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Electrical Engineering 20N - Fall 1998 - Final Exam

# Electrical Engineering 20N - Fall 1998 - Final Exam - EECS...

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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) = (ejn=2 +e jn=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) = (ejn + e jn )=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 zero-input 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 zero-state response is 8t; y (t) = 2. 3 (4) ...
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