text stringlengths 30 4k | source stringlengths 60 201 |
|---|---|
}
• Path to accepting state:
ε
a Æ a Æ a Æ e Æ f Æ g
1 0
0
0
Viewing computations as a tree
0,
1
a
ε
ε
b
e
0
1
c
f
1
0
d
g
Input w = 01
a
0
In general, accept if
there is a path labeled
by the entire input
string, possibly
interspersed with εs,
leading to an
accepting state.
Here, lea... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b6d125bc94344e53285a1cdbdb33371_MIT6_045JS11_lec04.pdf |
it isn’t accepted.
• L(M), the language recognized by NFA M, =
{ w | w is accepted by M}.
NFAs vs. FAs
NFAs vs. DFAs
•
DFA = Deterministic Finite Automaton, new name for
ordinary Finite Automata (FA).
– To emphasize the difference from NFAs.
• What languages are recognized by NFAs?
• Since DFAs are special ... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b6d125bc94344e53285a1cdbdb33371_MIT6_045JS11_lec04.pdf |
δ1( r, a ) )
• Starting from states in S, δ2( S, a ) gives all states M1 could reach
after a and possibly some ε-transitions.
– M2 recognizes L(M1): At any point in processing the
string, the state of M2 represents exactly the set of states
that M1 could be in.
Example: NFA Æ DFA
• M1:
0,1
0
a
b
1
c
• S... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b6d125bc94344e53285a1cdbdb33371_MIT6_045JS11_lec04.pdf |
ited
Closure under operations
• The last example suggests we retry proofs of
closure of FA languages under concatenation and
star, this time using NFAs.
• OK since they have the same expressive power
(recognize the same languages) as DFAs.
• We already proved closure under common set-
theoretic operations---u... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b6d125bc94344e53285a1cdbdb33371_MIT6_045JS11_lec04.pdf |
2), a) = (δ1(q1, a), δ2(q2, a))
– q03 = (q01, q02)
– F3 = { (q1,q2) | q1 ∈ F1 or q2 ∈ F2 }
Closure under union
• Theorem: FA-recognizable languages are closed
under union.
• New Proof:
– Start with NFAs M1 and M2.
– Get another NFA, M3, with L(M3) = L(M1) ∪ L(M2).
ε
Add new
start state
ε
M1
M2
Use f... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b6d125bc94344e53285a1cdbdb33371_MIT6_045JS11_lec04.pdf |
* = { x | x = y1 y2 … yk for some k ≥ 0, every y in L }
= L0 ∪ L1 ∪ L2 ∪ …
• Theorem: FA-recognizable languages are closed under
•
star.
Proof:
– Start with FA M1.
– Get an NFA, M2, with L(M2) = L(M1)*.
ε
M1
Add new start
state; it’s also
a final state,
since ε is in
L(M1)*.
ε
Use final states
from M1 an... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b6d125bc94344e53285a1cdbdb33371_MIT6_045JS11_lec04.pdf |
– ε,
– ∅,
– ( R1 ∪ R2 ), where R1 and R2 are smaller regular expressions,
– ( R1 ° R2 ), where R1 and R2 are smaller regular expressions, or
– ( R1* ), where R1 is a smaller regular expression.
• A recursive definition.
Regular expressions
• Definition: R is a regular expression over alphabet Σ
exactly if... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b6d125bc94344e53285a1cdbdb33371_MIT6_045JS11_lec04.pdf |
R1* ) = ( L(R1) )*
• Example: Expression ( ( 0 ∪ 1 ) ε )* ∪ 0 denotes
language { 0, 1 }* ∪ { 0 } = { 0, 1 }*, all strings.
• Example: ( 0 ∪ 1 )* 111 ( 0 ∪ 1 )* denotes { 0, 1 }*
{ 111 } { 0, 1 }*, all strings with substring 111.
More examples
• Definition:
– L(a) = { a }; one string, with one symbol a.
... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b6d125bc94344e53285a1cdbdb33371_MIT6_045JS11_lec04.pdf |
� 1 )*
Abbreviate:
Σ* 01 Σ* ∪ Σ* 10 Σ*
• Example: L = strings with neither substring 01 or
10.
– Can’t write complement.
– But can write: 0* ∪ 1*.
• Example: L = strings with no more than two
consecutive 0s or two consecutive 1s
– Would be easy if we could write complement.
( ε ∪ 1 ∪ 11 ) (( 0 ∪ 00 ) (1 ∪ ... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b6d125bc94344e53285a1cdbdb33371_MIT6_045JS11_lec04.pdf |
of R:
• Show for the three base cases.
• Show how to construct NFAs for more complex expressions
from NFAs for their subexpressions.
– Case 1: R = a
• L(R) = { a }
a
– Case 2: R = ε
• L(R) = { ε }
Accepts only a.
Accepts only
ε.
Theorem 1
• Theorem 1: If R is a regular expression, then L(R)
is a regul... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b6d125bc94344e53285a1cdbdb33371_MIT6_045JS11_lec04.pdf |
:
b:
a
• ab:
• a*:
a
ε
• ab ∪ a*:
ε
a
ε
ε
ε
b
a
ε
b
ε
a
ε
b
Theorem 2
• Theorem 2: If L is a regular language, then there
is a regular expression R with L = L(R).
• Proof:
– For each NFA M, define a regular expression R with
L(R) = L(M).
– Show with an example:
b
a
a
x
a
b
b
y
z
– Convert ... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b6d125bc94344e53285a1cdbdb33371_MIT6_045JS11_lec04.pdf |
.
Theorem 2
a ∪ bb* a
b*a
q0
b a*
y
qf
• Finally, remove y:
b*a (a ∪ bb* a)* b a*
q0
qf
• New label describes all strings that can move the machine
from q0 to qf, visiting (just) y any number of times.
• This final label is the needed regular expression.
Theorem 2
• Define a generalized NFA (gNFA).
... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b6d125bc94344e53285a1cdbdb33371_MIT6_045JS11_lec04.pdf |
Lemma
• Algorithms that answer questions about
FAs.
• Reading: Sipser, Section 1.4; some pieces
from 4.1
MIT OpenCourseWare
http://ocw.mit.edu
6.045J / 18.400J Automata, Computability, and Complexity
Spring 2011
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b6d125bc94344e53285a1cdbdb33371_MIT6_045JS11_lec04.pdf |
6.045: Automata, Computability, and
Complexity
Or, Great Ideas in Theoretical
Computer Science
Spring, 2010
Class 7
Nancy Lynch
Today
• Basic computability theory
• Topics:
– Decidable and recognizable languages
– Recursively enumerable languages
– Turing Machines that solve problems involving FAs
– Undecidability ... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b9f68b5081985b135f6ad521a93d81d_MIT6_045JS11_lec07.pdf |
Definition:
In fact, these two notions define different language classes:
– L is Turing-recognizable
– L is Turing-decidable
if there is some TM that recognizes L.
if there is some TM that decides L.
• The classes of Turing-recognizable and Turing-decidable
languages are different.
• Theorem 2: If L is Turing-decid... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b9f68b5081985b135f6ad521a93d81d_MIT6_045JS11_lec07.pdf |
Idea: Run both M1 and M2 on w.
• One must accept.
• If M1 accepts, then M accepts.
• If M2 accepts, then M rejects.
– But, we can’t run M1 and M2 one after the other because
the first one might never halt.
– Run them in parallel, until one accepts?
– How? We don’t have a parallel Turing Machine model.
Decidable and r... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b9f68b5081985b135f6ad521a93d81d_MIT6_045JS11_lec07.pdf |
to
renaming).
Examples
• Example: Every regular language L is
decidable.
– Let M be a DFA with L(M) = L.
– Design a Turing machine M′ that simulates M.
– If, after processing the input, the simulated M is
in an accepting state, M′ accepts; else M′
rejects.
Examples
• Example: Let X = be the set of binary
represen... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b9f68b5081985b135f6ad521a93d81d_MIT6_045JS11_lec07.pdf |
sequence of finite strings (possibly infinitely many)
•
•
•
on output tape.
It may print the same string more than once.
If E is an enumerator, then define
L(E) = { x | x is printed by E }.
If L = L(E) for some enumerator E, then we say that L is
recursively enumerable (r.e.).
Recursively enumerable languages
• Int... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b9f68b5081985b135f6ad521a93d81d_MIT6_045JS11_lec07.pdf |
steps for 1st, 2nd, and 3rd inputs, ε, 0 and 1.
• …
• Run more and more steps for more and more inputs.
– Eventually succeeds in reaching qacc for each accepting
computation of M, so enumerates all elements of L.
Recursively enumerable languages
• Theorem 8: L is recursively enumerable if and
only if L is Turing-rec... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b9f68b5081985b135f6ad521a93d81d_MIT6_045JS11_lec07.pdf |
Turing Machines that solve graph problems
• Consider the problem of whether a given graph
has a cycle of length > 2.
• Graph = (V,E), V = vertices, E = edges,
undirected.
• Write <G> for the encoding of G.
• Using this representation for the input, we can
write an algorithm to determine whether or not a
given graph... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b9f68b5081985b135f6ad521a93d81d_MIT6_045JS11_lec07.pdf |
inputs using standard encodings and
formalize the algorithms that we’ve already discussed,
using Turing machines.
Turing Machines that solve DFA problems
• Example: Equivalence for DFAs
L2 = { < M1, M2 > | L(M1) = L(M2) } is Turing-decidable.
• Elements of L2 are bit-string representations of
pairs of DFAs that rec... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b9f68b5081985b135f6ad521a93d81d_MIT6_045JS11_lec07.pdf |
DFA and L(M) = ∅ }, decidable by Turing machine that
searches M’s digraph.
{ < M, w > | M is a DFA, w is a word in M’s alphabet, and w ∈ L(M) },
decidable by a Turing machine that emulates M on w.
• Turing machines compute only on strings, but we can
regard them as computing on DFAs by encoding the DFAs
as strings (... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b9f68b5081985b135f6ad521a93d81d_MIT6_045JS11_lec07.pdf |
accepts exactly < M, w> encodings for which M accepts w.
• U is sometimes called a universal Turing machine because
it runs all TMs.
– Like an interpreter for a programming language.
The Acceptance Problem
• AccTM = { < M, w > | M is a TM and M accepts w }.
• U: On input < M, w >:
– Simulate M on input w.
– If M acce... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b9f68b5081985b135f6ad521a93d81d_MIT6_045JS11_lec07.pdf |
H(<M,w>) accepts if M accepts w, rejects if M rejects w
or if M loops on w.
– Use H to construct another TM H′ that decides a special
case of the same language.
– Instead of considering whether M halts on an arbitrary
w, just consider M on its own representation:
– H′(<M>):
• accepts if M accepts <M>,
• rejects if ... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b9f68b5081985b135f6ad521a93d81d_MIT6_045JS11_lec07.pdf |
, with TMs labeling rows and strings
that represent TMs labeling columns.
• The major diagonal describes results for M(<M>), for all M.
• D is a diagonal machine, constructed explicitly to differ
•
from the diagonal entries: D(<M>)’s result differs from
M(<M>)’s.
Implies that D itself can’t appear as a label for a r... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b9f68b5081985b135f6ad521a93d81d_MIT6_045JS11_lec07.pdf |
:
– Suppose for contradiction that HaltTM is Turing-
decidable, say by Turing machine R:
• R(<M,w>):
– accepts if M halts on (accepts or rejects) w,
– rejects if M loops (neither accepts nor rejects) on w.
– Using R, define new TM S to decide AccTM:
• S: On input <M,w>:
– Run R on <M,w>; R must either accept or reject... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b9f68b5081985b135f6ad521a93d81d_MIT6_045JS11_lec07.pdf |
Suppose HaltTM is Turing-decidable by TM R.
– S: On input <M,w>:
– Run R on <M,w>; R must either accept or reject;
can’t loop, by definition of R.
– If R accepts then M must halt (accept or reject)
on w. Then simulate M on w, knowing this must
terminate. If M accepts, accept. If rejects, reject.
– If R rejects, then... | https://ocw.mit.edu/courses/6-045j-automata-computability-and-complexity-spring-2011/5b9f68b5081985b135f6ad521a93d81d_MIT6_045JS11_lec07.pdf |
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
6.436J/15.085J
Lecture 3
Fall 2018
CONDITIONING AND INDEPENDENCE
Most of the material in this lecture is covered in [Bertsekas & Tsitsiklis] Sec-
tions 1.3-1.5 and Problem 48 (or problem 43, in the 1st edition), available at
http://athenasc.com/Prob-2nd-Ch1.pdf. Solutions to ... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/5ba99e3803701c145f1d06c26dcf61b8_MIT6_436JF18_lec03.pdf |
we have
∪∞P
i=1
�
Ai | B =
∞
X
i=1
P(Ai | B).
As a result, suppose PB : F → [0, 1] is defined by PB(A) = P(A | B).
Then, PB is a probability measure on (
, F).
(b) Let A be an event. If the events Bi, i ∈ N, form a partition of
P(Bi) > 0 for every i, then
, and
P(A) =
∞
X
i=1
P(A | Bi)P(Bi).
In parti... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/5ba99e3803701c145f1d06c26dcf61b8_MIT6_436JF18_lec03.pdf |
ned.
Proof.
(a) We have P(
| B) = P( ∩ B)/P(B) = P(B)/P(B) = 1. Also
Ω
Ω
∪∞P
i=1
�
P B ∩ (∪∞ Ai)
�
P(B)
Since the sets B ∩ Ai, i ∈ N are disjoint, countable additivity, applied to the
P(∪∞ (B ∩ Ai))
P(B)
Ai | B =
i=1
i=1
=
.
2
right-hand side, yields
∪∞
i=1
P
�
Ai | B =
P
∞
i=1
P(B ∩ Ai) ... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/5ba99e3803701c145f1d06c26dcf61b8_MIT6_436JF18_lec03.pdf |
and the result from part (c).
(d) Note that the sequence of events ∩n
is decreasing and converges to
∩∞ Ai. By the continuity property of probability measures, we have P ∩∞
i=1
i=1
Ai = limn→∞ P ∩i
n
=1 Ai . Note that
i=1Ai
�
�
∩n
i=1
P
�
Ai = P(A1) ·
P(A1)
·
P(A1 ∩ A2)
P(A1 ∩ A2) P(A1 ∩ A2 ∩ A3... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/5ba99e3803701c145f1d06c26dcf61b8_MIT6_436JF18_lec03.pdf |
condition is P(A) =
P(A | B).
⊥
(b) Let S be an index set (possibly infinite, or even uncountable), and let {As |
s ∈ S} be a family (set) of events. The events in this family are said to be
independent if for every finite subset S0 of S, we have
P ∩s∈S0 As =
�
P(As).
Y
s∈S0
(c) Let F1 ⊂ F and F2 ⊂ F be two ˙-... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/5ba99e3803701c145f1d06c26dcf61b8_MIT6_436JF18_lec03.pdf |
the ith toss resulted in a “1”. If i
6= j, the events Ai and Aj
are independent.
(b) The events in the (infinite) family {Ai | i ∈ N} are independent. This statement
captures the intuitive idea of “independent” coin tosses.
(c) Let Fn be the collection of all events whose occurrence can be decided by looking
at the... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/5ba99e3803701c145f1d06c26dcf61b8_MIT6_436JF18_lec03.pdf |
F2
turn out to be independent. This statement captures the intuitive idea that knowing
the results of the tosses at odd times provides no information on the results of the
tosses at even times.
4
2.1 How to check independence of ˙-algebras? p-systems.
How can one establish that two complicated ˙-fields (e.g., as ... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/5ba99e3803701c145f1d06c26dcf61b8_MIT6_436JF18_lec03.pdf |
,
A, B ∈ ⇒ A ∩ B ∈ ) is called a p-system.
Examples of p-systems are intervals (a, b) on R, rectangles (a, b) × (c, d) on
R2 , convex sets in Rd , etc.
Theorem 2. Let 1 and 2 be p-systems and Fi = ˙(i), i = 1, 2. If
P(A ∩ B) = P(A)P(B)
(1)
for every A ∈ 1, B ∈ 2, then F1 and F2 are independent.
Proof. F... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/5ba99e3803701c145f1d06c26dcf61b8_MIT6_436JF18_lec03.pdf |
P(B)P(A) ,
n→∞
n→∞
implying A ∈ LB. Similar argument holds for An ց A.
5
It turns out that 1 and 2 imply that LB contains
(1) (see Proposition to
follow). Thus by the monotone class theorem LB = F1. Thus (1) holds for all
A ∈ F1 and B ∈ 2. By symmetry it also holds for all A ∈ 1 and B ∈ F2.
And applying th... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/5ba99e3803701c145f1d06c26dcf61b8_MIT6_436JF18_lec03.pdf |
it is sufficient to prove D0 is a p-
system. Fix C ∈ and let
LC = {A ∈ D0 : A ∩ C ∈ D0}
On one hand, LC contains and
. On the other hand, LC is closed under
punching holes: For A ⊂ B we have (B \ A) ∩ C = (B ∩ C) \ (A ∩ C). Thus
LC = D0 by minimality of D0. Hence D0 is closed under intersections with
elements ... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/5ba99e3803701c145f1d06c26dcf61b8_MIT6_436JF18_lec03.pdf |
form)
(C) consists of Ø,
and all sets that can be written in the
Ω
α
(A1,1 ∩ . . . ∩ A1,n1 ) ∪ . . . ∪ (Am,1 ∩ . . . ∩ Am,nm ) ,
with Ai,j ∈ C or Ac ∈ C.
i,j
6
3 THE BOREL-CANTELLI LEMMA
The Borel-Cantelli lemma is a tool that is often used to establish that a certain
event has probability zero or... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/5ba99e3803701c145f1d06c26dcf61b8_MIT6_436JF18_lec03.pdf |
(b) If
P
Remark: The result in part (b) is not true without the independence assumption.
Indeed, consider an arbitrary event C such that 0 < P(C) < 1 and let An = C
for all n. Then P({An i.o.}) = P(C) < 1, even though
P(An) = ∞.
n
P
The following lemma is useful here and in may other contexts.
Lemma 1. Suppose ... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/5ba99e3803701c145f1d06c26dcf61b8_MIT6_436JF18_lec03.pdf |
This is true for every k. By taking the limit as k → ∞, we obtain log
pi) = −∞, and
(1 − pi) = 0.
For the converse statement, note that under pi < 1 we have
∞
i=1
Q
∞
i=1
(1 −
Q
∞
(1 − pi) = 0 ⇐⇒ ∀n
Y
i=1
∞
(1 − pi) = 0
Y
i=n
If pi 6→ 0 the result is automatic. Hence, we may also assume pi → 0. Then
takin... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/5ba99e3803701c145f1d06c26dcf61b8_MIT6_436JF18_lec03.pdf |
P
∞
n=1
∞
i=n
∞
P(A) ≤ P
�
∪∞
i=n
Ai ≤
P(Ai).
X
i=n
We take the limit of both sides as n → ∞. Since the right-hand side con-
verges to zero, P(A) must be equal to zero.
8
(b) Let Bn = ∪∞ Ai, and note that A = ∩∞ Bn. We claim that P(Bc ) = 0.
i=n
n=1
n
This will imply the desired result because
P(Ac) ... | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/5ba99e3803701c145f1d06c26dcf61b8_MIT6_436JF18_lec03.pdf |
m
Y �
i=n
1 − P(Ai) = 0,
where the second equality made use of the continuity property of probability
measures.
9
MIT OpenCourseWare
https://ocw.mit.edu
6.436J / 15.085J Fundamentals of Probability
Fall 2018
For information about citing these materials or our Terms of Use, visit: https://ocw.mit.edu/terms | https://ocw.mit.edu/courses/6-436j-fundamentals-of-probability-fall-2018/5ba99e3803701c145f1d06c26dcf61b8_MIT6_436JF18_lec03.pdf |
6.581 / 20.482 J
Foundations of Algorithms and Computational Techniques in Systems Biology
Adjoint Sensitivity Analysis for Optimization
From: Cao Y, Li ST, Petzold L, Serban R, Adjoint sensitivity
analysis or differential-algebraic equations: The adjoint DAE
system and its numerical solution, SIAM Journal on Scie... | https://ocw.mit.edu/courses/20-482j-foundations-of-algorithms-and-computational-techniques-in-systems-biology-spring-2006/5bb25cdfb8333d9ef2d0631fcffc8e01_adjoint.pdf |
&
x dp {
p
dp {
{ Easy { { Easy
+ F
x
dx&
dp
Hard
Hard
Hard
dt
2
6.581J / 20.482JFoundations of Algorithms and Computational Techniques in Systems BiologyProfessor Bruce TidorProfessor Jacob K. White
Trick 2: Integration By Parts
dG
dp
=
T
∫0
g
dx
x
dp
{
Hard
dx
+ g p +λ*(F
+ F
dt
)
x&
p
dp ... | https://ocw.mit.edu/courses/20-482j-foundations-of-algorithms-and-computational-techniques-in-systems-biology-spring-2006/5bb25cdfb8333d9ef2d0631fcffc8e01_adjoint.pdf |
⎝
dx
& dp
x
t T=
t =0
g −
x
T
d (λ Fx& ) +λ F = 0
dt
T
x
x
x
T
T
⎧ d (λ Fx& ) = λ F + g
⎪ dt
⎪
dG⎨
⎪
⎪ dp
⎩
( g p +λ F
p )
∫0 14243
Easy
=
T
T
dt
+
⎛
⎜λT F
x&
⎝
dx
dp
t T=
t =0
3
6.581J / 20.482JFoundations of Algorithms and Computational Techniques in Systems BiologyProfessor Bruce TidorProfessor Jac... | https://ocw.mit.edu/courses/20-482j-foundations-of-algorithms-and-computational-techniques-in-systems-biology-spring-2006/5bb25cdfb8333d9ef2d0631fcffc8e01_adjoint.pdf |
TidorProfessor Jacob K. White
What Do you Need To Know?
For Your Dynamic System:
( ,
f x t
,
( ,
f x t
x
f p x t
( ,
+
p A x A x x B u B2 x
) = 1 + 2 ⊗ +
1
) =
⊗ x
p A A I
,
+ ⊗ ) +
+ 2 (
x
I
⊗
1
( p) ⊗ + 1
p A x A x x B u B u (
) = 1
+ 1
B I
2
+ 1
u
⊗
( p)
,
p
(
)
u
For Your Objective Function... | https://ocw.mit.edu/courses/20-482j-foundations-of-algorithms-and-computational-techniques-in-systems-biology-spring-2006/5bb25cdfb8333d9ef2d0631fcffc8e01_adjoint.pdf |
Testing
6.170 Lecture 5
Fall 2005
Reading: Chapter 10 of Program Development in Java by Barbara Liskov
1
Program verification techniques and input space partitioning
The goal of testing — like that of all other program verification techniques — is to ensure that
a program functions correctly. Testing cannot prove ... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/5bb92f87c17d1cb32a055db3af9cb4cc_lec5.pdf |
check each output for correctness, then we are guaranteed that the program is correct:
it operates correctly no matter what input it is given. Checking the output is usually done
by recording the expected answer and comparing the actual answer against that, though the
checking may also be done procedurally. The desi... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/5bb92f87c17d1cb32a055db3af9cb4cc_lec5.pdf |
the equivalence group, and
if it works for any member of the equivalence group, then it works for all members. Testing
one member is equivalent, in terms of finding errors, to testing every member of the group.
Thus, a test suite that includes one (arbitrary) input from each equivalence group provides
complete testi... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/5bb92f87c17d1cb32a055db3af9cb4cc_lec5.pdf |
to partition the
input space. The end result is the same: we have guidance about how to build a test suite.
Three key heuristics for test suite selection are coverage, boundary cases, and duplicates.
• Coverage This is the key heuristic used in testing. Coverage criteria aims to make some
input partition (some test... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/5bb92f87c17d1cb32a055db3af9cb4cc_lec5.pdf |
domain requires having an element in each part of it. For a monolithic domain,
this requires only one test case. Some domains have a natural structure, and in such a case
2
all parts should be covered. For instance, given an integer output, it would be wise to supply
negative, positive, and zero inputs.
• Boundary... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/5bb92f87c17d1cb32a055db3af9cb4cc_lec5.pdf |
� 22� 23�
Actual partition�
As another example, you might reason that another program behaves one way for inputs less
than 10 and another way for inputs greater than or equal to 10; you presume that the test
suite consisting of 5 and 15 is exhaustive. Suppose that your reasoning was faulty and the
program has thre... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/5bb92f87c17d1cb32a055db3af9cb4cc_lec5.pdf |
) is sometimes called “stress testing.” It is particularly useful when the system
under test does not check its representation invariants, preconditions, or postconditions. If
something goes wrong in such a system, it may not be immediately apparent. A lengthy run
of the program gives the problem time to manifest it... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/5bb92f87c17d1cb32a055db3af9cb4cc_lec5.pdf |
you should take their cross-product; for instance, if one heuristic suggests two partitions
(say, over one input) and another suggests three partitions (say, over another input), then the test
suite should include at least six test cases.
3
Black-box testing
Black-box test cases are generated by examining the spec... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/5bb92f87c17d1cb32a055db3af9cb4cc_lec5.pdf |
. Supplying
Integer.MIN VALUE would uncover a bug in our implementation: no Java function can
satisfy the specification, because Integer.MIN VALUE = -2147483648 and Integer.MAX VALUE
= 2147483647.
– output: if the output is viewed as the ints, we would attempt to produce outputs
of Integer.MIN VALUE and Integer.MAX... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/5bb92f87c17d1cb32a055db3af9cb4cc_lec5.pdf |
// returns the length of the longest stutter in s
int stutterCount(String s)
We can apply the three heuristics:
• coverage
– input: include some test; the string input domain is monolithic. (Inputting the empty
string is a wise choice and implicitly suggested; it will also be covered by other heuristics,
below.) ... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/5bb92f87c17d1cb32a055db3af9cb4cc_lec5.pdf |
they don’t hurt.
• duplicates A string (of whatever length) containing only a single character achieves the goal
of making the stutter be the same as the whole string. Another kind of duplication is having
multiple stutters of the same length, either with the same character or with different char
acters. Such stutte... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/5bb92f87c17d1cb32a055db3af9cb4cc_lec5.pdf |
places.
Consider the following routine:
6
// returns the maximum of its arguments
int max3(int x, int y, int z) {
if (x>y) {
if (x>z) return x; else return z;
} else {
if (y>z) return y; else return z;
}
}
The declarative specification is much better than an operational (procedural) one, but it only
suggests three ... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/5bb92f87c17d1cb32a055db3af9cb4cc_lec5.pdf |
achieve branch coverage (and those two cases achieve
every variety of glass-box coverage). Likewise, branch coverage does not imply path coverage. Two
executions can ensure branch (and statement) coverage of
if (p1)
a;
else
b;
if (p2)
c;
else
d;
if the first sets p1 and p2 to true and the second sets them to false. ... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/5bb92f87c17d1cb32a055db3af9cb4cc_lec5.pdf |
may read about various types or levels of testing, such as unit, module, system, and
acceptance testing. These refer to testing larger and larger portions of an entire system and are
described in many software engineering books.
Testing can be performed either bottom-up, in which subpieces are verified before they ar... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/5bb92f87c17d1cb32a055db3af9cb4cc_lec5.pdf |
of this approach is that design errors can be identified early
rather than only when most of the system is built (and the pieces still don’t work together).
In general, the right approach to testing is to do the variety which suits your development style.
Testing as you code is more effective because you remember the ... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/5bb92f87c17d1cb32a055db3af9cb4cc_lec5.pdf |
are violated, then trouble is very likely later on.
The more quickly such an error is detected (in terms of how many lines of code have executed
between the occurrence of the error and its detection), the easier it is to locate. Locating an error
is frequently much, much harder than fixing it.
6
Static verification ... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/5bb92f87c17d1cb32a055db3af9cb4cc_lec5.pdf |
programmers to
realize that their implementation does not handle a particular case.
• ask a friend. Code inspections, in which a programmer who did not write the code reads and
reasons about it, are remarkably effective at finding errors. You aren’t able to do this in the
first half of 6.170 — that’s what the LAs and ... | https://ocw.mit.edu/courses/6-170-laboratory-in-software-engineering-fall-2005/5bb92f87c17d1cb32a055db3af9cb4cc_lec5.pdf |
2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 18 - Solution of F.E. equations
Prof. K.J. Bathe
In structures,
F (u, p) = R.
In heat transfer,
F (θ) = Q
In fluid flow,
F (v, p, θ) = R
In structures/solids
�
F (m) � �
F =
=
m
m
0V (m)
Elastic materials
Example p. 590 textbook
MIT O... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/5bbd789c4c008f43a78ffc97d314ed5d_MIT2_094S11_lec18.pdf |
0ρ 0L = tρ tL
⇒ 0
tL
tS11 = 0L
� �2
0L
tL
0L
tτ11 = tL
tτ11
∴
0
L tτ11 = E˜ ·
tL
1
2
��
t
u
1 + 0
L
�2
�
− 1
⇒ tτ11A =
tP =
��
1 +
E˜A
2
tu �2
0L
� �
− 1
1 +
tu �
0L
This is because of the material-law assumption (18.5) (okay for small strains . . . )
Hyperelasticity
tW = f (Green-Lagrange... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/5bbd789c4c008f43a78ffc97d314ed5d_MIT2_094S11_lec18.pdf |
of (18.1) (similarly (18.2) and (18.3))
Newton-Raphson Find U ∗ as the zero of f (U ∗)
f (U ∗) =
−
R
t+Δt
�
t+Δt
t+Δt
F
�
(i−1) +
U
= f
∂f
∂U
�
�
�
�
t+Δt
(i−1)
U
�
U ∗ − t+ΔtU (i−1) + H.O.T.
·
�
where t+ΔtU (i−1) is the value we just calculated and an approximation to U ∗.
Assume t+ΔtR is independent ... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/5bbd789c4c008f43a78ffc97d314ed5d_MIT2_094S11_lec18.pdf |
MIT 2.094
18. Solution of F.E. equations
Pictorially for a single degree of freedom system
i = 1;
i = 2;
tK Δu(1) = t+ΔtR − tF
t+ΔtK (1)Δu(2) = t+ΔtR
−
t+ΔtF (1)
Convergence Use
�ΔU (i)�2 < �
�a�2 =
��
(ai)2
i
But, if incremental displacements are small in every iteration, need to also use
�t+ΔtR − t+ΔtF... | https://ocw.mit.edu/courses/2-094-finite-element-analysis-of-solids-and-fluids-ii-spring-2011/5bbd789c4c008f43a78ffc97d314ed5d_MIT2_094S11_lec18.pdf |
Massachusetts Institute of Technology
Department of Electrical Engineering and Computer Science
6.245: MULTIVARIABLE CONTROL SYSTEMS
by A. Megretski �
Interpretations for Standard Optimization Setup1
This is the second lecture on standard feedback optimization setup. It describes a variety
of ways to come up with... | https://ocw.mit.edu/courses/6-245-multivariable-control-systems-spring-2004/5bca558808b8b652b14b2be6128d8e7d_lec2_6245_2004.pdf |
(transfer function 1/s) maps a signal f ≤ L1 to signals of the form
g(t) = c0 +
t
0
f (δ )dδ,
where c0 is an arbitrary constant. In general, a system’s output is not necessarily unique
because it may also depend on a set of auxiliary parameters (e.g. the initial states of
the system), as in Figure 2.1. In the c... | https://ocw.mit.edu/courses/6-245-multivariable-control-systems-spring-2004/5bca558808b8b652b14b2be6128d8e7d_lec2_6245_2004.pdf |
rate 1/M for every
positive integer M (though, indeed, it will not be the result of sampling f (t) = cos(�t)
at rate T = 1/M for M ≥= 1).
An alternative way of representing a discrete time signal f = f (t) with sampling rate
T > 0 is by specifying T and the sequence of its values f [k] at time instances tk = kT ,
... | https://ocw.mit.edu/courses/6-245-multivariable-control-systems-spring-2004/5bca558808b8b652b14b2be6128d8e7d_lec2_6245_2004.pdf |
f [k] into g[k] = f [k − 1], with the value of g[0]
being arbitrary.
∞� lm
[T ]
+
2.1.3 Finite Order CT LTI Models
A state-space model for a finite order CT LTI system H with input f = f (t), output
g = g(t), and state x = x(t) has the form
x˙ (t) = Ax(t) + Bf (t),
g(t) = Cx(t) + Df (t),
(2.1)
(2.2)
where A, ... | https://ocw.mit.edu/courses/6-245-multivariable-control-systems-spring-2004/5bca558808b8b652b14b2be6128d8e7d_lec2_6245_2004.pdf |
(2.3)
(2.4)
where A, B, C, D are constant matrices with real entries. Here f, g, x are DT signals
with same sampling rate T > 0. In terms of sampled value sequences f [k] = f (kT ),
x[k] = x(kT ), and g[k] = g(kT ), equations have the form
x[k + 1] = Ax[k] + Bf [k],
g[k] = Cx[k] + Df [k],
(2.5)
(2.6)
The trans... | https://ocw.mit.edu/courses/6-245-multivariable-control-systems-spring-2004/5bca558808b8b652b14b2be6128d8e7d_lec2_6245_2004.pdf |
A)−1B,
and, for an k-by-m complex matrix M ,
�max(M ) = max
u→Cm
, |u|=1
5
|M u|,
and |v| denotes the standard Hermitean norm (length) of vector v.
If H is a discrete time system, maximization over the imaginary axis is replaced with
maximization over the unit circle:
∈H∈� = sup �max(H(ej�)).
�→[−�,�]
... | https://ocw.mit.edu/courses/6-245-multivariable-control-systems-spring-2004/5bca558808b8b652b14b2be6128d8e7d_lec2_6245_2004.pdf |
. The definition for discrete time systems is similar, with integrals replaced by
sums:
N
inf
N �0
{π2|f [k]| − |g[k]|2} > −→,
2
�
k=0
6
where g = S(f ).
The informal rationale behind the definition is as follows: for “zero initial conditions”
(whatever this means), we expect the “energy” of the output to ... | https://ocw.mit.edu/courses/6-245-multivariable-control-systems-spring-2004/5bca558808b8b652b14b2be6128d8e7d_lec2_6245_2004.pdf |
≤ (0, →) (for gT this is
true since A is a Hurwitz matrix), and hence both have Fourier transforms fT and ˜T
respectively. In addition, by causality, g(t) = gT (t) for t < T . Hence
˜
g
|g(t)|2dt =
T
|g T (t)|2dt �
T
0
�
0
|g T (t)|2dt
0
�
=
1
2� −�
�
1
2� −�
=
|˜T
g (jσ)|2dσ =
|H(jσ)f˜T (jσ)|2d... | https://ocw.mit.edu/courses/6-245-multivariable-control-systems-spring-2004/5bca558808b8b652b14b2be6128d8e7d_lec2_6245_2004.pdf |
τ
�
0
|g1(t)|2dt.
To show that H-Infinity norm cannot be larger than L2 gain, consider zero initial
conditions and sinusoidal inputs f (t) = f0 cos(σt), where frequency σ and vector f0 are
appropriately chosen.
7
2.2.3 H2 norm and L2-to-L-Infinity gain
L2-to-L-Infinity gain of a stable state space model is defined... | https://ocw.mit.edu/courses/6-245-multivariable-control-systems-spring-2004/5bca558808b8b652b14b2be6128d8e7d_lec2_6245_2004.pdf |
[k] with zero mean, unit variance, and zero cross-corellation:
Ef [k] = 0, Ef [k1]f [k2]≥ = �(k1, k2)I,
where �(k1, k2) = 0 for k1 ≥= k2 and �(k, k) = 1 for all k.
The continuous time white noise is a slightly more complicated concept: it is a general
ized random process f = f (t) (something akin the Dirac delta in... | https://ocw.mit.edu/courses/6-245-multivariable-control-systems-spring-2004/5bca558808b8b652b14b2be6128d8e7d_lec2_6245_2004.pdf |
through an input, and will in turn be driven by an output
of an LTI system, as shown on Figure 2.2, left, where � is the nonlinear/time varying
block, and G is the LTI system.
� �
z1
G
� z2
w1
�
w2 �
�w1
w2 �
G
� z1
� z2
Figure 2.2: Small gain theorem
The following theorem, a version of the famous small... | https://ocw.mit.edu/courses/6-245-multivariable-control-systems-spring-2004/5bca558808b8b652b14b2be6128d8e7d_lec2_6245_2004.pdf |
for the standard setup shown on
Figure 2.3, where G is an unstable LTI plant with delay,
P (s) =
−� s
e
,
s − 1
and F is the controller to be designed to guarantee good tracking of the reference input
r at lower frequencies.
r
� � �
−�
F
�
P
�
q
Figure 2.3: Feedback design with delay
We begin by approxi... | https://ocw.mit.edu/courses/6-245-multivariable-control-systems-spring-2004/5bca558808b8b652b14b2be6128d8e7d_lec2_6245_2004.pdf |
P0=exp(-tau)/(s-1)+G0;
g=5;
d=150;
[A,B,C,D]=linmod2(’lec2_ex1a’);
p=pck(A,B,C,D);
k=hinfsyn(p,1,1,0,20,0.01);
2.3.3 A small gain theorem for H2
H2 norm, too, can serve as a measure of robustness to white noise perturbations of the
feedback loop gain.
Remember that the usual small gain bound on a CT system � with i... | https://ocw.mit.edu/courses/6-245-multivariable-control-systems-spring-2004/5bca558808b8b652b14b2be6128d8e7d_lec2_6245_2004.pdf |
every square integrable row vector function p = p(t). In the discrete time case, simply
require that
t2
t1
�
�
�
�
2
=
�
�
�
�
�
k2
p[k]g[k]
�
k=k1
k2
�
k=k1
|p[k]|2E|g[k]|
2
for every sequence p = p[k] of row vectors.
Theorem 2.4 Let G be a stable finite order LTI system with two inputs w1, w2 and two
ou... | https://ocw.mit.edu/courses/6-245-multivariable-control-systems-spring-2004/5bca558808b8b652b14b2be6128d8e7d_lec2_6245_2004.pdf |
, v = [v1; v2] is a strict sense white noise signal, i.e.
Evi[k]2 = 1, Evi[k] = 0,
and it is known that vi[k] is independent of y[t], vi[t − 1] for all t � k. What is the
asymptotic value of Ey[k]2 as k � →?
This question can be answered easily without relying on the H2 small gain theorem,
because
Ey[k + 1]2 = (a... | https://ocw.mit.edu/courses/6-245-multivariable-control-systems-spring-2004/5bca558808b8b652b14b2be6128d8e7d_lec2_6245_2004.pdf |
8.871 Lecture 3
M. Padi
November 14, 2004
Let’s use dimensional analysis to make some general remarks about the tensions of various branes. We’ll
start with the M2- and M5-branes. Note that the tension of an object that fills 2 space dimensions must
have units of [L−3] = [L2] . The only length scale we have is lp, ... | https://ocw.mit.edu/courses/8-871-selected-topics-in-theoretical-particle-physics-branes-and-gauge-theory-dynamics-fall-2004/5bcdff3a3071c28befd43ad3ef4fc350_lec3.pdf |
anes and D-branes using S-duality between
M-theory compactified on a circle and Type IIA string theory. We use the fact that the M2-brane compactified
over one of its dimensions becomes an F1-string in Type IIA. Thus we can write
TF 1 = =
1 R
l3
l2
p
s
.
(3)
An M2 which does not wrap the M theory circle becomes ... | https://ocw.mit.edu/courses/8-871-selected-topics-in-theoretical-particle-physics-branes-and-gauge-theory-dynamics-fall-2004/5bcdff3a3071c28befd43ad3ef4fc350_lec3.pdf |
is S-dual to itself at coupling g� = 1/gs. Under
this duality, the D1-brane becomes the F1-string, the D3-brane is self-dual and the D5-brane becomes the
NS5-brane. So now we have the following equalities:
s
TD1 =
TF 1 =
1
gsl2
s
1
l2
s
= T � =
F 1
= T � =
D1
1
l�2
s
1
g� l�2
s s
From these equations we ... | https://ocw.mit.edu/courses/8-871-selected-topics-in-theoretical-particle-physics-branes-and-gauge-theory-dynamics-fall-2004/5bcdff3a3071c28befd43ad3ef4fc350_lec3.pdf |
Frequency response: Resonance, Bandwidth, Q factor
Resonance.
Let’s continue the exploration of the frequency response of RLC circuits by investigating
the series RLC circuit shown on Figure 1.
I
R
C
Vs
R
+
VR
-
The magnitude of the transfer function when the output is taken across the resistor is
Figure 1
H
(
)
... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/5bcec4bfba5f2e99754b77509e9e7ab4_resonance_qfactr.pdf |
the current through it (
cos(
cos(
)
tω are
RV
RI
)
)
p t
( )
=
=
V
R
V I
R R
cos(
ω
t I
)
R
t
ω
2
cos (
6.071/22.071 Spring 2006, Chaniotakis and Cory
cos(
t
ω
)
)
(1.4)
2
And the average power becomes
P
(
)
ω =
=
1
2
1
2
V I
R R
2
I R
R
(1.5)
Notice that this power is a function o... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/5bcec4bfba5f2e99754b77509e9e7ab4_resonance_qfactr.pdf |
⎠
+
1
2
ω
0
ω
2
=
R
L
2
+
2
⎛
⎜
⎝
R
L
2
⎞
⎟
⎠
+
1
2
ω
0
6.071/22.071 Spring 2006, Chaniotakis and Cory
(1.8)
(1.9)
(1.10)
4
The bandwidth is the difference between the half power frequencies
Bandwidth B
=
=
−
1ω ω
2
(1.11)
By multiplying Equation (1.9) with Equation (1.10) ... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/5bcec4bfba5f2e99754b77509e9e7ab4_resonance_qfactr.pdf |
Vc
dt
=
ω
CA
cos(
t
ω
)
. The
SE
=
1
2
2
L C A
ω
2
2
2
cos (
t
ω
)
+
2
CA
2
sin (
ω
t
)
1
2
(1.15)
At the resonance frequency where
ω ω=
0
=
1
LC
the energy stored in the circuit
becomes
SE
=
2
CA
1
2
6.071/22.071 Spring 2006, Chaniotakis and Cory
(1.16)
5
... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/5bcec4bfba5f2e99754b77509e9e7ab4_resonance_qfactr.pdf |
the resonance characteristics of the parallel RLC circuit.
I s(t)
IR(t)
R
L
C
Here the impedance seen by the current source is
Figure 4
Z
//
=
j L
ω
(1
−
2
ω
LC
)
+
j L
ω
R
(1.20)
At the resonance frequency
resistive. The parallel combination of the capacitor and the inductor act as an open
circuit. Therefore at ... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/5bcec4bfba5f2e99754b77509e9e7ab4_resonance_qfactr.pdf |
⎜
⎝
1
RC
⎞
⎟
⎠
2
+
1
2
ω
0
ω
2
=
1
RC
2
+
2
⎛
⎜
⎝
1
RC
⎞
⎟
⎠
2
+
1
2
ω
0
And the bandwidth for the parallel RLC circuit is
Figure 5
The Q factor is
PB
−
ω ω=
2
1
=
1
RC
Q
=
ω
0
B
P
=
ω
0
RC
=
R
L
ω
0
(1.23)
(1.24)
(1.25)
(1.26)
(1.27)
6.071/22.071 Spring 2006, Chaniotakis and Cory
8
... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/5bcec4bfba5f2e99754b77509e9e7ab4_resonance_qfactr.pdf |
2
+
2
⎛
⎜
⎝
1
RC
⎞
⎟
⎠
2
+
1
2
ω
0
ω
2
=
1
RC
2
+
2
⎛
⎜
⎝
1
RC
⎞
⎟
⎠
2
+
1
2
ω
0
Bandwidth
B
S
−
ω ω=
2
1
=
R
L
PB
−
ω ω=
2
1
=
1
RC
Q factor
Q
=
0
L
ω ω
0
=
R
B
S
=
1
RC
ω
0
Q
=
ω
0
B
P
=
ω
0
RC
=
R
L
ω
0
6.071/22.071 Spring 2006, Chaniotakis and Cory
9
... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/5bcec4bfba5f2e99754b77509e9e7ab4_resonance_qfactr.pdf |
“notch” filter.
If we are interested in suppressing a 60 Hz noise signal then
2 60
π =
1
LC
6.071/22.071 Spring 2006, Chaniotakis and Cory
(1.29)
(1.30)
(1.31)
10
For L=47mH, the corresponding value of the capacitor is C=150µF.
The plot of the transfer function wit... | https://ocw.mit.edu/courses/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/5bcec4bfba5f2e99754b77509e9e7ab4_resonance_qfactr.pdf |
15.083J/6.859J Integer Optimization
Lecture 3: Methods to enhance formulations
1 Outline
• Polyhedral review
• Methods to generate valid inequalities
• Methods to generate facet defining inequalities
Slide 1
2 Polyhedral review
2.1 Dimension of polyhedra
• Definition: The vectors x1 , . . . , xk ∈ (cid:3)n
uni... | https://ocw.mit.edu/courses/15-083j-integer-programming-and-combinatorial-optimization-fall-2009/5be17dfa9a150e8db5b9fb76905d2408_MIT15_083JF09_lec03.pdf |
P .
• Let f (cid:2) x ≥ g be a valid inequality for a polyhedron P , and let F = {x ∈
P | f (cid:2) x = g}. Then, F is called a face of P and we say that f (cid:2) x ≥ g
represents F . A face is called proper if F = Ø(cid:6)
, P .
• A face F of P represented by the inequality f
(cid:2) x ≥ g, is called a facet of
P if... | https://ocw.mit.edu/courses/15-083j-integer-programming-and-combinatorial-optimization-fall-2009/5be17dfa9a150e8db5b9fb76905d2408_MIT15_083JF09_lec03.pdf |
of dimension one: −x1 + 2x2 ≤ 4, and 2x1 + x2 ≤ 8.
3 Methods to generate valid inequalities
3.1 Rounding
• Choose u = (u1, . . . , um)(cid:2) ≥ 0; Multiply ith constraint with ui and sum:
Slide 5
Slide 6
Slide 7
(cid:2)
n
(u
(cid:2)Aj )xj ≤ u
(cid:2)b.
j=1
• Since (cid:5)u (cid:2)Aj (cid:6) ≤ u (cid:2)Aj a... | https://ocw.mit.edu/courses/15-083j-integer-programming-and-combinatorial-optimization-fall-2009/5be17dfa9a150e8db5b9fb76905d2408_MIT15_083JF09_lec03.pdf |
e∈E(U )
• Round to (|U | is odd)
1
2
(cid:2)
e∈δ(U )
1
xe ≤ |U |.
2
(cid:2)
xe +
e∈E(U )
xe ≤ 1 |U |.
2
(cid:2)
xe ≤
e∈E(U )
(cid:9)
(cid:10)
1 |U | =
2
|U | − 1 ,
2
3.2 Superadditivity
3.2.1 Definition
A function F : D ⊂ (cid:3) (cid:8)n → (cid:3) is superadditive if for a1, a2 ∈ D,: a1 + a2 ∈ D :
... | https://ocw.mit.edu/courses/15-083j-integer-programming-and-combinatorial-optimization-fall-2009/5be17dfa9a150e8db5b9fb76905d2408_MIT15_083JF09_lec03.pdf |
F (Aj ) + F (Aj )(k − 1)
≤ F (Aj ) + F (Aj (k − 1))
≤ F (Aj + Aj (k − 1)),
by superadditivity, and the induction is complete.
Therefore,
n
(cid:2)
n
(cid:2)
F (Aj )xj ≤
F (Aj xj ).
Slide 12
j=1
j=1
3
By superadditivity,
⎛
n
(cid:2)
F (Aj xj ) ≤ F
⎝
n
(cid:2)
Aj xj
⎞
⎠ = F (Ax).
j=1
j=1
Since... | https://ocw.mit.edu/courses/15-083j-integer-programming-and-combinatorial-optimization-fall-2009/5be17dfa9a150e8db5b9fb76905d2408_MIT15_083JF09_lec03.pdf |
4
x1 + 4x2 + x3 + x4 ≥ 8 is valid for S.
(cid:5)
• For d = 1, and aj are not integers. In this case, since x ≥ 0, we obtain
n
(cid:5)aj (cid:6)xj ≤ (cid:5)a0(cid:6), and thus the following
j=1
(cid:5)aj (cid:6)xj ≤ a0. Since x ∈ Z,
(cid:5)
n
j=1
inequality is valid for S
(cid:2)
n
(aj − (cid:5)aj (cid:6))xj ... | https://ocw.mit.edu/courses/15-083j-integer-programming-and-combinatorial-optimization-fall-2009/5be17dfa9a150e8db5b9fb76905d2408_MIT15_083JF09_lec03.pdf |
is valid for S for some d ≥ 0, and the
aj xj + c(xk − α − 1) ≤ b is valid for S for some c ≥ 0, then the
aj xj ≤ b is valid for S.
n
j=1
n
j=1
(cid:5)
(cid:5)
(−x1 + x2) + (x2 − 3) ≤ 1, (−x1 + x2) − (x2 − 2) ≤ 1.
α = 2, −x1 + x2 ≤ 1 is valid.
3.5 Mixed integer rounding
3.5.1 Proposition
• For v ∈ (cid:3), f ... | https://ocw.mit.edu/courses/15-083j-integer-programming-and-combinatorial-optimization-fall-2009/5be17dfa9a150e8db5b9fb76905d2408_MIT15_083JF09_lec03.pdf |
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