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C
1
Image removed for copyright reasons.
Please see: Figure 6.2b, page 262, from Orlando, T., and K. Delin.
Foundations of Applied Superconductivity. Reading, MA:
Addison-Wesley, 1991. ISBN: 0201183234.
But along
mininum, so that J
the hexagonal path C
1 B
is a
vanishes along this path.
Therefore,
And exper... | https://ocw.mit.edu/courses/6-763-applied-superconductivity-fall-2005/29bb21252c2fd3293111508de473d2ac_lecture10.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
(cid:10) 6.642 Continuum Electromechanics
Fall 2008
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
(cid:13)
6.642, Continuum Electromechanics, Fall 2004
Prof. Markus Zahn
Lecture 7: Pressure–Velocity Relations for Inviscid,... | https://ocw.mit.edu/courses/6-642-continuum-electromechanics-fall-2008/29d24b6ff1988b3c0fc376b437fa7706_lec07_f08.pdf |
x
(
)
Δ =
(cid:3)
α
v
x
6.642, Continuum Electromechanics Lecture 7
Prof. Markus Zahn Page 1 of 5
Θ
(
' 0
)
β
(cid:108)
= Θ
(cid:3)
(
, v 0
x
)
=
(cid:3)
v
β
x
(cid:108) (
Θ
x
)
=
(cid:108)
Θ
α
sinh kx
(cid:108)
− Θ
β
(
sinh k x
− Δ
)
sinh k
Δ
(cid:3)
v
x
... | https://ocw.mit.edu/courses/6-642-continuum-electromechanics-fall-2008/29d24b6ff1988b3c0fc376b437fa7706_lec07_f08.pdf |
k U
z
) (cid:108)
Θ
⎡
⎢
⎢
⎣
(cid:108)
Θ
α
(cid:108)
Θ
β
⎤
⎥
⎥
⎦
=
1
k
⎡
⎢
⎢
⎢
⎢
⎣
−
coth k
Δ
−
1
sinh k
Δ
1
sinh k
coth k
⎤ ⎡
⎥
Δ ⎢
⎥
⎢
⎥ ⎢
⎥ ⎣
⎦
Δ
(cid:3)
v
(cid:3)
v
α
x
β
x
⎤
⎥
⎥
⎥
⎦
⎡
(cid:3)
p
⎢
⎢
(cid:3)
p
⎣
α
β
⎤
⎥
⎥
⎦
=
(
ρ ω −
j
k U
z
)
k
⎡
⎢
⎢
⎢
⎢
⎣
−
coth k
Δ
−
1
sinh k
Δ
1
inh k
s
coth k
⎤ ⎡
⎥
Δ ⎢
⎥
⎢
⎥ ⎢
... | https://ocw.mit.edu/courses/6-642-continuum-electromechanics-fall-2008/29d24b6ff1988b3c0fc376b437fa7706_lec07_f08.pdf |
0=
,
0ξ =
−ρ
b
gx P
+
0
x 0
<
Perturbations:
(cid:3)
⎡
p
1
⎢
(cid:3)
⎢
p
⎣
2
⎤
⎥ =
⎥
⎦
j
a
⎡
−⎢
ωρ ⎢
⎢
k
−⎢
⎣
coth ka
1
sinh ka
(cid:3)
⎡
p
3
⎢
(cid:3)
⎢
p
⎣
4
⎤
⎥ =
⎥
⎦
j
ωρ
b
k
⎡
−⎢
⎢
⎢
−⎢
⎣
coth kb
1
sinh kb
1
sinh ka
⎤
⎥
⎡
⎥ ⎢
⎥ ⎢
⎣
coth ka
⎥
⎦
(cid:3)
v
(cid:3)
v
x
1
x2
⎤
⎥
⎥
⎦
1
sinh kb
coth kb
⎤
⎥
⎡
⎥ ⎢
⎥ ⎢
⎣... | https://ocw.mit.edu/courses/6-642-continuum-electromechanics-fall-2008/29d24b6ff1988b3c0fc376b437fa7706_lec07_f08.pdf |
P
30
(
0
+ ξ
)
'
+ P
3
( )
ξ
= P
30
(
0
)
+
dP
30
dx
x 0
=
ξ +
(
'
P 0
3
)
'
P
3
( )
ξ
'
P
2
( )
ξ = −ρ ξ +
g
a
'
P
2
(
0
)
= - g
ρ ξ +
b
(
'
P 0
3
)
(cid:3)
g
−ρ ξ +
b
(cid:3)
P + g
(cid:3)
ρ ξ −
3
a
(cid:3)
P
2
= γ
k
2
(cid:3)
ξ
(cid:3)
P =
3
j
ωρ
b
k
(cid:3)
P =
2
j
ωρ
a
k
⎡
⎣
⎡
⎣
(cid:3)
coth kbv
−
⎤
⎦
x3
= +
2
ρ ω... | https://ocw.mit.edu/courses/6-642-continuum-electromechanics-fall-2008/29d24b6ff1988b3c0fc376b437fa7706_lec07_f08.pdf |
ρ
b
)a
Instability if:
2
k + g
γ
(
ρ − ρ
b
a
)
<
0
Rayleigh-Taylor Instability
(heavier fluid above)
if
aρ > ρb
2
λ
π
T
k
c
=
⎡
⎢
⎢
⎣
g
(
ρ − ρ
b
a
)
γ
1
2
⎤
⎥
⎥
⎦
⎡
λ = π ⎢
⎢
⎣
2
T
γ
ρ − ρ
b
a
)
g
(
⎤
⎥
⎥
⎦
=
1
2
(Taylor Wavelength)
λ > λ
λ < λ
T
T
Unstable
Stable
Stable if ρ >
b
aρ
Long Wavelength Limit:
ka... | https://ocw.mit.edu/courses/6-642-continuum-electromechanics-fall-2008/29d24b6ff1988b3c0fc376b437fa7706_lec07_f08.pdf |
Lecture 4
PN Junction and MOS Electrostatics(I)
Semiconductor Electrostatics in Thermal
Equilibrium
Outline
• Nonuniformly doped semiconductor in thermal
equilibrium
• Relationships between potential, φ(x) and
equilibrium carrier concentrations, po(x), no(x)
–Boltzmann relations & “60 mV Rule”
• Quasineutral situa... | https://ocw.mit.edu/courses/6-012-microelectronic-devices-and-circuits-spring-2009/2a02d79958a93f067bd1ae334492fb6a_MIT6_012S09_lec04.pdf |
balances Drift
J n(x) = J n
drift(x) + Jn
diff (x) = 0
What is n o(x) that satisfies this condition?
no, Nd
partially uncompensated �
donor charge
Nd(x) +
no(x)
net electron charge
-
Let us examine the electrostatics implications of
n o(x) ≠ Nd(x)
x
6.012 Spring 2009
Lecture 4
5
Space charge density
ρ... | https://ocw.mit.edu/courses/6-012-microelectronic-devices-and-circuits-spring-2009/2a02d79958a93f067bd1ae334492fb6a_MIT6_012S09_lec04.pdf |
4
8
2. Relationships between potential, φφφφ(x) and
equilibrium carrier concentrations, po(x), no(x)
(Boltzmann relations)
dn
o
dx
Jn = 0 = qnoµnE + qDn
µµµµn •
Dn
dφφφφ
1
=
dx no
dno
dx
•
Using Einstein relation:
q
dφφφφ
•
kT dx
=
d(ln no)
dx
Integrate:
q
kT
(φφφφ− φφφφref )= ln no − ln no,ref... | https://ocw.mit.edu/courses/6-012-microelectronic-devices-and-circuits-spring-2009/2a02d79958a93f067bd1ae334492fb6a_MIT6_012S09_lec04.pdf |
1:
)• ln 10
( ) •
n
log o
n i
n = 1018cm −3 ⇒ φφφφ= (60m )× 8 = 480 mV
o
6.012 Spring 2009
Lecture 4
11
“60 mV” Rule: contd.
With holes:
φφφφ= −(25m )• ln
p o = −(25m )• ln 10)
ni
(
• log
po
ni
Or
p
o
φφφφ≈ −(60 m )• log
ni
EXAMPLE 2:
18
−3
no = 10 cm ⇒ po = 10 cm
⇒ φφφφ= −(60m ) × −8 = 480mV
2
−3
... | https://ocw.mit.edu/courses/6-012-microelectronic-devices-and-circuits-spring-2009/2a02d79958a93f067bd1ae334492fb6a_MIT6_012S09_lec04.pdf |
1 102
104
106
108
1010
1012
1014
1016
1018 1019
p-type
intrinsic
n-type
n o, equilibrium electron concentration (cm−3)
Note: φ cannot exceed 550 mV or be smaller than -550 mV.
(Beyond this point different physics come into play.)
6.012 Spring 2009
Lecture 4
13
Example 3: Compute potential difference in ... | https://ocw.mit.edu/courses/6-012-microelectronic-devices-and-circuits-spring-2009/2a02d79958a93f067bd1ae334492fb6a_MIT6_012S09_lec04.pdf |
−3) ==== −−−−190 mV
6.012 Spring 2009
Lecture 4
14
3. Quasineutral situation
If Nd(x) changes slowly with x ⇒ n o(x) also changes
slowly with x. WHY?
Small dn o/dx implies a small diffusion current. We do
not need a large drift current to balance it.
Small drift current implies a small electric field and
th... | https://ocw.mit.edu/courses/6-012-microelectronic-devices-and-circuits-spring-2009/2a02d79958a93f067bd1ae334492fb6a_MIT6_012S09_lec04.pdf |
Maximum Likelihood Estimation
Parameter Estimation
Fitting Probability Distributions
Maximum Likelihood
MIT 18.443
Dr. Kempthorne
Spring 2015
MIT 18.443 Parameter EstimationFitting Probability DistributionsMaximum Likelihood
eliho od
1 Maximum Likelihood Estimation
Framework/Definitions
Outline
1 Maximum Likel... | https://ocw.mit.edu/courses/18-443-statistics-for-applications-spring-2015/2a08e2ffa8c87f3187d3969cc24c168b_MIT18_443S15_LEC4.pdf |
Estimation
Framework/Definitions
Likelihood Definition
Case B: Time-Series Model
X1, X2, . . . , Xn are observations of a time series
{Xt , t = 1, 2, . . .}
Joint density of X = (X1, X2, . . . , Xn) is given by:
f (x1, . . . , xn | θ) =
f (x1 | θ) × f (x2 | θ, x1) × f (x3 | θ, x1, x2) × · · ·
=
=⇒
lik(θ) =
×f... | https://ocw.mit.edu/courses/18-443-statistics-for-applications-spring-2015/2a08e2ffa8c87f3187d3969cc24c168b_MIT18_443S15_LEC4.pdf |
most likely”
ˆθMLE maximizes the log likelihood
n
n
£(θ) = log lik(θ) =
log[f (xi | θ)],
(Case A)
i=1
MIT 18.443 Parameter EstimationFitting Probability DistributionsMaximum Likelihood
elihood
5 Maximum Likelihood Estimation
Framework/Definitions
Specifying the MLE
Example 8.5.A: Poisson Distribution
X1, . . ... | https://ocw.mit.edu/courses/18-443-statistics-for-applications-spring-2015/2a08e2ffa8c87f3187d3969cc24c168b_MIT18_443S15_LEC4.pdf |
i=1
2
(x−µ)2
2 σ2
]
n
= −
2
ln(2π) − n ln(σ) −
1
2σ2
n
n
(xi − µ)2
i=1
MLE of θ = (µ, σ2):
ˆ
σ2
θMLE = (ˆµMLE , ˆMLE )
£(θˆMLE ) maximizes £(θ) = £(µ, σ)
ˆ
= 0 and
θMLE solves:
∂£(µ, σ2)
∂µ
∂£(µ, σ2)
∂σ2
= 0
MIT 18.443
Parameter EstimationFitting Probability DistributionsMaximum Likelihood
elihood
... | https://ocw.mit.edu/courses/18-443-statistics-for-applications-spring-2015/2a08e2ffa8c87f3187d3969cc24c168b_MIT18_443S15_LEC4.pdf |
n
i=1
= α/X
ˆ
0 =
∂£(α, λˆ)
∂α
= −n
Γ'(α)
Γ(α)
Γ'(α)
Γ(α)
+ ln(α) − ln(X ) + 1
n
MIT 18.443
−n
=
+ n ln(λˆ) +
ln(xi )
n
n
i=1
n
n
+ n ln(α) − n ln(X ) +
ln(xi )
(cid:80)
n
i=1
ln(xi )
i=1
=⇒ 0 =
Γ/
(α)
Γ(α)
Parameter EstimationFitting Probability DistributionsMaximum Likelihood
elihood
10
Maxim... | https://ocw.mit.edu/courses/18-443-statistics-for-applications-spring-2015/2a08e2ffa8c87f3187d3969cc24c168b_MIT18_443S15_LEC4.pdf |
18.156 Lecture Notes
Febrary 17, 2015
trans. Jane Wang
The main goal of this lecture is to prove Korn’s inequality, which as we recall is as follows:
Theorem 1 (Korn’s Inequality). If u ∈ C2
comp(Rn), and ∆u = f , then
[∂i∂ju]Cα ≤ C(n, α)[∆u]Cα.
First, let us recall the progress that we made last time. To start, we hav... | https://ocw.mit.edu/courses/18-156-differential-analysis-ii-partial-differential-equations-and-fourier-analysis-spring-2016/2a167573b1412e48588f173774f6e158_MIT18_156S16_lec5.pdf |
x2)| into pieces that look like behaviors that
we can understand.
Recall that last class, we examined a few examples.
1. f supported between x1 and x2.
1
2. f supported over x1.
Used that |K(x)| (cid:46) |x|−n.
3. f supported on B3d(x1), and (cid:15) < d. Note that as opposed to
can be (cid:29) dα.
the previous exampl... | https://ocw.mit.edu/courses/18-156-differential-analysis-ii-partial-differential-equations-and-fourier-analysis-spring-2016/2a167573b1412e48588f173774f6e158_MIT18_156S16_lec5.pdf |
:12)
(cid:12)
(cid:12)
(cid:12)
(cid:12)
(cid:12)
=
(cid:12)
(cid:90)
(cid:90)
f (y)K(cid:15)(x1 − y) dy −
(cid:90)
f (y)K(cid:15)(x2 − y) dy
(cid:12)
(cid:12)
(cid:12)
(cid:12)
(f (y) − A)K(cid:15)(x1 − y) dy
−
(
f (y)
−
B
)K(cid:15)(x2
−
y) dy
N1
+
(cid:90)
N
c
1
(f (y) − C)K(cid:15)(x1 − y) dy −
(cid:90)
N c
2
(f (y... | https://ocw.mit.edu/courses/18-156-differential-analysis-ii-partial-differential-equations-and-fourier-analysis-spring-2016/2a167573b1412e48588f173774f6e158_MIT18_156S16_lec5.pdf |
N2
(cid:12)
(cid:12)
(cid:12) +
(cid:12)
(cid:12)
(cid:12)
I2
(cid:90)
F
I3 − I4
(cid:12)
(cid:12)
(cid:12) +
(cid:12)
(cid:12)
(cid:12)
(cid:90)
N1\N2
(cid:12)
(cid:12)
I4 +
(cid:12)
(cid:12) (cid:90)
(cid:12)
(cid:12)
N2\N1
(cid:12)
(cid:12)
I3 .
(cid:12)
1.
The first two terms will behave like example 2 and the last ... | https://ocw.mit.edu/courses/18-156-differential-analysis-ii-partial-differential-equations-and-fourier-analysis-spring-2016/2a167573b1412e48588f173774f6e158_MIT18_156S16_lec5.pdf |
but K(cid:15) is discontinuous. However, the (cid:15) < d/10 means that in F we avoid this discontinuity.
We also note that we didn’t need to choose a to be the midpoint of x1 and x2. We just needed
something like |x1 − y| ∼ |a − y| ∼ |x2 − y| on F .
The following proposition then almost gives us Korn’s inequality, exc... | https://ocw.mit.edu/courses/18-156-differential-analysis-ii-partial-differential-equations-and-fourier-analysis-spring-2016/2a167573b1412e48588f173774f6e158_MIT18_156S16_lec5.pdf |
:15) ∈ Cc
∞, we have that u(cid:15) → u in C2.
|partiali∂ju(x1) − ∂i∂ju(x2)| =
(cid:90)
(cid:12)
(cid:12)
(cid:12)
(∆u(x1 − y) − ∆u(x2 − y))ϕ(cid:15)(y) dy
(cid:12)
(cid:12)
(cid:12)
(cid:46) lim inf[∆u(cid:15)]Cα.
→0
(cid:15)
Note that this isn’t quite good enough, since we could have something like the following dang... | https://ocw.mit.edu/courses/18-156-differential-analysis-ii-partial-differential-equations-and-fourier-analysis-spring-2016/2a167573b1412e48588f173774f6e158_MIT18_156S16_lec5.pdf |
/2) ≤ C(α, n, B)(cid:107)u(cid:107)C2(B1).
As a step toward proving Schauder’s inequality, let us change one of the conditions in this lemma.
Proposition 7 (Baby Schauder). If 0 < λ ≤ aij ≤ Λ, [aij]Cα(B1) ≤ B, Lu = 0 on B1, then
(cid:107)u(cid:107)C2,α(B ) ≤ C(α, n, B, λ, Λ)(cid:107)u(cid:107)C2(B1).
1/2
Proof. First, ... | https://ocw.mit.edu/courses/18-156-differential-analysis-ii-partial-differential-equations-and-fourier-analysis-spring-2016/2a167573b1412e48588f173774f6e158_MIT18_156S16_lec5.pdf |
6.034 Artificial Intelligence. Copyright © 2004 by Massachusetts Institute of Technology.
6.034 Notes: Section 7.1
Slide 7.1.1
We have been using this simulated bankruptcy data set to illustrate the different learning
algorithms that operate on continuous data. Recall that R is supposed to be the ratio of earnings ... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a180cde9d8748aa97cbf51950ecb096_ch7_mach3.pdf |
training data. The linear separator is a very simple
hypothesis class, not nearly as powerful as either 1-NN or decision trees. However, as simple as this
class is, in general, there will be many possible linear separators to choose from.
Also, note that, once again, this decision boundary disagrees with that drawn ... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a180cde9d8748aa97cbf51950ecb096_ch7_mach3.pdf |
is, the equation of a linear separator. We will be illustrating everything in
two dimensions but all the equations hold for an arbitrary number of dimensions.
The equation of a linear separator in an n-dimensional feature space is (surprise!) a linear equation
which is determined by n+1 values, the components of an ... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a180cde9d8748aa97cbf51950ecb096_ch7_mach3.pdf |
will always be equal to 1. Then we can write a linear equation as a dot product. When we do this, we
will indicate it by using an overbar over the vectors.
6.034 Artificial Intelligence. Copyright © 2004 by Massachusetts Institute of Technology.
Slide 7.1.12
First a word on terminology: the equations we will be wr... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a180cde9d8748aa97cbf51950ecb096_ch7_mach3.pdf |
the hyperplane.
Looking at the right triangle defined by the w-hat and the x vector, both emanating from the origin,
we see that the projection of x onto w-hat is the length of the base of the triangle, where x is the
hypotenuse and the base angle is theta.
Now, if we subtract out the perpendicular distance to the ... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a180cde9d8748aa97cbf51950ecb096_ch7_mach3.pdf |
far we've talked about how to represent a linear hypothesis but not how to find one. In this slide
is the perceptron algorithm, developed by Rosenblatt in the mid 50's. This is not exactly the
original form of the algorithm but it is equivalent and it will help us later to see it in this form.
This is a greedy, "mis... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a180cde9d8748aa97cbf51950ecb096_ch7_mach3.pdf |
. In general, we will have to go around multiple times.
The remarkable fact is that the algorithm is guaranteed to terminate with the weights for a separating
hyperplane as long as the data is linearly separable. The proof of this fact is beyond our scope.
Notice that if the data is not separable, then this algorith... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a180cde9d8748aa97cbf51950ecb096_ch7_mach3.pdf |
calculate the slope of the function at that input value and we take a step that is proportional to the
slope. Note that the sign of the slope will tell us whether an increase of the input variable will
increase or decrease the value of the output. The magnitude of the slope will tell us how fast the
function is chan... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a180cde9d8748aa97cbf51950ecb096_ch7_mach3.pdf |
th search.
In more sophisticated search algorithms one does a search along the specified direction looking for a
value of the step size that guarantees an increase in the function value.
Slide 7.2.8
Now we can see that our choice of increment in the perceptron algorithm is related to the gradient of
the sum of the... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a180cde9d8748aa97cbf51950ecb096_ch7_mach3.pdf |
.
The perceptron algorithm can be described as a gradient ascent algorithm, but its error criterion is
slightly unusual in that there are many separators that all have zero error.
6.034 Artificial Intelligence. Copyright © 2004 by Massachusetts Institute of Technology.
Slide 7.2.11
Recall that the perceptron algo... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a180cde9d8748aa97cbf51950ecb096_ch7_mach3.pdf |
separator. If the margin is
positive, the point is classified correctly, so do nothing. If the margin is negative, add that point into
the weights of the separator. We can do that simply by incrementing the associated alpha.
Finally, when all of the points are classified correctly, we return the weighted sum of the ... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a180cde9d8748aa97cbf51950ecb096_ch7_mach3.pdf |
(note
the sign is different from what we saw in our previous treatment but the idea is the same). In this
way, we can write the basic rule of operation as computing the weighted sum of all the inputs and
comparing to 0.
The key observation is that the decision boundary for a single perceptron unit is a hyperplane i... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a180cde9d8748aa97cbf51950ecb096_ch7_mach3.pdf |
) units into these
networks make them much more powerful: they are no longer limited to linearly separable
problems.
What these networks do is basically use the earlier layers (closer to the input) to transform the
problem into more tractable problems for the latter layers.
6.034 Artificial Intelligence. Copyrigh... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a180cde9d8748aa97cbf51950ecb096_ch7_mach3.pdf |
unit. The other point has negative
distance and produces a zero output. This is shown in the shaded column in the table.
Slide 7.3.10
On the lower right, we see that the problem has been mapped into a linearly separable problem in
the space of the outputs of the hidden units. We can now easily find a linear separat... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a180cde9d8748aa97cbf51950ecb096_ch7_mach3.pdf |
that way.
Slide 7.3.14
The classic "soft threshold" that is used in neural nets is referred to as a "sigmoid" (meaning S-like)
and is shown here. The variable z is the "total input" or "activation" of a neuron, that is, the
weighted sum of all of its inputs.
Note that when the input (z) is 0, the sigmoid's value i... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a180cde9d8748aa97cbf51950ecb096_ch7_mach3.pdf |
the logistic function we
saw in the last slide.
The output of this function (y) varies smoothly with changes in the input and, importantly, with
changes in the weights. In fact, the weights and inputs both play similar roles in the function.
Slide 7.3.17
Given a dataset of training points, each of which specifies ... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a180cde9d8748aa97cbf51950ecb096_ch7_mach3.pdf |
respect to the weights,
that is, the vector of changes in the output due to a change in each of the weights.
The output (y) of a single sigmoid unit is simply the output of the sigmoid function for the current
activation (that is, total weighted input) of the unit. So, this output depends both on the values of the
... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a180cde9d8748aa97cbf51950ecb096_ch7_mach3.pdf |
a very simple form when expressed in terms of the output of the
sigmoid. Then, it is just the output times 1 minus the output. We will use this fact liberally later.
Slide 7.3.22
Now, what happens if the input to our unit is not a direct input but the output of another unit and
we're interested in the rate of chang... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a180cde9d8748aa97cbf51950ecb096_ch7_mach3.pdf |
respect to any of the weights,
we get a term that measures the error at the output (y-y^i) times the change in the output which is
produced by the change in the weight (dy/dw).
Slide 7.3.27
Let's pick weight w13, that weights the output of unit 1 (y1) coming into the output unit (unit 3).
What is the change in the... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a180cde9d8748aa97cbf51950ecb096_ch7_mach3.pdf |
weights on the inputs to unit 1 change the output by
changing one input to unit 3 and so the final gradient depends on the behavior of unit 3. It is the
realization of this reuse of terms that leads to an efficient strategy for computing the error gradient.
Slide 7.3.29
The cases we have seen so far are not complet... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a180cde9d8748aa97cbf51950ecb096_ch7_mach3.pdf |
form of dy/dw, the form of
delta in the pink box should be plausible: the product of the slope of the output sigmoid times the
sum of the products of weights and other deltas. This is exactly the form of the dy/dw expressions
we saw before.
The clever part here is that by computing the deltas starting with that of ... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a180cde9d8748aa97cbf51950ecb096_ch7_mach3.pdf |
are stuck at 0 or 1 because the magnitude of the total
input is too large (positive or negative). If we get saturation, the slope of the sigmoid is 0 and there
will not be any meaningful information of which way to change the weight.
Slide 7.3.34
Now we pick a sample input feature vector. We will use this to define... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a180cde9d8748aa97cbf51950ecb096_ch7_mach3.pdf |
over the whole network, the
delta's capture the recursion that we observed earlier. It is for this reason, simplicity, locality and,
therefore, efficiency that backpropagation has become the dominant paradigm for training neural
nets.
As mentioned before, however, the difficult choice of the learning rate and relat... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a180cde9d8748aa97cbf51950ecb096_ch7_mach3.pdf |
.
6.034 Artificial Intelligence. Copyright © 2004 by Massachusetts Institute of Technology.
Slide 7.4.4
Then we perform the minimization of the training error, for example, using backpropagation. This
will generally involve going through the input data and making changes to the weights many times.
A common term u... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a180cde9d8748aa97cbf51950ecb096_ch7_mach3.pdf |
we can use them once on a held out test set to estimate the
expected behavior on new data. Note the emphasis on doing this once. If we change the weights to
improve this behavior, then we no longer have a held out set.
6.034 Artificial Intelligence. Copyright © 2004 by Massachusetts Institute of Technology.
Slide ... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a180cde9d8748aa97cbf51950ecb096_ch7_mach3.pdf |
Note, however, that during most of the time that the training error is dropping, the test error is
increasing. This indicates that the net is overfitting the data. If you look at the net output at the end
of training, you can see what is happening. The net has constructed a baroque decision boundary to
capture preci... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a180cde9d8748aa97cbf51950ecb096_ch7_mach3.pdf |
the points are densest, this actually
approximates the linear boundary but then deviates wildly in response to an outlier near (-2, 0). This
illustrates how the choice of kernel can affect the generalization ability of an SVM classifier.
Slide 7.4.13
We mentioned earlier that backpropagation is an on-line training ... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a180cde9d8748aa97cbf51950ecb096_ch7_mach3.pdf |
error), in the
presence of a momentum term, the change in the weights will not necessarily be zero. So, this may
cause the system to move through a shallow minimum, which may be good. However, it may also
lead to undesirable oscillations in some circumstances.
In practice, choosing a good value of momentum for a pr... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a180cde9d8748aa97cbf51950ecb096_ch7_mach3.pdf |
problems, one can have multiple output units, for example, each aimed at
recognizing one class, sharing the hidden units with the other classes.
One difficulty with this approach is that there may be ambiguous outputs, e.g. two values above the
0.5 threshold when using a unary encoding. How do we treat such a case? ... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a180cde9d8748aa97cbf51950ecb096_ch7_mach3.pdf |
be linear, that is, simply remove the sigmoid non
linearity and have the unit returned a weighted sum of its inputs.
Slide 7.4.21
One very interesting application of neural networks is the ALVINN project from CMU. The project
was the brainchild of Dean Pomerleau. ALVINN is an automatic steering system for a car bas... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a180cde9d8748aa97cbf51950ecb096_ch7_mach3.pdf |
't want to generate simulated images from scratch, as in a video game, since they are
insufficiently realistic. What they did, instead, is transform the real images and fill in the few
missing pixels by a form of "interpolation" on the actual pixels. The results were amazingly good.
However, it turned out that once ... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a180cde9d8748aa97cbf51950ecb096_ch7_mach3.pdf |
6.034 Artificial Intelligence. Copyright © 2004 by Massachusetts Institute of Technology.
6.034 Notes: Section 3.4
Slide 3.4.1
In this section, we will look at some of the basic approaches for building programs that play two-
person games such as tic-tac-toe, checkers and chess.
Much of the work in this area has be... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a2f347357dc1e683ece44aec2f81e5e_ch3_csp_games2.pdf |
to generate the descendants of each of
these positions, etc. Note that these trees are enormous and cannot be explicitly represented in their
entirety for any complex game.
6.034 Artificial Intelligence. Copyright © 2004 by Massachusetts Institute of Technology.
Slide 3.4.4
Here's a little piece of the game tree ... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a2f347357dc1e683ece44aec2f81e5e_ch3_csp_games2.pdf |
use our scoring function to see what the values are at the
leaves of this tree. These are called the "static evaluations". What we want is to compute a value for
each of the nodes above this one in the tree by "backing up" these static evaluations in the tree.
The player who is building the tree is trying to maximiz... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a2f347357dc1e683ece44aec2f81e5e_ch3_csp_games2.pdf |
ranking, we see a graph that looks something like this. The
earliest serious chess program (MacHack6), which had a ranking of 1200, searched on average to a
depth of 4. Belle, which was one of the first hardware-assisted chess programs doubled the depth
and gained about 800 points in ranking. Deep Blue, which search... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a2f347357dc1e683ece44aec2f81e5e_ch3_csp_games2.pdf |
we look at might bring an even nastier surprise, but it doesn't matter what it is: we
already know that this move is worse than the one to the left, so why bother looking any further? In
fact, it may be that this unknown position is a great one for the maximizer, but then the minimizer
would never choose it. So, no ... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a2f347357dc1e683ece44aec2f81e5e_ch3_csp_games2.pdf |
-Value is at the leftmost leaf, whose static value is 2 and so it returns that.
Slide 3.4.16
This first value, since it is less than infinity, becomes the new value of beta in Min-Value.
Slide 3.4.17
So, now we call Max-Value with the next successor, which is also a leaf whose value is 7.
Slide 3.4.18
7 is not le... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a2f347357dc1e683ece44aec2f81e5e_ch3_csp_games2.pdf |
would have needed under
pure Min-Max.
Slide 3.4.26
We can write alpha-beta in a more compact form that captures the symmetry between the Max-
Value and Min-Value procedures. This is sometimes called the NegaMax form (instead of the Min-
Max form). Basically, this exploits the idea that minimizing is the same as maxi... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a2f347357dc1e683ece44aec2f81e5e_ch3_csp_games2.pdf |
Slide 3.4.29
The Move Generator would seem to be an unremarkable component of a game program, and this
would be true if its only function were to list the legal moves. In fact, it is a crucial component of
the program because its goal is to produce ordered moves. We saw that if the moves are ordered
well, then alph... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a2f347357dc1e683ece44aec2f81e5e_ch3_csp_games2.pdf |
off between the complexity of the evaluator and the depth of the search.
6.034 Artificial Intelligence. Copyright © 2004 by Massachusetts Institute of Technology.
Slide 3.4.31
As one can imagine in an area that has received as much attention as game playing programs, there
are a million and one techniques that hav... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a2f347357dc1e683ece44aec2f81e5e_ch3_csp_games2.pdf |
as I mentioned earlier, some moves are searched to a depth of 30 ply because of this.
Obviously, Deep Blue makes extensive use of parallelization in its search. This turns out to be surprisingly hard to do effectively and was probably the most significant innovation
in Deep Blue.
Slide 3.4.32
In this section, we ha... | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a2f347357dc1e683ece44aec2f81e5e_ch3_csp_games2.pdf |
many years of gradual refinement to achieve human-level
competence even in well-defined activities such as chess. We should not expect immediate success
in attacking any of the grand challenges of AI. | https://ocw.mit.edu/courses/6-034-artificial-intelligence-spring-2005/2a2f347357dc1e683ece44aec2f81e5e_ch3_csp_games2.pdf |
Chapter 2
Abelian Gauge Symmetry
As I have already mentioned, local gauge symmetry is the major new principle, beyond
the generic implementation of special relativity and quantum mechanics in quantum
field theory, which arises in formulating the standard model. It is, however, a rather
abstract concept, and one whose... | https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/2a43685db9124bc04925067ae33c6f23_chap2.pdf |
can correspond to the same fields. Gauge symmetry is a family of functional
transformations among the potentials that leaves the field strength unchanged. The
charge and current distributions, which provide the source terms in the Maxwell
equations, are under gauge transformations.
In the general, classical continuum... | https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/2a43685db9124bc04925067ae33c6f23_chap2.pdf |
“vector” field Aµ(�x, t) – whose components we shall
−
from here on call potentials – we add to the action a term
corresponding to the Lagrangian
Sint. =
q Aµdxµ
−
�
The momentum is now, with L
Lint. =
dxj
dt
.
qA0 + qAj
−
Lf ree + Lint.,
≡
∂L
∂ ˙xj = m
pj =
vj
√1
v2
−
+ qAj
dpj
dt
=
d
dt
(
mvj
√... | https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/2a43685db9124bc04925067ae33c6f23_chap2.pdf |
.7)
(2.8)
(2.9)
with
and
Ej
∂A0
∂xj −
∂Aj
∂t
≡ −
Bj
�jlm ∂Al
∂xm
≡
5
(2.11)
(2.12)
Identifying �
E and �
B as electric and magnetic field strengths, we thereby arrive at
the Lorentz force law for a particle of mass m, charge q.
Note that with this identification two of the Maxwell equations, viz.
∂Bj... | https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/2a43685db9124bc04925067ae33c6f23_chap2.pdf |
simply integrate by parts in
defines local gauge symmetry. Since local gauge symmetry leaves the equation of
motion unchanged, it must leave �
B unchanged, as of course one can verify
directly.
E and �
�
Clearly, the requirement that the world-line of a charged particle should have no
ends is closely related to th... | https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/2a43685db9124bc04925067ae33c6f23_chap2.pdf |
=
v
2 −
p
(�
�
qA)2
2
−
m
(2.20)
1
−
H = m2 + (�
p
�
qA)2 + qA0.
(2.21)
�
−
The appearance of square root here leads to difficulties in quantization. In order to
implement the commutation relations, or (more heuristically) wave-particle duality,
one would like to make the substitution �
p
odinger
wave equation ... | https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/2a43685db9124bc04925067ae33c6f23_chap2.pdf |
)2
lating the Schr¨odinger wave equation, which becomes
−
2m
(Of course, the constant term m on the right hand side can be eliminated by
+ qA0)ψ
= (m +
(2.23)
i
∂ψ
∂t
i∂�
(
−
q �
A)2
absorbing a factor e−
imt into ψ.)
7
For the gauge transformation A� = Aµ + ∂µχ on the potentials to leave the
µ
Schr¨odin... | https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/2a43685db9124bc04925067ae33c6f23_chap2.pdf |
Then the condition that an interaction term of the
with charges q1, q2,
symmetry law as primary.
ψ1, ψ2,
general form
· · ·
· · ·
which may also contain derivations, should conserve charge is that
ΔL =
1 ψp2
ψp1
2
· · ·
p1q1 + p2q2 +
= 0
· · ·
(2.25)
(2.26)
for the term destroys p1 particles charge q1,... | https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/2a43685db9124bc04925067ae33c6f23_chap2.pdf |
containing derivatives. But in order to obtain
sensible equations of motion it is necessary to have derivative terms in the action.
So if we want to promote charge conservation to ]it local gauge symmetry, permit
ting transformations in which χ depends on space and time, we must modify the
derivatives.
(2.28)
A s... | https://ocw.mit.edu/courses/8-325-relativistic-quantum-field-theory-iii-spring-2003/2a43685db9124bc04925067ae33c6f23_chap2.pdf |
MIT 2.852
Manufacturing Systems Analysis
Lectures 18–19
Loops
Stanley B. Gershwin
Spring, 2007
Copyright c�2007 Stanley B. Gershwin.
Problem
Statement
B1
M2
B2
M3
B3
M1
M4
B6
M6
B5
M5
B4
• Finite buffers (0 � ni(t) � Ni).
• Single closed loop – fixed population (
• Focus is on the Buzacott model (deter... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/2a492e1e011aa53ffe87fefef19d6461_MIT2_852S10_loops.pdf |
.01
r2
.1
p2
.01
20N1
e
t
a
r
n
o
i
t
c
u
d
o
r
p
0.89
0.885
0.88
0.875
0.87
0.865
0.86
0
N2=10
N2=15
N2=20
N2=30
N2=40
10
20
30
40
50
60
population
Copyright �2007 Stanley B. Gershwin.
c
5
Expected
population
method
• Treat the loop as a line in which the first machine
and the last are th... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/2a492e1e011aa53ffe87fefef19d6461_MIT2_852S10_loops.pdf |
1
ei+1
− 2 −
pu(i + 1)
ru(i + 1)
�
, i = k, ..., 1
pb(i + 1)rd(i)
.
pd(i)E(i + 1)
This is 4k equations in 4k unknowns. But only 4k − 1 of them are independent
because the derivation uses E(i) = E(i + 1) for i = 1, ..., k. The first k − 1
are E(1) = E(2), E(2) = E(3), ..., E(k − 1) = E(k). But this implies
E(... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/2a492e1e011aa53ffe87fefef19d6461_MIT2_852S10_loops.pdf |
blockage are reduced and
probabilities of starvation are increased. (Similarly if it is almost empty.)
� Suppose the population is smaller than the smallest buffer. Then there
will be no blockage. The expected population method does not take this
into account.
Copyright �2007 Stanley B. Gershwin.
c
10
Loop Behavi... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/2a492e1e011aa53ffe87fefef19d6461_MIT2_852S10_loops.pdf |
3
Loop Behavior
Ranges
Range of starvation
7
B1
0
B6
M1
M2
M6
10
B2
0
B5
M3
M5
10
B3
10
B4
M4
• If M5 stays down for a long time, it will starve M1.
• Therefore M5 is in the range of starvation of M1.
• Similarly, M6 is in the range of starvation of M1.
Copyright c
�2007 Stanley B. Gershwin.
14
... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/2a492e1e011aa53ffe87fefef19d6461_MIT2_852S10_loops.pdf |
culty for decomposition
Range of blocking of M6
B1
M
2
B
2
M3
B3
L(1)
M u(1)
B(1)
M d(1)
M1
M4
Range of blocking of M
1
B6
M6
B5
M5
B4
Range of starvation of M1
M1
M4
B1
M
2
B
2
M3
B3
M (6)d
B(6)
u
M (6)
L(6)
B6
M6
B5
M5
B4
Range of starvation of M
2
Ranges of blocking and starvation of... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/2a492e1e011aa53ffe87fefef19d6461_MIT2_852S10_loops.pdf |
to the
failure mode of a real machine.
Copyright c
�2007 Stanley B. Gershwin.
20
Multiple Failure
Mode Line
Decomposition
1,2
3
4
5,6,7
8
9,10
1,2,
3,4
5,6,7,
8,9,10
• There is an observer in each buffer who is told that
he is actually in the buffer of a two-machine line.
Copyright c
�2007 Stanley B. Ge... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/2a492e1e011aa53ffe87fefef19d6461_MIT2_852S10_loops.pdf |
and probability of blocking in each down
mode, average inventory.
Copyright c
�2007 Stanley B. Gershwin.
24
Line Decomposition
Multiple Failure
Mode Line
Decomposition
• A set of decomposition equations are formulated.
• They are solved by a Dallery-David-Xie-like
algorithm.
• The results are a little more accur... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/2a492e1e011aa53ffe87fefef19d6461_MIT2_852S10_loops.pdf |
f (i)
E(i − 1)
rjf
where pjf (i) is the probability of failure of the upstream machine into mode
jf ; Ps,jf (i − 1) is the probability of starvation of line i − 1 due to mode jf ;
rjf is the probability of repair of the upstream machine from mode jf ; etc.
• Also, E(i − 1) = E(i).
• pjf (i), pjf (i) are used to e... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/2a492e1e011aa53ffe87fefef19d6461_MIT2_852S10_loops.pdf |
M (6)u
M1
M (6)d
10
B
2
2
B5
M3
M5
10
B3
0
B4
M4
• The B6 observer knows how many parts there are in his buffer.
• If there are 5, he knows that the modes he sees in M d(6) could
be those corresponding to the modes of M1, M2, M3, and M4.
Copyright c
�2007 Stanley B. Gershwin.
30
Thresholds
10
B1
8
... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/2a492e1e011aa53ffe87fefef19d6461_MIT2_852S10_loops.pdf |
8
13
M2
B 2
1
M1
1
B 4
15
M3
M4
B 3
threshold
population = 21
Copyright c
�2007 Stanley B. Gershwin.
• When M1 fails for a long time,
3 B4 and B3 fill up, and there
is one part in B2. Therefore
there is a threshold of 1 in B2.
• When M2 fails for a long time,
B1 fills up, and there is one
part in B4.... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/2a492e1e011aa53ffe87fefef19d6461_MIT2_852S10_loops.pdf |
4
B
3
1M
*
M4
*
M3
*
M2
M2
M3
M4
• Break up each buffer into a sequence of buffers of size 1 and
reliable machines.
• Count backwards from each real machine the number of
buffers equal to the population.
• Identify the reliable machine that the count ends at.
Copyright c
�2007 Stanley B. Gershwin.
36
... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/2a492e1e011aa53ffe87fefef19d6461_MIT2_852S10_loops.pdf |
average buffer level error 5% with a
maximum of 21%.
� Ten-machine cases: mean throughput error 1.4% with a
maximum of 4%; average buffer level error 6% with a
maximum of 44%.
Copyright �2007 Stanley B. Gershwin.
c
40
Numerical
Results
Other algorithm attributes
• Convergence reliability: almost always.
• Speed: ... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/2a492e1e011aa53ffe87fefef19d6461_MIT2_852S10_loops.pdf |
.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
• Production rate vs. r1.
• Usual saturating graph.
Copyright �2007 Stanley B. Gershwin.
c
44
Numerical
Results
B1
M2
B2
* M1
B4
M4
B3
10
9
8
7
6
5
4
3
2
1
0
0
M3
Behavior
b1 average
b2 average
b3 average
b4 ... | https://ocw.mit.edu/courses/2-852-manufacturing-systems-analysis-spring-2010/2a492e1e011aa53ffe87fefef19d6461_MIT2_852S10_loops.pdf |
Practical multitone architectures
Lecture 4
Vladimir Stojanović
6.973 Communication System Design – Spring 2006
Massachusetts Institute of Technology
Cite as: Vladimir Stojanovic, course materials for 6.973 Communication System Design, Spring 2006.
MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Te... | https://ocw.mit.edu/courses/6-973-communication-system-design-spring-2006/2a8950c0b8a77687bae83a70ff459b08_lecture_4.pdf |
http://ocw.mit.edu/), Massachusetts Institute of Technology.
Downloaded on [DD Month YYYY].
6.973 Communication System Design
3
Efficientizing example
1+0.9D-1 channel (Pe=10-6, gap=8.8dB, PAM/QAM)
PAM and single-sideband
QAM
bn
b +1n
Cite as: Vladimir Stojanovic, course materials for 6.973 Communication S... | https://ocw.mit.edu/courses/6-973-communication-system-design-spring-2006/2a8950c0b8a77687bae83a70ff459b08_lecture_4.pdf |
w.mit.edu/), Massachusetts Institute of Technology.
Downloaded on [DD Month YYYY].
6.973 Communication System Design
6
Dynamic rate adaptation
Change the loading when channel changes
LC is a natural candidate
Keep the ET bit distribution and perturb based on
channel changes
Bit is moved from channel n to... | https://ocw.mit.edu/courses/6-973-communication-system-design-spring-2006/2a8950c0b8a77687bae83a70ff459b08_lecture_4.pdf |
- kT)
k
n=1
pm(t) * pn(-t)
||pm|| . ||pn||
*
Qk = Q(kT)
Qk = Idk
Figure by MIT OpenCourseWare.
Generalized Nyquist criterion
No interference between symbols
No interference between sub-channels
Cite as: Vladimir Stojanovic, course materials for 6.973 Communication System Design, Spring 2006.
MIT OpenCourseWare... | https://ocw.mit.edu/courses/6-973-communication-system-design-spring-2006/2a8950c0b8a77687bae83a70ff459b08_lecture_4.pdf |
, course materials for 6.973 Communication System Design, Spring 20
06.
MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology.
Downloaded on [DD Month YYYY].
6.973 Communication System Design
9
Convergence of multitone to modal modulation
Modal modulation is optimal for finite symbol t... | https://ocw.mit.edu/courses/6-973-communication-system-design-spring-2006/2a8950c0b8a77687bae83a70ff459b08_lecture_4.pdf |
Technology.
Downloaded on [DD Month YYYY].
6.973 Communication System Design
11
Discrete time channel partitioning
Digital realization
X0
X1
XN-2
XN-1
m0
m1
mN-2
mN-1
1/T
x
+
D
A
C
(t)n
x(t)
(t)
(t)h
+
*
(-t)
(LPF)
y(t) (LPF)
1/T
A
D
C
y
y = Px + n
f *
0
f *
1
f *
N-2
f *
N-1
Y0
Y1
YN-2
YN-1
yN-1
yN-2
y0
=
p1
p0... | https://ocw.mit.edu/courses/6-973-communication-system-design-spring-2006/2a8950c0b8a77687bae83a70ff459b08_lecture_4.pdf |
guardband
Etot=(N+1)*E_dim=(8+1)*1=9
SVD on
Gives singular values
Sub-channel SNRs
Waterfilling shows only 7 dimensions can be used
Sub-channel energies
SNRs are then
Total SNR
VC capacity
Would get 1.55bits/dim if N-> inf
Cite as: Vladimir Stojanovic, course materials for 6.973 Communica... | https://ocw.mit.edu/courses/6-973-communication-system-design-spring-2006/2a8950c0b8a77687bae83a70ff459b08_lecture_4.pdf |
N
−
1
VC
0...0
x
0 ...
x
xν−
...
N
N
−
1
2
2
SVD can be replaced by eigen-decomposition (spectral
factorization)
A discrete form of modal modulation
While SNRs are unique, many choices for M and F
Cite as: Vladimir Stojanovic, course materials for 6.973 Communication System Design, Spring 2006.
MIT Open... | https://ocw.mit.edu/courses/6-973-communication-system-design-spring-2006/2a8950c0b8a77687bae83a70ff459b08_lecture_4.pdf |
973 Communication System Design, Spring 2006.
MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology.
Downloaded on [DD Month YYYY].
6.973 Communication System Design
18
Figure by MIT OpenCourseWare.
One-tap frequency equalizer
Need to compensate for channel attenuation
To recover the o... | https://ocw.mit.edu/courses/6-973-communication-system-design-spring-2006/2a8950c0b8a77687bae83a70ff459b08_lecture_4.pdf |
VC, N=8, SNR=8.1 dB, 7*8/9=6.2MAC/sample
DMT, N=8, SNR=7.6 dB, 8pt FFT/IFFT,
2.7MAC/sample
N=16, 3.8MAC/sample, SNR=8.8 dB
DFE needs 10FF taps, 1FB tap, SNR=8.4 dB, 11MAC/sample
Cite as: Vladimir Stojanovic, course materials for 6.973 Communication System Design, Spring 2006.
MIT OpenCourseWare (http://ocw... | https://ocw.mit.edu/courses/6-973-communication-system-design-spring-2006/2a8950c0b8a77687bae83a70ff459b08_lecture_4.pdf |
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