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Unformatted text preview: EECS 20. Final Exam Solution 9 December
1998
1. (a) Let = exp(2i 5=12). Then
n
n
X k 1 n+1
X
exp(2ik 5=12) = =
1
k=0
k=0
So this sum is zero if
1 n+1 = 0 , n+1 = exp[2(n+1)5=12] = 1 , 2(n+1)5=12 is a multiple of 2
The smallest n > 0 for which this holds is n = 11.
(b) Re[(1 + i) exp i = cos sin = 0 if = =4.
(c) Let A = exp i in polar coordinates. Then A exp(i!t) + A exp( i!t) = 2Re[ exp(i!t + )
= 2 cos(!t + ) = cos(!t + =4)
if = 1=2; = =4. So A = 1=2 exp(i=4).
2. The sets are shown in Figure 2 1 1 (a) (b) 1 (c) Figure 1: These are the required sets.
3. (a) The lists are F0 = f(a; a); (a; b); (b; c); (c; c); (c; d); (d; b)g
F1 = f(a; a); (b; b); (b; c); (d; a); (d; b); (d; d)g
F01 = f(a; a); (b; c); (d; b)g
F0or1 = f(a; a); (a; b); (b; b); (b; c); (c; c); (c; d); (d; a); (d; b); (d; d)g
(b) Both assertions are true.
(c) We have F00 = f(a; a); (a; b); (a; c); (b; c); (b; d); (c; c); (c; d); (c; b); (d; c)g
1 4. Observe that x0 (t) = x(t) x(t 1). By linearity and time invariance, it must therefore
be true that
8
> 1; if 0 t < 1
<
0 (t) = y (t) y (t 1) =
y
> 0;1; if 2 t < 3
:
otherwise
This is sketched below: y'(t) 2 3 t 1 Figure 2: The output signal.
5. (a) Observe that x(n) = cos(n=2) = (ejn=2 +e jn=2 )=2. Since H (=2) = H ( =2) =
=2, it follows that y(n) = cos(n=2).
2
j 0n . Since H (0) = 0, it follows that y (n) = 0.
(b) Observe that x(n) = 5e
(c) Observe that x(n) = cos(n) = (ejn + e jn )=2. Since H ( ) = H ( ) = , it
follows that
(
y(n) = cos(n) = +; n even
; n odd
6. The state transition diagram is shown in Figure 3. Note the names of the four states
indicate the pattern that is remembered. The output sequence is fffftfftft .
1 0 1 φ
0 1 10 f f f 101
t
0 1 1 0 Figure 3: Required state transition diagram.
7. (a) The general expression is 8t = 0; 1; y(t) = c0Atx0 + t 1
X 0 s
c A bu(t s 1) s=0 (b) We have 8t
t 1
0At x0 + X c0As b
1 = c
s=0 2 (1) 4 = 2c0At x0 + t 1
X 0 s
cA b s=0 (2)
(3) i. If we subtract (1) from (2) we get 8t
3 = c0At x0
so the zeroinput response is 8t; y (t) = 3.
ii. If we subtract (4) from (1) we get
2= X1 0 s
cA b t s s=0 So the zerostate response is 8t; y (t) = 2. 3 (4) ...
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This note was uploaded on 05/17/2009 for the course EE 20N taught by Professor Ayazifar during the Spring '08 term at University of California, Berkeley.
 Spring '08
 Ayazifar
 Electrical Engineering

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