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2.2. MAGNETIC FORCE ON A CURRENT-CARRYING WIRE 13 The net force on a current-carrying loop of wire in a uniform magnetic ﬁeld is zero. Note that this does not preclude the possibility that the rigid loop rotates; for example, the force on opposite sides of the loop may be equal and opposite. What we have found is merely that the force will not lead to a translational net force on the loop; e.g., force that would propel the loop away from its current position in space. The possibility of rotation without translation leads to the most rudimentary concept for an electric motor. Practical electric motors use variations on essentially this same idea; see “Additional Reading” for more information. The third case of practical interest is the force experienced by two parallel inﬁnitesimally-thin wires in free space, as shown in Figure 2.3. Here the wires are inﬁnite in length (we’ll return to that in a moment), lie in the x = 0 plane, are separated by distance d, and carry currents I1 and I2, respectively. The current in wire 1 gives rise to a magnetic ﬂux density B1. The force exerted on wire 2 by B1 is: F2 = Z C [I2dl (r) × B1 (r)] (2.9) where C is the path followed by I2, and dl (r) = ˆzdz. A simple way to determine B1 in this situation is as follows. First, if wire 1 had been aligned along the x = y = 0 line, then the magnetic ﬂux density everywhere would be ˆφµ0I1 2πρ In the present problem, wire 1 is displaced by d/2 in the −ˆy direction. Although this would seem to make the new expression more complicated, note that the only positions where values of B1 (r) are required are those corresponding to C; i.e., points on wire 2. For these points, B1 (r) = −ˆxµ0I1 2πd along C (2.10) That is, the relevant distance is d (not ρ), and the direction of B1 (r) for points along C is −ˆx (not ˆφ). Returning to Equation 2.9, we obtain: F2 = Z C I2 ˆzdz × −ˆxµ0I1 2πd = −ˆyµ0I1I2 2πd Z C dz (2.11) The remaining integral is simply the length of wire 2 that we wish to consider. Inﬁnitely-long wires will therefore result in inﬁnite force. This is not a very interesting or useful result. However, the force per unit length of wire is ﬁnite, and is obtained simply by dropping the integral in the previous equation. We obtain: F2 ∆l = −ˆyµ0I1I2 2πd (2.12) where ∆l is the length of the section of wire 2 being considered. Note that when the currents I1 and I2 ﬂow in the same direction (i.e., have the same sign), the magnetic force exerted by the current on wire 1 pulls wire 2 toward wire 1. The same process can be used to determine the magnetic force F1 exerted by the current in wire 1 on wire 2. The result is F1 ∆l = +ˆyµ0I1I2 2πd (2.13) c⃝Y. Zhao CC BY-SA 4.0 Figure 2.3: Parallel current-carrying wires.	Electromagnetics_Vol2_Page_26_Chunk2201
14 CHAPTER 2. MAGNETOSTATICS REDUX When the currents I1 and I2 ﬂow in the same direction (i.e., when the product I1I2 is positive), then the magnetic force exerted by the current on wire 2 pulls wire 1 toward wire 2. We are now able to summarize the results as follows: If currents in parallel wires ﬂow in the same di- rection, then the wires attract; whereas if the cur- rents ﬂow in opposite directions, then the wires repel. Also: The magnitude of the associated force is µ0I1I2/2πd for wires separated by distance d in non-magnetic media. If the wires are ﬁxed in position and not able to move, these forces represent stored (potential) energy. It is worth noting that this is precisely the energy which is stored by an inductor – for example, the two wire segments here might be interpreted as segments in adjacent windings of a coil-shaped inductor. Example 2.1. DC power cable. A power cable connects a 12 V battery to a load exhibiting an impedance of 10 Ω. The conductors are separated by 3 mm by a plastic insulating jacket. Estimate the force between the conductors. Solution. The current ﬂowing in each conductor is 12 V divided by 10 Ω, which is 1.2 A. In terms of the theory developed in this section, a current I1 = +1.2 A ﬂows from the positive terminal of the battery to the load on one conductor, and a current I2 = −1.2 A returns to the battery on the other conductor. The change in sign indicates that the currents at any given distance from the battery are ﬂowing in opposite directions. Also from the problem statement, d = 3 mm and the insulator is presumably non-magnetic. Assuming the conductors are approximately straight, the force between conductors is ≈µ0I1I2 2πd ∼= −96.0 µN with the negative sign indicating that the wires repel. Note in the above example that this force is quite small, which explains why it is not always observed. However, this force becomes signiﬁcant when the current is large or when many sets of conductors are mechanically bound together (amounting to a larger net current), as in a motor. Additional Reading: • “Electric motor” on Wikipedia.	Electromagnetics_Vol2_Page_27_Chunk2202
2.3. TORQUE INDUCED BY A MAGNETIC FIELD 15 2.3 Torque Induced by a Magnetic Field [m0024] A magnetic ﬁeld exerts a force on current. This force is exerted in a direction perpendicular to the direction of current ﬂow. For this reason, current-carrying structures in a magnetic ﬁeld tend to rotate. A convenient description of force associated with rotational motion is torque. In this section, we deﬁne torque and apply this concept to a closed loop of current. These concepts apply to a wide range of practical devices, including electric motors. Figure 2.4 illustrates the concept of torque. Torque depends on the following: • A local origin r0, • A point r which is connected to r0 by a perfectly-rigid mechanical structure, and • The force F applied at r. In terms of these parameters, the torque T is: T ≜d × F (2.14) where the lever arm d ≜r −r0 gives the location of r relative to r0. Note that T is a position-free vector c⃝C. Wang CC BY-SA 4.0 Figure 2.4: Torque associated with a single lever arm. which points in a direction perpendicular to both d and F. Note that T does not point in the direction of rotation. Nevertheless, T indicates the direction of rotation through a “right hand rule”: If you point the thumb of your right hand in the direction of T, then the curled ﬁngers of your right hand will point in the direction of torque-induced rotation. Whether rotation actually occurs depends on the geometry of the structure. For example, if T aligns with the axis of a perfectly-rigid mechanical shaft, then all of the work done by F will be applied to rotation of the shaft on this axis. Otherwise, torque will tend to rotate the shaft in other directions as well. If the shaft is not free to rotate in these other directions, then the effective torque – that is, the torque that contributes to rotation of the shaft – is reduced. The magnitude of T has SI base units of N·m and quantiﬁes the energy associated with the rotational force. As you might expect, the magnitude of the torque increases with increasing lever arm magnitude \|d\|. In other words, the torque resulting from a constant applied force increases with the length of the lever arm. Torque, like the translational force F, satisﬁes superposition. That is, the torque resulting from forces applied to multiple rigidly-connected lever arms is the sum of the torques applied to the lever arms individually. Now consider the current loop shown in Figure 2.5. The loop is perfectly rigid and is rigidly attached to a non-conducting shaft. The assembly consisting of the loop and the shaft may rotate without friction around the axis of the shaft. The loop consists of four straight segments that are perfectly-conducting and inﬁnitesimally-thin. A spatially-uniform and static impressed magnetic ﬂux density B0 = ˆxB0 exists throughout the domain of the problem. (Recall that an impressed ﬁeld is one that exists in the absence of any other structure in the problem.) What motion, if any, is expected? Recall that the net translational force on a current loop in a spatially-uniform and static magnetic ﬁeld is	Electromagnetics_Vol2_Page_28_Chunk2203
16 CHAPTER 2. MAGNETOSTATICS REDUX zero (Section 2.2). However, this does not preclude the possibility of different translational forces acting on each of the loop segments resulting in a rotation of the shaft. Let us ﬁrst calculate these forces. The force FA on segment A is FA = IlA × B0 (2.15) where lA is a vector whose magnitude is equal to the length of the segment and which points in the direction of the current. Thus, FA = I (ˆzL) × (ˆxB0) = ˆyILB0 (2.16) Similarly, the force FC on segment C is FC = I (−ˆzL) × (ˆxB0) = −ˆyILB0 (2.17) The forces FB and FD on segments B and D, respectively, are: FB = I (−ˆxL) × (ˆxB0) = 0 (2.18) and FD = I (+ˆxL) × (ˆxB0) = 0 (2.19) Thus, the force exerted on the current loop by the impressed magnetic ﬁeld will lead to rotation in the +ˆφ direction. z x A B C D L W I B0=x̂B0 current source wire loop non-conducting shaft c⃝C. Wang CC BY-SA 4.0 Figure 2.5: A rudimentary electric motor consisting of a single current loop. We calculate the associated torque T as T = TA + TB + TC + TD (2.20) where TA, TB, TC, and TD are the torques associated with segments A, B, C, and D, respectively. For example, the torque associated with segment A is TA = W 2 ˆx × FA = ˆzLW 2 IB0 (2.21) Similarly, TB = 0 since FB = 0 (2.22) TC = ˆzLW 2 IB0 (2.23) TD = 0 since FD = 0 (2.24) Summing these contributions, we ﬁnd T = ˆzLWIB0 (2.25) Note that T points in the +ˆz direction, indicating rotational force exerted in the +ˆφ direction, as expected. Also note that the torque is proportional to the area LW of the loop, is proportional to the current I, and is proportional to the magnetic ﬁeld magnitude B0. The analysis that we just completed was static; that is, it applies only at the instant depicted in Figure 2.5. If the shaft is allowed to turn without friction, then the loop will rotate in the +ˆφ direction. So, what will happen to the forces and torque? First, note that FA and FC are always in the +ˆy and −ˆy directions, respectively, regardless of the rotation of the loop. Once the loop rotates away from the position shown in Figure 2.5, the forces FB and FD become non-zero; however, they are always equal and opposite, and so do not affect the rotation. Thus, the loop will rotate one-quarter turn and then come to rest, perhaps with some damped oscillation around the rest position depending on the momentum of the loop. At the rest position, the lever arms for segments A and C are pointing in the same directions as FA and FC, respectively. Therefore, the cross product of the lever arm and translational force for each segment is zero and subsequently TA = TC = 0. Once stopped in this position, both the net translational force and the net torque are zero.	Electromagnetics_Vol2_Page_29_Chunk2204
2.3. TORQUE INDUCED BY A MAGNETIC FIELD 17 c⃝Abnormaal CC BY-SA 3.0 Figure 2.6: This DC electric motor uses brushes (here, the motionless leads labeled “+” and “−”) combined with the motion of the shaft to periodically alternate the direction of current between two coils, thereby cre- ating nearly constant torque. If such a device is to be used as a motor, it is necessary to ﬁnd a way to sustain the rotation. There are several ways in which this might be accomplished. First, one might make I variable in time. For example, the direction of I could be reversed as the loop passes the quarter-turn position. This reverses FA and FC, propelling the loop toward the half-turn position. The direction of I can be changed again as the loop passes half-turn position, propelling the loop toward the three-quarter-turn position. Continuing this periodic reversal of the current sustains the rotation. Alternatively, one may periodically reverse the direction of the impressed magnetic ﬁeld to the same effect. These methods can be combined or augmented using multiple current loops or multiple sets of time-varying impressed magnetic ﬁelds. Using an appropriate combination of current loops, magnetic ﬁelds, and waveforms for each, it is possible to achieve sustained torque throughout the rotation. An example is shown in Figure 2.6. Additional Reading: • “Torque” on Wikipedia. • “Electric motor” on Wikipedia.	Electromagnetics_Vol2_Page_30_Chunk2205
18 CHAPTER 2. MAGNETOSTATICS REDUX 2.4 The Biot-Savart Law [m0066] The Biot-Savart law (BSL) provides a method to calculate the magnetic ﬁeld due to any distribution of steady (DC) current. In magnetostatics, the general solution to this problem employs Ampere’s law; i.e., Z C H · dl = Iencl (2.26) in integral form or ∇× H = J (2.27) in differential form. The integral form is relatively simple when the problem exhibits a high degree of symmetry, facilitating a simple description in a particular coordinate system. An example is the magnetic ﬁeld due to a straight and inﬁnitely-long current ﬁlament, which is easily determined by solving the integral equation in cylindrical coordinates. However, many problems of practical interest do not exhibit the necessary symmetry. A commonly-encountered example is the magnetic ﬁeld due to a single loop of current, which will be addressed in Example 2.2. For such problems, the differential form of Ampere’s law is needed. BSL is the solution to the differential form of Ampere’s law for a differential-length current element, illustrated in Figure 2.7. The current element is I dl, where I is the magnitude of the current (SI base units of A) and dl is a differential-length vector indicating the direction of the current at the “source point” r′. The resulting contribution to the magnetic ﬁeld intensity at the “ﬁeld point” r is dH(r) = I dl 1 4πR2 × ˆR (2.28) where R = ˆRR ≜r −r′ (2.29) In other words, R is the vector pointing from the source point to the ﬁeld point, and dH at the ﬁeld point is given by Equation 2.28. The magnetic ﬁeld due to a current-carrying wire of any shape may be obtained by integrating over the length of the wire: H(r) = Z C dH(r) = I 4π Z C dl × ˆR R2 (2.30) In addition to obviating the need to solve a differential equation, BSL provides some useful insight into the behavior of magnetic ﬁelds. In particular, Equation 2.28 indicates that magnetic ﬁelds follow the inverse square law – that is, the magnitude of the magnetic ﬁeld due to a differential current element decreases in proportion to the inverse square of distance (R−2). Also, Equation 2.28 indicates that the direction of the magnetic ﬁeld due to a differential current element is perpendicular to both the direction of current ﬂow ˆl and the vector ˆR pointing from the source point to ﬁeld point. This observation is quite useful in anticipating the direction of magnetic ﬁeld vectors in complex problems. It may be helpful to note that BSL is analogous to Coulomb’s law for electric ﬁelds, which is a solution to the differential form of Gauss’ law, ∇· D = ρv. However, BSL applies only under magnetostatic conditions. If the variation in currents or magnetic ﬁelds over time is signiﬁcant, then the problem becomes signiﬁcantly more complicated. See “Jeﬁmenko’s Equations” in “Additional Reading” for more information. Example 2.2. Magnetic ﬁeld along the axis of a circular loop of current. Consider a ring of radius a in the z = 0 plane, centered on the origin, as shown in Figure 2.8. As indicated in the ﬁgure, the current I ﬂows in R R dH(r) dl @r' I c⃝C. Wang CC BY-SA 4.0 Figure 2.7: Use of the Biot-Savart law to calculate the magnetic ﬁeld due to a line current.	Electromagnetics_Vol2_Page_31_Chunk2206
2.4. THE BIOT-SAVART LAW 19 the ˆφ direction. Find the magnetic ﬁeld intensity along the z axis. Solution. The source current position is given in cylindrical coordinates as r′ = ˆρa (2.31) The position of a ﬁeld point along the z axis is r = ˆzz (2.32) Thus, ˆRR ≜r −r′ = −ˆρa + ˆzz (2.33) and R ≜\|r −r′\| = p a2 + z2 (2.34) Equation 2.28 becomes: dH(ˆzz) = I ˆφ adφ 4π [a2 + z2] × ˆzz −ˆρa √ a2 + z2 = Ia 4π ˆza −ˆρz [a2 + z2]3/2 dφ (2.35) Now integrating over the current: H(ˆzz) = Z 2π 0 Ia 4π ˆza −ˆρz [a2 + z2]3/2 dφ (2.36) = Ia 4π [a2 + z2]3/2 Z 2π 0 (ˆza −ˆρz) dφ (2.37) = Ia 4π [a2 + z2]3/2 ˆza Z 2π 0 dφ −z Z 2π 0 ˆρ dφ (2.38) The second integral is equal to zero. To see this, note that the integral is simply summing values of ˆρ for all possible values of φ. Since ˆρ(φ + π) = −ˆρ(φ), the integrand for any given value of φ is equal and opposite the integrand π radians later. (This is one example of a symmetry argument.) c⃝K. Kikkeri CC BY SA 4.0 (modiﬁed) Figure 2.8: Calculation of the magnetic ﬁeld along the z axis due to a circular loop of current centered in the z = 0 plane. The ﬁrst integral in the previous equation is equal to 2π. Thus, we obtain H(ˆzz) = ˆz Ia2 2 [a2 + z2]3/2 (2.39) Note that the result is consistent with the associated “right hand rule” of magnetostatics: That is, the direction of the magnetic ﬁeld is in the direction of the curled ﬁngers of the right hand when the thumb of the right hand is aligned with the location and direction of current. It is a good exercise to conﬁrm that this result is also dimensionally correct. Equation 2.28 extends straightforwardly to other distributions of current. For example, the magnetic ﬁeld due to surface current Js (SI base units of A/m) can be calculated using Equation 2.28 with I dl replaced by Js ds where ds is the differential element of surface area. This can be conﬁrmed by dimensional analysis: I dl has SI base units of A·m, as does JS ds. Similarly, the magnetic ﬁeld due to volume current J (SI base units of A/m2) can be calculated using Equation 2.28 with I dl replaced by J dv where dv is the differential element of volume. For a single particle with charge q (SI base units of C) and	Electromagnetics_Vol2_Page_32_Chunk2207
20 CHAPTER 2. MAGNETOSTATICS REDUX velocity v (SI base units of m/s), the relevant quantity is qv since C·m/s = (C/s)·m = A·m. In all of these cases, Equation 2.28 applies with the appropriate replacement for I dl. Note that the quantities qv, I dl, JS ds, and J dv, all having the same units of A·m, seem to be referring to the same physical quantity. This physical quantity is known as current moment. Thus, the “input” to BSL can be interpreted as current moment, regardless of whether the current of interest is distributed as a line current, a surface current, a volumetric current, or simply as moving charged particles. See “Additional Reading” at the end of this section for additional information on the concept of “moment” in classical physics. Additional Reading: • “Biot-Savart Law” on Wikipedia. • “Jeﬁmenko’s Equations” on Wikipedia. • “Moment (physics)” on Wikipedia. 2.5 Force, Energy, and Potential Difference in a Magnetic Field [m0059] The force Fm experienced by a particle at location r bearing charge q due to a magnetic ﬁeld is Fm = qv × B(r) (2.40) where v is the velocity (magnitude and direction) of the particle, and B(r) is the magnetic ﬂux density at r. Now we must be careful: In this description, the motion of the particle is not due to Fm. In fact the cross product in Equation 2.40 clearly indicates that Fm and v must be in perpendicular directions. Instead, the reverse is true: i.e., it is the motion of the particle that is giving rise to the force. The motion described by v may be due to the presence of an electric ﬁeld, or it may simply be that that charge is contained within a structure that is itself in motion. Nevertheless, the force Fm has an associated potential energy. Furthermore, this potential energy may change as the particle moves. This change in potential energy may give rise to an electrical potential difference (i.e., a “voltage”), as we shall now demonstrate. The change in potential energy can be quantiﬁed using the concept of work, W. The incremental work ∆W done by moving the particle a short distance ∆l, over which we assume the change in Fm is negligible, is ∆W ≈Fm ·ˆl∆l (2.41) where in this case ˆl is the unit vector in the direction of the motion; i.e., the direction of v. Note that the purpose of the dot product in Equation 2.41 is to ensure that only the component of Fm parallel to the direction of motion is included in the energy tally. Any component of v which is due to Fm (i.e., ultimately due to B) must be perpendicular to Fm, so ∆W for such a contribution must be, from Equation 2.41, equal to zero. In other words: In the absence of a mechanical force or an electric ﬁeld, the potential energy of a charged particle remains constant regardless of how it is moved by Fm. This surprising result may be summarized as follows:	Electromagnetics_Vol2_Page_33_Chunk2208
2.5. FORCE, ENERGY, AND POTENTIAL DIFFERENCE IN A MAGNETIC FIELD 21 The magnetic ﬁeld does no work. Instead, the change of potential energy associated with the magnetic ﬁeld must be completely due to a change in position resulting from other forces, such as a mechanical force or the Coulomb force. The presence of a magnetic ﬁeld merely increases or decreases this potential difference once the particle has moved, and it is this change in the potential difference that we wish to determine. We can make the relationship between potential difference and the magnetic ﬁeld explicit by substituting the right side of Equation 2.40 into Equation 2.41, yielding ∆W ≈q [v × B(r)] ·ˆl∆l (2.42) Equation 2.42 gives the work only for a short distance around r. Now let us try to generalize this result. If we wish to know the work done over a larger distance, then we must account for the possibility that v × B varies along the path taken. To do this, we may sum contributions from points along the path traced out by the particle, i.e., W ≈ N X n=1 ∆W(rn) (2.43) where rn are positions deﬁning the path. Substituting the right side of Equation 2.42, we have W ≈q N X n=1 [v × B(rn)] ·ˆl(rn)∆l (2.44) Taking the limit as ∆l →0, we obtain W = q Z C [v × B(r)] ·ˆl(r)dl (2.45) where C is the path (previously, the sequence of rn’s) followed by the particle. Now omitting the explicit dependence on r in the integrand for clarity: W = q Z C [v × B] · dl (2.46) where dl = ˆldl as usual. Now, we are able to determine the change in potential energy for a charged particle moving along any path in space, given the magnetic ﬁeld. At this point, it is convenient to introduce the electric potential difference V21 between the start point (1) and end point (2) of C. V21 is deﬁned as the work done by traversing C, per unit of charge; i.e., V21 ≜W q (2.47) This has units of J/C, which is volts (V). Substituting Equation 2.46, we obtain: V21 = Z C [v × B] · dl (2.48) Equation 2.48 is electrical potential induced by charge traversing a magnetic ﬁeld. Figure 2.9 shows a simple scenario that illustrates this concept. Here, a straight perfectly-conducting wire of length l is parallel to the y axis and moves at speed v in the +z direction through a magnetic ﬁeld B = ˆxB. Thus, v × B = ˆzv × ˆxB = ˆyBv (2.49) Taking endpoints 1 and 2 of the wire to be at y = y0 (2) (1) B v z y x l c⃝C. Wang CC BY 4.0 Figure 2.9: A straight wire moving through a mag- netic ﬁeld.	Electromagnetics_Vol2_Page_34_Chunk2209
22 CHAPTER 2. MAGNETOSTATICS REDUX and y = y0 + l, respectively, we obtain V21 = Z y0+l y0 [ˆyBv] · ˆydy = Bvl (2.50) Thus, we see that endpoint 2 is at an electrical potential of Bvl greater than that of endpoint 1. This “voltage” exists even though the wire is perfectly-conducting, and therefore cannot be attributed to the electric ﬁeld. This voltage exists even though the force required for movement must be the same on both endpoints, or could even be zero, and therefore cannot be attributed to mechanical forces. Instead, this change in potential is due entirely to the magnetic ﬁeld. Because the wire does not form a closed loop, no current ﬂows in the wire. Therefore, this scenario has limited application in practice. To accomplish something useful with this concept we must at least form a closed loop, so that current may ﬂow. For a closed loop, Equation 2.48 becomes: V = I C [v × B] · dl (2.51) Examination of this equation indicates one additional requirement: v × B must somehow vary over C. This is because if v × B does not vary over C, the result will be [v × B] · I C dl which is zero because the integral is zero. The following example demonstrates a practical application of this idea. Example 2.3. Potential induced in a time-varying loop. Figure 2.10 shows a modiﬁcation to the problem originally considered in Figure 2.9. Now, we have created a closed loop using perfectly-conducting and motionless wire to form three sides of a rectangle, and assigned the origin to the lower left corner. An inﬁnitesimally-small gap has been inserted in the left (z = 0) side of the loop and closed with an ideal resistor of value R. As before, B = ˆxB (spatially uniform and time invariant) and v B z x y VT + - R c⃝C. Wang CC BY-SA 4.0 Figure 2.10: A time-varying loop created by moving a “shorting bar” along rails comprising two adjacent sides of the loop. v = ˆzv (constant). What is the voltage VT across the resistor and what is the current in the loop? Solution. Since the gap containing the resistor is inﬁnitesimally small, VT = I C [v × B] · dl (2.52) where C is the perimeter formed by the loop, beginning at the “−” terminal of VT and returning to the “+” terminal of VT . Only the shorting bar is in motion, so v = 0 for the other three sides of the loop. Therefore, only the portion of C traversing the shorting bar contributes to VT . Thus, we ﬁnd VT = Z l y=0 [ˆzv × ˆxB] · ˆydy = Bvl (2.53) This potential gives rise to a current Bvl/R, which ﬂows in the counter-clockwise direction. Note in the previous example that the magnetic ﬁeld has induced VT , not the current. The current is simply a response to the existence of the potential, regardless of the source. In other words, the same potential VT would exist even if the gap was not closed by a resistor.	Electromagnetics_Vol2_Page_35_Chunk2210
2.5. FORCE, ENERGY, AND POTENTIAL DIFFERENCE IN A MAGNETIC FIELD 23 Astute readers will notice that this analysis seems to have a lot in common with Faraday’s law, V = −∂ ∂tΦ (2.54) which says the potential induced in a single closed loop is proportional to the time rate of change of magnetic ﬂux Φ, where Φ = Z S B · ds (2.55) and where S is the surface through which the ﬂux is calculated. From this perspective, we see that Equation 2.51 is simply a special case of Faraday’s law, pertaining speciﬁcally to “motional emf.” Thus, the preceding example can also be solved by Faraday’s law, taking S to be the time-varying surface bounded by C. [m0182]	Electromagnetics_Vol2_Page_36_Chunk2211
24 CHAPTER 2. MAGNETOSTATICS REDUX Image Credits Fig. 2.1: c⃝Stannered, https://en.wikipedia.org/wiki/File:Charged-particle-drifts.svg, CC BY 2.5 (https://creativecommons.org/licenses/by/2.5/deed.en). Modiﬁed by Maschen, author. Fig. 2.2: c⃝M. Biaek, https://en.wikipedia.org/wiki/File:Cyclotron motion.jpg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 2.3: c⃝YahuiZ (Y. Zhao), https://commons.wikimedia.org/wiki/File:Figure2.3Yahui.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Modiﬁed by author. Fig. 2.4: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Torque associated with a single lever arm.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 2.5: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Rudimentary electric motor.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 2.6: c⃝Abnormaal, https://en.wikipedia.org/wiki/File:Electric motor.gif, CC BY SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0/). Fig. 2.7: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Magnetic ﬁeld due to a line current.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 2.8: c⃝K. Kikkeri, https://commons.wikimedia.org/wiki/File:M0104 fRing.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Modiﬁed by author. Fig. 2.9: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Straight wire moving in a magnetic ﬁeld.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 2.10: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Moving shorting bar of a circuit in mag ﬁeld.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).	Electromagnetics_Vol2_Page_37_Chunk2212
Chapter 3 Wave Propagation in General Media 3.1 Poynting’s Theorem [m0073] Despite the apparent complexity of electromagnetic theory, there are in fact merely four ways that electromagnetic energy can be manipulated. Electromagnetic energy can be: • Transferred; i.e., conveyed by transmission lines or in waves; • Stored in an electric ﬁeld (capacitance); • Stored in a magnetic ﬁeld (inductance); or • Dissipated (converted to heat; i.e., resistance). Poynting’s theorem is an expression of conservation of energy that elegantly relates these various possibilities. Once recognized, the theorem has important applications in the analysis and design of electromagnetic systems. Some of these emerge from the derivation of the theorem, as opposed to the unsurprising result. So, let us now derive the theorem. We begin with a statement of the theorem. Consider a volume V that may contain materials and structures in any combination. This is crudely depicted in Figure 3.1. Also recall that power is the time rate of change of energy. Then: Pnet,in = PE + PM + PΩ (3.1) where Pnet,in is the net power ﬂow into V, PE is the power associated with energy storage in electric ﬁelds within V, PM is the power associated with energy storage in magnetic ﬁelds within V, and PΩis the power dissipated (converted to heat) in V. Some preliminary mathematical results. We now derive two mathematical relationships that will be useful later. Let E be the electric ﬁeld intensity (SI base units of V/m) and let D be the electric ﬂux density (SI base units of C/m2). We require that the constituents of V be linear and time-invariant; therefore, D = ǫE where the permittivity ǫ is constant with respect to time, but not necessarily with respect to position. Under these conditions, we ﬁnd ∂ ∂t (E · D) = 1 ǫ ∂ ∂t (D · D) (3.2) Note that E and D point in the same direction. Let ˆe be the unit vector describing this direction. Then, E = ˆeE and D = ˆeD where E and D are the scalar c⃝C. Wang CC BY-SA 4.0 Figure 3.1: Poynting’s theorem describes the fate of power entering a region V consisting of materials and structures capable of storing and dissipating energy. Electromagnetics Vol. 2. c⃝2020 S.W. Ellingson CC BY SA 4.0. https://doi.org/10.21061/electromagnetics-vol-2	Electromagnetics_Vol2_Page_38_Chunk2213
26 CHAPTER 3. WAVE PROPAGATION IN GENERAL MEDIA components of E and D, respectively. Subsequently, 1 ǫ ∂ ∂t (D · D) = 1 ǫ ∂ ∂tD2 = 1 ǫ 2D ∂ ∂tD = 2E · ∂ ∂tD (3.3) Summarizing: ∂ ∂t (E · D) = 2E · ∂ ∂tD (3.4) which is the expression we seek. It is worth noting that the expressions on both sides of the equation have the same units, namely, those of power. Using the same reasoning, we ﬁnd: ∂ ∂t (H · B) = 2H · ∂ ∂tB (3.5) where H is the magnetic ﬁeld intensity (SI base units of A/m) and B is the magnetic ﬂux density (SI base units of T). Derivation of the theorem. We begin with the differential form of Ampere’s law: ∇× H = J + ∂ ∂tD (3.6) Taking the dot product with E on both sides: E · (∇× H) = E · J + E · ∂ ∂tD (3.7) Let’s deal with the left side of this equation ﬁrst. For this, we will employ a vector identity (Equation B.27, Appendix B.3). This identity states that for any two vector ﬁelds F and G, ∇· (F × G) = G · (∇× F) −F · (∇× G) (3.8) Substituting E for F and H for G and rearranging terms, we ﬁnd E · (∇× H) = H · (∇× E) −∇· (E × H) (3.9) Next, we invoke the Maxwell-Faraday equation: ∇× E = −∂ ∂tB (3.10) Using this equation to replace the factor in the parentheses in the second term of Equation 3.9, we obtain: E·(∇× H) = H· −∂ ∂tB −∇·(E × H) (3.11) Substituting this expression for the left side of Equation 3.7, we have −H· ∂ ∂tB−∇·(E × H) = E·J+E· ∂ ∂tD (3.12) Next we invoke the identities of Equations 3.4 and 3.5 to replace the ﬁrst and last terms: −1 2 ∂ ∂t (H · B) −∇· (E × H) = E · J + 1 2 ∂ ∂t (E · D) (3.13) Now we move the ﬁrst term to the right-hand side and integrate both sides over the volume V: − Z V ∇· (E × H) dv = Z V E · J dv + Z V 1 2 ∂ ∂t (E · D) dv + Z V 1 2 ∂ ∂t (H · B) dv (3.14) The left side may be transformed into a surface integral using the divergence theorem: Z V ∇· (E × H) dv = I S (E × H) · ds (3.15) where S is the closed surface that bounds V and ds is the outward-facing normal to this surface, as indicated in Figure 3.1. Finally, we exchange the order of time differentiation and volume integration in the last two terms: − I S (E × H) · ds = Z V E · J dv + 1 2 ∂ ∂t Z V E · D dv + 1 2 ∂ ∂t Z V H · B dv (3.16)	Electromagnetics_Vol2_Page_39_Chunk2214
3.1. POYNTING’S THEOREM 27 Equation 3.16 is Poynting’s theorem. Each of the four terms has the particular physical interpretation identiﬁed in Equation 3.1, as we will now demonstrate. Power dissipated by ohmic loss. The ﬁrst term of the right side of Equation 3.16 is PΩ≜ Z V E · J dv (3.17) Equation 3.17 is Joule’s law. Joule’s law gives the power dissipated due to ﬁnite conductivity of material. The role of conductivity (σ, SI base units of S/m) can be made explicit using the relationship J = σE (Ohm’s law) in Equation 3.17, which yields: PΩ= Z V E · σE dv = Z V σ \|E\|2 dv (3.18) PΩis due to the conversion of the electric ﬁeld into conduction current, and subsequently into heat. This mechanism is commonly known as ohmic loss or joule heating. It is worth noting that this expression has the expected units. For example, in Equation 3.17, E (SI base units of V/m) times J (SI base units of A/m2) yields a quantity with units of W/m3; i.e., power per unit volume, which is power density. Then integration over volume yields units of W; hence, power. Energy storage in electric and magnetic ﬁelds. The second term on the right side of Equation 3.16 is: PE ≜1 2 ∂ ∂t Z V E · D dv (3.19) Recall D = ǫE, where ǫ is the permittivity (SI base units of F/m) of the material. Thus, we may rewrite the previous equation as follows: PE = 1 2 ∂ ∂t Z V E · ǫE dv = 1 2 ∂ ∂t Z V ǫ \|E\|2 dv = ∂ ∂t 1 2 Z V ǫ \|E\|2 dv (3.20) The quantity in parentheses is We, the energy stored in the electric ﬁeld within V. The third term on the right side of Equation 3.16 is: PM ≜1 2 ∂ ∂t Z V H · B dv (3.21) Recall B = µH, where µ is the permeability (SI base units of H/m) of the material. Thus, we may rewrite the previous equation as follows: PM = 1 2 ∂ ∂t Z V H · µH dv = 1 2 ∂ ∂t Z V µ \|H\|2 dv = ∂ ∂t 1 2 Z V µ \|H\|2 dv (3.22) The quantity in parentheses is Wm, the energy stored in the magnetic ﬁeld within V. Power ﬂux through S. The left side of Equation 3.16 is Pnet,in ≜− I S (E × H) · ds (3.23) The quantity E × H is the Poynting vector, which quantiﬁes the spatial power density (SI base units of W/m2) of an electromagnetic wave and the direction in which it propagates. The reader has likely already encountered this concept. Regardless, we’ll conﬁrm this interpretation of the quantity E × H in Section 3.2. For now, observe that integration of the Poynting vector over S as indicated in Equation 3.23 yields the total power ﬂowing out of V through S. The negative sign in Equation 3.23 indicates that the combined quantity represents power ﬂow in to V through S. Finally, note the use of a single quantity Pnet,in does not imply that power is entirely inward-directed or outward-directed. Rather, Pnet,in represents the net ﬂux; i.e., the sum of the inward- and outward-ﬂowing power. Summary. Equation 3.16 may now be stated concisely as follows: Pnet,in = PΩ+ ∂ ∂tWe + ∂ ∂tWm (3.24)	Electromagnetics_Vol2_Page_40_Chunk2215
28 CHAPTER 3. WAVE PROPAGATION IN GENERAL MEDIA Poynting’s theorem (Equation 3.24, with Equa- tions 3.23, 3.17, 3.19, and 3.21) states that the net electromagnetic power ﬂowing into a region of space may be either dissipated, or used to change the energy stored in electric and magnetic ﬁelds within that region. Since we derived this result directly from the general form of Maxwell’s equations, these three possibilities are all the possibilities allowed by classical physics, so this is a statement of conservation of power. Finally, note the theorem also works in reverse; i.e., the net electromagnetic power ﬂowing out of a region of space must have originated from some active source (i.e., the reverse of power dissipation) or released from energy stored in electric or magnetic ﬁelds. Additional Reading: • “Poynting vector” on Wikipedia. 3.2 Poynting Vector [m0122] In this section, we use Poynting’s theorem (Section 3.1) to conﬁrm the interpretation of the Poynting vector S ≜E × H (3.25) as the spatial power density (SI base units of W/m2) and the direction of power ﬂow. Figure 3.2 shows a uniform plane wave incident on a homogeneous and lossless region V deﬁned by the closed right cylinder S. The axis of the cylinder is aligned along the direction of propagation ˆk of the plane wave. The ends of the cylinder are planar and perpendicular to ˆk. Now let us apply Poynting’s theorem: Pnet,in = PΩ+ ∂ ∂tWe + ∂ ∂tWm (3.26) Since the region is lossless, PΩ= 0. Presuming no other electric or magnetic ﬁelds, We and Wm must also be zero. Consequently, Pnet,in = 0. However, this does not mean that there is no power ﬂowing into V. Instead, Pnet,in = 0 merely indicates that the net power ﬂowing into V is zero. Thus, we might instead express the result for the present scenario as follows: Pnet,in = Pin −Pout = 0 (3.27) c⃝C. Wang CC BY-SA 4.0 Figure 3.2: A uniform plane wave incident on a region bounded by the cylindrical surface S.	Electromagnetics_Vol2_Page_41_Chunk2216
3.2. POYNTING VECTOR 29 where Pin and Pout indicate power ﬂow explicitly into and explicitly out of V as separate quantities. Proceeding, let’s ignore what we know about power ﬂow in plane waves, and instead see where Poynting’s theorem takes us. Here, Pnet,in ≜− I S (E × H) · ds = 0 (3.28) The surface S consists of three sides: The two ﬂat ends, and the curved surface that connects them. Let us refer to these sides as S1, S2, and S3, from left to right as shown in Figure 3.2. Then, − Z S1 (E × H) · ds (3.29) − Z S2 (E × H) · ds − Z S3 (E × H) · ds = 0 On the curved surface S2, ds is everywhere perpendicular to E × H (i.e., perpendicular to ˆk). Therefore, the second integral is equal to zero. On the left end S1, the outward-facing differential surface element ds = −ˆkds. On the right end S3, ds = +ˆkds. We are left with: Z S1 (E × H) · ˆkds (3.30) − Z S3 (E × H) · ˆkds = 0 Compare this to Equation 3.27. Since ˆk is, by deﬁnition, the direction in which E × H points, the ﬁrst integral must be Pin and the second integral must be Pout. Thus, Pin = Z S1 (E × H) · ˆkds (3.31) and it follows that the magnitude and direction of E × H, which we recognize as the Poynting vector, are spatial power density and direction of power ﬂow, respectively. The Poynting vector is named after English physicist J.H. Poynting, one of the co-discoverers of the concept. The fact that this vector points in the direction of power ﬂow, and is therefore also a “pointing vector,” is simply a remarkable coincidence. Additional Reading: • “Poynting vector” on Wikipedia.	Electromagnetics_Vol2_Page_42_Chunk2217
30 CHAPTER 3. WAVE PROPAGATION IN GENERAL MEDIA 3.3 Wave Equations for Lossy Regions [m0128] The wave equations for electromagnetic propagation in lossless and source-free media, in differential phasor form, are: ∇2 eE + ω2µǫeE = 0 (3.32) ∇2 eH + ω2µǫ eH = 0 (3.33) The constant ω2µǫ is labeled β2, and β turns out to be the phase propagation constant. Now, we wish to upgrade these equations to account for the possibility of loss. First, let’s be clear on what we mean by “loss.” Speciﬁcally, we mean the possibility of conversion of energy from the propagating wave into current, and subsequently to heat. This mechanism is described by Ohm’s law: eJ = σeE (3.34) where σ is conductivity and eJ is conduction current density (SI base units of A/m2). In the lossless case, σ is presumed to be zero (or at least negligible), so J is presumed to be zero. To obtain wave equations for media exhibiting signiﬁcant loss, we cannot assume J = 0. To obtain equations that account for the possibility of signiﬁcant conduction current, hence loss, we return to the phasor forms of Maxwell’s equations: ∇· eE = eρv ǫ (3.35) ∇× eE = −jωµ eH (3.36) ∇· eH = 0 (3.37) ∇× eH = eJ + jωǫeE (3.38) where eρv is the charge density. If the region of interest is source-free, then eρv = 0. However, we may not similarly suppress eJ since σ may be non-zero. To make progress, let us identify the possible contributions to eJ as follows: eJ = eJimp + eJind (3.39) where eJimp represents impressed sources of current and eJind represents current which is induced by loss. An impressed current is one whose behavior is independent of other features, analogous to an independent current source in elementary circuit theory. In the absence of such sources, Equation 3.39 becomes: eJ = 0 + σeE (3.40) Equation 3.38 may now be rewritten as: ∇× eH = σeE + jωǫeE (3.41) = (σ + jωǫ) eE (3.42) = jωǫc eE (3.43) where we deﬁned the new constant ǫc as follows: ǫc ≜ǫ −j σ ω (3.44) This constant is known as complex permittivity. In the lossless case, ǫc = ǫ; i.e., the imaginary part of ǫc →0 so there is no difference between the physical permittivity ǫ and ǫc. The effect of the material’s loss is represented as a non-zero imaginary component of the permittivity. It is also common to express ǫc as follows: ǫc = ǫ′ −jǫ′′ (3.45) where the real-valued constants ǫ′ and ǫ′′ are in this case: ǫ′ = ǫ (3.46) ǫ′′ = σ ω (3.47) This alternative notation is useful for three reasons. First, some authors use the symbol ǫ to refer to both physical permittivity and complex permittivity. In this case, the “ǫ′ −jǫ′′” notation is helpful in mitigating confusion. Second, it is often more convenient to specify ǫ′′ at a frequency than it is to specify σ, which may also be a function of frequency. In fact, in some applications the loss of a material is most conveniently speciﬁed using the ratio ǫ′′/ǫ′ (known as loss tangent, for reasons explained elsewhere). Finally, it turns out that nonlinearity of permittivity can also be accommodated as an imaginary	Electromagnetics_Vol2_Page_43_Chunk2218
3.3. WAVE EQUATIONS FOR LOSSY REGIONS 31 component of the permittivity. The “ǫ′′” notation allows us to accommodate both effects – nonlinearity and conductivity – using common notation. In this section, however, we remain focused exclusively on conductivity. Complex permittivity ǫc (SI base units of F/m) describes the combined effects of permittivity and conductivity. Conductivity is represented as an imaginary-valued component of the permittiv- ity. Returning to Equations 3.35–3.38, we obtain: ∇· eE = 0 (3.48) ∇× eE = −jωµ eH (3.49) ∇· eH = 0 (3.50) ∇× eH = jωǫceE (3.51) These equations are identical to the corresponding equations for the lossless case, with the exception that ǫ has been replaced by ǫc. Similarly, we may replace the factor ω2µǫ in Equations 3.32 and 3.33, yielding: ∇2 eE + ω2µǫc eE = 0 (3.52) ∇2 eH + ω2µǫc eH = 0 (3.53) In the lossless case, ω2µǫc →ω2µǫ, which is β2 as expected. For the general (i.e., possibly lossy) case, we shall make an analogous deﬁnition γ2 ≜−ω2µǫc (3.54) such that the wave equations may now be written as follows: ∇2 eE −γ2 eE = 0 (3.55) ∇2 eH −γ2 eH = 0 (3.56) Note the minus sign in Equation 3.54, and the associated changes of signs in Equations 3.55 and 3.56. For the moment, this choice of sign may be viewed as arbitrary – we would have been equally justiﬁed in choosing the opposite sign for the deﬁnition of γ2. However, the choice we have made is customary, and yields some notational convenience that will become apparent later. In the lossless case, β is the phase propagation constant, which determines the rate at which the phase of a wave progresses with increasing distance along the direction of propagation. Given the similarity of Equations 3.55 and 3.56 to Equations 3.32 and 3.33, respectively, the constant γ must play a similar role. So, we are motivated to ﬁnd an expression for γ. At ﬁrst glance, this is simple: γ = p γ2. However, recall that every number has two square roots. When one declares β = p β2 = ω√µǫ, there is no concern since β is, by deﬁnition, positive; therefore, one knows to take the positive-valued square root. In contrast, γ2 is complex-valued, and so the two possible values of p γ2 are both potentially complex-valued. We are left without clear guidance on which of these values is appropriate and physically-relevant. So we proceed with caution. First, consider the special case that γ is purely imaginary; e.g., γ = jγ′′ where γ′′ is a real-valued constant. In this case, γ2 = −(γ′′)2 and the wave equations become ∇2 eE + (γ′′)2 eE = 0 (3.57) ∇2 eH + (γ′′)2 eH = 0 (3.58) Comparing to Equations 3.32 and 3.33, we see γ′′ plays the exact same role as β; i.e., γ′′ is the phase propagation constant for whatever wave we obtain as the solution to the above equations. Therefore, let us make that deﬁnition formally: β ≜Im {γ} (3.59) Be careful: Note that we are not claiming that γ′′ in the possibly-lossy case is equal to ω2µǫ. Instead, we are asserting exactly the opposite; i.e., that there is a phase propagation constant in the general (possibly lossy) case, and we should ﬁnd that this constant simpliﬁes to ω2µǫ in the lossless case. Now we make the following deﬁnition for the real component of γ: α ≜Re {γ} (3.60) Such that γ = α + jβ (3.61) where α and β are real-valued constants. Now, it is possible to determine γ explicitly in terms of ω and the constitutive properties of the material.	Electromagnetics_Vol2_Page_44_Chunk2219
32 CHAPTER 3. WAVE PROPAGATION IN GENERAL MEDIA First, note: γ2 = (α + jβ)2 = α2 −β2 + j2αβ (3.62) Expanding Equation 3.54 using Equations 3.45–3.47, we obtain: γ2 = −ω2µ ǫ −j σ ω = −ω2µǫ + jωµσ (3.63) The real and imaginary parts of Equations 3.62 and 3.63 must be equal. Enforcing this equality yields the following equations: α2 −β2 = −ω2µǫ (3.64) 2αβ = ωµσ (3.65) Equations 3.64 and 3.65 are independent simultaneous equations that may be solved for α and β. Sparing the reader the remaining steps, which are purely mathematical, we ﬁnd: α = ω    µǫ′ 2   s 1 + ǫ′′ ǫ′ 2 −1      1/2 (3.66) β = ω    µǫ′ 2   s 1 + ǫ′′ ǫ′ 2 + 1      1/2 (3.67) Equations 3.66 and 3.67 can be veriﬁed by conﬁrming that Equations 3.64 and 3.65 are satisﬁed. It is also useful to conﬁrm that the expected results are obtained in the lossless case. In the lossless case, σ = 0, so ǫ′′ = 0. Subsequently, Equation 3.66 yields α = 0 and Equation 3.67 yields β = ω√µǫ, as expected. The electromagnetic wave equations account- ing for the possibility of lossy media are Equa- tions 3.55 and 3.56 with γ = α + jβ, where α and β are the positive real-valued constants deter- mined by Equations 3.66 and 3.67, respectively. We conclude this section by pointing out a very useful analogy to transmission line theory. In the section “Wave Propagation on a TEM Transmission Line,”1 we found that the potential and current along a transverse electromagnetic (TEM) transmission line satisfy the same wave equations that we have developed in this section, having a complex-valued propagation constant γ = α + jβ, and the same physical interpretation of β as the phase propagation constant. As is explained in another section, α too has the same interpretation in both applications – that is, as the attenuation constant. Additional Reading: • “Electromagnetic Wave Equation” on Wikipedia. 1This section may appear in a different volume, depending on the version of this book.	Electromagnetics_Vol2_Page_45_Chunk2220
3.4. COMPLEX PERMITTIVITY 33 3.4 Complex Permittivity [m0134] The relationship between electric ﬁeld intensity E (SI base units of V/m) and electric ﬂux density D (SI base units of C/m2) is: D = ǫE (3.68) where ǫ is the permittivity (SI base units of F/m). In simple media, ǫ is a real positive value which does not depend on the time variation of E. That is, the response (D) to a change in E is observed instantaneously and without delay. In practical materials, however, the change in D in response to a change in E may depend on the manner in which E changes. The response may not be instantaneous, but rather might take some time to fully manifest. This behavior can be modeled using the following generalization of Equation 3.68: D = a0E+a1 ∂ ∂tE+a2 ∂2 ∂t2 E+a3 ∂3 ∂t3 E+... (3.69) where a0, a1, a2, and so on are real-valued constants, and the number of terms is inﬁnite. In practical materials, the importance of the terms tends to diminish with increasing order. Thus, it is common that only the ﬁrst few terms are signiﬁcant. In many applications involving common materials, only the ﬁrst term is signiﬁcant; i.e., a0 ≈ǫ and an ≈0 for n ≥1. As we shall see in a moment, this distinction commonly depends on frequency. In the phasor domain, differentiation with respect to time becomes multiplication by jω. Thus, Equation 3.69 becomes eD = a0 eE + a1 (jω) eE + a2 (jω)2 eE + a3 (jω)3 eE + ... = a0 eE + jωa1 eE −ω2a2 eE −jω3a3 eE + ... =	Electromagnetics_Vol2_Page_46_Chunk2221
34 CHAPTER 3. WAVE PROPAGATION IN GENERAL MEDIA c⃝K.A. Mauritz (modiﬁed) Figure 3.3: The relative contributions of the real and imaginary components of permittivity for a typical di- electric material (in this case, a polymer). Additional Reading: • “Permittivity” on Wikipedia. 3.5 Loss Tangent [m0132] In Section 3.3, we found that the effect of loss due to non-zero conductivity σ could be concisely quantiﬁed using the ratio ǫ′′ ǫ′ = σ ωǫ (3.76) where ǫ′ and ǫ′′ are the real and imaginary components of the complex permittivity ǫc, and ǫ′ ≜ǫ. In this section, we explore this relationship in greater detail. Recall Ampere’s law in differential (but otherwise general) form: ∇× H = J + ∂ ∂tD (3.77) The ﬁrst term on the right is conduction current, whereas the second term on the right is displacement current. In the phasor domain, differentiation with respect to time (∂/∂t) becomes multiplication by jω. Therefore, the phasor form of Equation 3.77 is ∇× eH = eJ + jω eD (3.78) Also recall that eJ = σeE and that eD = ǫeE. Thus, ∇× eH = σeE + jωǫeE (3.79) Interestingly, the total current is the sum of a real-valued conduction current and an imaginary-valued displacement current. This is shown graphically in Figure 3.4. Note that the angle δ indicated in Figure 3.4 is given by tan δ ≜σ ωǫ (3.80) The quantity tan δ is referred to as the loss tangent. Note that loss tangent is zero for a lossless (σ ≡0) material, and increases with increasing loss. Thus, loss tangent provides an alternative way to quantify the effect of loss on the electromagnetic ﬁeld within a material. Loss tangent presuming only ohmic (conduction) loss is given by Equation 3.80.	Electromagnetics_Vol2_Page_47_Chunk2222
3.5. LOSS TANGENT 35 Comparing Equation 3.80 to Equation 3.76, we see loss tangent can equivalently be calculated as tan δ = ǫ′′ ǫ (3.81) and subsequently interpreted as shown in Figure 3.5. The discussion in this section has assumed that ǫc is complex-valued solely due to ohmic loss. However, it is explained in Section 3.4 that permittivity may also be complex-valued as a way to model delay in the response of D to changing E. Does the concept of loss tangent apply also in this case? Since the math does not distinguish between permittivity which is complex due to loss and permittivity which is complex due to delay, subsequent mathematically-derived results apply in either case. On the other hand, there may be potentially signiﬁcant differences in the physical manifestation of these effects. For example, a material having large loss tangent due to ohmic loss might become hot when a large electric ﬁeld is applied, whereas a material having large loss tangent due to delayed response might not. Summarizing: The expression for loss tangent given by Equa- tion 3.81 and Figure 3.5 does not distinguish be- tween ohmic loss and delayed response. c⃝C. Wang CC BY-SA 4.0 Figure 3.4: In the phasor domain, the total current is the sum of a real-valued conduction current and an imaginary-valued displacement current. c⃝C. Wang CC BY-SA 4.0 Figure 3.5: Loss tangent deﬁned in terms of the real and imaginary components of the complex permittiv- ity ǫc. Additional Reading: • “Dielectric loss” on Wikipedia.	Electromagnetics_Vol2_Page_48_Chunk2223
36 CHAPTER 3. WAVE PROPAGATION IN GENERAL MEDIA 3.6 Plane Waves in Lossy Regions [m0130] The electromagnetic wave equations for source-free regions consisting of possibly-lossy material are (see Section 3.3): ∇2 eE −γ2 eE = 0 (3.82) ∇2 eH −γ2 eH = 0 (3.83) where γ2 ≜−ω2µǫc (3.84) We now turn our attention to the question, what are the characteristics of waves that propagate in these conditions? As in the lossless case, these equations permit waves having a variety of geometries including plane waves, cylindrical waves, and spherical waves. In this section, we will consider the important special case of uniform plane waves. To obtain the general expression for the uniform plane wave solution, we follow precisely the same procedure described in the section “Uniform Plane Waves: Derivation.”2 Although that section presumed lossless media, the only difference in the present situation is that the real-valued constant +β2 is replaced with the complex-valued constant −γ2. Thus, we obtain the desired solution through a simple modiﬁcation of the solution for the lossless case. For a wave exhibiting uniform magnitude and phase in planes of constant z, we ﬁnd that the electric ﬁeld is: eE = ˆx eEx + ˆy eEy (3.85) where eEx = E+ x0e−γz + E− x0e+γz (3.86) eEy = E+ y0e−γz + E− y0e+γz (3.87) where the complex-valued coefﬁcients E+ x0, E− x0, E+ y0, and E− y0 are determined by boundary conditions (possibly including sources) outside the region of interest. This result can be conﬁrmed by verifying that Equations 3.86 and 3.87 each satisfy Equation 3.82. Also, it may be helpful to note that these expressions are identical to those obtained for 2Depending on the version of this book, this section may appear in a different volume. the voltage and current in lossy transmission lines, as described in the section “Wave Equation for a TEM Transmission Line.”3 Let’s consider the special case of an ˆx-polarized plane wave propagating in the +ˆz direction: eE = ˆxE+ x0e−γz (3.88) We established in Section 3.3 that γ may be written explicitly in terms of its real and imaginary components as follows: γ = α + jβ (3.89) where α and β are positive real-valued constants depending on frequency (ω) and constitutive properties of the medium; i.e., permittivity, permeability, and conductivity. Thus: eE = ˆxE+ x0e−(α+jβ)z = ˆxE+ x0e−αze−jβz (3.90) Observe that the variation of phase with distance is determined by β through the factor e−jβz; thus, β is the phase propagation constant and plays precisely the same role as in the lossless case. Observe also that the variation in magnitude is determined by α through the real-valued factor e−αz. Speciﬁcally, magnitude is reduced inverse-exponentially with increasing distance along the direction of propagation. Thus, α is the attenuation constant and plays precisely the same role as the attenuation constant for a lossy transmission line. The presence of loss in material gives rise to a real-valued factor e−αz which describes the at- tenuation of the wave with propagation (in this case, along z) in the material. We may continue to exploit the similarity of the potentially-lossy and lossless plane wave results to quickly ascertain the characteristics of the magnetic ﬁeld. In particular, the plane wave relationships apply exactly as they do in the lossless case. These relationships are: eH = 1 η ˆk × eE (3.91) 3Depending on the version of this book, this section may appear in a different volume.	Electromagnetics_Vol2_Page_49_Chunk2224
3.7. WAVE POWER IN A LOSSY MEDIUM 37 eE = −ηˆk × eH (3.92) where ˆk is the direction of propagation and η is the wave impedance. In the lossless case, η = p µ/ǫ; however, in the possibly-lossy case we must replace ǫ = ǫ′ with ǫc = ǫ′ −jǫ′′. Thus: η →ηc = r µ ǫc = r µ ǫ′ −jǫ′′ = r µ ǫ′ s 1 1 −j (ǫ′′/ǫ′) (3.93) Thus: ηc = r µ ǫ′ · 1 −j ǫ′′ ǫ′ −1/2 (3.94) Remarkably, we ﬁnd that the wave impedance for a lossy material is equal to p µ/ǫ – the wave impedance we would calculate if we neglected loss (i.e., assumed σ = 0) – times a correction factor that accounts for the loss. This correction factor is complex-valued; therefore, E and H are not in phase when propagating through lossy material. We now see that in the phasor domain: eH = 1 ηc ˆk × eE (3.95) eE = −ηcˆk × eH (3.96) The plane wave relationships in media which are possibly lossy are given by Equations 3.95 and 3.96, with the complex-valued wave impedance given by Equation 3.94. 3.7 Wave Power in a Lossy Medium [m0133] In this section, we consider the power associated with waves propagating in materials which are potentially lossy; i.e., having conductivity σ signiﬁcantly greater than zero. This topic has previously been considered in the section “Wave Power in a Lossless Medium” for the case in loss is not signiﬁcant.4 A review of that section may be useful before reading this section. Recall that the Poynting vector S ≜E × H (3.97) indicates the power density (i.e., W/m2) of a wave and the direction of power ﬂow. This is “instantaneous” power, applicable to waves regardless of the way they vary with time. Often we are interested speciﬁcally in waves which vary sinusoidally, and which subsequently may be represented as phasors. In this case, the time-average Poynting vector is Save ≜1 2Re n eE × eH∗o (3.98) Further, we have already used this expression to ﬁnd that the time-average power density for a sinusoidally-varying uniform plane wave in a lossless medium is simply Save = \|E0\|2 2η (lossless case) (3.99) where \|E0\| is the peak (as opposed to RMS) magnitude of the electric ﬁeld intensity phasor, and η is the wave impedance. Let us now use Equation 3.98 to determine the expression corresponding to Equation 3.99 in the case of possibly-lossy media. We may express the electric and magnetic ﬁeld intensities of a uniform plane wave as eE = ˆxE0e−αze−jβz (3.100) and eH = ˆyE0 ηc e−αze−jβz (3.101) 4Depending on the version of this book, this section may appear in another volume.	Electromagnetics_Vol2_Page_50_Chunk2225
38 CHAPTER 3. WAVE PROPAGATION IN GENERAL MEDIA where α and β are the attenuation constant and phase propagation constant, respectively, and ηc is the complex-valued wave impedance. As written, these expressions describe a wave which is +ˆx-polarized and propagates in the +ˆz direction. We make these choices for convenience only – as long as the medium is homogeneous and isotropic, we expect our ﬁndings to apply regardless of polarization and direction of propagation. Applying Equation 3.98: Save = 1 2Re n eE × eH∗o = 1 2 ˆz Re ( \|E0\|2 η∗c e−2αz ) = ˆz\|E0\|2 2 Re 1 η∗c e−2αz (3.102) Because ηc is complex-valued when the material is lossy, we must proceed with caution. First, let us write ηc explicitly in terms of its magnitude \|ηc\| and phase ψη: η ≜\|η\| ejψη (3.103) Then: η∗ c = \|ηc\| e−jψη (3.104) (η∗ c)−1 = \|ηc\|−1 e+jψη (3.105) Re n (η∗ c)−1o = \|ηc\|−1 cos ψη (3.106) Then Equation 3.102 may be written: Save = ˆz\|E0\|2 2 \|ηc\| e−2αz cos ψη (3.107) The time-average power density of the plane wave described by Equation 3.100 in a possibly- lossy material is given by Equation 3.107. As a check, note that this expression gives the expected result for lossless media; i.e., α = 0, \|ηc\| = η, and ψη = 0. We now see that the effect of loss is that power density is now proportional to (e−αz)2, so, as expected, power density is proportional to the square of either \|E\| or \|H\|. The result further indicates a one-time scaling of the power density by a factor of \|η\| \|ηc\| cos ψη < 1 (3.108) relative to a medium without loss. The reduction in power density due to non- zero conductivity is proportional to a distance- dependent factor e−2αz and an additional factor that depends on the magnitude and phase of ηc.	Electromagnetics_Vol2_Page_51_Chunk2226
3.8. DECIBEL SCALE FOR POWER RATIO 39 3.8 Decibel Scale for Power Ratio [m0154] In many disciplines within electrical engineering, it is common to evaluate the ratios of powers and power densities that differ by many orders of magnitude. These ratios could be expressed in scientiﬁc notation, but it is more common to use the logarithmic decibel (dB) scale in such applications. In the conventional (linear) scale, the ratio of power P1 to power P0 is simply G = P1 P0 (linear units) (3.109) Here, “G” might be interpreted as “power gain.” Note that G < 1 if P1 < P0 and G > 1 if P1 > P0. In the decibel scale, the ratio of power P1 to power P0 is G ≜10 log10 P1 P0 (dB) (3.110) where “dB” denotes a unitless quantity which is expressed in the decibel scale. Note that G < 0 dB (i.e., is “negative in dB”) if P1 < P0 and G > 0 dB if P1 > P0. The power gain P1/P0 in dB is given by Equa- tion 3.110. Alternatively, one might choose to interpret a power ratio as a loss L with L ≜1/G in linear units, which is L = −G when expressed in dB. Most often, but not always, engineers interpret a power ratio as “gain” if the output power is expected to be greater than input power (e.g., as expected for an ampliﬁer) and as “loss” if output power is expected to be less than input power (e.g., as expected for a lossy transmission line). Power loss L is the reciprocal of power gain G. Therefore, L = −G when these quantities are expressed in dB. Example 3.1. Power loss from a long cable. A 2 W signal is injected into a long cable. The power arriving at the other end of the cable is 10 µW. What is the power loss in dB? In linear units: G = 10 µW 2 W = 5 × 10−6 (linear units) In dB: G = 10 log10	Electromagnetics_Vol2_Page_52_Chunk2227
40 CHAPTER 3. WAVE PROPAGATION IN GENERAL MEDIA However, note that this is not true if R1 ̸= R0. A power ratio in dB is equal to 20 log10 of the voltage ratio only if the associated impedances are equal. Adding to the potential for confusion on this point is the concept of voltage gain Gv: Gv ≜20 log10 V1 V0 (dB) (3.114) which applies regardless of the associated impedances. Note that Gv = G only if the associated impedances are equal, and that these ratios are different otherwise. Be careful! The decibel scale simpliﬁes common calculations. Here’s an example. Let’s say a signal having power P0 is injected into a transmission line having loss L. Then the output power P1 = P0/L in linear units. However, in dB, we ﬁnd: 10 log10 P1 = 10 log10 P0 L = 10 log10 P0 −10 log10 L Division has been transformed into subtraction; i.e., P1 = P0 −L (dB) (3.115) This form facilitates easier calculation and visualization, and so is typically preferred. Finally, note that the units of P1 and P0 in Equation 3.115 are not dB per se, but rather dB with respect to the original power units. For example, if P1 is in mW, then taking 10 log10 of this quantity results in a quantity having units of dB relative to 1 mW. A power expressed in dB relative to 1 mW is said to have units of “dBm.” For example, “0 dBm” means 0 dB relative to 1 mW, which is simply 1 mW. Similarly +10 dBm is 10 mW, −10 dBm is 0.1 mW, and so on. Additional Reading: • “Decibel” on Wikipedia. • “Scientiﬁc notation” on Wikipedia. 3.9 Attenuation Rate [m0155] Attenuation rate is a convenient way to quantify loss in general media, including transmission lines, using the decibel scale. Consider a transmission line carrying a wave in the +z direction. Let P0 be the power at z = 0. Let P1 be the power at z = l. Then the power at z = 0 relative to the power at z = l is: P0 P1 = e−2α·0 e−2α·l = e2αl (linear units) (3.116) where α is the attenuation constant; that is, the real part of the propagation constant γ = α + jβ. Expressed in this manner, the power ratio is a loss; that is, a number greater than 1 represents attenuation. In the decibel scale, the loss is 10 log10 P0 P1 = 10 log10 e2αl = 20αl log10 e ∼= 8.69αl dB (3.117) Attenuation rate is deﬁned as this quantity per unit length. Dividing by l, we obtain: attenuation rate ∼= 8.69α (3.118) This has units of dB/length, where the units of length are the same length units in which α is expressed. For example, if α is expressed in units of m−1, then attenuation rate has units of dB/m. Attenuation rate ∼= 8.69α is the loss in dB, per unit length. The utility of the attenuation rate concept is that it allows us to quickly calculate loss for any distance of wave travel: This loss is simply attenuation rate (dB/m) times length (m), which yields loss in dB. Example 3.2. Attenuation rate in a long cable. A particular coaxial cable has an attenuation constant α ∼= 8.5 × 10−3 m−1. What is the attenuation rate and the loss in dB for 100 m of this cable?	Electromagnetics_Vol2_Page_53_Chunk2228
3.10. POOR CONDUCTORS 41 The attenuation rate is ∼= 8.69α ∼= 0.0738 dB/m The loss in 100 m of this cable is ∼= (0.0738 dB/m) (100 m) ∼= 7.4 dB Note that it would be entirely appropriate, and equivalent, to state that the attenuation rate for this cable is 7.4 dB/(100 m). The concept of attenuation rate is used in precisely the same way to relate ratios of spatial power densities for unguided waves. This works because spatial power density has SI base units of W/m2, so the common units of m−2 in the numerator and denominator cancel in the power density ratio, leaving a simple power ratio. 3.10 Poor Conductors [m0156] A poor conductor is a material for which conductivity is low, yet sufﬁcient to exhibit signiﬁcant loss. To be clear, the loss we refer to here is the conversion of the electric ﬁeld to current through Ohm’s law. The threshold of signiﬁcance depends on the application. For example, the dielectric spacer separating the conductors in a coaxial cable might be treated as lossless (σ = 0) for short lengths at low frequencies; whereas the loss of the cable for long lengths and higher frequencies is typically signiﬁcant, and must be taken into account. In the latter case, the material is said to be a poor conductor because the loss is signiﬁcant yet the material can still be treated in most other respects as an ideal dielectric. A quantitative but approximate criterion for identiﬁcation of a poor conductor can be obtained from the concept of complex permittivity ǫc, which has the form: ǫc = ǫ′ −jǫ′′ (3.119) Recall that ǫ′′ quantiﬁes loss, whereas ǫ′ exists independently of loss. In fact, ǫc = ǫ′ = ǫ for a perfectly lossless material. Therefore, we may quantify the lossiness of a material using the ratio ǫ′′/ǫ′, which is sometimes referred to as loss tangent (see Section 3.5). Using this quantity, we deﬁne a poor conductor as a material for which ǫ′′ is very small relative to ǫ′. Thus, ǫ′′ ǫ′ ≪1 (poor conductor) (3.120) A poor conductor is a material having loss tan- gent much less than 1, such that it behaves in most respects as an ideal dielectric except that ohmic loss may not be negligible. An example of a poor conductor commonly encountered in electrical engineering includes the popular printed circuit board substrate material FR4 (ﬁberglass epoxy), which has ǫ′′/ǫ′ ∼0.008 over the frequency range it is most commonly used. Another example, already mentioned, is the dielectric spacer	Electromagnetics_Vol2_Page_54_Chunk2229
42 CHAPTER 3. WAVE PROPAGATION IN GENERAL MEDIA material (for example, polyethylene) typically used in coaxial cables. The loss of these materials may or may not be signiﬁcant, depending on the particulars of the application. The imprecise deﬁnition of Equation 3.120 is sufﬁcient to derive some characteristics exhibited by all poor conductors. To do so, ﬁrst recall that the propagation constant γ is given in general as follows: γ2 = −ω2µǫc (3.121) Therefore: γ = p −ω2µǫc (3.122) In general a number has two square roots, so some caution is required here. In this case, we may proceed as follows: γ = jω√µ p ǫ′ −jǫ′′ = jω p µǫ′ r 1 −j ǫ′′ ǫ′ (3.123) The requirement that ǫ′′/ǫ′ ≪1 for a poor conductor allows this expression to be “linearized.” For this, we invoke the binomial series representation: (1 + x)n = 1 + nx + n(n −1) 2! x2 + ... (3.124) where x and n are, for our purposes, any constants; and “...” indicates the remaining terms in this inﬁnite series, with each term containing the factor xn with n > 2. If x ≪1, then all terms containing xn with n ≥2 will be very small relative to the ﬁrst two terms of the series. Thus, (1 + x)n ≈1 + nx for x ≪1 (3.125) Applying this to the present problem: 1 −j ǫ′′ ǫ′ 1/2 ≈1 −j ǫ′′ 2ǫ′ (3.126) where we have used n = 1/2 and x = −jǫ′′/ǫ′. Applying this approximation to Equation 3.123, we obtain: γ ≈jω p µǫ′ 1 −j ǫ′′ 2ǫ′ ≈jω p µǫ′ + ω p µǫ′ ǫ′′ 2ǫ′ (3.127) At this point, we are able to identify an expression for the phase propagation constant: β ≜Im {γ} ≈ω p µǫ′ (poor conductor) (3.128) Remarkably, we ﬁnd that β for a poor conductor is approximately equal to β for an ideal dielectric. For the attenuation constant, we ﬁnd α ≜Re {γ} ≈ω p µǫ′ ǫ′′ 2ǫ′ (poor conductor) (3.129) Alternatively, this expression may be written in the following form: α ≈1 2β ǫ′′ ǫ′ (poor conductor) (3.130) Presuming that ǫc is determined entirely by ohmic loss, then ǫ′′ ǫ′ = σ ωǫ (3.131) Under this condition, Equation 3.129 may be rewritten: α ≈ω p µǫ′ σ 2ωǫ (poor conductor) (3.132) Since ǫ′ = ǫ under these assumptions, the expression simpliﬁes to α ≈σ 2 rµ ǫ = 1 2ση (poor conductor) (3.133) where η ≜ p µ/ǫ′ is the wave impedance presuming lossless material. This result is remarkable for two reasons: First, factors of ω have been eliminated, so there is no dependence on frequency separate from the frequency dependence of the constitutive parameters σ, µ, and ǫ. These parameters vary slowly with frequency, so the value of α for a poor conductor also varies slowly with frequency. Second, we see α is proportional to σ and η. This makes it quite easy to anticipate how the attenuation constant is affected by changes in conductivity and wave impedance in poor conductors. Finally, what is the wave impedance in a poor conductor? In contrast to η, ηc is potentially complex-valued and may depend on σ. First, recall: ηc = r µ ǫ′ · 1 −j ǫ′′ ǫ′ −1/2 (3.134)	Electromagnetics_Vol2_Page_55_Chunk2230
3.11. GOOD CONDUCTORS 43 Applying the same approximation applied to γ earlier, this may be written ηc ≈ r µ ǫ′ · 1 −j ǫ′′ 2ǫ′ (poor conductor) (3.135) We see that for a poor conductor, Re{ηc} ≈η and that Im{ηc} ≪Re {ηc}. The usual approximation in this case is simply ηc ≈η (poor conductor) (3.136) Additional Reading: • “Binomial series” on Wikipedia. 3.11 Good Conductors [m0157] A good conductor is a material which behaves in most respects as a perfect conductor, yet exhibits signiﬁcant loss. Now, we have to be very careful: The term “loss” applied to the concept of a “conductor” means something quite different from the term “loss” applied to other types of materials. Let us take a moment to disambiguate this term. Conductors are materials which are intended to efﬁciently sustain current, which requires high conductivity σ. In contrast, non-conductors are materials which are intended to efﬁciently sustain the electric ﬁeld, which requires low σ. “Loss” for a non-conductor (see in particular “poor conductors,” Section 3.10) means the conversion of energy in the electric ﬁeld into current. In contrast, “loss” for a conductor refers to energy already associated with current, which is subsequently dissipated in resistance. Summarizing: A good (“low-loss”) conductor is a material with high conductivity, such that power dissipated in the resistance of the material is low. A quantitative criterion for a good conductor can be obtained from the concept of complex permittivity ǫc, which has the form: ǫc = ǫ′ −jǫ′′ (3.137) Recall that ǫ′′ is proportional to conductivity (σ) and so ǫ′′ is very large for a good conductor. Therefore, we may identify a good conductor using the ratio ǫ′′/ǫ′, which is sometimes referred to as “loss tangent” (see Section 3.5). Using this quantity we deﬁne a good conductor as a material for which: ǫ′′ ǫ′ ≫1 (good conductor) (3.138) This condition is met for most materials classiﬁed as “metals,” and especially for metals exhibiting very high conductivity such as gold, copper, and aluminum. A good conductor is a material having loss tan- gent much greater than 1.	Electromagnetics_Vol2_Page_56_Chunk2231
44 CHAPTER 3. WAVE PROPAGATION IN GENERAL MEDIA The imprecise deﬁnition of Equation 3.138 is sufﬁcient to derive some characteristics that are common to materials over a wide range of conductivity. To derive these characteristics, ﬁrst recall that the propagation constant γ is given in general as follows: γ2 = −ω2µǫc (3.139) Therefore: γ = p −ω2µǫc (3.140) In general, a number has two square roots, so some caution is required here. In this case, we may proceed as follows: γ = jω√µ p ǫ′ −jǫ′′ = jω p µǫ′ r 1 −j ǫ′′ ǫ′ (3.141) Since ǫ′′/ǫ′ ≫1 for a good conductor, γ ≈jω p µǫ′ r −j ǫ′′ ǫ′ ≈jω p µǫ′′p −j (3.142) To proceed, we must determine the principal value of √−j. The answer is that √−j = (1 −j) / √ 2.5 Continuing: γ ≈jω r µǫ′′ 2 + ω r µǫ′′ 2 (3.143) We are now able to identify expressions for the attenuation and phase propagation constants: α ≜Re {γ} ≈ω r µǫ′′ 2 (good conductor) (3.144) β ≜Im {γ} ≈α (good conductor) (3.145) Remarkably, we ﬁnd that α ≈β for a good conductor, and neither α nor β depend on ǫ′. In the special case that ǫc is determined entirely by conductivity loss (i.e., σ > 0) and is not accounting 5You can conﬁrm this simply by squaring this result. The easiest way to derive this result is to work in polar form, in which −j is 1 at an angle of −π/2, and the square root operation consists of taking the square root of the magnitude and dividing the phase by 2. for delayed polarization response (as described in Section 3.4), then ǫ′′ = σ ω (3.146) Under this condition, Equation 3.144 may be rewritten: α ≈ω rµσ 2ω = rωµσ 2 (good conductor) (3.147) Since ω = 2πf, another possible form for this expression is α ≈ p πfµσ (good conductor) (3.148) The conductivity of most materials changes very slowly with frequency, so this expression indicates that α (and β) increases approximately in proportion to the square root of frequency for good conductors. This is commonly observed in electrical engineering applications. For example, the attenuation rate of transmission lines increases approximately as √f. This is so because the principal contribution to the attenuation is resistance in the conductors comprising the line. The attenuation rate for signals conveyed by transmission lines is approximately proportional to the square root of frequency. Let us now consider the wave impedance ηc in a good conductor. Recall: ηc = r µ ǫ′ · 1 −j ǫ′′ ǫ′ −1/2 (3.149) Applying the same approximation applied to γ earlier in this section, the previous expression may be written ηc ≈ r µ ǫ′ · −j ǫ′′ ǫ′ −1/2 (good conductor) ≈ r µ ǫ′′ · 1 √−j (3.150) We’ve already established that √−j = (1 −j) / √ 2. Applying that result here: ηc ≈ r µ ǫ′′ · √ 2 1 −j (3.151)	Electromagnetics_Vol2_Page_57_Chunk2232
3.11. GOOD CONDUCTORS 45 Now multiplying numerator and denominator by 1 + j, we obtain ηc ≈ r µ 2ǫ′′ · (1 + j) (3.152) In the special case that ǫc is determined entirely by conductivity loss and is not accounting for delayed polarization response, then ǫ′′ = σ/ω, and we ﬁnd: ηc ≈ rµω 2σ · (1 + j) (3.153) There are at least two other ways in which this expression is commonly written. First, we can use ω = 2πf to obtain: ηc ≈ r πfµ σ · (1 + j) (3.154) Second, we can use the fact that α ≈√πfµσ for good conductors to obtain: ηc ≈α σ · (1 + j) (3.155) In any event, we see that the magnitude of the wave impedance bears little resemblance to the wave impedance for a poor conductor. In particular, there is no dependence on the physical permittivity ǫ′ = ǫ, as we saw also for α and β. In this sense, the concept of permittivity does not apply to good conductors, and especially so for perfect conductors. Note also that ψη, the phase of ηc, is always ≈π/4 for a good conductor, in contrast to ≈0 for a poor conductor. This has two implications that are useful to know. First, since ηc is the ratio of the magnitude of the electric ﬁeld intensity to the magnitude of the magnetic ﬁeld intensity, the phase of the magnetic ﬁeld will be shifted by ≈π/4 relative to the phase of the electric ﬁeld in a good conductor. Second, recall from Section 3.7 that the power density for a wave is proportional to cos ψη. Therefore, the extent to which a good conductor is able to “extinguish” a wave propagating through it is determined entirely by α, and speciﬁcally is proportional to e−αl where l is distance traveled through the material. In other words, only a perfect conductor (σ →∞) is able to completely suppress wave propagation, whereas waves are always able to penetrate some distance into any conductor which is merely “good.” A measure of this distance is the skin depth of the material. The concept of skin depth is presented in Section 3.12. The dependence of β on conductivity leads to a particularly surprising result for the phase velocity of the beleaguered waves that do manage to propagate within a good conductor. Recall that for both lossless and low-loss (“poor conductor”) materials, the phase velocity vp is either exactly or approximately c/√ǫr, where ǫr ≜ǫ′/ǫ0, resulting in typical phase velocities within one order of magnitude of c. For a good conductor, we ﬁnd instead: vp = ω β ≈ ω √πfµσ (good conductor) (3.156) and since ω = 2πf: vp ≈ s 4πf µσ (good conductor) (3.157) Note that the phase velocity in a good conductor increases with frequency and decreases with conductivity. In contrast to poor conductors and non-conductors, the phase velocity in good conductors is usually a tiny fraction of c. For example, for a non-magnetic (µ ≈µ0) good conductor with typical σ ∼106 S/m, we ﬁnd vp ∼100 km/s at 1 GHz – just ∼0.03% of the speed of light in free space. This result also tells us something profound about the nature of signals that are conveyed by transmission lines. Regardless of whether we analyze such signals as voltage and current waves associated with the conductors or in terms of guided waves between the conductors, we ﬁnd that the phase velocity is within an order of magnitude or so of c. Thus, the information conveyed by signals propagating along transmission lines travels primarily within the space between the conductors, and not within the conductors. Information cannot travel primarily in the conductors, as this would then result in apparent phase velocity which is orders of magnitude less than c, as noted previously. Remarkably, classical transmission line theory employing the R′, G′, C′, L′ equivalent circuit model6 gets this right, even though that approach does not explicitly consider the possibility of guided waves traveling between the conductors. 6Depending on the version of this book, this topic may appear in another volume.	Electromagnetics_Vol2_Page_58_Chunk2233
46 CHAPTER 3. WAVE PROPAGATION IN GENERAL MEDIA 3.12 Skin Depth [m0158] The electric and magnetic ﬁelds of a wave are diminished as the wave propagates through lossy media. The magnitude of these ﬁelds is proportional to e−αl where α ≜Re {γ} is the attenuation constant (SI base units of m−1), γ is the propagation constant, and l is the distance traveled. Although the rate at which magnitude is reduced is completely described by α, particular values of α typically do not necessarily provide an intuitive sense of this rate. An alternative way to characterize attenuation is in terms of skin depth δs, which is deﬁned as the distance at which the magnitude of the electric and magnetic ﬁelds is reduced by a factor of 1/e. In other words: e−αδs = e−1 ∼= 0.368 (3.158) Skin depth δs is the distance over which the mag- nitude of the electric or magnetic ﬁeld is reduced by a factor of 1/e ∼= 0.368. Since power is proportional to the square of the ﬁeld magnitude, δs may also be interpreted as the distance at which the power in the wave is reduced by a factor of (1/e)2 ∼= 0.135. In yet other words: δs is the distance at which ∼= 86.5% of the power in the wave is lost. This deﬁnition for skin depth makes δs easy to compute: From Equation 3.158, it is simply δs = 1 α (3.159) The concept of skin depth is most commonly applied to good conductors. For a good conductor, α ≈√πfµσ (Section 3.11), so δs ≈ 1 √πfµσ (good conductors) (3.160) Example 3.3. Skin depth of aluminum. Aluminum, a good conductor, exhibits σ ≈3.7 × 107 S/m and µ ≈µ0 over a broad range of radio frequencies. Using Equation 3.160, we ﬁnd δs ∼26 µm at 10 MHz. Aluminum sheet which is 1/16-in (∼= 1.59 mm) thick can also be said to have a thickness of ∼61δs at 10 MHz. The reduction in the power density of an electromagnetic wave after traveling through this sheet will be ∼ e−α(61δs)2 = e−α(61/α)2 = e−122 which is effectively zero from a practical engineering perspective. Therefore, 1/16-in aluminum sheet provides excellent shielding from electromagnetic waves at 10 MHz. At 1 kHz, the situation is signiﬁcantly different. At this frequency, δs ∼2.6 mm, so 1/16-in aluminum is only ∼0.6δs thick. In this case the power density is reduced by only ∼ e−α(0.6δs)2 = e−α(0.6/α)2 = e−1.2 ≈0.3 This is a reduction of only ∼70% in power density. Therefore, 1/16-in aluminum sheet provides very little shielding at 1 kHz. [m0183]	Electromagnetics_Vol2_Page_59_Chunk2234
3.12. SKIN DEPTH 47 Image Credits Fig. 3.1: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Poynting%E2%80%99s theorem illustration.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 3.2: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Uniform plane wave incident cylindrical surface.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 3.3: c⃝K.A. Mauritz, https://commons.wikimedia.org/wiki/File:Dielectric responses.svg, Used with permission (see URL) and modiﬁed by author. Fig. 3.4: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Total current in phasor domain.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 3.5: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Loss tangent deﬁnition.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).	Electromagnetics_Vol2_Page_60_Chunk2235
Chapter 4 Current Flow in Imperfect Conductors 4.1 AC Current Flow in a Good Conductor [m0069] In this section, we consider the distribution of current in a conductor which is imperfect (i.e., a “good conductor”) and at frequencies greater than DC. To establish context, consider the simple DC circuit shown in Figure 4.1. In this circuit, the current source provides a steady current which ﬂows through a cylinder-shaped wire. As long as the conductivity σ (SI base units of S/m) of the wire is uniform throughout the wire, the current density J (SI base units of A/m2) is uniform throughout the wire. Now let us consider the AC case. Whereas the electric ﬁeld intensity E is constant in the DC case, E exists as a wave in the AC case. In a good conductor, the c⃝C. Wang CC BY-SA 4.0 Figure 4.1: Current ﬂow in cylinder at DC. magnitude of E decreases in proportion to e−αd where α is the attenuation constant and d is distance traversed by the wave. The attenuation constant increases with increasing σ, so the rate of decrease of the magnitude of E increases with increasing σ. In the limiting case of a perfect conductor, α →∞ and so E →0 everywhere inside the material. Any current within the wire must be the result of either an impressed source or it must be a response to E. Without either of these, we conclude that in the AC case, J →0 everywhere inside a perfect conductor.1 But if J = 0 in the material, then how does current pass through the wire? We are forced to conclude that the current must exist as a surface current; i.e., entirely outside the wire, yet bound to the surface of the wire. Thus: In the AC case, the current passed by a perfectly- conducting material lies entirely on the surface of the material. The perfectly-conducting case is unobtainable in practice, but the result gives us a foothold from which we may determine what happens when σ is not inﬁnite. If σ is merely ﬁnite, then α is also ﬁnite and subsequently wave magnitude may be non-zero over ﬁnite distances. Let us now consider the direction in which this putative wave propagates. The two principal directions in the present problem are parallel to the 1You might be tempted to invoke Ohm’s law (J = σE) to argue against this conclusion. However, Ohm’s law provides no useful information about the current in this case, since σ →∞at the same time E →0. What Ohm’s law is really saying in this case is that E = J/σ →0 because J must be ﬁnite and σ →∞. Electromagnetics Vol. 2. c⃝2020 S.W. Ellingson CC BY SA 4.0. https://doi.org/10.21061/electromagnetics-vol-2	Electromagnetics_Vol2_Page_61_Chunk2236
4.1. AC CURRENT FLOW IN A GOOD CONDUCTOR 49 axis of the wire and perpendicular to the axis of the wire. Waves propagating in any other direction may be expressed as a linear combination of waves traveling in the principal directions, so we need only consider the principal directions to obtain a complete picture. Consider waves propagating in the perpendicular direction ﬁrst. In this case, we presume a boundary condition in the form of non-zero surface current, which we infer from the perfectly-conducting case considered previously. We also note that E deep inside the wire must be weaker than E closer to the surface, since a wave deep inside the wire must have traversed a larger quantity of material than a wave measured closer to the surface. Applying Ohm’s law (J = σE), the current deep inside the wire must similarly diminish for a good conductor. We conclude that a wave traveling in the perpendicular direction exists and propagates toward the center of the wire, decreasing in magnitude with increasing distance from the surface. We are unable to infer the presence of a wave traveling in the other principal direction – i.e., along the axis of the wire – since there is no apparent boundary condition to be satisﬁed on either end of the wire. Furthermore, the presence of such a wave would mean that different cross-sections of the wire exhibit different radial distributions of current. This is not consistent with physical observations. We conclude that the only relevant wave is one which travels from the surface of the wire inward. Since the current density is proportional to the electric ﬁeld magnitude, we conclude: In the AC case, the current passed by a wire com- prised of a good conductor is distributed with maximum current density on the surface of the wire, and the current density decays exponen- tially with increasing distance from the surface. This phenomenon is known as the skin effect, referring to the notion of current forming a skin-like layer below the surface of the wire. The effect is illustrated in Figure 4.2. Since α increases with increasing frequency, we see that the speciﬁc distribution of current within the wire by Biezl (modiﬁed) (public domain) Figure 4.2: Distribution of AC current in a wire of cir- cular cross-section. Shading indicates current density. depends on frequency. In particular, the current will be concentrated close to the surface at high frequencies, uniformly distributed throughout the wire at DC, and in an intermediate state for intermediate frequencies. Additional Reading: • “Skin effect” on Wikipedia.	Electromagnetics_Vol2_Page_62_Chunk2237
50 CHAPTER 4. CURRENT FLOW IN IMPERFECT CONDUCTORS 4.2 Impedance of a Wire [m0159] The goal of this section is to determine the impedance – the ratio of potential to current – of a wire. The answer to this question is relatively simple in the DC (“steady current”) case: The impedance is found to be equal to the resistance of the wire, which is given by R = l σA (DC) (4.1) where l is the length of the wire and A is the cross-sectional area of the wire. Also, the impedance of a wire comprised of a perfect conductor at any frequency is simply zero, since there is no mechanism in the wire that can dissipate or store energy in this case. However, all practical wires are comprised of good – not perfect – conductors, and of course many practical signals are time-varying, so the two cases above do not address a broad category of practical interest. The more general case of non-steady currents in imperfect conductors is complicated by the fact that the current in an imperfect conductor is not uniformly distributed in the wire, but rather is concentrated near the surface and decays exponentially with increasing distance from the surface (this is determined in Section 4.1). We are now ready to consider the AC case for a wire comprised of a good but imperfect conductor. What is the impedance of the wire if the current source is sinusoidally-varying? Equation 4.1 for the DC case was determined by ﬁrst obtaining expressions for the potential (V ) and net current (I) for a length-l section of wire in terms of the electric ﬁeld intensity E in the wire. To follow that same approach here, we require an expression for I in terms of E that accounts for the non-uniform distribution of current. This is quite difﬁcult to determine in the case of a cylindrical wire. However, we may develop an approximate solution by considering a surface which is not cylindrical, but rather planar. Here we go: Consider the experiment described in Figure 4.3. Here, a semi-inﬁnite region ﬁlled with a homogeneous good conductor meets a semi-inﬁnite region of free space along a planar interface at z = 0, with increasing z corresponding to increasing depth into the material. A plane wave propagates in the +ˆz direction, beginning from just inside the structure’s surface. The justiﬁcation for presuming the existence of this wave was presented in Section 4.1. The electric ﬁeld intensity is given by eE = ˆxE0e−αze−jβz (4.2) where E0 is an arbitrary complex-valued constant. The current density is given by Ohm’s law of electromagnetics:2 eJ = σeE = ˆxσE0e−αze−jβz (4.3) Recall that α = δ−1 s where δs is skin depth (see Section 3.12). Also, for a good conductor, we know that β ≈α (Section 3.11). Using these relationships, we may rewrite Equation 4.3 as follows: eJ ≈ˆxσE0e−z/δse−jz/δs = ˆxσE0e−(1+j)z/δs (4.4) The net current eI is obtained by integrating eJ over any cross-section S through which all the current ﬂows; i.e., eI = Z S eJ · ds (4.5) Here, the simplest solution is obtained by choosing S to be a rectangular surface that is perpendicular to the direction of eJ at x = 0. This is shown in Figure 4.4. c⃝C. Wang CC BY-SA 4.0 Figure 4.3: Experiment used to determine current density and resistance in the AC case.	Electromagnetics_Vol2_Page_63_Chunk2238
4.2. IMPEDANCE OF A WIRE 51 c⃝Y. Zhao CC BY-SA 4.0 Figure 4.4: Choice of S for calculating net current I. The dimensions of S are width W in the y dimension and extending to inﬁnity in the z direction. Then we have eI ≈ Z W y=0 Z ∞ z=0 ˆxσE0e−(1+j)z/δs · (ˆx dy dz) = σE0W Z ∞ z=0 e−(1+j)z/δsdz (4.6) For convenience, let us deﬁne the constant K ≜(1 + j)/δs. Since K is constant with respect to z, the remaining integral is straightforward to evaluate: Z ∞ 0 e−Kzdz = −1 K e−Kz ∞ 0 = + 1 K (4.7) Incorporating this result into Equation 4.6, we obtain: eI ≈σE0W δs 1 + j (4.8) We calculate eV for a length l of the wire as follows: eV = − Z 0 x=l eE · dl (4.9) 2To be clear, this is the “point form” of Ohm’s law, as opposed to the circuit theory form (V = IR). where we have determined that x = 0 corresponds to the “+” terminal and x = l corresponds to the “−” terminal.3 The path of integration can be any path that begins and ends at the proscribed terminals. The simplest path to use is one along the surface, parallel to the x axis. Along this path, z = 0 and thus eE = ˆxE0. For this path: eV = − Z 0 x=l (ˆxE0) · (ˆxdx) = E0l (4.10) The impedance Z measured across terminals at x = 0 and x = l is now determined to be: Z ≜ eV eI ≈1 + j σδs · l W (4.11) The resistance is simply the real part, so we obtain R ≈ l σ(δsW) (AC case) (4.12) The quantity R in this case is referred to speciﬁcally as the ohmic resistance, since it is due entirely to the limited conductivity of the material as quantiﬁed by Ohm’s law.4 Note the resemblance to Equation 4.1 (the solution for the DC case): In the AC case, the product δsW, having units of area, plays the role of the physical cross-section S. Thus, we see an interesting new interpretation of the skin depth δs: It is the depth to which a uniform (DC) current would need to ﬂow in order to produce a resistance equal to the observed (AC) resistance. Equations 4.11 and 4.12 were obtained for a good conductor ﬁlling an inﬁnite half-space, having a ﬂat surface. How well do these results describe a cylindrical wire? The answer depends on the radius of the wire, a. For δs ≪a, Equations 4.11 and 4.12 are excellent approximations, since δs ≪a implies that most of the current lies in a thin shell close to the surface of the wire. In this case, the model used to develop the equations is a good approximation for any given radial slice through the wire, and we are 3If this is not clear, recall that the electric ﬁeld vector must point away from positive charge (thus, the + terminal). 4This is in contrast to other ways that voltage and current can be related; for example, the non-linear V -I characteristic of a diode, which is not governed by Ohm’s law.	Electromagnetics_Vol2_Page_64_Chunk2239
52 CHAPTER 4. CURRENT FLOW IN IMPERFECT CONDUCTORS justiﬁed in replacing W with the circumference 2πa. Thus, we obtain the following expressions: Z ≈1 + j σδs · l 2πa (δs ≪a) (4.13) and so R ≈ l σ(δs2πa) (δs ≪a) (4.14) The impedance of a wire of length l and radius a ≫δs is given by Equation 4.13. The resistance of such a wire is given by Equation 4.14. If, on the other hand, a < δs or merely ∼δs, then current density is signiﬁcant throughout the wire, including along the axis of the wire. In this case, we cannot assume that the current density decays smoothly to zero with increasing distance from the surface, and so the model leading to Equation 4.13 is a poor approximation. The frequency required for validity of Equation 4.13 can be determined by noting that δs ≈1/√πfµσ for a good conductor; therefore, we require 1 √πfµσ ≪a (4.15) for the derived expressions to be valid. Solving for f, we ﬁnd: f ≫ 1 πµσa2 (4.16) For commonly-encountered wires comprised of typical good conductors, this condition applies at frequencies in the MHz regime and above. These results lead us to one additional interesting ﬁnding about the AC resistance of wires. Since δs ≈1/√πfµσ for a good conductor, Equation 4.14 may be rewritten in the following form: R ≈1 2 r µf πσ · l a (4.17) We have found that R is approximately proportional to √f. For example, increasing frequency by a factor of 4 increases resistance by a factor of 2. This frequency dependence is evident in all kinds of practical wires and transmission lines. Summarizing: The resistance of a wire comprised of a good but imperfect conductor is proportional to the square root of frequency. At this point, we have determined that resistance is given approximately by Equation 4.17 for δs ≪a, corresponding to frequencies in the MHz regime and above, and by Equation 4.1 for δs ≫a, typically corresponding to frequencies in the kHz regime and below. We have also found that resistance changes slowly with frequency; i.e., in proportion to √f. Thus, it is often possible to roughly estimate resistance at frequencies between these two frequency regimes by comparing the DC resistance from Equation 4.1 to the AC resistance from Equation 4.17. An example follows. Example 4.1. Resistance of the inner conductor of RG-59. Elsewhere we have considered RG-59 coaxial cable (see the section “Coaxial Line,” which may appear in another volume depending on the version of this book). We noted that it was not possible to determine the AC resistance per unit length R′ for RG-59 from purely electrostatic and magnetostatic considerations. We are now able to consider the resistance per unit length of the inner conductor, which is a solid wire of the type considered in this section. Let us refer to this quantity as R′ ic. Note that R′ = R′ ic + R′ oc (4.18) where R′ oc is the resistance per unit length of the outer conductor. R′ oc remains a bit too complicated to address here. However, R′ ic is typically much greater than R′ oc, so R′ ∼R′ ic. That is, we get a pretty good idea of R′ for RG-59 by considering the inner conductor alone. The relevant parameters of the inner conductor are µ ≈µ0, σ ∼= 2.28 × 107 S/m, and	Electromagnetics_Vol2_Page_65_Chunk2240
4.2. IMPEDANCE OF A WIRE 53 a ∼= 0.292 mm. Using Equation 4.17, we ﬁnd: R′ ic ≜Ric l = 1 2 r µf πσ · 1 a ∼= 227 µΩ· m−1 · Hz−1/2 p f (4.19) Using Expression 4.16, we ﬁnd this is valid only for f ≫130 kHz. So, for example, we may be conﬁdent that R′ ic ≈0.82 Ω/m at 13 MHz. At the other extreme (f ≪130 kHz), Equation 4.1 (the DC resistance) is a better estimate. In this low frequency case, we estimate that R′ ic ≈0.16 Ω/m and is approximately constant with frequency. We now have a complete picture: As frequency is increased from DC to 13 MHz, we expect that R′ ic will increase monotonically from ≈0.16 Ω/m to ≈0.82 Ω/m, and will continue to increase in proportion to √f from that value. Returning to Equation 4.11, we see that resistance is not the whole story here. The impedance Z = R + jX also has a reactive component X equal to the resistance R; i.e., X ≈R ≈ l σ(δs2πa) (4.20) This is unique to good conductors at AC; that is, we see no such reactance at DC. Because this reactance is positive, it is often referred to as an inductance. However, this is misleading since inductance refers to the ability of a structure to store energy in a magnetic ﬁeld, and energy storage is decidedly not what is happening here. The similarity to inductance is simply that this reactance is positive, as is the reactance associated with inductance. As long as we keep this in mind, it is reasonable to model the reactance of the wire as an equivalent inductance: Leq ≈ 1 2πf · l σ(δs2πa) (4.21) Now substituting an expression for skin depth: Leq ≈ 1 2πf · r πfµ σ · l 2πa = 1 4π3/2 r µ σf · l a (4.22) for a wire having a circular cross-section with δs ≪a. The utility of this description is that it facilitates the modeling of wire reactance as an inductance in an equivalent circuit. Summarizing: A practical wire may be modeled using an equiv- alent circuit consisting of an ideal resistor (Equa- tion 4.17) in series with an ideal inductor (Equa- tion 4.22). Whereas resistance increases with the square root of frequency, inductance decreases with the square root of frequency. If the positive reactance of a wire is not due to physical inductance, then to what physical mechanism shall we attribute this effect? A wire has reactance because there is a phase shift between potential and current. This is apparent by comparing Equation 4.6 to Equation 4.10. This is the same phase shift that was found to exist between the electric and magnetic ﬁelds propagating in a good conductor, as explained in Section 3.11. Example 4.2. Equivalent inductance of the inner conductor of RG-59. Elsewhere in the book we worked out that the inductance per unit length L′ of RG-59 coaxial cable was about 370 nH/m. We calculated this from magnetostatic considerations, so the reactance associated with skin effect is not included in this estimate. Let’s see how L′ is affected by skin effect for the inner conductor. Using Equation 4.22 with µ = µ0, σ ∼= 2.28 × 107 S/m, and a ∼= 0.292 mm, we ﬁnd Leq ≈	Electromagnetics_Vol2_Page_66_Chunk2241
54 CHAPTER 4. CURRENT FLOW IN IMPERFECT CONDUCTORS associated with skin effect is as important as the magnetostatic inductance in the kHz regime, and becomes gradually less important with increasing frequency. Recall that the phase velocity in a low-loss transmission line is approximately 1/ √ L′C′. This means that skin effect causes the phase velocity in such lines to decrease with decreasing frequency. In other words: Skin effect in the conductors comprising com- mon transmission lines leads to a form of disper- sion in which higher frequencies travel faster than lower frequencies. This phenomenon is known as chromatic dispersion, or simply “dispersion,” and leads to signiﬁcant distortion for signals having large bandwidths. Additional Reading: • “Skin effect” on Wikipedia. 4.3 Surface Impedance [m0160] In Section 4.2, we derived the following expression for the impedance Z of a good conductor having width W, length l, and which is inﬁnitely deep: Z ≈1 + j σδs · l W (AC case) (4.25) where σ is conductivity (SI base units of S/m) and δs is skin depth. Note that δs and σ are constitutive parameters of material, and do not depend on geometry; whereas l and W describe geometry. With this in mind, we deﬁne the surface impedance ZS as follows: ZS ≜1 + j σδs (4.26) so that Z ≈ZS l W (4.27) Unlike the terminal impedance Z, ZS is strictly a materials property. In this way, it is like the intrinsic or “wave” impedance η, which is also a materials property. Although the units of ZS are those of impedance (i.e., ohms), surface impedance is usually indicated as having units of “Ω/□” (“ohms per square”) to prevent confusion with the terminal impedance. Summarizing: Surface impedance ZS (Equation 4.26) is a ma- terials property having units of Ω/□, and which characterizes the AC impedance of a material in- dependently of the length and width of the mate- rial. Surface impedance is often used to specify sheet materials used in the manufacture of electronic and semiconductor devices, where the real part of the surface impedance is more commonly known as the surface resistance or sheet resistance. Additional Reading: • “Sheet resistance” on Wikipedia. [m0184]	Electromagnetics_Vol2_Page_67_Chunk2242
4.3. SURFACE IMPEDANCE 55 Image Credits Fig. 4.1: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Current ﬂow in cylinder new.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 4.2: Biezl, https://commons.wikimedia.org/wiki/File:Skin depth.svg, public domain. Modiﬁed from original. Fig. 4.3: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Experiment for current density and resistance.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 4.4: c⃝YahuiZ (Y. Zhao), https://commons.wikimedia.org/wiki/File:Figure4.4.svg, CC BY SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Modiﬁed by author.	Electromagnetics_Vol2_Page_68_Chunk2243
Chapter 5 Wave Reﬂection and Transmission 5.1 Plane Waves at Normal Incidence on a Planar Boundary [m0161] When a plane wave encounters a discontinuity in media, reﬂection from the discontinuity and transmission into the second medium is possible. In this section, we consider the scenario illustrated in Figure 5.1: a uniform plane wave which is normally incident on the planar boundary between two semi-inﬁnite material regions. By “normally-incident” we mean the direction of propagation ˆk is perpendicular to the boundary. We shall assume that the media are “simple” and lossless (i.e., the imaginary component of permittivity ǫ′′ is equal to zero) and therefore the media are completely deﬁned by a real-valued permittivity and a real-valued permeability. Figure 5.1 shows the wave incident on the boundary, which is located at the z = 0 plane. The electric ﬁeld intensity eEi of this wave is given by eEi(z) = ˆxEi 0e−jβ1z , z ≤0 (5.1) where β1 = ω√µ1ǫ1 is the phase propagation constant in Region 1 and Ei 0 is a complex-valued constant. eEi serves as the “stimulus” in this problem. That is, all other contributions to the total ﬁeld may be expressed in terms of eEi. In fact, all other contributions to the total ﬁeld may be expressed in terms of Ei 0. From the plane wave relationships, we determine that the associated magnetic ﬁeld intensity is eHi(z) = ˆyEi 0 η1 e−jβ1z , z ≤0 (5.2) where η1 = p µ1/ǫ1 is the wave impedance in Region 1. The possibility of a reﬂected plane wave propagating in exactly the opposite direction is inferred from two pieces of evidence: The general solution to the wave equation, which includes terms corresponding to waves traveling in both +ˆz or −ˆz; and the geometrical symmetry of the problem, which precludes waves traveling in any other directions. The symmetry of the problem also precludes a change of polarization, so the reﬂected wave should have no ˆy component. Therefore, we may be conﬁdent that the reﬂected electric ﬁeld has the form eEr(z) = ˆxBe+jβ1z , z ≤0 (5.3) c⃝C. Wang CC BY-SA 4.0 Figure 5.1: A uniform plane wave normally incident on the planar boundary between two semi-inﬁnite ma- terial regions. Electromagnetics Vol. 2. c⃝2020 S.W. Ellingson CC BY SA 4.0. https://doi.org/10.21061/electromagnetics-vol-2	Electromagnetics_Vol2_Page_69_Chunk2244
5.1. PLANE WAVES AT NORMAL INCIDENCE ON A PLANAR BOUNDARY 57 where B is a complex-valued constant that remains to be determined. Since the direction of propagation for the reﬂected wave is −ˆz, we have from the plane wave relationships that eHr(z) = −ˆy B η1 e+jβ1z , z ≤0 (5.4) Similarly, we infer the existence of a “transmitted” plane wave propagating on the z > 0 side of the boundary. The symmetry of the problem precludes any direction of propagation other than +z, and with no possibility of a wave traveling in the −ˆz direction for z > 0. Therefore, we may be conﬁdent that the transmitted electric ﬁeld has the form: eEt(z) = ˆxCe−jβ2z , z ≥0 (5.5) and an associated magnetic ﬁeld having the form eHt(z) = ˆy C η2 e−jβ2z , z ≥0 (5.6) where β2 = ω√µ2ǫ2 and η2 = p µ2/ǫ2 are the phase propagation constant and wave impedance, respectively, in Region 2. The constant C, like B, is a complex-valued constant that remains to be determined. At this point, the only unknowns in this problem are the complex-valued constants B and C. Once these values are known, the problem is completely solved. These values can be determined by the application of boundary conditions at z = 0. First, recall that the tangential component of the total electric ﬁeld intensity must be continuous across a material boundary. To apply this boundary condition, let us deﬁne eE1 and eE2 to be the total electric ﬁeld intensities in Regions 1 and 2, respectively. The total ﬁeld in Region 1 is the sum of incident and reﬂected ﬁelds, so eE1(z) = eEi(z) + eEr(z) (5.7) The ﬁeld in Region 2 is simply eE2(z) = eEt(z) (5.8) Also, we note that all ﬁeld components are already tangent to the boundary. Thus, continuity of the tangential component of the electric ﬁeld across the boundary requires eE1(0) = eE2(0), and therefore eEi(0) + eEr(0) = eEt(0) (5.9) Now employing Equations 5.1, 5.3, and 5.5, we obtain: Ei 0 + B = C (5.10) Clearly a second equation is required to determine both B and C. This equation can be obtained by enforcing the boundary condition on the magnetic ﬁeld. Recall that any discontinuity in the tangential component of the total magnetic ﬁeld intensity must be supported by a current ﬂowing on the surface. There is no impressed current in this problem, and there is no reason to suspect a current will arise in response to the ﬁelds present in the problem. Therefore, the tangential components of the magnetic ﬁeld must be continuous across the boundary. This becomes the same boundary condition that we applied to the total electric ﬁeld intensity, so the remaining steps are the same. We deﬁne eH1 and eH2 to be the total magnetic ﬁeld intensities in Regions 1 and 2, respectively. The total ﬁeld in Region 1 is eH1(z) = eHi(z) + eHr(z) (5.11) The ﬁeld in Region 2 is simply eH2(z) = eHt(z) (5.12) The boundary condition requires eH1(0) = eH2(0), and therefore eHi(0) + eHr(0) = eHt(0) (5.13) Now employing Equations 5.2, 5.4, and 5.6, we obtain: Ei 0 η1 −B η1 = C η2 (5.14) Equations 5.10 and 5.14 constitute a linear system of two simultaneous equations with two unknowns. A straightforward method of solution is to ﬁrst eliminate C by substituting the left side of Equation 5.10 into Equation 5.14, and then to solve for B. One obtains: B = Γ12Ei 0 (5.15) where Γ12 ≜η2 −η1 η2 + η1 (5.16) Γ12 is known as a reﬂection coefﬁcient. The subscript “12” indicates that this coefﬁcient applies for	Electromagnetics_Vol2_Page_70_Chunk2245
58 CHAPTER 5. WAVE REFLECTION AND TRANSMISSION incidence from Region 1 toward Region 2. We may now solve for C by substituting Equation 5.15 into Equation 5.10. We ﬁnd: C = (1 + Γ12) Ei 0 (5.17) Now summarizing the solution: eEr(z) = ˆxΓ12Ei 0e+jβ1z , z ≤0 (5.18) eEt(z) = ˆx (1 + Γ12) Ei 0e−jβ2z , z ≥0 (5.19) Equations 5.18 and 5.19 are the reﬂected and transmitted ﬁelds, respectively, in response to the incident ﬁeld given in Equation 5.1 in the normal incidence scenario shown in Figure 5.1. Expressions for eHr and eHt may be obtained by applying the plane wave relationships to the preceding expressions. It is useful to check this solution by examining some special cases. First: If the material in Region 2 is identical to the material in Region 1, then there should be no reﬂection and eEt = eEi. In this case, η2 = η1, so Γ12 = 0, and we obtain the expected result. A second case of practical interest is when Region 2 is a perfect conductor. First, note that this may seem at ﬁrst glance to be a violation of the “lossless” assumption made at the beginning of this section. While it is true that we did not explicitly account for the possibility of a perfect conductor in Region 2, let’s see what the present analysis has to say about this case. If the material in Region 2 is a perfect conductor, then there should be no transmission since the electric ﬁeld is zero in a perfect conductor. In this case, η2 = 0 since the ratio of electric ﬁeld intensity to magnetic ﬁeld intensity is zero in Region 2, and subsequently Γ12 = −1 and 1 + Γ12 = 0. As expected, eEt is found to be zero, and the reﬂected electric ﬁeld experiences a sign change as required to enforce the boundary condition eEi(0) + eEr(0) = 0. Thus, we obtained the correct answer because we were able to independently determine that η2 = 0 in a perfect conductor. When Region 2 is a perfect conductor, the reﬂec- tion coefﬁcient Γ12 = −1 and the solution de- scribed in Equations 5.18 and 5.19 applies. It may be helpful to note the very strong analogy between reﬂection of the electric ﬁeld component of a plane wave from a planar boundary and the reﬂection of a voltage wave in a transmission line from a terminating impedance. In a transmission line, the voltage reﬂection coefﬁcient Γ is given by Γ = ZL −Z0 ZL + Z0 (5.20) where ZL is the load impedance and Z0 is the characteristic impedance of the transmission line. Comparing this to Equation 5.16, we see η1 is analogous to Z0 and η2 is analogous to ZL. Furthermore, we see the special case η2 = η1, considered previously, is analogous to a matched load, and the special case η2 = 0, also considered previously, is analogous to a short-circuit load. In the case of transmission lines, we were concerned about what fraction of the power was delivered into a load and what fraction of the power was reﬂected from a load. A similar question applies in the case of plane wave reﬂection. In this case, we are concerned about what fraction of the power density is transmitted into Region 2 and what fraction of the power density is reﬂected from the boundary. The time-average power density Si ave associated with the incident wave is: Si ave = \|Ei 0\|2 2η1 (5.21) presuming that Ei 0 is expressed in peak (as opposed to rms) units. Similarly, the time-average power density Sr ave associated with the reﬂected wave is: Sr ave = Γ12Ei 0 2 2η1 = \|Γ12\|2 Ei 0 2 2η1 = \|Γ12\|2 Si ave (5.22) From the principle of conservation of power, the power density St ave transmitted into Region 2 must be	Electromagnetics_Vol2_Page_71_Chunk2246
5.1. PLANE WAVES AT NORMAL INCIDENCE ON A PLANAR BOUNDARY 59 equal to the incident power density minus the reﬂected power density. Thus: St ave = Si ave −Sr ave =	Electromagnetics_Vol2_Page_72_Chunk2247
60 CHAPTER 5. WAVE REFLECTION AND TRANSMISSION 5.2 Plane Waves at Normal Incidence on a Material Slab [m0162] In Section 5.1, we considered what happens when a uniform plane wave is normally incident on the planar boundary between two semi-inﬁnite media. In this section, we consider the problem shown in Figure 5.2: a uniform plane wave normally incident on a “slab” sandwiched between two semi-inﬁnite media. This scenario arises in many practical engineering problems, including the design and analysis of ﬁlters and impedance-matching devices at RF and optical frequencies, the analysis of RF propagation through walls, and the design and analysis of radomes. The ﬁndings of Section 5.1 are an important stepping stone to the solution of this problem, so a review of Section 5.1 is recommended before reading this section. For consistency of terminology, let us refer to the problem considered in Section 5.1 as the “single-boundary” problem and the present (slab) problem as the “double-boundary” problem. Whereas there are only two regions (“Region 1” and “Region 2”) in the single-boundary problem, in the double-boundary problem there is a third region that we shall refer to as “Region 3.” We assume that the media comprising each slab are “simple” and lossless (i.e., the imaginary component of permittivity ǫ′′ is equal to zero) and therefore the media are completely deﬁned by a real-valued permittivity and a real-valued c⃝C. Wang CC BY-SA 4.0 Figure 5.2: A uniform plane wave normally incident on a slab. permeability. The boundary between Regions 1 and 2 is at z = −d, and the boundary between Regions 2 and 3 is at z = 0. Thus, the thickness of the slab is d. In both problems, we presume an incident wave in Region 1 incident on the boundary with Region 2 having electric ﬁeld intensity eEi(z) = ˆxEi 0e−jβ1z (Region 1) (5.28) where β1 = ω√µ1ǫ1 is the phase propagation constant in Region 1. eEi serves as the “stimulus” in this problem. That is, all other contributions to the total ﬁeld may be expressed in terms of eEi. From the plane wave relationships, we determine that the associated magnetic ﬁeld intensity is eHi(z) = ˆyEi 0 η1 e−jβ1z (Region 1) (5.29) where η1 = p µ1/ǫ1 is the wave impedance in Region 1. The symmetry arguments of the single-boundary problem apply in precisely the same way to the double-boundary problem. Therefore, we presume that the reﬂected electric ﬁeld intensity is: eEr(z) = ˆxBe+jβ1z (Region 1) (5.30) where B is a complex-valued constant that remains to be determined; and subsequently the reﬂected magnetic ﬁeld intensity is: eHr(z) = −ˆy B η1 e+jβ1z (Region 1) (5.31) Similarly, we infer the existence of a transmitted plane wave propagating in the +ˆz direction in Region 2. The electric and magnetic ﬁeld intensities of this wave are given by: eEt2(z) = ˆxCe−jβ2z (Region 2) (5.32) and an associated magnetic ﬁeld having the form: eHt2(z) = ˆy C η2 e−jβ2z (Region 2) (5.33) where β2 = ω√µ2ǫ2 and η2 = p µ2/ǫ2 are the phase propagation constant and wave impedance, respectively, in Region 2. The constant C, like B, is a	Electromagnetics_Vol2_Page_73_Chunk2248
5.2. PLANE WAVES AT NORMAL INCIDENCE ON A MATERIAL SLAB 61 complex-valued constant that remains to be determined. Now let us consider the boundary between Regions 2 and 3. Note that eEt2 is incident on this boundary in precisely the same manner as eEi is incident on the boundary between Regions 1 and 2. Therefore, we infer a reﬂected electric ﬁeld intensity in Region 2 as follows: eEr2(z) = ˆxDe+jβ2z (Region 2) (5.34) where D is a complex-valued constant that remains to be determined; and an associated magnetic ﬁeld eHr2(z) = −ˆy D η2 e+jβ2z (Region 2) (5.35) Subsequently, we infer a transmitted electric ﬁeld intensity in Region 3 as follows: eEt(z) = ˆxFe−jβ3z (Region 3) (5.36) and an associated magnetic ﬁeld having the form eHt(z) = ˆy F η3 e−jβ3z (Region 3) (5.37) where β3 = ω√µ3ǫ3 and η3 = p µ3/ǫ3 are the phase propagation constant and wave impedance, respectively, in Region 3. The constant F, like D, is a complex-valued constant that remains to be determined. We infer no wave traveling in the (−ˆz) in Region 3, just as we inferred no such wave in Region 2 of the single-boundary problem. For convenience, Table 5.1 shows a complete summary of the ﬁeld components we have just identiﬁed. Note that the total ﬁeld intensity in each region is the sum of the ﬁeld components in that region. For example, the total electric ﬁeld intensity in Region 2 is eEt2 + eEr2. Now observe that there are four unknown constants remaining; namely B, C, D, and F. The double-boundary problem is completely solved once we have expressions for these constants in terms of the “given” quantities in the problem statement. Solutions for the unknown constants can be obtained by enforcing boundary conditions on the electric and magnetic ﬁelds at each of the two boundaries. As in the single-boundary case, the relevant boundary condition is that the total ﬁeld should be continuous across each boundary. Applying this condition to the boundary at z = 0, we obtain: eEt2(0) + eEr2(0) = eEt(0) (5.38) eHt2(0) + eHr2(0) = eHt(0) (5.39) Applying this condition to the boundary at z = −d, we obtain: eEi(−d) + eEr(−d) = eEt2(−d) + eEr2(−d) (5.40) eHi(−d) + eHr(−d) = eEt2(−d) + eEr2(−d) (5.41) Making substitutions from Table 5.1 and dividing out common factors, Equation 5.38 becomes: C + D = F (5.42) Equation 5.39 becomes: C η2 −D η2 = F η3 (5.43) Equation 5.40 becomes: Ei 0e+jβ1d + Be−jβ1d = Ce+jβ2d + De−jβ2d (5.44) Equation 5.41 becomes: Ei 0 η1 e+jβ1d −B η1 e−jβ1d = C η2 e+jβ2d −D η2 e−jβ2d (5.45) Equations 5.42–5.45 are recognizable as a system of 4 simultaneous linear equations with the number of unknowns equal to the number of equations. We could simply leave it at that, however, some very useful insights are gained by solving this system of equations in a particular manner. First, note the resemblance between the situation at the z = 0 boundary in this problem and the situation at z = 0 boundary in the single-boundary problem. In fact, the two are the same problem, with the following transformation of variables (single-boundary → double-boundary): Ei 0 →C i.e., the independent variable (5.46) B →D i.e., reﬂection (5.47) C →F i.e., transmission (5.48)	Electromagnetics_Vol2_Page_74_Chunk2249
62 CHAPTER 5. WAVE REFLECTION AND TRANSMISSION Electric Field Intensity Magnetic Field Intensity Region of Validity Region 1 eEi(z) = ˆxEi 0e−jβ1z eHi(z) = +ˆy	Electromagnetics_Vol2_Page_75_Chunk2250
5.2. PLANE WAVES AT NORMAL INCIDENCE ON A MATERIAL SLAB 63 yielding: ηeq = η2 1 + Γ23e−j2β2d 1 −Γ23e−j2β2d (5.55) Equation 5.55 is the wave impedance in the re- gion to the right of the boundary in the equiva- lent scenario shown in Figure 5.3. “Equivalent” in this case means that the incident and reﬂected ﬁelds in Region 1 are identical to those in the original (slab) problem. Two comments on this expression before proceeding. First: Note that if the materials in Regions 2 and 3 are identical, then η2 = η3, so Γ23 = 0, and thus ηeq = η2, as expected. Second: Note that ηeq is, in general, complex-valued. This may initially seem troubling, since the imaginary component of the wave impedance is normally associated with general (e.g., possibly lossy) material. We speciﬁcally precluded this possibility in the problem statement, so clearly the complex value of the wave impedance is not indicating loss. Instead, the non-zero phase of ηeq represents the ability of the standing wave inside the slab to impart a phase shift between the electric and magnetic ﬁelds. This is precisely the same effect that one observes at the input of a transmission line: The input impedance Zin is, in general, complex-valued even if the line is lossless and the characteristic impedance and load impedance are real-valued.2 In fact, the impedance looking into a transmission line is given by an expression of precisely the same form as Equation 5.55. This striking analogy between plane waves at planar boundaries and voltage and current waves in transmission lines applies broadly. We can now identify an “equivalent reﬂection coefﬁcient” Γ1,eq for the scenario shown in Figure 5.3: Γ1,eq ≜ηeq −η1 ηeq + η1 (5.56) The quantity Γ1,eq may now be used precisely in the same way as Γ12 was used in the single-boundary problem to ﬁnd the reﬂected ﬁelds and reﬂected power density in Region 1. 2See the section “Input Impedance of a Terminated Lossless Transmission Line” for a reminder. This section may appear in a different volume depending on the version of this book. Example 5.2. Reﬂection of WiFi from a glass pane. A WiFi (wireless LAN) signal at a center frequency of 2.45 GHz is normally incident on a glass pane which is 1 cm thick and is well-characterized in this application as a lossless dielectric with ǫr = 4. The source of the signal is sufﬁciently distant from the window that the incident signal is well-approximated as a plane wave. Determine the fraction of power reﬂected from the pane. Solution. In this case, we identify Regions 1 and 3 as approximately free space, and Region 2 as the pane. Thus η1 = η3 = η0 ∼= 376.7 Ω (5.57) and η2 = η0 √ǫr ∼= 188.4 Ω (5.58) The reﬂection coefﬁcient from Region 2 to Region 3 is Γ23 = η3 −η2 η3 + η2 = η0 −η2 η0 + η2 ∼= 0.3333 (5.59) Given f = 2.45 GHz, the phase propagation constant in the glass is β2 = 2π λ = 2πf√ǫr c ∼= 102.6 rad/m (5.60) Given d = 1 cm, the equivalent wave impedance is ηeq = η2 1 + Γ23e−j2β2d 1 −Γ23e−j2β2d ∼= 117.9 −j78.4 Ω (5.61) Next we calculate Γ1,eq = ηeq −η1 ηeq + η1 ∼= −0.4859 −j0.2354 (5.62)	Electromagnetics_Vol2_Page_76_Chunk2251
64 CHAPTER 5. WAVE REFLECTION AND TRANSMISSION The ratio of reﬂected power density to incident power density is simply the squared magnitude of this reﬂection coefﬁcient, i.e.: Sr ave Siave = \|Γ1,eq\|2 = 0.292 ∼= 29.2% (5.63) where Sr ave and Si ave are the reﬂected and incident power densities, respectively. Since B = Γ1,eqEi 0, we have: eEr(z) = ˆxΓ1,eqEi 0e+jβ1z , z ≤−d (5.64) and eHr(z) can be obtained from the plane wave relationships. If desired, it is now quite simple to obtain solutions for the electric and magnetic ﬁelds in Regions 2 and 3. However, it is the usually the power density transmitted into Region 3 that is of greatest interest. This power density is easily determined from the principle of conservation of power. If the loss in Region 2 is negligible, then no power can be dissipated there. In this case, all power not reﬂected from the z = −d interface must be transmitted into Region 3. In other words: St ave Siave = 1 −\|Γ1,eq\|2 (5.65) where St ave is the transmitted power density. Example 5.3. Transmission of WiFi through a glass pane. Continuing Example 5.2: What fraction of incident power passes completely through the glass pane? Solution. St ave Siave = 1 −\|Γ1,eq\|2 ∼= 70.8% (5.66) Additional Reading: • “Radome” on Wikipedia. 5.3 Total Transmission Through a Slab [m0163] Section 5.2 details the solution of the “single-slab” problem. This problem is comprised of three material regions: A semi-inﬁnite Region 1, from which a uniform plane wave is normally incident; Region 2, the slab, deﬁned by parallel planar boundaries separated by distance d; and a semi-inﬁnite Region 3, through which the plane wave exits. The solution that was developed assumes simple media with negligible loss, so that the media in each region is characterized entirely by permittivity and permeability. We now focus on a particular class of applications involving this structure. In this class of applications, we seek total transmission through the slab. By “total transmission” we mean 100% of power incident on the slab is transmitted through the slab into Region 3, and 0% of the power is reﬂected back into Region 1. There are many applications for such a structure. One application is the radome, a protective covering which partially or completely surrounds an antenna, but nominally does not interfere with waves being received by or transmitted from the antenna. Another application is RF and optical wave ﬁltering; that is, passing or rejecting waves falling within a narrow range of frequencies. In this section, we will ﬁrst describe the conditions for total transmission, and then shall provide some examples of these applications. We begin with characterization of the media. Region 1 is characterized by its permittivity ǫ1 and permeability µ1, such that the wave impedance in Region 1 is η1 = p µ1/ǫ1. Similarly, Region 2 is characterized by its permittivity ǫ2 and permeability µ2, such that the wave impedance in Region 2 is η2 = p µ2/ǫ2. Region 3 is characterized by its permittivity ǫ3 and permeability µ3, such that the wave impedance in Region 3 is η3 = p µ3/ǫ3. The analysis in this section also depends on β2, the phase propagation constant in Region 2, which is given by ω√µ2ǫ2. Recall that reﬂection from the slab is quantiﬁed by	Electromagnetics_Vol2_Page_77_Chunk2252
5.3. TOTAL TRANSMISSION THROUGH A SLAB 65 the reﬂection coefﬁcient Γ1,eq = ηeq −η1 ηeq + η1 (5.67) where ηeq is given by ηeq = η2 1 + Γ23e−j2β2d 1 −Γ23e−j2β2d (5.68) and where Γ23 is given by Γ23 = η3 −η2 η3 + η2 (5.69) Total transmission requires that Γ1,eq = 0. From Equation 5.67 we see that Γ1,eq is zero when η1 = ηeq. Now employing Equation 5.68, we see that total transmission requires: η1 = η2 1 + Γ23e−j2β2d 1 −Γ23e−j2β2d (5.70) For convenience and clarity, let us deﬁne the quantity P ≜e−j2β2d (5.71) Using this deﬁnition, Equation 5.70 becomes η1 = η2 1 + Γ23P 1 −Γ23P (5.72) Also, let us substitute Equation 5.69 for Γ23, and multiply numerator and denominator by η3 + η2 (the denominator of Equation 5.69). We obtain: η1 = η2 η3 + η2 + (η3 −η2)P η3 + η2 −(η3 −η2)P (5.73) Rearranging the numerator and denominator, we obtain: η1 = η2 (1 + P)η3 + (1 −P)η2 (1 −P)η3 + (1 + P)η2 (5.74) The parameters η1, η2, η3, β2, and d deﬁning any single-slab structure that exhibits total transmis- sion must satisfy Equation 5.74. Our challenge now is to identify combinations of parameters that satisfy this condition. There are two general categories of solutions. These categories are known as half-wave matching and quarter-wave matching. Half-wave matching applies when we have the same material on either side of the slab; i.e., η1 = η3. Let us refer to this common value of η1 and η3 as ηext. Then the condition for total transmission becomes: ηext = η2 (1 + P)ηext + (1 −P)η2 (1 −P)ηext + (1 + P)η2 (5.75) For the above condition to be satisﬁed, we need the fraction on the right side of the equation to be equal to ηext/η2. From Equation 5.71, the magnitude of P is always 1, so the ﬁrst value of P you might think to try is P = +1. In fact, this value satisﬁes Equation 5.75. Therefore, e−j2β2d = +1. This new condition is satisﬁed when 2β2d = 2πm, where m = 1, 2, 3, ... (We do not consider m ≤0 to be valid solutions since these would represent zero or negative values of d.) Thus, we ﬁnd d = πm β2 = λ2 2 m , where m = 1, 2, 3, ... (5.76) where λ2 = 2π/β2 is the wavelength inside the slab. Summarizing: Total transmission through a slab embedded in regions of material having equal wave impedance (i.e., η1 = η3) may be achieved by setting the thickness of the slab equal to an integer number of half-wavelengths at the frequency of interest. This is known as half-wave matching. A remarkable feature of half-wave matching is that there is no restriction on the permittivity or permeability of the slab, and the only constraint on the media in Regions 1 and 3 is that they have equal wave impedance. Example 5.4. Radome design by half-wave matching. The antenna for a 60 GHz radar is to be protected from weather by a radome panel positioned directly in front of the radar. The panel is to be constructed from a low-loss material having µr ≈1 and ǫr = 4. To have sufﬁcient mechanical integrity, the panel must be at least 3 mm thick. What thickness should be	Electromagnetics_Vol2_Page_78_Chunk2253
66 CHAPTER 5. WAVE REFLECTION AND TRANSMISSION used? Solution. This is a good application for half-wave matching because the material on either side of the slab is the same (presumably free space) whereas the material used for the slab is unspeciﬁed. The phase velocity in the slab is vp = c √ǫr ∼= 1.5 × 108 m/s (5.77) so the wavelength in the slab is λ2 = vp f ∼= 2.5 mm (5.78) Thus, the minimum thickness of the slab that satisﬁes the half-wave matching condition is d = λ2/2 ∼= 1.25 mm. However, this does not meet the 3 mm minimum-thickness requirement. Neither does the next available thickness, d = λ2 ∼= 2.5 mm. The next-thickest option, d = 3λ2/2 ∼= 3.75 mm, does meet the requirement. Therefore, we select d ∼= 3.75 mm. It should be emphasized that designs employing half-wave matching will be narrowband – that is, total only for the design frequency. As frequency increases or decreases from the design frequency, there will be increasing reﬂection and decreasing transmission. Quarter-wave matching requires that the wave impedances in each region are different and related in a particular way. The quarter-wave solution is obtained by requiring P = −1, so that η1 = η2 (1 + P)η3 + (1 −P)η2 (1 −P)η3 + (1 + P)η2 = η2 η2 η3 (5.79) Solving this equation for wave impedance of the slab material, we ﬁnd η2 = √η1η3 (5.80) Note that P = −1 is obtained when 2β2d = π + 2πm where m = 0, 1, 2, .... Thus, we ﬁnd d = π 2β2 + π β2 m = λ2 4 + λ2 2 m , where m = 0, 1, 2, ... (5.81) Summarizing: Total transmission is achieved through a slab by selecting η2 = √η1η3 and making the slab one- quarter wavelength at the frequency of interest, or some integer number of half-wavelengths thicker if needed. This is known as quarter-wave match- ing. Example 5.5. Radome design by quarter-wave matching. The antenna for a 60 GHz radar is to be protected from weather by a radome panel positioned directly in front of the radar. In this case, however, the antenna is embedded in a lossless material having µr ≈1 and ǫr = 2, and the radome panel is to be placed between this material and the outside, which we presume is free space. The material in which the antenna is embedded, and against which the radome panel is installed, is quite rigid so there is no minimum thickness requirement. However, the radome panel must be made from a material which is lossless and non-magnetic. Design the radome panel. Solution. The radome panel must be comprised of a material having η2 = √η1η3 = r η0 √ 2 · η0 ∼= 317 Ω (5.82) Since the radome panel is required to be non-magnetic, the relative permittivity must be given by η2 = η0 √ǫr ⇒ǫr = η0 η2 2 ∼= 1.41 (5.83) The phase velocity in the slab will be vp = c √ǫr ∼= 3 × 108 m/s √ 1.41 ∼= 2.53 × 108 m/s (5.84) so the wavelength in the slab is λ2 = vp f ∼= 2.53 × 108 m/s 60 × 109 Hz ∼= 4.20 mm (5.85)	Electromagnetics_Vol2_Page_79_Chunk2254
5.4. PROPAGATION OF A UNIFORM PLANE WAVE IN AN ARBITRARY DIRECTION 67 Thus, the minimum possible thickness of the radome panel is d = λ2/4 ∼= 1.05 mm, and the relative permittivity of the radome panel must be ǫr ∼= 1.41. Additional Reading: • “Radome” on Wikipedia. 5.4 Propagation of a Uniform Plane Wave in an Arbitrary Direction [m0165] An example of a uniform plane wave propagating in a lossless medium is shown in Figure 5.4. This wave is expressed in the indicated coordinate system as follows: eE = ˆxE0e−jβz (5.86) This is a phasor-domain expression for the electric ﬁeld intensity, so E0 is a complex-valued number representing the magnitude and reference phase of the sinusoidally-varying wave. The term “reference phase” is deﬁned as the phase of eE at the origin of the coordinate system. Since the phase propagation constant β is real and positive, this wave is traveling in the +ˆz direction through simple lossless media. Note that electric ﬁeld intensity vector is linearly polarized in a direction parallel to ˆx. Depending on the position in space (and, for the physical time-domain waveform, time), eE points either in the +ˆx direction or the −ˆx direction. Let us be a bit more speciﬁc about the direction of the vector eE. To do this, let us deﬁne the reference polarization to be the direction in which eE points when Re E0e−jβz ≥0; i.e., when the phase of eE is between −π/2 and +π/2 radians. Thus, the reference polarization of eE in Equation 5.86 is always +ˆx. Note that Equation 5.86 indicates a speciﬁc combination of reference polarization and direction of propagation. However, we may obtain any other combination of reference polarization and direction of c⃝C. Wang CC BY-SA 4.0 Figure 5.4: The plane wave described by Equa- tion 5.86.	Electromagnetics_Vol2_Page_80_Chunk2255
68 CHAPTER 5. WAVE REFLECTION AND TRANSMISSION c⃝C. Wang CC BY-SA 4.0 Figure 5.5: The same plane wave described in a ro- tated coordinate system, yielding Equation 5.87. c⃝C. Wang CC BY-SA 4.0 Figure 5.6: The same plane wave described in yet an- other rotation of the coordinate system, yielding Equa- tion 5.88. propagation by rotation of the Cartesian coordinate system. For example, if we rotate the +x axis of the coordinate system into the position originally occupied by the +y axis, then the very same wave is expressed as eE = −ˆyE0e−jβz (5.87) This is illustrated in Figure 5.5. At ﬁrst glance, it appears that the reference polarization has changed; however, this is due entirely to our choice of coordinate system. That is, the reference polarization is precisely the same; it is only the coordinate system used to describe the reference polarization that has changed. Let us now rotate the +z axis of the coordinate system around the x axis into the position originally occupied by the −z axis. Now the very same wave is expressed as eE = +ˆyE0e+jβz (5.88) This is illustrated in Figure 5.6. At ﬁrst glance, it appears that the direction of propagation has reversed; but, again, it is only the coordinate system that has changed, and not the direction of propagation. Summarizing: Equations 5.86, 5.87, and 5.88 all represent the same wave. They only appear to be c⃝C. Wang CC BY-SA 4.0 Figure 5.7: A plane wave described in ray-ﬁxed coor- dinates, yielding Equation 5.89. different due to our choice for the orientation of the coordinate system in each case. Now consider the same thought experiment for the inﬁnite number of cases in which the wave does not propagate in one of the three basis directions of the Cartesian coordinate system. One situation in which we face this complication is when the wave is obliquely incident on a surface. In this case, it is impossible to select a single orientation of the coordinate system in which the directions of propagation, reference polarization, and surface normal can all be described in terms of one basis vector each. To be clear: There is no fundamental limitation imposed by this slipperiness of the coordinate system. However, practical problems such as the oblique-incidence scenario described above are much easier to analyze if we are able to express waves in a system of coordinates which always yields the same expressions. Fortunately, this is easily accomplished using ray-ﬁxed coordinates. Ray-ﬁxed coordinates are a unique set of coordinates that are determined from the characteristics of the wave, as opposed to being determined arbitrarily and separately from the characteristics of the wave. The ray-ﬁxed representation of a uniform plane wave is: eE(r) = ˆeE0e−jk·r (5.89) This is illustrated in Figure 5.7. In this representation, r is the position at which eE is evaluated, ˆe is the reference polarization expressed as a unit vector, and k is the unit vector ˆk in the direction of propagation, times β; i.e.: k ≜ˆkβ (5.90) Consider Equation 5.86 as an example. In this case, ˆe = ˆx, k = ˆzβ, and (as always) r = ˆxx + ˆyy + ˆzz (5.91)	Electromagnetics_Vol2_Page_81_Chunk2256
5.4. PROPAGATION OF A UNIFORM PLANE WAVE IN AN ARBITRARY DIRECTION 69 Thus, k · r = βz, as expected. In ray-ﬁxed coordinates, a wave can be represented by one – and only one – expression, which is the same expression regardless of the orientation of the “global” coordinate system. Moreover, only two basis directions (namely, ˆk and ˆe) must be deﬁned. Should a third coordinate be required, either ˆk × ˆe or ˆe × ˆk may be selected as the additional basis direction. Note that the ﬁrst choice has the possibly-useful feature that is the reference polarization of the magnetic ﬁeld intensity eH. A general procedure for recasting the ray-ﬁxed representation into a “coordinate-bound” representation is as follows. First, we represent k in the new ﬁxed coordinate system; e.g.: k = kxˆx + kyˆy + kzˆz (5.92) where kx ≜βˆk · ˆx (5.93) ky ≜βˆk · ˆy (5.94) kz ≜βˆk · ˆz (5.95) Then, k · r = kxx + kyy + kzz (5.96) With this expression in hand, Equation 5.89 may be rewritten as: eE = ˆeE0e−jkxxe−jkyye−jkzz (5.97) If desired, one can similarly decompose ˆe into its Cartesian components as follows: ˆe = (ˆe · ˆx) ˆx + (ˆe · ˆy) ˆy + (ˆe · ˆz) ˆz (5.98) Thus, we see that the ray-ﬁxed representation of Equation 5.89 accommodates all possible combinations of direction of propagation and reference polarization. Example 5.6. Plane wave propagating away from the z-axis. A uniform plane wave exhibiting a reference polarization of ˆz propagates away from the z-axis. Develop representations of this wave in ray-ﬁxed and global Cartesian coordinates. Solution. As always, the ray-ﬁxed representation is given by Equation 5.89. Since the reference polarization is ˆz, ˆe = ˆz. Propagating away from the z-axis means ˆk = ˆx cos φ + ˆy sin φ (5.99) where φ indicates the speciﬁc direction of propagation. For example, φ = 0 yields ˆk = +ˆx, φ = π/2 yields ˆk = +ˆy, and so on. Therefore, k ≜βˆk = β (ˆx cos φ + ˆy sin φ) (5.100) For completeness, note that the following factor appears in the phase-determining exponent in Equation 5.89: k · r = β (x cos φ + y sin φ) (5.101) In this case, we see kx = β cos φ, ky = β sin φ, and kz = 0. Thus, the wave may be expressed in Cartesian coordinates as follows: eE = ˆzE0e−jβx cos φe−jβy sin φ (5.102)	Electromagnetics_Vol2_Page_82_Chunk2257
70 CHAPTER 5. WAVE REFLECTION AND TRANSMISSION 5.5 Decomposition of a Wave into TE and TM Components [m0166] A broad range of problems in electromagnetics involve scattering of a plane wave by a planar boundary between dissimilar media. Section 5.1 (“Plane Waves at Normal Incidence on a Planar Boundary Between Lossless Media”) addressed the special case in which the wave arrives in a direction which is perpendicular to the boundary (i.e., “normal incidence”). Analysis of the normal incidence case is simpliﬁed by the fact that the directions of ﬁeld vectors associated with the reﬂected and transmitted are the same (except possibly with a sign change) as those of the incident wave. For the more general case in which the incident wave is obliquely incident (i.e., not necessarily normally-incident), the directions of the ﬁeld vectors will generally be different. This added complexity is easily handled if we take the effort to represent the incident wave as the sum of two waves having particular polarizations. These polarizations are referred to as transverse electric (TE) and transverse magnetic (TM). This section describes these polarizations and the method for decomposition of a plane wave into TE and TM components. We will then be prepared to address the oblique incidence case in a later section. To begin, we deﬁne the ray-ﬁxed coordinate system shown in Figure 5.8. In this ﬁgure, ˆki is a unit vector indicating the direction in which the incident wave propagates. The unit normal ˆn is perpendicular to the boundary, and points into the region from which the wave is incident. We now make the following deﬁnition: The plane of incidence is the plane in which both the normal to the surface (ˆn) and the direction of propagation (ˆki) lie. The TE-TM decomposition consists of ﬁnding the components of the electric and magnetic ﬁelds which are perpendicular (“transverse”) to the plane of incidence. Of the two possible directions that are perpendicular to the plane of incidence, we choose ˆe⊥, deﬁned as shown in Figure 5.8. From the ﬁgure, we see that: ˆe⊥≜ ˆki × ˆn ˆki × ˆn (5.103) Deﬁned in this manner, ˆe⊥is a unit vector which is perpendicular to both ˆki, and so may serve as a basis vector of a coordinate system which is attached to the incident ray. The remaining basis vector for this ray-ﬁxed coordinate system is chosen as follows: ˆei ∥≜ˆe⊥× ˆki (5.104) Deﬁned in this manner, ˆei ∥is a unit vector which is perpendicular to both ˆe⊥and ˆki, and parallel to the plane of incidence. Let us now examine an incident uniform plane wave in the new, ray-ﬁxed coordinate system. We begin with the following phasor representation of the electric ﬁeld intensity: eEi = ˆeiEi 0e−jki·r (5.105) where ˆei is a unit vector indicating the reference polarization, Ei 0 is a complex-valued scalar, and r is a vector indicating the position at which eEi is evaluated. We may express ˆei in the ray-ﬁxed coordinate system of Figure 5.8 as follows: ˆei =	Electromagnetics_Vol2_Page_83_Chunk2258
5.5. DECOMPOSITION OF A WAVE INTO TE AND TM COMPONENTS 71 The electric ﬁeld vector is always perpendicular to the direction of propagation, so ˆei · ˆki = 0. This leaves: ˆei =	Electromagnetics_Vol2_Page_84_Chunk2259
72 CHAPTER 5. WAVE REFLECTION AND TRANSMISSION 5.6 Plane Waves at Oblique Incidence on a Planar Boundary: TE Case [m0167] In this section, we consider the problem of reﬂection and transmission from a planar boundary between semi-inﬁnite media for a transverse electric (TE) uniform plane wave. Before attempting this section, a review of Sections 5.1 (“Plane Waves at Normal Incidence on a Planar Boundary Between Lossless Media”) and 5.5 (“Decomposition of a Wave into TE and TM Components”) is recommended. Also, note that this section has much in common with Section 5.7 (“Plane Waves at Oblique Incidence on a Planar Boundary: TM Case”), although it is recommended to attempt the TE case ﬁrst. The TE case is illustrated in Figure 5.9. The boundary between the two semi-inﬁnite and lossless regions is located at the z = 0 plane. The wave is incident from Region 1. The electric ﬁeld intensity eEi T E of this wave is given by eEi T E(r) = ˆyEi T Ee−jki·r (5.111) In this expression, r is the position at which eEi T E is evaluated, and ki = ˆkiβ1 (5.112) c⃝C. Wang CC BY-SA 4.0 Figure 5.9: A TE uniform plane wave obliquely inci- dent on the planar boundary between two semi-inﬁnite material regions. where ˆki is the unit vector indicating the direction of propagation and β1 = ω√µ1ǫ1 is the phase propagation constant in Region 1. eEi T E serves as the “stimulus” in this problem, so all other contributions to the total ﬁeld will be expressed in terms of quantities appearing in Equation 5.111. The presence of reﬂected and transmitted uniform plane waves is inferred from our experience with the normal incidence scenario (Section 5.1). There, as here, the symmetry of the problem indicates that the reﬂected and transmitted components of the electric ﬁeld will have the same polarization as that of the incident electric ﬁeld. This is because there is nothing present in the problem that could account for a change in polarization. Thus, the reﬂected and transmitted ﬁelds will also be TE. Therefore, we postulate the following expression for the reﬂected wave: eEr(r) = ˆyBe−jkr·r (5.113) where B is an unknown, possibly complex-valued constant to be determined and kr = ˆkrβ1 (5.114) indicates the direction of propagation, which is also currently unknown. Similarly, we postulate the following expression for the transmitted wave: eEt(r) = ˆyCe−jkt·r (5.115) where C is an unknown, possibly complex-valued constant to be determined; kt = ˆktβ2 (5.116) where ˆkt is the unit vector indicating the direction of propagation; and β2 = ω√µ2ǫ2 is the phase propagation constant in Region 2. At this point, the unknowns in this problem are the constants B and C, as well as the directions ˆkr and ˆkt. We may establish a relationship between Ei T E, B, and C by application of boundary conditions at z = 0. First, recall that the tangential component of the total electric ﬁeld intensity must be continuous across material boundaries. To apply this boundary condition, let us deﬁne eE1 and eE2 to be the total electric ﬁelds in Regions 1 and 2, respectively. The	Electromagnetics_Vol2_Page_85_Chunk2260
5.6. PLANE WAVES AT OBLIQUE INCIDENCE ON A PLANAR BOUNDARY: TE CASE 73 total ﬁeld in Region 1 is the sum of incident and reﬂected ﬁelds, so eE1(r) = eEi T E(r) + eEr(r) (5.117) The total ﬁeld in Region 2 is simply eE2(r) = eEt(r) (5.118) Next, note that all electric ﬁeld components are already tangent to the boundary. Thus, continuity of the tangential component of the electric ﬁeld across the boundary requires eE1(0) = eE2(0), and therefore eEi T E(r0) + eEr(r0) = eEt(r0) (5.119) where r = r0 ≜ˆxx + ˆyy since z = 0 on the boundary. Now employing Equations 5.111, 5.113, and 5.115, we obtain: ˆyEi T Ee−jki·r0 +ˆyBe−jkr·r0 = ˆyCe−jkt·r0 (5.120) Dropping the vector (ˆy) since it is the same in each term, we obtain: Ei T Ee−jki·r0 + Be−jkr·r0 = Ce−jkt·r0 (5.121) For Equation 5.121 to be true at every point r0 on the boundary, it must be true that ki · r0 = kr · r0 = kt · r0 (5.122) Essentially, we are requiring the phases of each ﬁeld in Regions 1 and 2 to be matched at every point along the boundary. Any other choice will result in a violation of boundary conditions at some point along the boundary. This phase matching criterion will determine the directions of propagation of the reﬂected and transmitted ﬁelds, which we shall do later. First, let us use the phase matching criterion to complete the solution for the coefﬁcients B and C. Enforcing Equation 5.122, we observe that Equation 5.121 reduces to: Ei T E + B = C (5.123) A second equation is needed since we currently have only one equation (Equation 5.123) and two unknowns (B and C). The second equation is c⃝C. Wang CC BY-SA 4.0 Figure 5.10: Incident, reﬂected, and transmitted mag- netic ﬁeld components associated with the TE electric ﬁeld components shown in Figure 5.9. obtained by applying the appropriate boundary conditions to the magnetic ﬁeld. The magnetic ﬁeld associated with each of the electric ﬁeld components is identiﬁed in Figure 5.10. Note the orientations of the magnetic ﬁeld vectors may be conﬁrmed using the plane wave relationships: Speciﬁcally, the cross product of the electric and magnetic ﬁelds should point in the direction of propagation. Expressions for each of the magnetic ﬁeld components is determined formally below. From the plane wave relationships, we determine that the incident magnetic ﬁeld intensity is eHi(r) = 1 η1 ˆki × eEi T E (5.124) where η1 = p µ1/ǫ1 is the wave impedance in Region 1. To make progress requires that we express ˆki in the global ﬁxed coordinate system. Here it is: ˆki = ˆx sin ψi + ˆz cos ψi (5.125) Thus: eHi(r) =	Electromagnetics_Vol2_Page_86_Chunk2261
74 CHAPTER 5. WAVE REFLECTION AND TRANSMISSION In the global coordinate system: ˆkr = ˆx sin ψr −ˆz cos ψr (5.128) Thus: eHr(r) = (ˆz sin ψr + ˆx cos ψr) B η1 e−jkr·r (5.129) The transmitted magnetic ﬁeld has the form: eHt(r) = 1 η2 ˆkt × eEt (5.130) In the global coordinate system: ˆkt = ˆx sin ψt + ˆz cos ψt (5.131) Thus: eHt(r) =	Electromagnetics_Vol2_Page_87_Chunk2262
5.6. PLANE WAVES AT OBLIQUE INCIDENCE ON A PLANAR BOUNDARY: TE CASE 75 simply state the result, and in Section 5.8 we shall perform this part of the derivation in detail and with greater attention to the implications. One ﬁnds: ψr = ψi (5.145) and ψt = arcsin β1 β2 sin ψi (5.146) Equation 5.145 is the unsurprising result that angle of reﬂection equals angle of incidence. Equation 5.146 – addressing angle of transmission – is a bit more intriguing. Astute readers may notice that there is something ﬁshy about this equation: It seems possible for the argument of arcsin to be greater than one. This oddity is addressed in Section 5.8. Finally, note that Equation 5.145 allows us to eliminate ψr from Equation 5.141, yielding: ΓT E = η2 cos ψi −η1 cos ψt η2 cos ψi + η1 cos ψt (5.147) Thus, we obtain what is perhaps the most important ﬁnding of this section: The electric ﬁeld reﬂection coefﬁcient for oblique TE incidence, ΓT E, is given by Equation 5.147. The following example demonstrates the utility of this result. Example 5.7. Power transmission at an air-to-glass interface (TE case). Figure 5.11 illustrates a TE plane wave incident from air onto the planar boundary with glass. The glass exhibits relative permittivity of 2.1. Determine the power reﬂected and transmitted relative to power incident on the boundary. Solution. The power reﬂected relative to power incident is \|ΓT E\|2 whereas the power transmitted relative to power incident is 1 −\|ΓT E\|2. ΓT E may be calculated using Equation 5.147. Calculating the quantities that enter into this expression: η1 ≈η0 ∼= 376.7 Ω(air) (5.148) c⃝C. Wang CC BY-SA 4.0 Figure 5.11: A TE uniform plane wave incident from air to glass. η2 ≈ η0 √ 2.1 ∼= 260.0 Ω(glass) (5.149) ψi = 30◦ (5.150) Note β1 β2 ≈ ω√µ0ǫ0 ω√µ0 · 2.1ǫ0 ∼= 0.690 (5.151) so ψt = arcsin β1 β2 sin ψi ∼= 20.2◦ (5.152) Now substituting these values into Equation 5.147, we obtain ΓT E ∼= −0.2220 (5.153) Subsequently, the fraction of power reﬂected relative to power incident is \|ΓT E\|2 ∼= 0.049; i.e., about 4.9%. 1 −\|ΓT E\|2 ∼= 95.1% of the power is transmitted into the glass.	Electromagnetics_Vol2_Page_88_Chunk2263
76 CHAPTER 5. WAVE REFLECTION AND TRANSMISSION 5.7 Plane Waves at Oblique Incidence on a Planar Boundary: TM Case [m0164] In this section, we consider the problem of reﬂection and transmission from a planar boundary between semi-inﬁnite media for a transverse magnetic (TM) uniform plane wave. Before attempting this section, a review of Sections 5.1 (“Plane Waves at Normal Incidence on a Planar Boundary Between Lossless Media”) and 5.5 (“Decomposition of a Wave into TE and TM Components”) is recommended. Also, note that this section has much in common with Section 5.6 (“Plane Waves at Oblique Incidence on a Planar Boundary: TE Case”), and it is recommended to attempt the TE case ﬁrst. In this section, we consider the scenario illustrated in Figure 5.12. The boundary between the two semi-inﬁnite and lossless regions is located at the z = 0 plane. The wave is incident from Region 1. The magnetic ﬁeld intensity eHi T M of this wave is given by eHi T M(r) = ˆyHi T Me−jki·r (5.154) In this expression, r is the position at which eHi T M is c⃝C. Wang CC BY-SA 4.0 Figure 5.12: A TM uniform plane wave obliquely incident on the planar boundary between two semi- inﬁnite material regions. evaluated. Also, ki = ˆkiβ1 (5.155) where ˆki is the unit vector indicating the direction of propagation and β1 = ω√µ1ǫ1 is the phase propagation constant in Region 1. eHi T M serves as the “stimulus” in this problem, and all other contributions to the total ﬁeld may be expressed in terms of parameters associated with Equation 5.154. The presence of reﬂected and transmitted uniform plane waves is inferred from our experience with the normal incidence scenario (Section 5.1). There, as here, the symmetry of the problem indicates that the reﬂected and transmitted components of the magnetic ﬁeld will have the same polarization as that of the incident electric ﬁeld. This is because there is nothing present in the problem that could account for a change in polarization. Thus, the reﬂected and transmitted ﬁelds will also be TM. So we postulate the following expression for the reﬂected wave: eHr(r) = −ˆyBe−jkr·r (5.156) where B is an unknown, possibly complex-valued constant to be determined and kr = ˆkrβ1 (5.157) indicates the direction of propagation. The reader may wonder why we have chosen −ˆy, as opposed to +ˆy, as the reference polarization for eHr. In fact, either +ˆy or −ˆy could be used. However, the choice is important because the form of the results we obtain in this section – speciﬁcally, the reﬂection coefﬁcient – will be determined with respect to this speciﬁc convention, and will be incorrect with respect to the opposite convention. We choose −ˆy because it has a particular advantage which we shall point out at the end of this section. Continuing, we postulate the following expression for the transmitted wave: eHt(r) = ˆyCe−jkt·r (5.158) where C is an unknown, possibly complex-valued constant to be determined and kt = ˆktβ2 (5.159)	Electromagnetics_Vol2_Page_89_Chunk2264
5.7. PLANE WAVES AT OBLIQUE INCIDENCE ON A PLANAR BOUNDARY: TM CASE 77 where ˆkt is the unit vector indicating the direction of propagation and β2 = ω√µ2ǫ2 is the phase propagation constant in Region 2. At this point, the unknowns in this problem are the constants B and C, as well as the unknown directions ˆkr and ˆkt. We may establish a relationship between Hi T M, B, and C by application of boundary conditions at z = 0. First, we presume no impressed current at the boundary. Thus, the tangential component of the total magnetic ﬁeld intensity must be continuous across the boundary. To apply this boundary condition, let us deﬁne eH1 and eH2 to be the total magnetic ﬁelds in Regions 1 and 2, respectively. The total ﬁeld in Region 1 is the sum of incident and reﬂected ﬁelds, so eH1(r) = eHi T M(r) + eHr(r) (5.160) The ﬁeld in Region 2 is simply eH2(r) = eHt(r) (5.161) Also, we note that all magnetic ﬁeld components are already tangent to the boundary. Thus, continuity of the tangential component of the magnetic ﬁeld across the boundary requires eH1(r0) = eH2(r0), where r0 ≜ˆxx + ˆyy since z = 0 on the boundary. Therefore, eHi T M(r0) + eHr(r0) = eHt(r0) (5.162) Now employing Equations 5.154, 5.156, and 5.158, we obtain: ˆyHi T Me−jki·r0 −ˆyBe−jkr·r0 = ˆyCe−jkt·r0 (5.163) Dropping the vector (ˆy) since it is the same in each term, we obtain: Hi T Me−jki·r0 −Be−jkr·r0 = Ce−jkt·r0 (5.164) For this to be true at every point r0 on the boundary, it must be true that ki · r0 = kr · r0 = kt · r0 (5.165) Essentially, we are requiring the phases of each ﬁeld in Regions 1 and 2 to be matched at every point along the boundary. Any other choice will result in a violation of boundary conditions at some point along the boundary. This expression allows us to solve for the directions of propagation of the reﬂected and transmitted ﬁelds, which we shall do later. Our priority for now shall be to solve for the coefﬁcients B and C. Enforcing Equation 5.165, we observe that Equation 5.164 reduces to: Hi T M −B = C (5.166) A second equation is needed since we currently have only one equation (Equation 5.166) and two unknowns (B and C). The second equation is obtained by applying the appropriate boundary conditions to the electric ﬁeld. The electric ﬁeld associated with each of the magnetic ﬁeld components is identiﬁed in Figure 5.12. Note the orientations of the electric ﬁeld vectors may be conﬁrmed using the plane wave relationships: Speciﬁcally, the cross product of the electric and magnetic ﬁelds should point in the direction of propagation. Expressions for each of the electric ﬁeld components is determined formally below. From the plane wave relationships, we determine that the incident electric ﬁeld intensity is eEi(r) = −η1ˆki × eHi T M (5.167) where η1 = p µ1/ǫ1 is the wave impedance in Region 1. To make progress requires that we express ˆki in the global ﬁxed coordinate system. Here it is: ˆki = ˆx sin ψi + ˆz cos ψi (5.168) Thus: eEi(r) =	Electromagnetics_Vol2_Page_90_Chunk2265
78 CHAPTER 5. WAVE REFLECTION AND TRANSMISSION The transmitted magnetic ﬁeld has the form: eEt(r) = −η2ˆkt × eHt (5.173) In the global coordinate system: ˆkt = ˆx sin ψt + ˆz cos ψt (5.174) Thus: eEt(r) =	Electromagnetics_Vol2_Page_91_Chunk2266
5.7. PLANE WAVES AT OBLIQUE INCIDENCE ON A PLANAR BOUNDARY: TM CASE 79 can be found using Equation 5.165. Here we shall simply state the result, and in Section 5.8 we shall perform this part of the derivation in detail and with greater attention to the implications. One ﬁnds: ψr = ψi (5.189) i.e., angle of reﬂection equals angle of incidence. Also, ψt = arcsin β1 β2 sin ψi (5.190) Astute readers may notice that there is something ﬁshy about Equation 5.190. Namely, it seems possible for the argument of arcsin to be greater than one. This oddity is addressed in Section 5.8. Now let us return to the following question, raised near the beginning of this section: Why choose −ˆy, as opposed to +ˆy, as the reference polarization for Hr, as shown in Figure 5.12? To answer this question, ﬁrst note that ΓT M (Equation 5.184) becomes the reﬂection coefﬁcient for normal (TEM) incidence when ψi = ψt = 0. If we had chosen +ˆy as the reference polarization for Hr, we would have instead obtained an expression for ΓT M that has the opposite sign for TEM incidence.3 There is nothing wrong with this answer, but it is awkward to have different values of the reﬂection coefﬁcient for the same physical scenario. By choosing −ˆy, the reﬂection coefﬁcient for the oblique incidence case computed for ψi = 0 converges to the reﬂection coefﬁcient previously computed for the normal-incidence case. It is important to be aware of this issue, as one occasionally encounters work in which the opposite (“+ˆy”) reference polarization has been employed. Finally, note that Equation 5.189 allows us to eliminate ψr from Equation 5.184, yielding: ΓT M = −η1 cos ψi + η2 cos ψt +η1 cos ψi + η2 cos ψt (5.191) Thus, we obtain what is perhaps the most important ﬁnding of this section: The electric ﬁeld reﬂection coefﬁcient for oblique TM incidence, ΓT M, is given by Equation 5.191. 3Obtaining this result is an excellent way for the student to con- ﬁrm their understanding of the derivation presented in this section. The following example demonstrates the utility of this result. Example 5.8. Power transmission at an air-to-glass interface (TM case). Figure 5.13 illustrates a TM plane wave incident from air onto the planar boundary with a glass region. The glass exhibits relative permittivity of 2.1. Determine the power reﬂected and transmitted relative to power incident on the boundary. Solution. The power reﬂected relative to power incident is \|ΓT M\|2 whereas the power transmitted relative to power incident is 1 −\|ΓT M\|2. ΓT M may be calculated using Equation 5.191. Calculating the quantities that enter into this expression: η1 ≈η0 ∼= 376.7 Ω(air) (5.192) η2 ≈ η0 √ 2.1 ∼= 260.0 Ω(glass) (5.193) ψi = 30◦ (5.194) Note β1 β2 ≈ ω√µ0ǫ0 ω√µ0 · 2.1ǫ0 ∼= 0.690 (5.195) so ψt = arcsin β1 β2 sin ψi ∼= 20.2◦ (5.196) Now substituting these values into Equation 5.191, we obtain ΓT M ∼= −0.1442 (5.197) (Did you get an answer closer to −0.1323? If so, you probably did not use sufﬁcient precision to represent intermediate results. This is a good example of a problem in which three signiﬁcant ﬁgures for results that are used in subsequent calculations is not sufﬁcient.) The fraction of power reﬂected relative to power incident is now determined to be	Electromagnetics_Vol2_Page_92_Chunk2267
80 CHAPTER 5. WAVE REFLECTION AND TRANSMISSION c⃝C. Wang CC BY-SA 4.0 Figure 5.13: A TM uniform plane wave incident from air to glass. \|ΓT M\|2 ∼= 0.021; i.e., about 2.1%. 1 −\|ΓT M\|2 ∼= 97.9% of the power is transmitted into the glass. Note that the result obtained in the preceding example is different from the result for a TE wave incident from the same direction (Example 5.7). In other words: The fraction of power reﬂected and transmitted from the planar boundary between dissimilar me- dia depends on the polarization of the incident wave relative to the boundary, as well as the an- gle of incidence. 5.8 Angles of Reﬂection and Refraction [m0168] Consider the situation shown in Figure 5.14: A uniform plane wave obliquely incident on the planar boundary between two semi-inﬁnite material regions. Let a point on the boundary be represented as the position vector r0 = ˆxx + ˆyy (5.198) In both Sections 5.6 (“Plane Waves at Oblique Incidence on a Planar Boundary: TE Case”) and 5.7 (“Plane Waves at Oblique Incidence on a Planar Boundary: TM Case”), it is found that ki · r0 = kr · r0 = kt · r0 (5.199) In this expression, ki = β1ˆki (5.200) kr = β1ˆkr (5.201) kt = β2ˆkt (5.202) where ˆki, ˆkr, and ˆkt are unit vectors in the direction of incidence, reﬂection, and transmission, c⃝C. Wang CC BY-SA 4.0 Figure 5.14: A uniform plane wave obliquely inci- dent on the planar boundary between two semi-inﬁnite material regions.	Electromagnetics_Vol2_Page_93_Chunk2268
5.8. ANGLES OF REFLECTION AND REFRACTION 81 respectively; and β1 and β2 are the phase propagation constants in Region 1 (from which the wave is incident) and Region 2, respectively. Equation 5.199 is essentially a boundary condition that enforces continuity of the phase of the electric and magnetic ﬁelds across the boundary, and is sometimes referred to as the “phase matching” requirement. Since the same requirement emerges independently in the TE and TM cases, and since any plane wave may be decomposed into TE and TM components, the requirement must apply to any incident plane wave regardless of polarization. Equation 5.199 is the key to ﬁnding the direction of reﬂection ψr and direction of transmission ψt. First, observe: ˆki = ˆx sin ψi + ˆz cos ψi (5.203) ˆkr = ˆx sin ψr −ˆz cos ψr (5.204) ˆkt = ˆx sin ψt + ˆz cos ψt (5.205) Therefore, we may express Equation 5.199 in the following form β1	Electromagnetics_Vol2_Page_94_Chunk2269
82 CHAPTER 5. WAVE REFLECTION AND TRANSMISSION c⃝Z. S´andor CC BY-SA 3.0 Figure 5.15: Angles of reﬂection and refraction for a light wave incident from air onto glass. As expected, the angle of reﬂection ψr is observed to be equal to ψi. The angle of refraction ψt is observed to be 35◦. What is the relative permittivity of the glass? Solution. Since the permittivity of glass is greater than that of air, we observe the expected result ψt < ψi. Since glass is non-magnetic, the expected relationship between these angles is given by Equation 5.213. Solving that equation for the relative permittivity of the glass, we obtain: ǫr2 = ǫr1 sin ψi sin ψt 2 ∼= 2.28 (5.216) When a wave travels in the reverse direction – i.e., from a non-magnetic medium of higher permittivity to a medium of lower permittivity – one ﬁnds ψt > ψr. In other words, the refraction is away from the surface normal. Figure 5.16 shows an example from common experience. In non-magnetic media, when ǫr1 < ǫr2, ψt < ψi (refraction toward the surface normal). When ǫr1 > ǫr2, ψt > ψi (refraction away from the surface normal). c⃝G. Saini CC BY-SA 4.0 Figure 5.16: Refraction accounts for the apparent dis- placement of an underwater object from the perspec- tive of an observer above the water. Under certain conditions, the ǫr2 < ǫr1 case leads to the following surprising observation: When calculating ψt using, for example, Equation 5.212, one ﬁnds that p ǫr1/ǫr2 sin ψi can be greater than 1. Since the sin function yields values between −1 and +1, the result of the arcsin function is undeﬁned. This odd situation is addressed in Section 5.11. For now, we will simply note that this condition leads to the phenomenon of total internal reﬂection. For now, all we can say is that when ǫr2 < ǫr1, ψt is able to reach π/2 radians, which corresponds to propagation parallel to the boundary. Beyond that threshold, we must account for the unique physical considerations associated with total internal reﬂection. We conclude this section with a description of the common waveguiding device known as the prism, shown in Figure 5.17. This particular device uses refraction to change the direction of light waves (similar devices can be used to manipulate radio waves as well). Many readers are familiar with the use of prisms to separate white light into its constituent colors (frequencies), as shown in Figure 5.18. The separation of colors is due to frequency dependence of the material comprising the prism. Speciﬁcally, the permittivity of the material is a function of frequency, and therefore the angle of refraction is a function of frequency. Thus, each frequency is refracted by a different amount. Conversely, a prism comprised of a material whose permittivity exhibits negligible variation with frequency will not separate incident	Electromagnetics_Vol2_Page_95_Chunk2270
5.9. TE REFLECTION IN NON-MAGNETIC MEDIA 83 c⃝D-Kuru CC BY-SA 3.0 Figure 5.17: A typical triangular prism. white light into its constituent colors since each color will be refracted by the same amount. Additional Reading: • “Prism” on Wikipedia. • “Refraction” on Wikipedia. • “Refractive index” on Wikipedia. • “Snell’s law” on Wikipedia. • “Total internal reﬂection” on Wikipedia. c⃝Suidroot CC BY-SA 4.0 Figure 5.18: A color-separating prism. 5.9 TE Reﬂection in Non-magnetic Media [m0171] Figure 5.19 shows a TE uniform plane wave incident on the planar boundary between two semi-inﬁnite material regions. In this case, the reﬂection coefﬁcient is given by: ΓT E = η2 cos ψi −η1 cos ψt η2 cos ψi + η1 cos ψt (5.217) where ψi and ψt are the angles of incidence and transmission (refraction), respectively; and η1 and η2 are the wave impedances in Regions 1 and 2, respectively. Many materials of practical interest are non-magnetic; that is, they have permeability that is not signiﬁcantly different from the permeability of free space. In this section, we consider the behavior of the reﬂection coefﬁcient for this class of materials. To begin, recall the general form of Snell’s law: sin ψt = β1 β2 sin ψi (5.218) In non-magnetic media, the permeabilities µ1 and µ2 are assumed equal to µ0. Thus: β1 β2 = ω√µ1ǫ1 ω√µ2ǫ2 = rǫ1 ǫ2 (5.219) c⃝C. Wang CC BY-SA 4.0 Figure 5.19: A TE uniform plane wave obliquely inci- dent on the planar boundary between two semi-inﬁnite regions of lossless non-magnetic material.	Electromagnetics_Vol2_Page_96_Chunk2271
84 CHAPTER 5. WAVE REFLECTION AND TRANSMISSION Since permittivity ǫ can be expressed as ǫ0 times the relative permittivity ǫr, we may reduce further to: β1 β2 = rǫr1 ǫr2 (5.220) Now Equation 5.218 reduces to: sin ψt = rǫr1 ǫr2 sin ψi (5.221) Next, note that for any value ψ, one may write cosine in terms of sine as follows: cos ψ = q 1 −sin2 ψ (5.222) Therefore, cos ψt = r 1 −ǫr1 ǫr2 sin2 ψi (5.223) Also we note that in non-magnetic media η1 = rµ1 ǫ1 = r µ0 ǫr1ǫ0 = η0 √ǫr1 (5.224) η2 = rµ2 ǫ2 = r µ0 ǫr2ǫ0 = η0 √ǫr2 (5.225) where η0 is the wave impedance in free space. Making substitutions into Equation 5.217, we obtain: ΓT E =	Electromagnetics_Vol2_Page_97_Chunk2272
5.10. TM REFLECTION IN NON-MAGNETIC MEDIA 85 5.10 TM Reﬂection in Non-magnetic Media [m0172] Figure 5.21 shows a TM uniform plane wave incident on the planar boundary between two semi-inﬁnite material regions. In this case, the reﬂection coefﬁcient is given by: ΓT M = −η1 cos ψi + η2 cos ψt +η1 cos ψi + η2 cos ψt (5.229) where ψi and ψt are the angles of incidence and transmission (refraction), respectively; and η1 and η2 are the wave impedances in Regions 1 and 2, respectively. Many materials of practical interest are non-magnetic; that is, they have permeability that is not signiﬁcantly different from the permeability of free space. In this section, we consider the behavior of the reﬂection coefﬁcient for this class of materials. To begin, recall the general form of Snell’s law: sin ψt = β1 β2 sin ψi (5.230) In non-magnetic media, the permeabilities µ1 and µ2 are assumed equal to µ0. Thus: β1 β2 = ω√µ1ǫ1 ω√µ2ǫ2 = rǫ1 ǫ2 (5.231) c⃝C. Wang CC BY-SA 4.0 Figure 5.21: A transverse magnetic uniform plane wave obliquely incident on the planar boundary be- tween two semi-inﬁnite material regions. Since permittivity ǫ can be expressed as ǫ0 times the relative permittivity ǫr, we may reduce further to: β1 β2 = rǫr1 ǫr2 (5.232) Now Equation 5.230 reduces to: sin ψt = rǫr1 ǫr2 sin ψi (5.233) Next, note that for any value ψ, one may write cosine in terms of sine as follows: cos ψ = q 1 −sin2 ψ (5.234) Therefore, cos ψt = r 1 −ǫr1 ǫr2 sin2 ψi (5.235) Also we note that in non-magnetic media η1 = rµ1 ǫ1 = r µ0 ǫr1ǫ0 = η0 √ǫr1 (5.236) η2 = rµ2 ǫ2 = r µ0 ǫr2ǫ0 = η0 √ǫr2 (5.237) where η0 is the wave impedance in free space. Making substitutions into Equation 5.229, we obtain: ΓT M = −	Electromagnetics_Vol2_Page_98_Chunk2273
86 CHAPTER 5. WAVE REFLECTION AND TRANSMISSION -1 -0.5 0 0.5 1 0 10 20 30 40 50 60 70 80 90 2 10 100 Reflection Coefficient angle of incidence [deg] Figure 5.22: The reﬂection coefﬁcient ΓT M as a func- tion of angle of incidence ψi for various media combi- nations, parameterized as ǫr2/ǫr1. Using Equation 5.241, we can easily see how different combinations of material affect the reﬂection coefﬁcient. First, we note that when ǫr1 = ǫr2 (i.e., same media on both sides of the boundary), ΓT M = 0 as expected. When ǫr1 > ǫr2 (e.g., wave traveling in glass toward air), we see that it is possible for ǫr2/ǫr1 −sin2 ψi to be negative, which makes ΓT M complex-valued. This results in total internal reﬂection, and is addressed in another section. When ǫr1 < ǫr2 (e.g., wave traveling in air toward glass), we see that ǫr2/ǫr1 −sin2 ψi is always positive, so ΓT M is always real-valued. Let us continue with the ǫr1 < ǫr2 condition. Figure 5.22 shows ΓT M plotted for various combinations of media over all possible angles of incidence from 0 (normal incidence) to π/2 (grazing incidence). We observe: In non-magnetic media with ǫr1 < ǫr2, ΓT M is real-valued and increases from a negative value for normal incidence to +1 as ψi approaches grazing incidence. Note that at any particular angle of incidence, ΓT M trends toward −1 as ǫr2/ǫr1 →∞. In this respect, the behavior of the TM component is similar to that of the TE component. In other words: As c⃝C. Wang CC BY-SA 4.0 Figure 5.23: Reﬂection of a plane wave with angle of incidence ψi equal to the polarizing angle ψi B. Here the media are non-magnetic with ǫr1 < ǫr2. ǫr2/ǫr1 →∞, the result for both the TE and TM components are increasingly similar to the result we would obtain for a perfect conductor in Region 2. Also note that when ǫr1 < ǫr2, ΓT M changes sign from negative to positive as angle of incidence increases from 0 to π/2. This behavior is quite different from that of the TE component, which is always negative for ǫr1 < ǫr2. The angle of incidence at which ΓT M = 0 is referred to as Brewster’s angle, which we assign the symbol ψi B. Thus: ψi B ≜ψi at which ΓT M = 0 (5.242) In the discussion that follows, here is the key point to keep in mind: Brewster’s angle ψi B is the angle of incidence at which ΓT M = 0. Brewster’s angle is also referred to as the polarizing angle. The motivation for the term “polarizing angle” is demonstrated in Figure 5.23. In this ﬁgure, a plane wave is incident with ψi = ψi B. The wave may contain TE and TM components in any combination. Applying the principle of superposition, we may consider these components separately. The TE component of the incident wave will scatter as reﬂected and transmitted waves which are also TE. However, ΓT M = 0 when ψi = ψi B, so the TM	Electromagnetics_Vol2_Page_99_Chunk2274
5.10. TM REFLECTION IN NON-MAGNETIC MEDIA 87 component of the transmitted wave will be TM, but the TM component of the reﬂected wave will be zero. Thus, the total (TE+TM) reﬂected wave will be purely TE, regardless of the TM component of the incident wave. This principle can be exploited to suppress the TM component of a wave having both TE and TM components. This method can be used to isolate the TE and TM components of a wave. Derivation of a formula for Brewster’s angle. Brewster’s angle for any particular combination of non-magnetic media may be determined as follows. ΓT M = 0 when the numerator of Equation 5.241 equals zero, so: −R cos ψi B + q R −sin2 ψi B = 0 (5.243) where we have made the substitution R ≜ǫr2/ǫr1 to improve clarity. Moving the second term to the right side of the equation and squaring both sides, we obtain: R2 cos2 ψi B = R −sin2 ψi B (5.244) Now employing a trigonometric identity on the left side of the equation, we obtain: R2	Electromagnetics_Vol2_Page_100_Chunk2275
88 CHAPTER 5. WAVE REFLECTION AND TRANSMISSION 5.11 Total Internal Reﬂection [m0169] Total internal reﬂection refers to a particular condition resulting in the complete reﬂection of a wave at the boundary between two media, with no power transmitted into the second region. One way to achieve complete reﬂection with zero transmission is simply to require the second material to be a perfect conductor. However, total internal reﬂection is a distinct phenomenon in which neither of the two media are perfect conductors. Total internal reﬂection has a number of practical applications; notably, it is the enabling principle of ﬁber optics. Consider the situation shown in Figure 5.25: A uniform plane wave is obliquely incident on the planar boundary between two semi-inﬁnite material regions. In Section 5.8, it is found that ψr = ψi (5.255) i.e., angle of reﬂection equals angle of incidence. Also, from Snell’s law: √µr1ǫr1 sin ψi = √µr2ǫr2 sin ψt (5.256) where “r” in the subscripts indicates the relative (unitless) quantities. The associated formula for ψt c⃝C. Wang CC BY-SA 4.0 (modiﬁed) Figure 5.25: A uniform plane wave obliquely inci- dent on the planar boundary between two semi-inﬁnite material regions. Here µr1ǫr1 > µr2ǫr2, so ψt > ψi. explicitly is: ψt = arcsin rµr1ǫr1 µr2ǫr2 sin ψi (5.257) From Equation 5.257, it is apparent that when µr1ǫr1 > µr2ǫr2, ψt > ψr; i.e., the transmitted wave appears to bend away from the surface normal, as shown in Figure 5.25. In fact, ψt can be as large as π/2 (corresponding to propagation parallel to the boundary) for angles of incidence which are less than π/2. What happens if the angle of incidence is further increased? When calculating ψt using Equation 5.257, one ﬁnds that the argument of the arcsine function becomes greater than 1. Since the possible values of the sine function are between −1 and +1, the arcsine function is undeﬁned. Clearly our analysis is inadequate in this situation. To make sense of this, let us begin by identifying the threshold angle of incidence ψi c at which the trouble begins. From the analysis in the previous paragraph, rµr1ǫr1 µr2ǫr2 sin ψi c = 1 (5.258) therefore, ψi c = arcsin rµr2ǫr2 µr1ǫr1 (5.259) This is known as the critical angle. When ψi < ψi c, our existing theory applies. When ψi ≥ψi c, the situation is, at present, unclear. For non-magnetic materials, Equation 5.259 simpliﬁes to ψi c = arcsin rǫr2 ǫr1 (5.260) Now let us examine the behavior of the reﬂection coefﬁcient. For example, the reﬂection coefﬁcient for TE component and non-magnetic materials is ΓT E = cos ψi − q ǫr2/ǫr1 −sin2 ψi cos ψi + q ǫr2/ǫr1 −sin2 ψi (5.261) From Equation 5.260, we see that sin2 ψi c = ǫr2 ǫr1 (5.262)	Electromagnetics_Vol2_Page_101_Chunk2276
5.11. TOTAL INTERNAL REFLECTION 89 So, when ψi > ψi c, we see that ǫr2/ǫr1 −sin2 ψi < 0 (ψi > ψi c) (5.263) and therefore q ǫr2/ǫr1 −sin2 ψi = jB (ψi > ψi c) (5.264) where B is a positive real-valued number. Now we may write Equation 5.261 as follows: ΓT E = A −jB A + jB (ψi > ψi c) (5.265) where A ≜cos ψi is also a positive real-valued number. Note that the numerator and denominator in Equation 5.265 have equal magnitude. Therefore, the magnitude of \|ΓT E\| = 1 and ΓT E = ejζ (ψi > ψi c) (5.266) where ζ is a real-valued number indicating the phase of ΓT E. In other words, When the angle of incidence ψi exceeds the criti- cal angle ψi c, the magnitude of the reﬂection coef- ﬁcient is 1. In this case, all power is reﬂected, and no power is transmitted into the second medium. This is total internal reﬂection. Although we have obtained this result for TE component, the identical conclusion is obtained for TM component as well. This is left as an exercise for the student. Example 5.11. Total internal reﬂection in glass. Figure 5.15 (Section 5.8) shows a demonstration of refraction of a beam of light incident from air onto a planar boundary with glass. Analysis of that demonstration revealed that the relative permittivity of the glass was ∼= 2.28. Figure 5.26 shows a modiﬁcation of the demonstration: In this case, a beam of light incident from glass onto a planar boundary with air. Conﬁrm that total internal reﬂection is the expected result in this demonstration. Solution. Assuming the glass is non-magnetic, the critical angle is given by Equation 5.260. In the present example, ǫr1 ∼= 2.28 (glass), ǫr2 ∼= 1 (air). Therefore, the critical angle ψi c ∼= 41.5◦. c⃝Z. S´andor CC BY-SA 3.0 Figure 5.26: Total internal reﬂection of a light wave incident on a planar boundary between glass and air. c⃝Timwether CC BY-SA 3.0 Figure 5.27: Laser light in a dielectric rod exhibiting the Goos-H¨anchen effect. The angle of incidence in Figure 5.26 is seen to be about ∼= 50◦, which is greater than the critical angle. Therefore, total internal reﬂection is expected. Note also that the phase ζ of the reﬂection coefﬁcient is precisely zero unless ψi > ψi c, at which point it is both non-zero and varies with ψi. This is known as the Goos-H¨anchen effect. This leads to the startling phenomenon shown in Figure 5.27. In this ﬁgure, the Goos-H¨anchen phase shift is apparent as a displacement between the point of incidence and the point of reﬂection.	Electromagnetics_Vol2_Page_102_Chunk2277
90 CHAPTER 5. WAVE REFLECTION AND TRANSMISSION The presence of an imaginary component in the reﬂection coefﬁcient is odd for two reasons. First, we are not accustomed to seeing a complex-valued reﬂection coefﬁcient emerge when the wave impedances of the associated media are real-valued. Second, the total reﬂection of the incident wave seems to contradict the boundary condition that requires the tangential components of the electric and magnetic ﬁelds to be continuous across the boundary. That is, how can these components of the ﬁelds be continuous across the boundary if no power is transmitted across the boundary? These considerations suggest that there is a ﬁeld on the opposite side of the boundary, but – somehow – it must have zero power. This is exactly the case, as is explained in Section 5.12. Additional Reading: • “Goos-H¨anchen effect” on Wikipedia. • “Snell’s law” on Wikipedia. • “Total internal reﬂection” on Wikipedia. 5.12 Evanescent Waves [m0170] Consider the situation shown in Figure 5.28: A uniform plane wave obliquely incident on the planar boundary between two semi-inﬁnite material regions, and total internal reﬂection occurs because the angle of incidence ψi is greater than the critical angle ψi c = arcsin rµr2ǫr2 µr1ǫr1 (5.267) Therefore, the reﬂection coefﬁcient is complex-valued with magnitude equal to 1 and phase that depends on polarization and the constitutive parameters of the media. The total reﬂection of the incident wave seems to contradict the boundary conditions that require the tangential components of the electric and magnetic ﬁelds to be continuous across the boundary. How can these components of the ﬁelds be continuous across the boundary if no power is transmitted across the boundary? There must be a ﬁeld on the opposite side of the boundary, but – somehow – it must have zero power. To make sense of this, let us attempt to ﬁnd a solution for the transmitted ﬁeld. c⃝C. Wang CC BY-SA 4.0 Figure 5.28: A uniform plane wave obliquely inci- dent on the planar boundary between two semi-inﬁnite material regions. Here µr1ǫr1 > µr2ǫr2 and ψi > ψi c.	Electromagnetics_Vol2_Page_103_Chunk2278
5.12. EVANESCENT WAVES 91 We begin by postulating a complex-valued angle of transmission ψtc. Although the concept of a complex-valued angle may seem counterintuitive, there is mathematical support for this concept. For example, consider the well-known trigonometric identities: sin θ = 1 j2	Electromagnetics_Vol2_Page_104_Chunk2279
92 CHAPTER 5. WAVE REFLECTION AND TRANSMISSION components in this region is described by the factor e−jkt·r where kt = β2ˆkt = β2	Electromagnetics_Vol2_Page_105_Chunk2280
5.12. EVANESCENT WAVES 93 Image Credits Fig. 5.1: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Upw incident on planar boundary.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 5.2: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Upw incident on a slab.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 5.3: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Double-boundary problem equivalent.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 5.4: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Plane wave in basic coord.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 5.5: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Plane wave in rotated coord.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 5.6: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Plane wave in another rotation coord.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 5.7: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Plane wave in ray-ﬁxed coord.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 5.8: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Coord system for TE-TM decomposition.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 5.9: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:TE-polarized upw incident on planar boundary.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 5.10: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Incident reﬂected and transmitted magnetic ﬁeld components.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 5.11: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:TE-polarized upw incident from air to glass.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 5.12: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:TM-polarized upw incident on planar boundary.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 5.13: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:TM-polarized upw incident from air to glass.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 5.14: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Upw incident obliquely on planar boundary.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).	Electromagnetics_Vol2_Page_106_Chunk2281
94 CHAPTER 5. WAVE REFLECTION AND TRANSMISSION Fig. 5.15: c⃝Z. S´andor, https://commons.wikimedia.org/wiki/File:F%C3%A9nyt%C3%B6r%C3%A9s.jpg, CC BY-SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0/). Fig. 5.16: c⃝G. Saini, https://kids.kiddle.co/Image:Refractionn.jpg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 5.17: c⃝D-Kuru, https://commons.wikimedia.org/wiki/File:Prism-side-fs PNr%C2%B00117.jpg, CC BY-SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0/). Fig. 5.18: c⃝Suidroot, https://commons.wikimedia.org/wiki/File:Prism-rainbow.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 5.19: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:TE-polarized upw incident on planar boundary.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 5.21: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:TM-polarized upw incident on planar boundary.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 5.23: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Reﬂection of plane wave incidence angle equals polarizing angle.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 5.24: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Brewster%E2%80%99s angle for non-magnetic media.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/). Fig. 5.25: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Upw incident on planar boundary angle reﬂection equals incidence.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/), modiﬁed. Fig. 5.26: c⃝Z. S´andor, https://en.wikipedia.org/wiki/File:Teljes f%C3%A9nyvisszaver%C5%91d%C3%A9s.jpg, CC BY-SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0/). Fig. 5.27: c⃝Timwether, https://en.wikipedia.org/wiki/File:Laser in ﬁbre.jpg, CC BY-SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0/). Fig. 5.28: c⃝Sevenchw (C. Wang), https://commons.wikimedia.org/wiki/File:Upw incident on planar boundary total internal reﬂection.svg, CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/).	Electromagnetics_Vol2_Page_107_Chunk2282
Chapter 6 Waveguides 6.1 Phase and Group Velocity [m0176] Phase velocity is the speed at which a point of constant phase travels as the wave propagates.1 For a sinusoidally-varying wave, this speed is easy to quantify. To see this, consider the wave: A cos (ωt −βz + ψ) (6.1) where ω = 2πf is angular frequency, z is position, and β is the phase propagation constant. At any given time, the distance between points of constant phase is one wavelength λ. Therefore, the phase velocity vp is vp = λf (6.2) Since β = 2π/λ, this may also be written as follows: vp = ω β (6.3) Noting that β = ω√µǫ for simple matter, we may also express vp in terms of the constitutive parameters µ and ǫ as follows: vp = 1 √µǫ (6.4) Since vp in this case depends only on the constitutive properties µ and ǫ, it is reasonable to view phase velocity also as a property of matter. 1Formally, “velocity” is a vector which indicates both the di- rection and rate of motion. It is common practice to use the terms “phase velocity” and “group velocity” even though we are actually referring merely to rate of motion. The direction is, of course, in the direction of propagation. Central to the concept of phase velocity is uniformity over space and time. Equations 6.2–6.4 presume a wave having the form of Equation 6.1, which exhibits precisely the same behavior over all possible time t from −∞to +∞and over all possible z from −∞to +∞. This uniformity over all space and time precludes the use of such a wave to send information. To send information, the source of the wave needs to vary at least one parameter as a function of time; for example A (resulting in amplitude modulation), ω (resulting in frequency modulation), or ψ (resulting in phase modulation). In other words, information can be transmitted only by making the wave non-uniform in some respect. Furthermore, some materials and structures can cause changes in ψ or other combinations of parameters which vary with position or time. Examples include dispersion and propagation within waveguides. Regardless of the cause, varying the parameters ω or ψ as a function of time means that the instantaneous distance between points of constant phase may be very different from λ. Thus, the instantaneous frequency of variation as a function of time and position may be very different from f. In this case Equations 6.2–6.4 may not necessarily provide a meaningful value for the speed of propagation. Some other concept is required to describe the speed of propagation of such waves. That concept is group velocity, vg, deﬁned as follows: Group velocity, vg, is the ratio of the apparent change in frequency ω to the associated change in the phase propagation constant β; i.e., ∆ω/∆β. Electromagnetics Vol. 2. c⃝2020 S.W. Ellingson CC BY SA 4.0. https://doi.org/10.21061/electromagnetics-vol-2	Electromagnetics_Vol2_Page_108_Chunk2283
96 CHAPTER 6. WAVEGUIDES Letting ∆β become vanishingly small, we obtain vg ≜∂ω ∂β (6.5) Note the similarity to the deﬁnition of phase velocity in Equation 6.3. Group velocity can be interpreted as the speed at which a disturbance in the wave propagates. Information may be conveyed as meaningful disturbances relative to a steady-state condition, so group velocity is also the speed of information in a wave. Note Equation 6.5 yields the expected result for waves in the form of Equation 6.1: vg = ∂β ∂ω −1 = ∂ ∂ω ω√µǫ −1 = 1 √µǫ = vp (6.6) In other words, the group velocity of a wave in the form of Equation 6.1 is equal to its phase velocity. To observe a difference between vp and vg, β must somehow vary as a function of something other than just ω and the constitutive parameters. Again, modulation (introduced by the source of the wave) and dispersion (frequency-dependent constitutive parameters) are examples in which vg is not necessarily equal to vp. Here’s an example involving dispersion: Example 6.1. Phase and group velocity for a material exhibiting square-law dispersion. A broad class of non-magnetic dispersive media exhibit relative permittivity ǫr that varies as the square of frequency over a narrow range of frequencies centered at ω0. For these media we presume ǫr = K ω ω0 2 (6.7) where K is a real-valued positive constant. What is the phase and group velocity for a sinusoidally-varying wave in this material? Solution. First, note β = ω√µ0ǫ = ω√µ0ǫ0 √ǫr = √ K · ω2 ω0 √µ0ǫ0 (6.8) The phase velocity is: vp = ω β = ω0 √ K · ω√µ0ǫ0 (6.9) Whereas the group velocity is: vg = ∂ω ∂β = ∂β ∂ω −1 (6.10) = ∂ ∂ω √ K · ω2 ω0 √µ0ǫ0 !−1 (6.11) = 2 √ K · ω ω0 √µ0ǫ0 !−1 (6.12) Now simplifying using Equation 6.8: vg = 2β ω −1 (6.13) = 1 2 ω β (6.14) = 1 2vp (6.15) Thus, we see that in this case the group velocity is always half the phase velocity. Another commonly-encountered example for which vg is not necessarily equal to vp is the propagation of guided waves; e.g., waves within a waveguide. In fact, such waves may exhibit phase velocity greater than the speed of light in a vacuum, c. However, the group velocity remains less than c, which means the speed at which information may propagate in a waveguide is less than c. No physical laws are violated, since the universal “speed limit” c applies to information, and not simply points of constant phase. (See “Additional Reading” at the end of this section for more on this concept.) Additional Reading: • “Group velocity” on Wikipedia.	Electromagnetics_Vol2_Page_109_Chunk2284
6.2. PARALLEL PLATE WAVEGUIDE: INTRODUCTION 97 • “Phase velocity” on Wikipedia. • “Speed of light” on Wikipedia. 6.2 Parallel Plate Waveguide: Introduction [m0173] A parallel plate waveguide is a device for guiding the propagation of waves between two perfectly-conducting plates. Our primary interest in this structure is as a rudimentary model applicable to a broad range of engineering problems. Examples of such problems include analysis of the ﬁelds within microstrip line and propagation of radio waves in the ionosphere. Figure 6.1 shows the geometry of interest. Here the plates are located at z = 0 and z = a. The plates are assumed to be inﬁnite in extent, and therefore there is no need to consider ﬁelds in the regions z < 0 or z > a. For this analysis, the region between the plates is assumed to consist of an ideal (lossless) material exhibiting real-valued permeability µ and real-valued permittivity ǫ. Let us limit our attention to a region within the waveguide which is free of sources. Expressed in phasor form, the electric ﬁeld intensity is governed by the wave equation ∇2 eE + β2 eE = 0 (6.16) where β = ω√µǫ (6.17) Equation 6.16 is a partial differential equation. This equation, combined with boundary conditions c⃝C. Wang CC BY-SA 4.0 (modiﬁed) Figure 6.1: Geometry for analysis of ﬁelds in a paral- lel plate waveguide.	Electromagnetics_Vol2_Page_110_Chunk2285
98 CHAPTER 6. WAVEGUIDES imposed by the perfectly-conducting plates, is sufﬁcient to determine a unique solution. This is most easily done in Cartesian coordinates, as we shall now demonstrate. First we express eE in Cartesian coordinates: eE = ˆx eEx + ˆy eEy + ˆz eEz (6.18) This facilitates the decomposition of Equation 6.16 into separate equations governing the ˆx, ˆy, ˆz components of eE: ∇2 eEx + β2 eEx = 0 (6.19) ∇2 eEy + β2 eEy = 0 (6.20) ∇2 eEz + β2 eEz = 0 (6.21) Next we observe that the operator ∇2 may be expressed in Cartesian coordinates as follows: ∇2 = ∂2 ∂x2 + ∂2 ∂y2 + ∂2 ∂z2 (6.22) so the equations governing the Cartesian components of eE may be written as follows: ∂2 ∂x2 eEx + ∂2 ∂y2 eEx + ∂2 ∂z2 eEx = −β2 eEx (6.23) ∂2 ∂x2 eEy + ∂2 ∂y2 eEy + ∂2 ∂z2 eEy = −β2 eEy (6.24) ∂2 ∂x2 eEz + ∂2 ∂y2 eEz + ∂2 ∂z2 eEz = −β2 eEz (6.25) Let us restrict our attention to scenarios that can be completely described in two dimensions; namely x and z, and for which there is no variation in y. This is not necessarily required, however, it is representative of a broad class of relevant problems, and allows Equations 6.23–6.25 to be considerably simpliﬁed. If there is no variation in y, then partial derivatives with respect to y are zero, yielding: ∂2 ∂x2 eEx + ∂2 ∂z2 eEx = −β2 eEx (6.26) ∂2 ∂x2 eEy + ∂2 ∂z2 eEy = −β2 eEy (6.27) ∂2 ∂x2 eEz + ∂2 ∂z2 eEz = −β2 eEz (6.28) The solution to the parallel plate waveguide problem may now be summarized as follows: The electric ﬁeld eE (Equation 6.18) is the solution to simultaneous Equations 6.26–6.28 subject to the boundary conditions that apply at the PEC surfaces at z = 0 and z = d; namely, that the tangent component of eE is zero at these surfaces. At this point, the problem has been reduced to a routine exercise in the solution of partial differential equations. However, a somewhat more useful approach is to ﬁrst decompose the total electric ﬁeld into transverse electric (TE) and transverse magnetic (TM) components,2 and determine the solutions to these components separately. The TE component of the electric ﬁeld is parallel to the plates, and therefore transverse (perpendicular) to the plane shown in Figure 6.1. Thus, the TE component has eEx = eEz = 0, and only eEy may be non-zero. The TM component of the magnetic ﬁeld intensity ( eH) is parallel to the plates, and therefore transverse to the plane shown in Figure 6.1. Thus, the TM component has eHx = eHz = 0, and only eHy may be non-zero. As always, the total ﬁeld is the sum of the TE and TM components. The TE and TM solutions are presented in Sections 6.3 (Electric component of the TE solution), 6.4 (Magnetic component of the TE solution), and 6.5 (Electric component of the TM solution). The magnetic component of the TM solution can be determined via a straightforward variation of the preceding three cases, and so is not presented here. Presentation of the solution for the ﬁelds in a parallel plate waveguide under the conditions described in the section continues in Section 6.3. 2For a refresher on TE-TM decomposition, see Section 5.5.	Electromagnetics_Vol2_Page_111_Chunk2286
6.3. PARALLEL PLATE WAVEGUIDE: TE CASE, ELECTRIC FIELD 99 6.3 Parallel Plate Waveguide: TE Case, Electric Field [m0174] In Section 6.2, the parallel plate waveguide was introduced. At the end of that section, we described the decomposition of the problem into its TE and TM components. In this section, we ﬁnd the electric ﬁeld component of the TE ﬁeld in the waveguide. Figure 6.2 shows the problem addressed in this section. (Additional details and assumptions are addressed in Section 6.2.) Since eEx = eEz = 0 for the TE component of the electric ﬁeld, Equations 6.26 and 6.28 are irrelevant, leaving only: ∂2 ∂x2 eEy + ∂2 ∂z2 eEy = −β2 eEy (6.29) The general solution to this partial differential equation is: eEy = e−jkzz Ae−jkxx + Be+jkxx + e+jkzz Ce−jkxx + De+jkxx (6.30) where A, B, C, and D are complex-valued constants and kx and kz are real-valued constants. We have assigned variable names to these constants with advance knowledge of their physical interpretation; however, at this moment they remain simply unknown constants whose values must be determined by enforcement of boundary conditions. c⃝C. Wang CC BY-SA 4.0 Figure 6.2: TE component of the electric ﬁeld in a parallel plate waveguide. Note that Equation 6.30 consists of two terms. The ﬁrst term includes the factor e−jkzz, indicating a wave propagating in the +ˆz direction, and the second term includes the factor e+jkzz, indicating a wave propagating in the −ˆz direction. If we impose the restriction that sources exist only on the left (z < 0) side of Figure 6.2, and that there be no structure capable of wave scattering (in particular, reﬂection) on the right (z > 0) side of Figure 6.2, then there can be no wave components propagating in the −ˆz direction. In this case, C = D = 0 and Equation 6.30 simpliﬁes to: eEy = e−jkzz Ae−jkxx + Be+jkxx (6.31) Before proceeding, let’s make sure that Equation 6.31 is actually a solution to Equation 6.29. As we shall see in a moment, performing this check will reveal some additional useful information. First, note: ∂eEy ∂x = e−jkzz −Ae−jkxx + Be+jkxx (jkx) (6.32) So: ∂2 eEy ∂x2 = e−jkzz Ae−jkxx + Be+jkxx	Electromagnetics_Vol2_Page_112_Chunk2287
100 CHAPTER 6. WAVEGUIDES This conﬁrms that kx and kz are in fact the components of the propagation vector k ≜βˆk = ˆxkx + ˆyky + ˆzkz (6.39) where ˆk is the unit vector pointing in the direction of propagation, and ky = 0 in this particular problem. The solution has now been reduced to ﬁnding the constants A, B, and either kx or kz. This is accomplished by enforcing the relevant boundary conditions. In general, the component of eE that is tangent to a perfectly-conducting surface is zero. Applied to the present problem, this means eEy = 0 at x = 0 and eEy = 0 at x = a. Referring to Equation 6.31, the boundary condition at x = 0 means e−jkzz [A · 1 + B · 1] = 0 (6.40) The factor e−jkzz cannot be zero; therefore, A + B = 0. Since B = −A, we may rewrite Equation 6.31 as follows: eEy = e−jkzzB e+jkxx −e−jkxx (6.41) This expression is simpliﬁed using a trigonometric identity: 1 2j e+jkxx −e−jkxx = sin kxx (6.42) Let us now make the deﬁnition Ey0 ≜j2B. Then: eEy = Ey0e−jkzz sin kxx (6.43) Now applying the boundary condition at x = a: Ey0e−jkzz sin kxa = 0 (6.44) The factor e−jkzz cannot be zero, and Ey0 = 0 yields only trivial solutions; therefore: sin kxa = 0 (6.45) This in turn requires that kxa = mπ (6.46) where m is an integer. Note that m = 0 is not of interest since this yields kx = 0, which according to Equation 6.43 yields the trivial solution eEy = 0. Also each integer value of m that is less than zero is excluded because the associated solution is different from the solution for the corresponding positive value of m in sign only, which can be absorbed in the arbitrary constant Ey0. At this point we have uncovered a family of solutions given by Equation 6.43 and Equation 6.46 with m = 1, 2, .... Each solution associated with a particular value of m is referred to as a mode, which (via Equation 6.46) has a particular value of kx. The value of kz for mode m is obtained using Equation 6.38 as follows: kz = p β2 −k2x = r β2 − mπ a 2 (6.47) Since kz is speciﬁed to be real-valued, we require: β2 − mπ a 2 > 0 (6.48) This constrains β; speciﬁcally: β > mπ a (6.49) Recall that β = ω√µǫ and ω = 2πf where f is frequency. Solving for f, we ﬁnd: f > m 2a√µǫ (6.50) Therefore, each mode exists only above a certain frequency, which is different for each mode. This cutoff frequency fc for mode m is given by f (m) c ≜ m 2a√µǫ (6.51) At frequencies below the cutoff frequency for mode m, modes 1 through m −1 exhibit imaginary-valued kz. The propagation constant must have a real-valued component in order to propagate; therefore, these modes do not propagate and may be ignored. Let us now summarize the solution. For the scenario depicted in Figure 6.2, the electric ﬁeld component of the TE solution is given by: ˆy eEy = ˆy ∞ X m=1 eE(m) y (6.52)	Electromagnetics_Vol2_Page_113_Chunk2288
6.3. PARALLEL PLATE WAVEGUIDE: TE CASE, ELECTRIC FIELD 101 where eE(m) y ≜ ( 0, f < f (m) c E(m) y0 e−jk(m) z z sin k(m) x x, f ≥f (m) c (6.53) where m enumerates modes (m = 1, 2, ...), k(m) z ≜ r β2 − h k(m) x i2 (6.54) and k(m) x ≜mπ/a (6.55) Finally, E(m) y0 is a complex-valued constant that depends on sources or boundary conditions to the left of the region of interest. For the scenario depicted in Figure 6.2, the elec- tric ﬁeld component of the TE solution is given by Equation 6.52 with modal components deter- mined as indicated by Equations 6.51–6.55. This solution presumes all sources lie to the left of the region of interest, and no scattering occurs to the right of the region of interest. To better understand this result, let us examine the lowest-order mode, m = 1. For this mode f (1) c = 1/2a√µǫ, so this mode can exist if f > 1/2a√µǫ. Also k(1) x = π/a, so k(1) z = r β2 − π a 2 (6.56) Subsequently, eE(1) y = E(1) y0 e−jk(1) z z sin πx a (6.57) Note that this mode has the form of a plane wave. The plane wave propagates in the +ˆz direction with phase propagation constant k(1) z . Also, we observe that the apparent plane wave is non-uniform, exhibiting magnitude proportional to sin πx/a within the waveguide. This is shown in Figure 6.3 (left image). In particular, we observe that the magnitude of the wave is zero at the perfectly-conducting surfaces – as is necessary to satisfy the boundary conditions – and is maximum in the center of the waveguide. Now let us examine the m = 2 mode. For this mode, f (2) c = 1/a√µǫ, so this mode can exist if c⃝C. Wang CC BY-SA 4.0 Figure 6.3: Magnitude of the two lowest-order TE modes in a parallel plate waveguide. Left: m = 1, Right: m = 2. f > 1/a√µǫ. This frequency is higher than f (1) c , so the m = 1 mode can exist at any frequency at which the m = 2 mode exists. Also k(2) x = 2π/a, so k(2) z = s β2 − 2π a 2 (6.58) Subsequently, eE(2) y = E(2) y0 e−jk(2) z z sin 2πx a (6.59) In this case, the apparent plane wave propagates in the +ˆz direction with phase propagation constant k(2) z , which is less than k(1) z . For m = 2, we ﬁnd magnitude is proportional to sin 2πx/a within the waveguide (Figure 6.3, right image). As in the m = 1 case, we observe that the magnitude of the wave is zero at the PEC surfaces; however, for m = 2, there are two maxima with respect to x, and the magnitude in the center of the waveguide is zero. This pattern continues for higher-order modes. In particular, each successive mode exhibits higher cutoff frequency, smaller propagation constant, and increasing integer number of sinusoidal half-periods in magnitude. Example 6.2. Single-mode TE propagation in a parallel plate waveguide. Consider an air-ﬁlled parallel plate waveguide consisting of plates separated by 1 cm. Determine the frequency range for which one	Electromagnetics_Vol2_Page_114_Chunk2289
102 CHAPTER 6. WAVEGUIDES (and only one) propagating TE mode is assured. Solution. Single-mode TE propagation is assured by limiting frequency f to greater than the cutoff frequency for m = 1, but lower than the cutoff frequency for m = 2. (Any frequency higher than the cutoff frequency for m = 2 allows at least 2 modes to exist.) Calculating the applicable cutoff frequencies, we ﬁnd: f (1) c = 1 2a√µ0ǫ0 ∼= 15.0 GHz (6.60) f (2) c = 2 2a√µ0ǫ0 ∼= 30.0 GHz (6.61) Therefore, 15.0 GHz ≤f ≤30.0 GHz. Finally, let us consider the phase velocity vp within the waveguide. For lowest-order mode m = 1, this is vp = ω k(1) z = ω q ω2µǫ −	Electromagnetics_Vol2_Page_115_Chunk2290
6.4. PARALLEL PLATE WAVEGUIDE: TE CASE, MAGNETIC FIELD 103 and ∂eE(m) y ∂x = ∂ ∂x E(m) y0 e−jk(m) z z sin k(m) x x = E(m) y0 e−jk(m) z z cos k(m) x x +k(m) x (6.68) We may now assemble a solution for the magnetic ﬁeld as follows: ˆx eHx + ˆz eHz = ˆx ∞ X m=1 eH(m) x + ˆz ∞ X m=1 eH(m) z (6.69) where eH(m) x = −k(m) z ωµ E(m) y0 e−jk(m) z z sin k(m) x x (6.70) eH(m) z = +j k(m) x ωµ E(m) y0 e−jk(m) z z cos k(m) x x (6.71) and modes may only exist at frequencies greater than the associated cutoff frequencies. Summarizing: The magnetic ﬁeld component of the TE solu- tion is given by Equation 6.69 with modal com- ponents as indicated by Equations 6.70 and 6.71. Caveats pertaining to the cutoff frequencies and the locations of sources and structures continue to apply. This result is quite complex, yet some additional insight is possible. At the perfectly-conducting (PEC) surface at x = 0, we see eH(m) x (x = 0) = 0 eH(m) z (x = 0) = +j k(m) x ωµ E(m) y0 e−jk(m) z z (6.72) Similarly, on the PEC surface at x = a, we see eH(m) x (x = a) = 0 eH(m) z (x = a) = −j k(m) x ωµ E(m) y0 e−jk(m) z z (6.73) Thus, we see the magnetic ﬁeld vector at the PEC surfaces is non-zero and parallel to the PEC surfaces. Recall that the magnetic ﬁeld is identically zero inside a PEC material. Also recall that boundary conditions require that discontinuity in the component of H tangent to a surface must be supported by a surface current. We conclude that Current ﬂows on the PEC surfaces of the waveg- uide. If this seems surprising, note that essentially the same thing happens in a coaxial transmission line. That is, signals in a coaxial transmission line can be described equally well in terms of either potentials and currents on the inner and outer conductors, or the electromagnetic ﬁelds between the conductors. The parallel plate waveguide is only slightly more complicated because the ﬁeld in a properly-designed coaxial cable is a single transverse electromagnetic (TEM) mode, whereas the ﬁelds in a parallel plate waveguide are combinations of TE and TM modes. Interestingly, we also ﬁnd that the magnetic ﬁeld vector points in different directions depending on position relative to the conducting surfaces. We just determined that the magnetic ﬁeld is parallel to the conducting surfaces at those surfaces. However, the magnetic ﬁeld is perpendicular to those surfaces at m locations between x = 0 and x = a. These locations correspond to maxima in the electric ﬁeld.	Electromagnetics_Vol2_Page_116_Chunk2291
104 CHAPTER 6. WAVEGUIDES 6.5 Parallel Plate Waveguide: TM Case, Electric Field [m0177] In Section 6.2, the parallel plate waveguide shown in Figure 6.4 was introduced. At the end of that section, we decomposed the problem into its TE and TM components. In this section, we ﬁnd the TM component of the ﬁelds in the waveguide. “Transverse magnetic” means the magnetic ﬁeld vector is perpendicular to the plane of interest, and is therefore parallel to the conducting surfaces. Thus, eH = ˆy eHy, with no component in the ˆx or ˆz directions. Following precisely the same reasoning employed in Section 6.2, we ﬁnd the governing equation for the magnetic component of TM ﬁeld is: ∂2 ∂x2 eHy + ∂2 ∂z2 eHy = −β2 eHy (6.74) The general solution to this partial differential equation is: eHy = e−jkzz Ae−jkxx + Be+jkxx + e+jkzz Ce−jkxx + De+jkxx (6.75) where A, B, C, and D are complex-valued constants; and kx and kz are real-valued constants. We have assigned variable names to these constants with advance knowledge of their physical interpretation; however, at this moment they remain simply unknown constants whose values must be determined by enforcement of boundary conditions. c⃝C. Wang CC BY-SA 4.0 Figure 6.4: TM component of the electric ﬁeld in a parallel plate waveguide. Note that Equation 6.75 consists of two terms. The ﬁrst term includes the factor e−jkzz, indicating a wave propagating in the +ˆz direction, and the second term includes the factor e+jkzz, indicating a wave propagating in the −ˆz direction. If we impose the restriction that sources exist only on the left (z < 0) side of Figure 6.4, and that there be no structure capable of wave scattering (in particular, reﬂection) on the right (z > 0) side of Figure 6.4, then there can be no wave components propagating in the −ˆz direction. In this case, C = D = 0 and Equation 6.75 simpliﬁes to: eHy = e−jkzz Ae−jkxx + Be+jkxx (6.76) Before proceeding, let’s make sure that Equation 6.76 is actually a solution to Equation 6.74. As in the TE case, this check yields a constraint (in fact, the same constraint) on the as-yet undetermined parameters kx and kz. First, note: ∂eHy ∂x = e−jkzz −Ae−jkxx + Be+jkxx (jkx) (6.77) So: ∂2 eHy ∂x2 = e−jkzz Ae−jkxx + Be+jkxx	Electromagnetics_Vol2_Page_117_Chunk2292
6.5. PARALLEL PLATE WAVEGUIDE: TM CASE, ELECTRIC FIELD 105 This is precisely the same constraint identiﬁed in the TE case, and conﬁrms that kx and ky are in fact the components of the propagation vector k ≜βˆk = ˆxkx + ˆyky + ˆzkz (6.83) where ˆk is the unit vector pointing in the direction of propagation, and ky = 0 in this particular problem. Our objective in this section is to determine the electric ﬁeld component of the TM ﬁeld. The electric ﬁeld may be obtained from the magnetic ﬁeld using Ampere’s law: ∇× eH = jωǫeE (6.84) Thus: eE = 1 jωǫ ∇× eH = 1 jωǫ ∇× ˆy eHy (6.85) The relevant form of the curl operator is Equation B.16 (Appendix B.2). Although the complete expression consists of 6 terms, all but 2 of these terms are zero because the ˆx and ˆz components of eH are zero. The two remaining terms are −ˆx∂eHy/∂z and +ˆz∂eHy/∂x. Thus: eE = 1 jωǫ −ˆx∂eHy ∂z + ˆz∂eHy ∂x ! (6.86) We may further develop this expression using Equations 6.77 and 6.79. We ﬁnd the ˆx component of eE is: eEx = kz ωǫ e−jkzz Ae−jkxx + Be+jkxx (6.87) and the ˆz component of eE is: eEz = kx ωǫ e−jkzz −Ae−jkxx + Be+jkxx (6.88) The solution has now been reduced to the problem of ﬁnding the constants A, B, and either kx or kz. This is accomplished by enforcing the relevant boundary conditions. In general, the component of the electric ﬁeld which is tangent to a perfectly-conducting surface is zero. Applied to the present (TM) case, this means eEz (x = 0) = 0 and eEz (x = a) = 0. Referring to Equation 6.88, the boundary condition at x = 0 means kx ωǫ e−jkzz [−A (1) + B (1)] = 0 (6.89) The factor e−jkzz always has unit magnitude, and so cannot be zero. We could require kx to be zero, but this is unnecessarily restrictive. Instead, we require A = B and we may rewrite Equation 6.88 as follows: eEz = Bkx ωǫ e−jkzz e+jkxx −e−jkxx (6.90) This expression is simpliﬁed using a trigonometric identity: sin kxa = 1 2j e+jkxa −e−jkxa (6.91) Thus: eEz = j2Bkx ωǫ e−jkzz sin kxx (6.92) Now following up with eEx, beginning from Equation 6.87: eEx = kz ωǫ e−jkzz Ae−jkxx + Be+jkxx = Bkz ωǫ e−jkzz e−jkxx + e+jkxx = 2Bkz ωǫ e−jkzz cos kxx (6.93) For convenience we deﬁne the following complex-valued constant: Ex0 ≜2Bkz ωǫ (6.94) This yields the following simpler expression: eEx = Ex0 e−jkzz cos kxx (6.95) Now let us apply the boundary condition at x = a to eEz: j2Bkz ωǫ e−jkzz sin kxa = 0 (6.96) Requiring B = 0 or kz = 0 yields only trivial solutions, therefore, it must be true that sin kxa = 0 (6.97) This in turn requires that kxa = mπ (6.98)	Electromagnetics_Vol2_Page_118_Chunk2293
106 CHAPTER 6. WAVEGUIDES where m is an integer. Note that this is precisely the same relationship that we identiﬁed in the TE case. There is an important difference, however. In the TE case, m = 0 was not of interest because this yields kx = 0, and the associated ﬁeld turned out to be identically zero. In the present (TM) case, m = 0 also yields kx = 0, but the associated ﬁeld is not necessarily zero. That is, for m = 0, eEz = 0 but eEx is not necessarily zero. Therefore, m = 0 as well as m = 1, m = 2, and so on are of interest in the TM case. At this point, we have uncovered a family of solutions with m = 0, 1, 2, .... Each solution is referred to as a mode, and is associated with a particular value of kx. In the discussion that follows, we shall ﬁnd that the consequences are identical to those identiﬁed in the TE case, except that m = 0 is now also allowed. Continuing: The value of kz for mode m is obtained using Equation 6.82 as follows: kz = p β2 −k2x = r β2 − mπ a 2 (6.99) Since kz is speciﬁed to be real-valued, we require: β2 − mπ a 2 > 0 (6.100) This constrains β; speciﬁcally: β > mπ a (6.101) Recall that β = ω√µǫ and ω = 2πf where f is frequency. Solving for f, we ﬁnd: f > m 2a√µǫ (6.102) Thus, each mode exists only above a certain frequency, which is different for each mode. This cutoff frequency fc for mode m is given by f (m) c ≜ m 2a√µǫ (6.103) At frequencies below the cutoff frequency for mode m, modes 0 through m −1 exhibit imaginary-valued kz and therefore do no propagate. Also, note that the cutoff frequency for m = 0 is zero, and so this mode is always able to propagate. That is, the m = 0 mode may exist for any a > 0 and any f > 0. Once again, this is a remarkable difference from the TE case, for which m = 0 is not available. Let us now summarize the solution. With respect to Figure 6.4, we ﬁnd that the electric ﬁeld component of the TM ﬁeld is given by: eE = ∞ X m=0 h ˆx eE(m) x + ˆz eE(m) z i (6.104) where eE(m) x ≜ ( 0, f < f (m) c E(m) x0 e−jk(m) z z cos k(m) x x, f ≥f (m) c (6.105) and eE(m) z ≜ ( 0, f < f (m) c j k(m) x k(m) z E(m) x0 e−jk(m) z z sin k(m) x x, f ≥f (m) c (6.106) where m enumerates modes (m = 0, 1, 2, ...) and k(m) z ≜ r β2 − h k(m) x i2 (6.107) k(m) x ≜mπ/a (6.108) Finally, the coefﬁcients E(m) x0 depend on sources and/or boundary conditions to the left of the region of interest. For the scenario depicted in Figure 6.4, the elec- tric ﬁeld component of the TM solution is given by Equation 6.104 with modal components de- termined as indicated by Equations 6.103–6.108. This solution presumes all sources lie to the left of the region of interest, with no additional sources or boundary conditions to the right of the region of interest. The m = 0 mode, commonly referred to as the “TM0” mode, is of particular importance in the analysis of microstrip transmission line, and is addressed in Section 6.6.	Electromagnetics_Vol2_Page_119_Chunk2294
6.6. PARALLEL PLATE WAVEGUIDE: THE TM0 MODE 107 6.6 Parallel Plate Waveguide: The TM0 Mode [m0220] In Section 6.2, the parallel plate waveguide (also shown in Figure 6.5) was introduced. At the end of that section we decomposed the problem into its constituent TE and TM ﬁelds. In Section 6.5, we determined the electric ﬁeld component of the TM ﬁeld, which was found to consist of a set of discrete modes. In this section, we address the lowest-order (m = 0) mode of the TM ﬁeld, which has special relevance in a number of applications including microstrip transmission lines. This mode is commonly referred to as the “TM0” mode. The TM electric ﬁeld intensity in the waveguide is given by Equation 6.104 with modal components determined as indicated by Equations 6.103–6.108 (Section 6.5). Recall that the m = 0 mode can only exist only in the TM case, and does not exist in the TE case. We also noted that the cutoff frequency for this mode is zero, so it may exist at any frequency, and within any non-zero plate separation a. For this mode k(0) x = 0, k(0) z = β, and we ﬁnd eE = ˆxE(0) x0 e−jβz (TM0 mode) (6.109) Remarkably, we ﬁnd that this mode has the form of a uniform plane wave which propagates in the +ˆz direction; i.e., squarely between the plates. The phase and group velocities of this wave are equal to each other and ω/β, precisely as expected for a uniform c⃝C. Wang CC BY-SA 4.0 (modiﬁed) Figure 6.5: TM0 mode in a parallel plate waveguide. plane wave in unbounded media; i.e., as if the plates did not exist. This observation allows us to easily determine the associated magnetic ﬁeld: Using the plane wave relationship, eH = 1 η ˆk × eE (6.110) = 1 η ˆz × ˆxE(0) x0 e−jβz (6.111) = ˆyE(0) x0 η e−jβz (TM0 mode) (6.112) Example 6.3. Guided waves in a printed circuit board (PCB). A very common form of PCB consists of a 1.575 mm-thick slab of low-loss dielectric having relative permittivity ≈4.5 sandwiched between two copper planes. Characterize the electromagnetic ﬁeld in a long strip of this material. Assume a single source exists at one end of the strip, and operates at frequencies below 10 GHz. Solution. Let us assume that the copper planes exhibit conductivity that is sufﬁciently high that the inward-facing surfaces may be viewed as perfectly-conducting. Also, let us limit scope to the ﬁeld deep within the “sandwich,” and neglect the region near the edges of the PCB. Under these conditions, the PCB is well-modeled as a parallel-plate waveguide. Thus, the electromagnetic ﬁeld consists of a combination of TE and TM modes. The active (non-zero) modes depend on the source (a mode must be “stimulated” by the source in order to propagate) and modal cutoff frequencies. The cutoff frequency for mode m is f (m) c = m 2a√µǫ (6.113) In this case, a = 1.575 mm, µ ≈µ0, and ǫ ≈4.5ǫ0. Therefore: f (m) c ≈(44.9 GHz) m (6.114) Since the cutoff frequency for m = 1 is much greater than 10 GHz, we may rest assured that	Electromagnetics_Vol2_Page_120_Chunk2295
108 CHAPTER 6. WAVEGUIDES the only mode that can propagate inside the PCB is TM0. Therefore, the ﬁeld deep inside the PCB may be interpreted as a single plane wave having the TM0 structure shown in Figure 6.5, propagating away from the source end of the PCB. The phase velocity is simply that of the apparent plane wave: vp = 1 √µǫ ≈1.41 × 108 m/s ≈0.47c (6.115) The scenario described in this example is essentially a very rudimentary form of microstrip transmission line. 6.7 General Relationships for Unidirectional Waves [m0222] Analysis of electromagnetic waves in enclosed spaces – and in waveguides in particular – is quite difﬁcult. The task is dramatically simpliﬁed if it can be assumed that the wave propagates in a single direction; i.e., is unidirectional. This does not necessarily entail loss of generality. For example: Within a straight waveguide, waves can travel either “forward” or “backward.” The principle of superposition allows one to consider these two unidirectional cases separately, and then to simply sum the results. In this section, the equations that relate the various components of a unidirectional wave are derived. The equations will be derived in Cartesian coordinates, anticipating application to rectangular waveguides. However, the underlying strategy is generally applicable. We begin with Maxwell’s curl equations: ∇× eE = −jωµ eH (6.116) ∇× eH = +jωǫeE (6.117) Let us consider Equation 6.117 ﬁrst. Solving for eE, we have: eE = 1 jωǫ∇× eH (6.118) This is actually three equations; that is, one each for the ˆx, ˆy, and ˆz components of eE, respectively. To extract these equations, let us deﬁne the components as follows: eE = ˆx eEx + ˆy eEy + ˆz eEz (6.119) eH = ˆx eHx + ˆy eHy + ˆz eHz (6.120) Now applying the equation for curl in Cartesian coordinates (Equation B.16 in Appendix B.2), we	Electromagnetics_Vol2_Page_121_Chunk2296
6.7. GENERAL RELATIONSHIPS FOR UNIDIRECTIONAL WAVES 109 ﬁnd: eEx = 1 jωǫ ∂eHz ∂y −∂eHy ∂z ! (6.121) eEy = 1 jωǫ ∂eHx ∂z −∂eHz ∂x ! (6.122) eEz = 1 jωǫ ∂eHy ∂x −∂eHx ∂y ! (6.123) Without loss of generality, we may assume that the single direction in which the wave is traveling is in the +ˆz direction. If this is the case, then eH, and subsequently each component of eH, contains a factor of e−jkzz where kz is the phase propagation constant in the direction of travel. The remaining factors are independent of z; i.e., depend only on x and y. With this in mind, we further decompose components of eH as follows: eHx = ehx(x, y)e−jkzz (6.124) eHy = ehy(x, y)e−jkzz (6.125) eHz = ehz(x, y)e−jkzz (6.126) where ehx(x, y), ehy(x, y), and ehz(x, y) represent the remaining factors. One advantage of this decomposition is that partial derivatives with respect to z reduce to algebraic operations; i.e.: ∂eHx ∂z = −jkzehx(x, y)e−jkzz = −jkz eHx (6.127) ∂eHy ∂z = −jkzehy(x, y)e−jkzz = −jkz eHy (6.128) ∂eHz ∂z = −jkzehz(x, y)e−jkzz = −jkz eHz (6.129) We now substitute Equations 6.124–6.126 into Equations 6.121–6.123 and then use Equations 6.127–6.129 to eliminate partial derivatives with respect to z. This yields: eEx = 1 jωǫ ∂eHz ∂y + jkz eHy ! (6.130) eEy = 1 jωǫ −jkz eHx −∂eHz ∂x ! (6.131) eEz = 1 jωǫ ∂eHy ∂x −∂eHx ∂y ! (6.132) Applying the same procedure to the curl equation for eH (Equation 6.116), one obtains: eHx = 1 −jωµ ∂eEz ∂y + jkz eEy ! (6.133) eHy = 1 −jωµ −jkz eEx −∂eEz ∂x ! (6.134) eHz = 1 −jωµ ∂eEy ∂x −∂eEx ∂y ! (6.135) Equations 6.130–6.135 constitute a set of simultaneous equations that represent Maxwell’s curl equations in the special case of a unidirectional (speciﬁcally, a +ˆz-traveling) wave. With just a little bit of algebraic manipulation of these equations, it is possible to obtain expressions for the ˆx and ˆy components of eE and eH which depend only on the ˆz components of eE and eH. Here they are:3 eEx = −j k2ρ +kz ∂eEz ∂x + ωµ∂eHz ∂y ! (6.136) eEy = +j k2ρ −kz ∂eEz ∂y + ωµ∂eHz ∂x ! (6.137) eHx = +j k2ρ ωǫ∂eEz ∂y −kz ∂eHz ∂x ! (6.138) eHy = −j k2ρ ωǫ∂eEz ∂x + kz ∂eHz ∂y ! (6.139) where k2 ρ ≜β2 −k2 z (6.140) Why deﬁne a parameter called “kρ”? Note from the deﬁnition that β2 = k2 ρ + k2 z. Further, note that this equation is an expression of the Pythagorean theorem, which relates the lengths of the sides of a right triangle. Since β is the “overall” phase propagation constant, and kz is the phase propagation constant for propagation in the ˆz direction in the waveguide, kρ must be associated with variation in ﬁelds in directions perpendicular to ˆz. In the cylindrical coordinate system, this is the ˆρ direction, hence the subscript “ρ.” We shall see later that kρ plays a special 3Students are encouraged to derive these for themselves – only algebra is required. Tip to get started: To get e Ex, begin with Equa- tion 6.130, eliminate e Hy using Equation 6.134, and then solve for e Ex.	Electromagnetics_Vol2_Page_122_Chunk2297
110 CHAPTER 6. WAVEGUIDES role in determining the structure of ﬁelds within the waveguide, and this provides additional motivation to identify this quantity explicitly in the ﬁeld equations. Summarizing: If you know the wave is unidirectional, then knowledge of the components of eE and eH in the direction of propagation is sufﬁcient to determine each of the remaining components of eE and eH. There is one catch, however: For this to yield sensible results, kρ may not be zero. So, ironically, these expressions don’t work in the case of a uniform plane wave, since such a wave would have eEz = eHz = 0 and kz = β (so kρ = 0), yielding values of zero divided by zero for each remaining ﬁeld component. The same issue arises for any other transverse electromagnetic (TEM) wave. For non-TEM waves – and, in particular, unidirectional waves in waveguides – either eEz or eHz must be non-zero and kρ will be non-zero. In this case, Equations 6.136–6.139 are both usable and useful since they allow determination of all ﬁeld components given just the z components. In fact, we may further exploit this simplicity by taking one additional step: Decomposition of the unidirectional wave into transverse electric (TE) and transverse magnetic (TM) components. In this case, TE means simply that eEz = 0, and TM means simply that eHz = 0. For a wave which is either TE or TM, Equations 6.136–6.139 reduce to one term each. 6.8 Rectangular Waveguide: TM Modes [m0223] A rectangular waveguide is a conducting cylinder of rectangular cross section used to guide the propagation of waves. Rectangular waveguide is commonly used for the transport of radio frequency signals at frequencies in the SHF band (3–30 GHz) and higher. The ﬁelds in a rectangular waveguide consist of a number of propagating modes which depends on the electrical dimensions of the waveguide. These modes are broadly classiﬁed as either transverse magnetic (TM) or transverse electric (TE). In this section, we consider the TM modes. Figure 6.6 shows the geometry of interest. Here the walls are located at x = 0, x = a, y = 0, and y = b; thus, the cross-sectional dimensions of the waveguide are a and b. The interior of the waveguide is presumed to consist of a lossless material exhibiting real-valued permeability µ and real-valued permittivity ǫ, and the walls are assumed to be perfectly-conducting. Let us limit our attention to a region within the waveguide which is free of sources. Expressed in phasor form, the electric ﬁeld intensity within the waveguide is governed by the wave equation ∇2 eE + β2 eE = 0 (6.141) where β = ω√µǫ (6.142) Figure 6.6: Geometry for analysis of ﬁelds in a rect- angular waveguide.	Electromagnetics_Vol2_Page_123_Chunk2298
6.8. RECTANGULAR WAVEGUIDE: TM MODES 111 Equation 6.141 is a partial differential equation. This equation, combined with boundary conditions imposed by the perfectly-conducting plates, is sufﬁcient to determine a unique solution. This solution is most easily determined in Cartesian coordinates, as we shall now demonstrate. First we express eE in Cartesian coordinates: eE = ˆx eEx + ˆy eEy + ˆz eEz (6.143) This facilitates the decomposition of Equation 6.141 into separate equations governing the ˆx, ˆy, and ˆz components of eE: ∇2 eEx + β2 eEx = 0 (6.144) ∇2 eEy + β2 eEy = 0 (6.145) ∇2 eEz + β2 eEz = 0 (6.146) Next we observe that the operator ∇2 may be expressed in Cartesian coordinates as follows: ∇2 = ∂2 ∂x2 + ∂2 ∂y2 + ∂2 ∂z2 (6.147) so the equations governing the Cartesian components of eE may be written as follows: ∂2 ∂x2 eEx + ∂2 ∂y2 eEx + ∂2 ∂z2 eEx + β2 eEx = 0 (6.148) ∂2 ∂x2 eEy + ∂2 ∂y2 eEy + ∂2 ∂z2 eEy + β2 eEy = 0 (6.149) ∂2 ∂x2 eEz + ∂2 ∂y2 eEz + ∂2 ∂z2 eEz + β2 eEz = 0 (6.150) In general, we expect the total ﬁeld in the waveguide to consist of unidirectional waves propagating in the +ˆz and −ˆz directions. We may analyze either of these waves; then the other wave is easily derived via symmetry, and the total ﬁeld is simply a linear combination (superposition) of these waves. With this in mind, we limit our focus to the wave propagating in the +ˆz direction. In Section 6.7, it is shown that all components of the electric and magnetic ﬁelds can be easily calculated once eEz and eHz are known. The problem is further simpliﬁed by decomposing the unidirectional wave into TM and TE components. In this decomposition, the TM component is deﬁned by the property that eHz = 0; i.e., is transverse (perpendicular) to the direction of propagation. Thus, the TM component is completely determined by eEz. Equations 6.136–6.139 simplify to become: eEx = −j kz k2ρ ∂eEz ∂x (6.151) eEy = −j kz k2ρ ∂eEz ∂y (6.152) eHx = +j ωµ k2ρ ∂eEz ∂y (6.153) eHy = −j ωµ k2ρ ∂eEz ∂x (6.154) where k2 ρ ≜β2 −k2 z (6.155) and kz is the phase propagation constant; i.e., the wave is assumed to propagate according to e−jkzz. Now let us address the problem of ﬁnding eEz, which will then completely determine the TM ﬁeld. As in Section 6.7, we recognize that eEz can be represented as the propagation factor e−jkzz times a factor that describes variation with respect to the remaining spatial dimensions x and y: eEz = eez(x, y)e−jkzz (6.156) Substitution of this expression into Equation 6.150 and dividing out the common factor of e−jkzz yields: ∂2 ∂x2 eez + ∂2 ∂y2 eez −k2 zez + β2eez = 0 (6.157) The last two terms may be combined using Equation 6.155, yielding: ∂2 ∂x2 eez + ∂2 ∂y2 eez + k2 ρeez = 0 (6.158) This is a partial differential equation for ez in the variables x and y. This equation may be solved using the technique of separation of variables. In this technique, we recognize that eez(x, y) can be written as the product of a function X(x) which depends only on x, and a function Y (y) that depends only on y. That is, eez(x, y) = X(x)Y (y) (6.159)	Electromagnetics_Vol2_Page_124_Chunk2299
112 CHAPTER 6. WAVEGUIDES Substituting this expression into Equation 6.158, we obtain: Y ∂2 ∂x2 X + X ∂2 ∂y2 Y + k2 ρXY = 0 (6.160) Next dividing through by XY , we obtain: 1 X ∂2 ∂x2 X + 1 Y ∂2 ∂y2 Y + k2 ρ = 0 (6.161) Note that the ﬁrst term depends only on x, the second term depends only on y, and the remaining term is a constant. Therefore, the sum of the ﬁrst and second terms is a constant; namely −k2 ρ. Since these terms depend on either x or y, and not both, the ﬁrst term must equal some constant, the second term must equal some constant, and these constants must sum to −k2 ρ. Therefore, we are justiﬁed in separating the equation into two equations as follows: 1 X ∂2 ∂x2 X + k2 x = 0 (6.162) 1 Y ∂2 ∂y2 Y + k2 y = 0 (6.163) where the new constants k2 x and k2 y must satisfy k2 x + k2 y = k2 ρ (6.164) Now multiplying Equations 6.162 and 6.163 by X and Y , respectively, we ﬁnd: ∂2 ∂x2 X + k2 xX = 0 (6.165) ∂2 ∂y2 Y + k2 yY = 0 (6.166) These are familiar one-dimensional differential equations. The solutions are:4 X = A cos (kxx) + B sin (kxx) (6.167) Y = C cos (kyy) + D sin (kyy) (6.168) where A, B, C, and D – like kx and ky – are constants to be determined. At this point, we observe that the wave we seek can be expressed as follows: eEz = eez(x, y)e−jkzz = X(x) Y (y) e−jkzz (6.169) 4Students are encouraged to conﬁrm that these are correct by conﬁrming that they are solutions to Equations 6.165 and 6.165, re- spectively. The solution is essentially complete except for the values of the constants A, B, C, D, kx, and ky. The values of these constants are determined by applying the relevant electromagnetic boundary condition. In this case, it is required that the component of eE that is tangent to a perfectly-conducting wall must be zero. Note that the ˆz component of eE is tangent to all four walls; therefore: eEz (x = 0) = 0 (6.170) eEz (x = a) = 0 (6.171) eEz (y = 0) = 0 (6.172) eEz (y = b) = 0 (6.173) Referring to Equation 6.169, these boundary conditions in turn require: X (x = 0) = 0 (6.174) X (x = a) = 0 (6.175) Y (y = 0) = 0 (6.176) Y (y = b) = 0 (6.177) Evaluating these conditions using Equations 6.167 and 6.168 yields: A · 1 + B · 0 = 0 (6.178) A cos (kxa) + B sin (kxa) = 0 (6.179) C · 1 + D · 0 = 0 (6.180) C cos (kyb) + D sin (kyb) = 0 (6.181) Equations 6.178 and 6.180 can be satisﬁed only if A = 0 and C = 0, respectively. Subsequently, Equations 6.179 and 6.181 reduce to: sin (kxa) = 0 (6.182) sin (kyb) = 0 (6.183) This in turn requires: kx = mπ a , m = 0, 1, 2... (6.184) ky = nπ b , n = 0, 1, 2... (6.185) Each positive integer value of m and n leads to a valid expression for eEz known as a mode. Solutions for which m = 0 or n = 0 yield kx = 0 or ky = 0, respectively. These correspond to zero-valued ﬁelds, and therefore are not of interest. The most general	Electromagnetics_Vol2_Page_125_Chunk2300

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