CPSC 421 Homework 6: Solutions
Fall 2009
7.17 of Sipser: Suppose that P = N P , and let A P be some languge that is not or . Clearly, P = N P and A P implies that A N P . It remains to be shown that every B N P is polynomial time reducible to A. Consider
CPSC 421 Homework 5: Solutions
Fall 2009
7.21 of Sipser: If f (x1 , . . . , xn ) is a formula function, then the function g dened on one more variable, g (x1 , . . . , xn , xn+1 ) = f (x1 , . . . , xn ), has twice as many satisfying assignments as f . Thu
RW324: Formal Languages, Automata Theory, Computability and Complexity, 2010 Week 8
L. van Zijl
Department of Computer Science University of Stellenbosch
2010
CS324-W8 (1)
RW324: Formal Languages, Automata Theory, Computability and Complexity, 2010 Week 8
6.045J/18.400J: Automata, Computability and Complexity
Prof. Nancy Lynch
Recitation 6: Decidability and Undecidability
March 15, 2007 Problem 1: These are the key concepts from lecture this week: 1. Undecidability - p. 172-176 have great example proofs. E
Winter 2009
1. (Problem 8-5) (a)
G( s) H ( s) =
o
K ( s + 8) s( s + 5)( s + 6)
and 270
o
Asymptotes: K > 0:
90
K < 0:
0
o
and 180
o
Intersect of Asymptotes:
1 =
Breakaway-point Equation: Breakaway Points:
3
0 5 6 ( 8) 3 1
2
= 15 .
2s + 35s + 176s + 240 =
Winter 2008
EECS 460 HW #5 Solutions 1. (Problem 7-19)
Closed-Loop Transfer Function:
Y (s) R( s) = 25 K s + ( 5 + 500 K t ) s + 25 K
2
Characteristic Equation:
s + ( 5 + 5000 K t ) s + 25 K = 0
2
= 0.6
2 n = 5 + 500 Kt = 1.2 n
From Eq. (7-118), settling
EECS 460 Homework #4 Solutions
1. (Problem 3-10) Use Masons Rule to obtain the transfer function from Y1(s) to Y3(s) as follows: (a) Direct Paths Single Loops P1 = G L1 = -GH
= 1 + GH Y ( s) G 3 = Y1 ( s ) 1 + GH
(b) The direct path and loop gain is the
Winter 2009 EECS 460 Homework #3 Solutions
1. (a)
+ 4 y = yy u
_
4
(b)
+ 4 y = 6u + 9u yy
2 + 4 y =. We know that Y ( s ) = 9s + 6 and yy u U (s) s3 s 2 + 4
Let us start off with
Y (s) 1 . Therefore, Y= Y ( s ) ( 9 s 2 + 6 ) 9 s 2Y ( s ) + 6Y ( s ) or =