text stringlengths 16 3.88k | source stringlengths 60 201 |
|---|---|
other plate
of the MOS capacitor.
Etching deposited materials to create the
appropriate geometric patterns.
Figures by MIT OCW.
Adapted from Maly, W.
Atlas of IC Technologies: An Introduction to VLSI Processes.
(Ignore dimensions in figures)
6.884 – Spring 2005
2/07/2005
L03 – CMOS Technology 8
UV Light
ETCHING... | https://ocw.mit.edu/courses/6-884-complex-digital-systems-spring-2005/5db68ba6abd100ecc232604ccfeb9f46_l03_cmos_gates.pdf |
fet
under polysi icon gate
l
(
)
6.884 – Spring 2005
Figure by MIT OCW.
2/07/2005
L03 – CMOS Technology 11
Design Rules
Surround rule
Exclusion rule
Extension
rules
Width
rules
Spacing rules
• Design rules are an abstraction of the fabrication process that specify
various geometric constraints on how diff... | https://ocw.mit.edu/courses/6-884-complex-digital-systems-spring-2005/5db68ba6abd100ecc232604ccfeb9f46_l03_cmos_gates.pdf |
DD
Supply Voltage = VDD
G
G
PFET only good
at pulling up
NFET only good
at pulling down
Ground = GND = 0V
6.884 – Spring 2005
2/07/2005
L03 – CMOS Technology 15
Generic Static CMOS Gate
VDD
IN1
IN2
INn
Pullup network,
connects output to
DD, contains only
V
PMOS
VOUT
Pulldown network,
connects output to
G... | https://ocw.mit.edu/courses/6-884-complex-digital-systems-spring-2005/5db68ba6abd100ecc232604ccfeb9f46_l03_cmos_gates.pdf |
a logic function f(x1, x2, ...)
– must be inverting for single level of CMOS logic
Pull up network should connect output to VDD when
f(x1, x2, ...) = 1
Pull down network should connect output to GND
when f(x1, x2, ...) = 1
Because PMOS is conducting with low inputs, useful
to write pullup as function of inverted ... | https://ocw.mit.edu/courses/6-884-complex-digital-systems-spring-2005/5db68ba6abd100ecc232604ccfeb9f46_l03_cmos_gates.pdf |
µm PMOS, in 0.25µm technology
Gate capacitance scales linearly with W
− ~2fF/µm
Diffusion capacitance scales linearly with W
− sum contributions from perimeter and area, ~2fF/µm
L
W
6.884 – Spring 2005
2/07/2005
L03 – CMOS Technology 23
Transistor Delay
When one gate drives another, all capacitance on the ... | https://ocw.mit.edu/courses/6-884-complex-digital-systems-spring-2005/5db68ba6abd100ecc232604ccfeb9f46_l03_cmos_gates.pdf |
gates rather than one large
complex gate
Only want to design and characterize a
small library of gates
What’s the best way to implement a
given logic function?
Figure by MIT OCW.
Figure by MIT OCW.
6.884 – Spring 2005
2/07/2005
L03 – CMOS Technology 27
Method of Logical Effort
(Sutherland and Sproul)
Fig... | https://ocw.mit.edu/courses/6-884-complex-digital-systems-spring-2005/5db68ba6abd100ecc232604ccfeb9f46_l03_cmos_gates.pdf |
1
1
Inverter
NAND
NOR
Input Cap = 3 units
Input Cap = 4 units
Input Cap = 5 units
L.E.=1 (definition)
L.E.=4/3
L.E.=5/3
6.884 – Spring 2005
2/07/2005
L03 – CMOS Technology 31
Electrical Effort
Cin
Logic Gate
Cout
Ratio of output load capacitance over input capacitance:
Electrical Effort = Cout/Cin
U... | https://ocw.mit.edu/courses/6-884-complex-digital-systems-spring-2005/5db68ba6abd100ecc232604ccfeb9f46_l03_cmos_gates.pdf |
F1/N
6.884 – Spring 2005
2/07/2005
L03 – CMOS Technology 35
Optimal Number of Stages
Cin
Cout
Minimum delay when:
stage effort = logical effort x electrical effort ~= 3.4-3.8
– Some derivations have e = 2.718.. as best stage effort – this ignores
parasitics
– Broad optimum, stage efforts of 2.4-6.0 within 15... | https://ocw.mit.edu/courses/6-884-complex-digital-systems-spring-2005/5db68ba6abd100ecc232604ccfeb9f46_l03_cmos_gates.pdf |
6.252 NONLINEAR PROGRAMMING
LECTURE 7: ADDITIONAL METHODS
LECTURE OUTLINE
• Least-Squares Problems and Incremental Gra-
dient Methods
• Conjugate Direction Methods
• The Conjugate Gradient Method
• Quasi-Newton Methods
• Coordinate Descent Methods
• Recall the least-squares problem:
minimize f (x) = 1
2
(cid:1)g... | https://ocw.mit.edu/courses/6-252j-nonlinear-programming-spring-2003/5db87e8d64f415730c362e7abd4042a6_6252_slides07.pdf |
itz condition on ∇gigi)
• Convergence to a “neighborhood” for a constant
stepsize
CONJUGATE DIRECTION METHODS
• Aim to improve convergence rate of steepest
descent, without incurring the overhead of New-
ton’s method
• Analyzed for a quadratic model. They require n
iterations to minimize f (x) = (1/2)x(cid:1)Qx ... | https://ocw.mit.edu/courses/6-252j-nonlinear-programming-spring-2003/5db87e8d64f415730c362e7abd4042a6_6252_slides07.pdf |
= ξi+1
(cid:1)
Qdj +
(cid:4)
(cid:1)
i
(cid:5)(cid:1)
c(i+1)mdm
Qdj = 0.
m=0
d2= ξ2 + c20d0 + c21d1
d1= ξ1 + c10d0
ξ1
ξ2
0
- c10d0
ξ0 = d0
0
d1
d0
CONJUGATE GRADIENT METHOD
• Apply Gram-Schmidt to the vectors ξk = gk =
∇f (xk), k = 0, 1, . . . , n − 1
dk = −gk +
k−1
(cid:1)
(cid:1)
gk
(cid:1)
dj
Qdj
... | https://ocw.mit.edu/courses/6-252j-nonlinear-programming-spring-2003/5db87e8d64f415730c362e7abd4042a6_6252_slides07.pdf |
≈ ∇2f (xk+1)pk,
pk = xk+1 − xk,
qk = ∇f (xk+1) − ∇f (xk).
(cid:7)−1
(cid:7)(cid:6)
(cid:6)
∇2f (xn) ≈ q0 · · · qn−1 p0 · · · pn−1
• Most popular Quasi-Newton method is a clever
way to implement this idea
Dk+1 = Dk +
(cid:1)
pkpk
(cid:1) qk
pk
−
(cid:1)
Dk
Dkqkqk
(cid:1)
qk
Dkqk
+ ξkτ kvkvk
(cid:1) ,
vk =... | https://ocw.mit.edu/courses/6-252j-nonlinear-programming-spring-2003/5db87e8d64f415730c362e7abd4042a6_6252_slides07.pdf |
linearly independent.
True for k = 1. Suppose no termination after k
steps, and g0, . . . , gk−1 are linearly independent.
Then, Span(d0, . . . , dk−1) = Span(g0, . . . , gk−1)
and there are two possibilities:
− gk = 0, and the method terminates.
− gk (cid:6)= 0, in which case from the expanding
manifold propert... | https://ocw.mit.edu/courses/6-252j-nonlinear-programming-spring-2003/5db87e8d64f415730c362e7abd4042a6_6252_slides07.pdf |
Physics 8.02T
For now, please sit anywhere, 9 to a table
P01 -
1
Class 1: Outline
Hour 1:
Why Physics?
Why Studio Physics? (& How?)
Vector and Scalar Fields
Hour 2:
Gravitational fields
Electric fields
P01 -
2
Why Physics?
P01 -
3
Why Study Physics?
Understand/appreciate nature
• Lightning
• Soap Films
• Butterfly W... | https://ocw.mit.edu/courses/8-02t-electricity-and-magnetism-spring-2005/5df2d2b1ad70d8dbd050e0e3ac1341cc_presentati_w01d1.pdf |
Overview of TEAL/Studio
Collaborative Learning
Groups of 3, Tables of 9
You teach, you discuss, you learn
In-Class Problem Solving
Desktop Experiments
Teacher-Student Interaction
Visualizations
PRS Questions
P01 -
11
Personal Response System
(PRS) Question:
Physics Experience
Pick up the nearest PRS
(under the table i... | https://ocw.mit.edu/courses/8-02t-electricity-and-magnetism-spring-2005/5df2d2b1ad70d8dbd050e0e3ac1341cc_presentati_w01d1.pdf |
Exam dates & times are online
Do NOT schedule early vacation departures, etc.
without consulting these times!
Any Questions?
P01 -
18
Physics is not Math…
P01 -
19
…but we use concepts from 18.02
(cid:71)
E
•Gradients
•Path Integrals
•Surface Integrals
B
V
∆ ≡ −
= −∇
V
(cid:71)
(cid:71)
∫ E s
d
⋅
(cid:71)
(cid:71)
Q... | https://ocw.mit.edu/courses/8-02t-electricity-and-magnetism-spring-2005/5df2d2b1ad70d8dbd050e0e3ac1341cc_presentati_w01d1.pdf |
B
d
Φ
dt
(cid:71)
(cid:71)
d
B s
⋅
=
µ
0
I
enc
+
µε
0 0
E
d
Φ
dt
(cid:118)
∫
C
(cid:118)
∫
C
Lorentz Force:
(cid:71)
F
q=
(
(cid:71)
(cid:71)
(cid:71)
E v B
+ ×
)
P01 -
25
Today: Fields
In General, then
Gravitational & Electric
P01 -
26
Scalar Fields
e.g. Temperature: Every location has
associated value (number with... | https://ocw.mit.edu/courses/8-02t-electricity-and-magnetism-spring-2005/5df2d2b1ad70d8dbd050e0e3ac1341cc_presentati_w01d1.pdf |
to field direction
All methods illustrated in
http://ocw.mit.edu/ans7870/8/8.02T/f04/visualization
s/electrostatics/39-pcharges/39-
twocharges320.html
P01 -
35
Vector Fields – Field Lines
• Direction of field line at any point is
tangent to field at that point
• Field lines never cross each other
P01 -
36
PRS Questi... | https://ocw.mit.edu/courses/8-02t-electricity-and-magnetism-spring-2005/5df2d2b1ad70d8dbd050e0e3ac1341cc_presentati_w01d1.pdf |
(cid:71)
g
Bonus: Where would
you put another mass
(cid:71)
g
m to make the field
become 0 at P?
NOTE: Solutions will be posted within one day of class
P01 -
44
From Gravitational to
Electric Fields
P01 -
45
Electric Charge (~Mass)
Two types of electric charge: positive and negative
Unit of charge is the coulomb [... | https://ocw.mit.edu/courses/8-02t-electricity-and-magnetism-spring-2005/5df2d2b1ad70d8dbd050e0e3ac1341cc_presentati_w01d1.pdf |
:71)
r
3
r
=
(
=
9 10 N m C 3C 3C
×
9
2
)(
2
)(
ˆ
i
(
1
2
−
)
ˆ3 m
j
3
(
1m
)
)
9
81 10 ˆ
×
(
2
−i
)
ˆ3 N
j
P01 -
49
The Superposition Principle
Many Charges Present:
Net force on any charge is vector sum of forces
from other individual charges
Example:
(cid:71)
F
3
(cid:71)
F
13
(cid:71)
F
23
+
=
In general:
(cid:71... | https://ocw.mit.edu/courses/8-02t-electricity-and-magnetism-spring-2005/5df2d2b1ad70d8dbd050e0e3ac1341cc_presentati_w01d1.pdf |
(cid:71)
E
This is easiest way to picture field
P01 -
53
PRS Question:
Electric Field
P01 -
54 | https://ocw.mit.edu/courses/8-02t-electricity-and-magnetism-spring-2005/5df2d2b1ad70d8dbd050e0e3ac1341cc_presentati_w01d1.pdf |
Determining n and μ: The Hall Effect
Vx, Ex
+ + + + + + + + + + +
- - - - - - - - -
Ey
I, Jx
r
F
r
+ qv r
= qE
r
× B
Bz
In steady state,
Fy = −evDBz
Fy = −eE y
EY = vDBZ = EH , the Hall Field
Since vD=-Jx/en,
EH = −
1
ne
J xBZ = RH J X BZ
RH = −
1
ne
σ = neμ
Experimental Hall Results on Metals
• V... | https://ocw.mit.edu/courses/3-225-electronic-and-mechanical-properties-of-materials-fall-2007/5e027de7290273530da1882cacc902ca_lecture_2.pdf |
dt
= −
p(t)
τ
− eE0 e
−iωt
B=0 in conductor,
r r
r r
and F (E) >> F (B)
− iωp0 = −
try p(t) = p0e−iωt
p0 − eE0
τ
− eE
0
1
τ
p =
0
− iω
ω>>1/τ, p out of phase with E
p0 =
eE0
iω
ω→ ∞, p → 0
ω<<1/τ, p in phase with E
p0 = −eE0τ
What if ωτ>>1?
When will J = σE break down? It depends on electrons u... | https://ocw.mit.edu/courses/3-225-electronic-and-mechanical-properties-of-materials-fall-2007/5e027de7290273530da1882cacc902ca_lecture_2.pdf |
0 = − nev =
− nep0
m
=
ne2
1
τ
m( − iω )
E0
σ 0
σ =
,σ =
1− iωτ 0
ne2τ
m
Response of e- to AC Electric Fields
•
Low frequency (ω<<1/τ)
– electron has many collisions before
direction change
– Ohm’s Law: J follows E, σ real
• High frequency (ω>>1/τ)
– electron has nearly 1 collision or less
when ... | https://ocw.mit.edu/courses/3-225-electronic-and-mechanical-properties-of-materials-fall-2007/5e027de7290273530da1882cacc902ca_lecture_2.pdf |
B = 0
r
∇ xE
= −
r
∇ xH =
r
r
r
D = E + 4πP
r
r
r
B = H + 4πM
r
B
1
∂
t
c
∂
r
4π r 1 ∂ D
J +
c ∂ t
c
Waves in Materials
• Non-magnetic material, μ =μ 0
• Polarization non-existent or swamped by free electrons,
P=0
r
∇ xE
= −
r
∂ B
∂ t
r
r
∇ xB = μ 0 J + μ 0ε 0
r
∇ x(∇ xE) = −
r
∂ E
∂ t
r... | https://ocw.mit.edu/courses/3-225-electronic-and-mechanical-properties-of-materials-fall-2007/5e027de7290273530da1882cacc902ca_lecture_2.pdf |
))
• Frequency dependence in ε(ω)
ε(ω) = 1+
iσ
ε0ω
= 1+
iσ0
ε0ω(1− iωτ)
ε(ω) = 1+
iω2τ
p
2ω− iωτ
ω2
p =
2
ne
ε0 m
Plasma Frequency
For ωτ>>>1, ε(ω) goes to 1
For an excellent conductor (σ0 large), ignore 1, look at case for ωτ<<1
ε(ω) ≈
iω2τ
p
iω2τ
p
2 ≈
ω− iωτ ω
Waves in Materials
k =
ω
c
ε (ω ... | https://ocw.mit.edu/courses/3-225-electronic-and-mechanical-properties-of-materials-fall-2007/5e027de7290273530da1882cacc902ca_lecture_2.pdf |
Success and Failure of Free e- Picture
• Success
– Metal conductivity
– Hall effect valence=1
– Skin Depth
– Wiedmann-Franz law
• Examples of Failure
– Insulators, Semiconductors
– Hall effect valence>1
– Thermoelectric effect
– Colors of metals
K/σ=thermal conduct./electrical conduct.~CT
Κ =
2
cvvthermτ
1... | https://ocw.mit.edu/courses/3-225-electronic-and-mechanical-properties-of-materials-fall-2007/5e027de7290273530da1882cacc902ca_lecture_2.pdf |
m K2)
2.22 x 108
2.12
2.23
2.42
2.20
2.31
2.32
2.36
2.14
2.90
2.61
2.28
2.49
2.14
2.58
2.75
2.48
2.64
3.53
2.57
373K
k
(watt cm-K)
0.73
k s T
(watt-ohm K2)
2.43 x 108
3.82
4.17
3.1
1.7
1.5
0.54
0.73
1.1
1.0
2.30
0.80
0.45
0.60
0.35
0.08
0.17
2.29
2.38
2.36
2.42
2.25
2.78
2.88
2.30
2.19
2.60
2.75
2.54
2.53
3.35
2.69
Exp... | https://ocw.mit.edu/courses/3-225-electronic-and-mechanical-properties-of-materials-fall-2007/5e027de7290273530da1882cacc902ca_lecture_2.pdf |
Class 5 – Project Choice Discussion
• Everyone should have a copy of the Project System Title
Table Handout
• Write your (readable) name on the line near the bottom
• Identify where in the list your “Title” (as shortened in many
cases by the faculty) appears
•
• Add the following late additions (anyone not appearing)
... | https://ocw.mit.edu/courses/esd-342-advanced-system-architecture-spring-2006/5e18ea1129b7d261f0046332fe676306_lec5.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
6.334 Power Electronics
Spring 2007
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/6-334-power-electronics-spring-2007/5e2f970d6c7c0ad911c62677e1b9fb82_ch6.pdf |
6.895 Essential Coding Theory
September 15, 2004
Lecturer: Madhu Sudan
Scribe: Adi Akavia
Lecture 3
Today’s Plan
• Converse coding theorem
• Shannon vs. Hamming theories
• Goals for the rest of the course
• Tools we use in this course
Shannon’s Converse Theorem
We complete our exposition of Shannon’s theory ... | https://ocw.mit.edu/courses/6-895-essential-coding-theory-fall-2004/5e3930912bd217554cd58c0573ca7a39_lect03.pdf |
decode to m is 2n−k , since D : {0, 1} ≥ {0, 1}k , for k = Rn. Now, since 2H(p)n >> 2n−k ,
then each corrupted word is mapped back to its original with only a very small probability.
n
More formally, fix some encoding and decoding mappings E, D. Let Im,� be a correct decoding
indicator, namely, Im,� = 1 if D(E(m) + �... | https://ocw.mit.edu/courses/6-895-essential-coding-theory-fall-2004/5e3930912bd217554cd58c0573ca7a39_lect03.pdf |
entries]
= exp(−n) +
�
�
� /�B(p� n,n) m
Pr[message m, error �, Im,� = 1]
3-1
As the choices of message, error and decoding algorithm are independent, and Pr[m] = 2−k, Pr[�|� /≤
B(p�n, n)] �
)n, we have: Pr[message m, error �, Im,� = 1] � 2−k · 2−H(p
� )n . Therefore,
⊆ 2−H(p
�
1
( n
� n)p
Pr[co... | https://ocw.mit.edu/courses/6-895-essential-coding-theory-fall-2004/5e3930912bd217554cd58c0573ca7a39_lect03.pdf |
Shannon also considers more general probability distributions on the error of the channel. In particu
lar, Shannon considers Markovian error model. Markovian models can capture situations where there are
bursts of huge amounts of error, by considering a finite set of correlated states. This models influenced
not only ... | https://ocw.mit.edu/courses/6-895-essential-coding-theory-fall-2004/5e3930912bd217554cd58c0573ca7a39_lect03.pdf |
encoding and decoding functions? Or, better yet, linear-time functions?
3-2
Shannon tells us that for any rate R < 1 − H(p) we can encode/decode. However, wjem we are
guaranteed 1 − H(p) − R = η, we are also interested in studying the complexity as a function of η. In
fact, a lot of the current research is conce... | https://ocw.mit.edu/courses/6-895-essential-coding-theory-fall-2004/5e3930912bd217554cd58c0573ca7a39_lect03.pdf |
be minimized of maximized. Nonetheless,
we general assume that we want to minimize q, as empirical observation indicate that it’s easier
to design good codes with smaller alphabet.
To simplify the parameters, we usually consider normalized parameters: R = n (to be maximized),
� = d (to be minimized), and study R vs... | https://ocw.mit.edu/courses/6-895-essential-coding-theory-fall-2004/5e3930912bd217554cd58c0573ca7a39_lect03.pdf |
group on the elements of F \ {0}, and
• there is a distribution-law of multiplication over addition.
The important thing for us is that finite fields exists.
Theorem 3. For any prime p, and m ≤ Z +, there exists a field Fpm of size p .
m
Where m = 1 the field of p elements is Fp = {0, ..., p − 1} with addition and mult... | https://ocw.mit.edu/courses/6-895-essential-coding-theory-fall-2004/5e3930912bd217554cd58c0573ca7a39_lect03.pdf |
defined below) is a family of codes that has two of these
properties: succinct representation, efficient encoding.
Let us first define what is a linear subspace.
Definition 5 (Linear Subspace). Let V be a vector space over a field F. A subspace L � V is a linear
subspace if every v1, v2 ≤ L and � ≤ F satisfy: v1 + v2 ≤ L ... | https://ocw.mit.edu/courses/6-895-essential-coding-theory-fall-2004/5e3930912bd217554cd58c0573ca7a39_lect03.pdf |
Therefore the study of
linear codes will be a big emphasis of this course. In particular we’ll consider codes based on polynomials
over finite fields.
1 Irreducible polynomial over Fp is one that can’t be factored over Fp
3-4
Polynomials over finite fields
F[X] denotes the ring of polynomials over F. The ele... | https://ocw.mit.edu/courses/6-895-essential-coding-theory-fall-2004/5e3930912bd217554cd58c0573ca7a39_lect03.pdf |
MIT 6.035
MIT 6 035
Introduction to Shift-Reduce Parsing
Martin Rinard
Laboratory for Computer Science
Massachusetts Institute of Technology
Orientation
• Specify Syntax Using
Context-Free Grammar
• Nonterminals
• Terminals
• Productions
• Given a grammar, Parser
Generator produces a
Generator produces a ... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/5e52815ef894b1842e678e6eb5c0a7b9_MIT6_035S10_lec03.pdf |
Op→ -
Op→ *
)
Expr
E
(
Expr
Op
Expr
num
*
num
+
num
Expr
Op
Expr
E
C
U
U
D
E
R
R
Shift -R e duce P a rser Exam p lep
Expr
Op
Expr
)
Expr
E
(
Expr → Expr Op Expr
Expr → ( Expr)
Expr → - Expr
Expr → num
Op → +
p
Op → -
Op → *
Expr
Op
Expr
num
*
num
+
num
Expr
E
C
U
U
D
E
R
R
Shift -R e duce P a... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/5e52815ef894b1842e678e6eb5c0a7b9_MIT6_035S10_lec03.pdf |
Happens if
Choose
Reduce
Expr
num
-
num
Expr
Expr
E
C
U
U
D
E
R
R
Shift/Reduce/Reduce Conflict
Expr → ExprOp Expr
Expr → Expr- Expr
Expr→ (Expr)
Expr→ Expr -
Expr→ num
Op→ +
Op→ -
Op →
Op → *
*
p
What Happens if
What Happens if
Choose
Reduce
Expr
num
-
num
Expr
Expr
!
S
S
L
I
A
F
F
... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/5e52815ef894b1842e678e6eb5c0a7b9_MIT6_035S10_lec03.pdf |
S
Shift/Reduce/Reduce Conflict
Expr → ExprOp Expr
Expr → Expr- Expr
Expr→ (Expr)
Expr→ Expr -
Expr→ num
Op→ +
Op→ -
Op →
Op → *
*
p
What Happens if
What Happens if
Choose
Reduce
num
-
num
Expr
Op
Expr
E
C
U
U
D
E
R
R
Shift/Reduce/Reduce Conflict
Expr → ExprOp Expr
Expr → Expr- Expr
Expr→ (E... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/5e52815ef894b1842e678e6eb5c0a7b9_MIT6_035S10_lec03.pdf |
licts
What Happens if
What Happens if
Choose
Shift
num
num
Expr → ExprOp Expr
Expr → Expr- Expr
Expr→ (Expr)
Expr→ Expr -
Expr→ num
Op→ +
Op→ -
Op →
Op → *
*
p
Expr
-
Expr
E
C
U
U
D
E
R
Conflicts
What Happens if
What Happens if
Choose
Shift
Expr
Expr
num
-
num
Expr
E
C
U
U
D
E
R
Expr → ExprOp ... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/5e52815ef894b1842e678e6eb5c0a7b9_MIT6_035S10_lec03.pdf |
Expr → Expr- Expr
Expr→ (Expr)
Expr→ Expr -
Expr→ num
Op→ +
Op→ -
Op →
Op → *
*
p
Parser Generator Should
Handle It
num
num
Constructing a Parser
• We will construct version with no lookahead
• Key Decisions
Key Decisions
• Shift or Reduce
• Which Production to Reduce
Which Production to Reduce
• Basic I... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/5e52815ef894b1842e678e6eb5c0a7b9_MIT6_035S10_lec03.pdf |
out the action
Parse Table Example
State
s0
s1
s2
s3
3
s4
s5
(
shift to s2
error
shift to s2
error
reduce (2)
reduce (3)
ACTION
)
error
error
shift to s5
shift to s4
4
reduce (2)
reduce (3)
hift t
$
error
accept
error
error
reduce (2)
reduce (3)
Goto
X
goto s1
goto s3
State Stack Symbol Stack
Input
(())
s0
s0
X... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/5e52815ef894b1842e678e6eb5c0a7b9_MIT6_035S10_lec03.pdf |
s3
3
s4
s5
(
shift to s2
error
shift to s2
error
reduce (2)
reduce (3)
ACTION
)
error
error
shift to s5
shift to s4
4
reduce (2)
reduce (3)
hift t
$
error
accept
error
error
reduce (2)
reduce (3)
Goto
X
goto s1
goto s3
• Reduce (n) (continued)
• Look up
t k]
T bl [t
• Table[top of the state stack][top of symbol stack... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/5e52815ef894b1842e678e6eb5c0a7b9_MIT6_035S10_lec03.pdf |
to s2
error
reduce (2)
reduce (3)
ACTION
)
error
error
shift to s5
shift to s4
4
reduce (2)
reduce (3)
hift t
$
error
accept
error
error
reduce (2)
reduce (3)
Goto
X
goto s1
goto s3
State Stack Symbol Stack
Input
Grammar
(())$
S → X$ (1)
X→ (X ) (2)
X→ ( ) (3)
X→ (
) (3)
s0
Parse Table In Action
State
s0
s1
s2
... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/5e52815ef894b1842e678e6eb5c0a7b9_MIT6_035S10_lec03.pdf |
s2
error
shift to s2
error
reduce (2)
reduce (3)
ACTION
)
error
error
shift to s5
shift to s4
4
reduce (2)
reduce (3)
hift t
$
error
accept
error
error
reduce (2)
reduce (3)
Goto
X
goto s1
goto s3
State Stack Symbol Stack
Input
Grammar
s2
s2
s2
s0
(
(
(
))$
S → X$ (1)
X→ (X ) (2)
X→ ( ) (3)
X→ (
) (3)
Parse Tab... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/5e52815ef894b1842e678e6eb5c0a7b9_MIT6_035S10_lec03.pdf |
Parse Table In Action
State
s0
s1
s2
s3
3
s4
s5
(
shift to s2
error
shift to s2
error
reduce (2)
reduce (3)
ACTION
)
error
error
shift to s5
shift to s4
4
reduce (2)
reduce (3)
hift t
$
error
accept
error
error
reduce (2)
reduce (3)
Goto
X
goto s1
goto s3
State Stack Symbol Stack
Input
Grammar
s5
s2
s2
s2
s0
)
(
(
(
)... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/5e52815ef894b1842e678e6eb5c0a7b9_MIT6_035S10_lec03.pdf |
s2
s0
(
)$
S → X$ (1)
X→ (X ) (2)
X→ ( ) (3)
X→ (
) (3)
Step Two: Push Nonterminal
p
State
s0
s1
s2
s3
3
s4
s5
(
shift to s2
error
shift to s2
error
reduce (2)
reduce (3)
ACTION
)
error
error
shift to s5
shift to s4
4
reduce (2)
reduce (3)
hift t
$
error
accept
error
error
reduce (2)
reduce (3)
Goto
X
goto s1
g... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/5e52815ef894b1842e678e6eb5c0a7b9_MIT6_035S10_lec03.pdf |
reduce (2)
reduce (3)
Goto
X
goto s1
goto s3
State Stack Symbol Stack
Input
Grammar
s2
s2
s0
X
X
(
)$
S → X$ (1)
X→ (X ) (2)
X→ ( ) (3)
X→ (
) (3)
Step Three: Use Goto, Push New State
p
,
State
s0
s1
s2
s3
3
s4
s5
(
shift to s2
error
shift to s2
error
reduce (2)
reduce (3)
ACTION
)
error
error
shift to s5
shift... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/5e52815ef894b1842e678e6eb5c0a7b9_MIT6_035S10_lec03.pdf |
)
ACTION
)
error
error
shift to s5
shift to s4
4
reduce (2)
reduce (3)
hift t
$
error
accept
error
error
reduce (2)
reduce (3)
Goto
X
goto s1
goto s3
State Stack Symbol Stack
Input
Grammar
s4
s3
s2
s2
s0
)
X
X
(
$
S → X$ (1)
X→ (X ) (2)
X→ ( ) (3)
X→ (
) (3)
Parse Table In Action
State
s0
s1
s2
s3
3
s4
s5
(
shi... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/5e52815ef894b1842e678e6eb5c0a7b9_MIT6_035S10_lec03.pdf |
One: Pop Stacks
p
p
State
s0
s1
s2
s3
3
s4
s5
(
shift to s2
error
shift to s2
error
reduce (2)
reduce (3)
ACTION
)
error
error
shift to s5
shift to s4
4
reduce (2)
reduce (3)
hift t
$
error
accept
error
error
reduce (2)
reduce (3)
Goto
X
goto s1
goto s3
State Stack Symbol Stack
Input
Grammar
$
S → X$ (1)
X→ (X ) (2)
X... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/5e52815ef894b1842e678e6eb5c0a7b9_MIT6_035S10_lec03.pdf |
Step Three: Use Goto, Push New State
p
,
State
s0
s1
s2
3
s3
s4
s5
(
shift to s2
error
shift to s2
error
reduce (2)
reduce (3)
ACTION
)
error
error
shift to s5
shift to s4
4
reduce (2)
reduce (3)
hift t
$
error
accept
error
error
reduce (2)
reduce (3)
Goto
X
goto s1
goto s3
State Stack Symbol Stack
Input
Grammar
$
S →... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/5e52815ef894b1842e678e6eb5c0a7b9_MIT6_035S10_lec03.pdf |
$ (1)
X→ (X ) (2)
X→ ( ) (3)
X→ (
) (3)
Key Concepts
t
•
•
t t
parser a
St k Fi i
t
c
controlling parser actions
• Pushdown automaton for parsing
• Stack, Finite state control
l
• Parse actions: shift, reduce, accept
Parse table for
Parse table for
• Indexed by parser state and input symbol
•
•
• Use sta... | https://ocw.mit.edu/courses/6-035-computer-language-engineering-spring-2010/5e52815ef894b1842e678e6eb5c0a7b9_MIT6_035S10_lec03.pdf |
6.801/6.866: Machine Vision, Lecture 12
Professor Berthold Horn, Ryan Sander, Tadayuki Yoshitake
MIT Department of Electrical Engineering and Computer Science
Fall 2020
These lecture summaries are designed to be a review of the lecture. Though I do my best to include all main topics from the
lecture, the lectures ... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e90d5693d5d378d3f19bf67913295aa_MIT6_801F20_lec12.pdf |
greyscale images - only black and white
– Arcane grammar is used for legal purposes - “comprises”, “apparatus”, “method”, etc.
– References of other patents are often included - sometimes these are added by the patent examiner, rather than the
patent authors
– Most patents end with something along the lines of “thi... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e90d5693d5d378d3f19bf67913295aa_MIT6_801F20_lec12.pdf |
higher-level machine vision tasks, such as:
• Attitude (pose estimation) of an object
• Object recognition
• Determining to position
We will touch more on these topics in later lectures.
1.2.1 Finding Edge with Derivatives
Recall that we find a proposed edge by finding an inflection point of the brightness E(x, y). ... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e90d5693d5d378d3f19bf67913295aa_MIT6_801F20_lec12.pdf |
maximum of ru(x):
2
Gradient2 of “Soft” Unit Step Function, ru(x)
0.1
5 · 10−2
)
x
(
u
2
r
0
−5 · 10−2
−0.1
−6
−4
−2
2
4
6
0
x
For those curious, here is the math behind this specific function, assuming a sigmoid for u(x):
1. u(x) = 1+exp (−x)
1
2. ru(x) =
du
dx
=
d
dx 1+exp −x
1
=
exp(−x)
(1+exp(... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e90d5693d5d378d3f19bf67913295aa_MIT6_801F20_lec12.pdf |
and 3, the
best point for estimating derivatives lies halfway between the two pixels.
How do we analyze the efficacy of this approach?
1. Taylor Series: From previous lectures we saw that we could use averaging to reduce the error terms from 2nd order
derivatives to third order derivatives. This is useful for analyti... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e90d5693d5d378d3f19bf67913295aa_MIT6_801F20_lec12.pdf |
2
1 −2
1
1
For deriving the sign here and understanding why we have symmetry, remember that convolution “flips” one of the two filters/-
operators!
Sanity Check: Let us apply this to some functions we already know the 2nd derivative of.
• f (x) = x :
2
2
f (x) = x
f 0(x) = 2x
f 00(x) = 2
Applying the 2... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e90d5693d5d378d3f19bf67913295aa_MIT6_801F20_lec12.pdf |
00(x) = 0
Applying the 2nd derivative estimator above to this function:
1 −2 1 ⊗
1
f (−1) = 1 f (0) = 1 f (1) = 1 =
1
2
((1 ∗ 1) + (−2 ∗ 1) + (1 ∗ 1)) =
1
2
(1 + −2 + 1) = 0
Where we note that = 1 due to the pixel spacing. This is equivalent to f 00(x) = 0.
In Practice: As demonstrated in the ... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e90d5693d5d378d3f19bf67913295aa_MIT6_801F20_lec12.pdf |
0(x + δ), for some δ ∈ R
Derivative of shifted function = Derivative equivalently shifted by same amount
• Linear :
d
dx
(af1(x) + bf2(x)) = af1
0 (x) + bf2
0 (x) for some a, b ∈ R
Derivative of scaled sum of two functions = Scaled sum of derivatives of both functions
We will exploit this linear, shift-invaria... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e90d5693d5d378d3f19bf67913295aa_MIT6_801F20_lec12.pdf |
.
5
1.2.4 Laplacian Estimators in 2D
Δ
2
∂2
∂x2
∂2
∂y2
The Laplacian r = Δ =
is another important estimator in machine vision, and, as we discussed last lecture, is the
lowest-order rotationally-symmetric derivative operator. Therefore, our finite difference/computational molecule estimates
should reflect this p... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e90d5693d5d378d3f19bf67913295aa_MIT6_801F20_lec12.pdf |
⎦ + 1
0
1
22
⎡
1
⎣0
1
⎤
1
0⎦ =
1
1
62
⎡
1
⎣4
1
0
−4
0
1
−4
1
4
−20
4
⎤
1
4⎦
1
Using Taylor Series, we can show that this estimator derived from this linear combination of estimators above results in an error
term that is one derivative higher than suing either of the individual estimat... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e90d5693d5d378d3f19bf67913295aa_MIT6_801F20_lec12.pdf |
but the maximum estimated gradient
(note that this is quantized at the octant level). Note that the quantized gradient direction is perpendicular to the edge. In this
case, for a candidate gradient point G0 and the adjacent pixels G− and G+, we must have:
G0 > G−, G0 ≥ G+
This forces − 1 ≤ s ≤ 1 . Note that we have... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e90d5693d5d378d3f19bf67913295aa_MIT6_801F20_lec12.pdf |
2 sign(s)s
0
2
• b = 2 → s = 4 sign(s)s
0
3
Where different interpolation methods give us different values of b.
1.2.8 Edge Transition and Defocusing Compensation
Another point of motivation: most edge detection results depend on the actual edge transition. Why are edges fuzzy (note that
some degree of fuzziness i... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e90d5693d5d378d3f19bf67913295aa_MIT6_801F20_lec12.pdf |
bright things appear. We can use this technique to see how accurately the algorithm plots the edge position - this allows for
7
error calculation since we have ground truth results that we can compute using the area of the circle. Our area of interest is
given by the area enclosed by the chord whose radial points i... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e90d5693d5d378d3f19bf67913295aa_MIT6_801F20_lec12.pdf |
(a)(b − x) + f (b)(x − a)
b − a
We can also leverage more sophisticated interpolation methods, such as cubic spline.
Why did the authors not leverage this interpolation strategy?
• this requires the spacing of any level, i.e. not just pixel and 2 pixel spacing, but everything in between.
√
• Since you interpolate,... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e90d5693d5d378d3f19bf67913295aa_MIT6_801F20_lec12.pdf |
Ex
(i)
Ey
#
=
Gradients at next step = Rotation R by θi × Gradients at current step
8
How do we select {θi}n
iterative update above if each time the candidate angle reduces |Ey| and increases |Ex|.
? We can select progressively smaller angles. We can accept the candidate angle and invoke the
i=1,2,···
The aggr... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e90d5693d5d378d3f19bf67913295aa_MIT6_801F20_lec12.pdf |
6.801/6.866: Machine Vision, Lecture 20
Professor Berthold Horn, Ryan Sander, Tadayuki Yoshitake
MIT Department of Electrical Engineering and Computer Science
Fall 2020
These lecture summaries are designed to be a review of the lecture. Though I do my best to include all main topics from the
lecture, the lectures ... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e98d8a6c0edce797859526dece67aea_MIT6_801F20_lec20.pdf |
Each of the tessellations from the platonic solids results in equal area projections on the sphere, but the division is
somewhat coarse.
For greater granularity, we can look at the 14 Archimedean solids. This allows for having multiple polygons in each
polyhedra (e.g. squares and triangles), resulting in unequal are... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e98d8a6c0edce797859526dece67aea_MIT6_801F20_lec20.pdf |
the performance of relative orientation
systems, and understanding their geometry can enable us to avoid using strategies that rely on these types of surfaces in order
to find the 2D transformation, for instance, between two cameras.
We will discuss more about these at the end of today’s lecture. For now, let’s intro... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e98d8a6c0edce797859526dece67aea_MIT6_801F20_lec20.pdf |
z
c
2
x
a2
2
y
b2
Note that this quadric surface has a linear, rather than quadratic dependence, on z.
Figure 6: 3D geometric depiction of a elliptic paraboloid, another member of the quadric family and a special case of the
hyperboloid of one sheet.
x − y
7. Hyperbolic Paraboloid: = 2
b2
a
z
c
2
2
Note that ... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e98d8a6c0edce797859526dece67aea_MIT6_801F20_lec20.pdf |
in te problems we solve: we are able to apply the same machine
vision framework regardless of “if the camera moved of it the world moved”.
• Therefore, the dual problem to binocular stereo is structure from motion, also known as Visual Odometry (VO).
This involves finding transformations over time from camera (i.e. o... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e98d8a6c0edce797859526dece67aea_MIT6_801F20_lec20.pdf |
process for the left and right cameras in our binocular stereo
system.
Figure 10: After finding the epipolar planes, we intersect these planes with the image plane, which gives us a set of lines in both
the left and righthand cameras/coordinate systems of our binocular stereo system.
A few notes on this process:
• ... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e98d8a6c0edce797859526dece67aea_MIT6_801F20_lec20.pdf |
question: How many correspondences do we
need to solve the binocular stereo/relative orientation problem?. To answer this, let us consider what happens when
we are given different numbers of correspondences:
• With 1 correspondence, we have many degrees of freedom left - i.e. not enough constraints for the number of ... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e98d8a6c0edce797859526dece67aea_MIT6_801F20_lec20.pdf |
Are there really 128 solutions? There are methods that have gotten this down to 20. With more correspondences and more
background knowledge of your system, it becomes easier to determine what the true solution is.
1.3.3 Determining Baseline and Rotation From Correspondences
To determine baseline and rotation for thi... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e98d8a6c0edce797859526dece67aea_MIT6_801F20_lec20.pdf |
and righthand
systems no longer intersect, as depicted in the diagram below:
6
Figure 12: With measurement noise, our rays do not line up exactly. We will use this idea to formulate our optimization problem
to solve for optimal baseline and rotation.
We can write that the error is perpendicular to the cross produ... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e98d8a6c0edce797859526dece67aea_MIT6_801F20_lec20.pdf |
·(r
l × (rl × rr
)): This yields the following:
Lefthand side :
Righthand side :
Combining :
0
0
0
l + γ(rl × rr
(αr
)) · ((r
l × (rl × rr)) = β||rl × rr
0
0
(b + βrr) · ((r
l × (rl × rr
0
2
β||r
l × rr
2
2
2||
0
0
)
)) = (b × r
l) · (rl × rr
0
0
)
= (b × r
l) · (rl × rr
||
0
0
Taking this dot... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e98d8a6c0edce797859526dece67aea_MIT6_801F20_lec20.pdf |
the direction along the rays in which we (almost) get an intersection between the left and right coordinate
systems. Typically, a negative α and/or β results in intersections behind the camera, which is often not physically feasible.
It turns out that one of the ways discard some of the 20 solutions produced by this ... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e98d8a6c0edce797859526dece67aea_MIT6_801F20_lec20.pdf |
1
2
0
How do we solve this? Because wi will change as our candidate solution changes, we will solve this problem iteratively and in
an alternating fashion - we will alternate between updated our conversion weights wi and solving for a new objective given the
recent update of weights. Therefore, we can write this opt... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e98d8a6c0edce797859526dece67aea_MIT6_801F20_lec20.pdf |
rb · qrlq
o
o
o o
= rr(bq) · qrl
o
o
o o
o
= rdd · qrl, where d = bq, which we can think of as a product of baseline and rotation.
o
o
Δ o
Recall that our goal here is to find the baseline b - this can be found by solving for our quantity d and multiplying it on the
∗ o
righthand side by q :
o
o
o
∗
o
o
o... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e98d8a6c0edce797859526dece67aea_MIT6_801F20_lec20.pdf |
we can interchange the order of how d and q interact with our measurements to produce the same
result.
o
o
oo
o o
t = rrd · qrl
o
o o
o
= rrq · drl
Given these symmetries, we can come up with some sample solutions:
oo
1. {q, d}
oo
2. {−q, d} (×2)
o
3. {q, −d} (×2)
o
o o
4. {d, q} (×2)
o
How do we av... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e98d8a6c0edce797859526dece67aea_MIT6_801F20_lec20.pdf |
of Hamilton’s quaternions by treating the
redundancies of this representation as extra constraints. Recall that Gibbs vectors, which are given as (cos
ωˆ), have
a singularity at θ = π.
θ
2
θ
2
, sin
1.3.6 When Does This Fail? Critical Surfaces
Like all the other machine vision approaches we have studied... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e98d8a6c0edce797859526dece67aea_MIT6_801F20_lec20.pdf |
of one sheet. We
need to ensure sections of surfaces you are looking at do not closely resemble sections of hyperboloids of one sheet.
There are also other types of critical surfaces that are far more common that we need to be mindful of when con-
sidering relative orientation problems - for instance, the intersectio... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e98d8a6c0edce797859526dece67aea_MIT6_801F20_lec20.pdf |
))−1J(θ)T e(θ)
2. LM: (J(θ)T J(θ) + λI)−1θ = J(θ)T e(θ) =⇒ θ = (J(θ)T J(θ) + λI)−1J(θ)T e(θ)
Where:
• θ is the vector of parameters and our solution point to this nonlinear optimization problem.
• J(θ) is the Jacobian of the nonlinear objective we seek to optimize.
• e(θ) is the residual function of the objective ... | https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/5e98d8a6c0edce797859526dece67aea_MIT6_801F20_lec20.pdf |
Dynamics of a Single Particle (Review) (continued)
1
2.003J/1.053J Dynamics and Control I, Spring 2007
Professor Thomas Peacock
2/12/2007
Lecture 2
Work-Energy Principle
Dynamics of a Single Particle (Review) (continued)
Reading: Williams 4–1, 4–2, 4–3 (Momentum Principles) - End of lecture last
time
Work-Energy... | https://ocw.mit.edu/courses/2-003j-dynamics-and-control-i-spring-2007/5eaa7b1b85eff45f6a92b5a6d6052665_lec02.pdf |
� W12 = T2 − T1 = (Work done is the change in the kinetic energy)
2 m|v|2
Consider the case:
F =
−∂V
∂r
For example,
⇒ W12 =
r2
�
r
1
F · dr =
Thus, in this special case:
F = (
∂V
∂x
r2 −∂V
∂r
�
r
1
ˆı −
∂V
∂y
jˆ)
· dr = V1 − V2(Potential Energy)
Or Conservation of Mechanical Energy
T2 − T1 = V1 ... | https://ocw.mit.edu/courses/2-003j-dynamics-and-control-i-spring-2007/5eaa7b1b85eff45f6a92b5a6d6052665_lec02.pdf |
.
Downloaded on [DD Month YYYY].
Dynamics of a Single Particle (Review) (continued)
3
Figure 2: Free body diagram. The only force acting on the mass is the force
due to gravity mg. Figure by MIT OCW.
Figure 3: Mass attached to a spring. Figure by MIT OCW.
Figure 4: Mass attached to linear dashpot. Figure ... | https://ocw.mit.edu/courses/2-003j-dynamics-and-control-i-spring-2007/5eaa7b1b85eff45f6a92b5a6d6052665_lec02.pdf |
a circular path. Determine
the angle θ0 to the position where the vehicle leaves the path and becomes a
projectile.
(Neglect friction and treat vehicle as a particle.)
Problem relates to driving a car down a hill.
Whenever you solve a problem, make sure that you draw a good diagram.
Cite as: Thomas Peacock and Ni... | https://ocw.mit.edu/courses/2-003j-dynamics-and-control-i-spring-2007/5eaa7b1b85eff45f6a92b5a6d6052665_lec02.pdf |
�eˆt
�
From Newton II:
F = p˙
�
N − mg cos θ = −mR θ˙2
Normal
mg sin θ = mR θ ¨
Tangential
(1)
(2)
We have:
−mg cos θ + N = −mR θ˙2
Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and
Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachus... | https://ocw.mit.edu/courses/2-003j-dynamics-and-control-i-spring-2007/5eaa7b1b85eff45f6a92b5a6d6052665_lec02.pdf |
48.2
2
3
◦
Note: This solution is independent of R and m.
Linear momentum principle used to solve problem.
Next time, use angular momentum principle to solve a problem.
Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and
Control I, Spring 2007. MIT OpenCourseWare ... | https://ocw.mit.edu/courses/2-003j-dynamics-and-control-i-spring-2007/5eaa7b1b85eff45f6a92b5a6d6052665_lec02.pdf |
18.657: Mathematics of Machine Learning
Lecturer: Philippe Rigollet
Scribe: Ali Makhdoumi
Lecture
6
Sep. 28, 2015
5. LEARNING WITH A GENERAL LOSS FUNCTION
In the previous lectures we have focused on binary losses for the classification problem and
developed VC theory for it. In particular, the risk for a classification f... | https://ocw.mit.edu/courses/18-657-mathematics-of-machine-learning-fall-2015/5ebb42429b252cbd2f711cd03e01b97f_MIT18_657F15_L6.pdf |
ification when needed.
[
−
[
−
∈
5.1 Empirical Risk Minimization
5.1.1 Notations
Loss function: In binary classification the loss function was 1I(h(X) = Y ). Here, we
replace this loss function by ℓ(Y, f (X)) which we assume is symmetric, where f
,
f :
1, 1] is the regression functions. Examples of loss function include
... | https://ocw.mit.edu/courses/18-657-mathematics-of-machine-learning-fall-2015/5ebb42429b252cbd2f711cd03e01b97f_MIT18_657F15_L6.pdf |
�
argmin R(f ).
f ∈F
Note that this is an oracle as in order to find it one need to have access to PXY and then
optimize R(f ) (we only observe the data Dn). Since f is the minimizer of the empirical
ˆ
risk minimizer, we have that Rn(f )
≤
ˆ
ˆR(f )
ˆ
Rn(f ) + Rn(f )
¯
ˆ
Rn(f ), which leads to
ˆ
¯
ˆ
Rn(f ) + Rn(f )
−
¯
ˆ... | https://ocw.mit.edu/courses/18-657-mathematics-of-machine-learning-fall-2015/5ebb42429b252cbd2f711cd03e01b97f_MIT18_657F15_L6.pdf |
n(f )
|
f ∈F
−
R(f )
| ≤
E
I
ˆ
sup Rn(f )
|
"
∈
f
F
−
R(f )
|#
+
r
log (1/delta)
2n
, w.p. 1
.
δ
−
As a result we only need to bound the expectation IE[supf ∈F |
ˆ
Rn(f )
].
R(f )
|
−
2
6
5.1.2 Symmetrization and Rademacher Complexity
Similar to the binary loss case we first use symmetrization technique and then intro-... | https://ocw.mit.edu/courses/18-657-mathematics-of-machine-learning-fall-2015/5ebb42429b252cbd2f711cd03e01b97f_MIT18_657F15_L6.pdf |
"|
n
i=1
X
n
ℓ(Yi, f (X
i))
1
− n
ℓ(Yi, f (X
i))
i=1
X
ℓ(Yi, f (Xi))
1
n
−
1
n
−
n
i=1
X
n
i=1
X
n
i=1
X
n
ℓ(Y ′
i , f (X ′
i)) Dn
|
ℓ(Yi , f (X ′
′
i)) Dn
|
# |#
# |#
′
ℓ(Y , f (X ′
i
i))
Dn
| |
##
i=1
X
ℓ(Y ′
i , f (X ′
i))
|#
|#
i=1
X
n
1
sup
f ∈F | n
(c)
≤
2IE
"
i=1
X
1
sup
f ∈F n
|
2 sup IE
Dn
≤
"
σi
ℓ(Yi, f (Xi))... | https://ocw.mit.edu/courses/18-657-mathematics-of-machine-learning-fall-2015/5ebb42429b252cbd2f711cd03e01b97f_MIT18_657F15_L6.pdf |
f ∈F n
|
"
n
i=1
X
ℓ(Yi, f (Xi))
−
IE[ℓ(Yi, f (Xi))]
|# ≤
n(ℓ
2
R
◦ F
)
and we only require to bound the Rademacher complexity.
5.1.3 Finite Class of functions
Suppose that the class of functions
F
is finite. We have the following bound.
3
Theorem: Assume that
F
is finite and that ℓ takes values in [0, 1]. We have
n(ℓ
)... | https://ocw.mit.edu/courses/18-657-mathematics-of-machine-learning-fall-2015/5ebb42429b252cbd2f711cd03e01b97f_MIT18_657F15_L6.pdf |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.