ECE 308 SIGNALS AND SYSTEMS FALL 2017 Answers to selected problems on prior years examinations

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1 ECE 308 SIGNALS AND SYSTEMS FALL 07 Answers to selected problems on prior years examinations Answers to problems on Midterm Examination #, Spring 009. x(t) = r(t + ) r(t ) u(t ) r(t ) + r(t 3) + u(t + 3). x[n] = δ[n + ] + δ[n] δ[n ] + δ[n 3] 3. Memoryless Linear Causal Time-invariant System YES NO YES NO System NO YES NO YES 5. (a) y[n] = y(t) =, n = 0 4, n = 6, n =, 3, 4, 5 4, n = 6, n = 7 0, otherwise ( e t ) u(t) One way to express the answer is y(t) = ( ) e (t+) u(t + ) ( ) e (t 3) u(t 3) and another is y(t) = 0, t < e e t ), t < 3 (e6 e ) e t, t 3

2 Answers to problems on Midterm Examination #, Spring 009. (a) X(ω) = x(t)e jω t dt x(t) = π X(ω)ejω t dω. x(t) = + 6 k= sin ( π k) cos ( k πt) π k V (Ω) = 6 ω sin ( 3 ω) e j(9/)ω (a) x(t) = 3 cos ( 6t + π 4 5. (a) V (ω) = (jω+3)(jω+) v(t) = [ e 3t + e t ] u(t) )

3 Answers to problems on Midterm Examination #3, Spring 009. W (Ω) = ( e jωn ) e jω ( e jω ). 3. H(ω) is given by y(t) = 3 + cos (3t) H(ω) 3 ( sin π ) 3t ω 5. y[n] = 3 ( )n

4 Answers to problems on the Final Examination, Spring 009. x(t) = u(t + ) r(t + ) + r(t ) r(t 3) u(t 3) r(t 4) + r(t 5). 3. (a) X k = { 0, k = 0,, 4, 6, k =, 3, 5, 7 x[n] = ( π ) cos 4 n + ( ) 3π cos 4 n y(t) = 3 cos(t) 5. y(t) = (e t e t )u(t) + (e (t ) e (t ) )u(t ) 6. y(t) = (( t 3 ) e t + 7 ) e 3t u(t)

5 Answers to problems on Midterm Examination #, Fall 009. (a) x[n] = u[n] + 4 u[n ] 3 u[n 5] 3 u[n 6] x[n] = δ[n] + 6 (δ[n ] + δ[n ] + δ[n 3] + δ[n 4]) + 3 δ[n 5]. (a) IS MEMORYLESS HAS MEMORY (c) LINEAR (d) LINEAR (e) NONCAUSAL (f) CAUSAL (g) TIME-INVARIANT (h) TIME-VARYING 3. y(t) y[n] =, n = 3, n = 6, n = 3 4, n = 4 0, n = 5 6, n = 6 5, n = 7 3, n = 8 0, otherwise t 5. (a) y(t) = e t ( e t ) u(t) y(t) = e (t ) ( e (t )) u(t )

6 Answers to problems on Midterm Examination #, Fall 009. Sum / Sinusoids / Periodic / Finite. x(t) = π + k= 4 π( 4k ) cos(kt) 3. x(t) = 3 π cos ( 5t + 3π 4 ) x(t) = 3 e t u(t) 3 e t u( t) 5. α = /, =

7 Answers to problems on Midterm Examination #3, Fall 009. or v[n] = v[n] = ( ) n u[n] + ( ) n u[ n] ( ) n u[n] + n u[ n]. (a) X 0 = 4, X =, X = 0, X 3 = x[n] = + cos ( π n) 3. (a) X(Ω) = + e jωn + e j3ωn (c) or v[n] = x[n] + x[n 4] + x[n 8] + + x[n 76] 9 v[n] = x[n 4k] k=0 j38ω sin(40ω) V (Ω) = X(Ω)e sin(ω) ( y(t) = 8 4 cos 3t π ) + 4 cos(000t π) 5. Set H(ω) to be H(ω) 0 ω

8 Answers to problems on the Final Examination, Fall 009 n y[n] n y[n] n y[n] (a) y (t) y (t) Y (ω) = e j(ω+) X(ω + ) + e j(ω ) X(ω ) Y (ω) = e jω [X(ω + ) + X(ω )] (c) Both systems are linear. (d) Both systems are time-varying. 3. y[n] = cos(π n) 3 y(t) = [ e 5(t ) + e 3(t )] u(t ) 5. y(t) = + e t sin(t), t 0

9 Answers to problems on the Midterm Examination #, Spring 0. x(t) = r(t + ) + r(t ) + r(t ) r(t 4) + u(t 4) u(t 6). 3. (a) (c) (d) 5. y[n] = 0, n 0, n = 4, n = 3, n = 3, n = 4, n = 5 0, n = 6, n = 7, n = 8 0, n = 9 y(t) = e t u(t) y(t) = e (t ) u(t ) y(t) = ( e t ) u(t) ) y(t) = 3 ( e (t ) u(t ) y(t) = ( e 6(t 3)) u(t 3) ( e 6(t 6)) u(t 6) y[n] = ( ) n u[n] 3

10 Answers to problems on the Midterm Examination #, Spring 0. x (t) = + ( ) ( ) ( ) k cos(kπt) (kπ) k= ( x (t) = ) ( ) k sin(kπt) kπ k=. (a) 3. sgn(t) = u(t) u( t) ( ) SGN(ω) = jω + πδ(ω) = jω t jπsgn( ω) ( ) jω + πδ( ω) Y (ω) = jaπδ(ω ω 0 ) + jaπδ(ω + ω 0 ) y(t) = A sin(ω 0 t) 5. X(4π) = X(Ω) = e jω ( e jω) 6. Even

11 Answers to problems on the Final Examination, Spring 0. (a) y(t) = (e t e t )u(t) y(t) = (e (t ) e (t ) )u(t ) (c) y(t) = (e t e t )u(t) e 4 (e (t ) e (t ) )u(t ). y[n] = + cos ( π n ) π 4 3. (a) y(t) = a 0 + ( a cos ( ) 4t π 4 + b sin ( )) 4t π 4 y(t) = cos ( ) 4t π π 4 Y (ω) = 5. y(t) = e t + e 4t for t 0 3e j4 (ω ) e j4 (ω + ) + 9

12 Answers to problems on the Midterm Examination #, Fall 0. (a) TRUE TRUE (c) TRUE (d) TRUE (e) TRUE (f) TRUE. x[n] n x(t) = r(t + 3) u(t + ) r(t + ) + u(t + ) + r(t ) r(t 3) u(t 4) y[n] = ( ( ) n ) u[n] 5. (a) y(t) = ( e t )u(t) ( e (t 3) )u(t 3) y(t) = 3( e (t ) )u(t ) (c) y(t) = ( e t )u(t)

13 Answers to problems on the Midterm Examination #, Fall 0. x(t) = k= ( cos(kπ )) sin(kπt) kπ. (a) X(ω) = /( + ω ) 3. y(t) = /( + t ) Y (ω) π - - π ω x(t) = π cos(ω 0t + θ) 5. (a) y(t + ) = x(t + ) cos(π(t + )) = x(t) cos(πt + 4π) = x(t) cos(πt) = y(t) d k = (c k + c k+ )

14 Answers to problems on the Midterm Examination #3, Fall 0. (a) y[n] = ( ) n cos ( π n) u[n]. (a) X 5 = Y (Ω) = e jω ( e jω) For k = 0,,..., 4 and for k = 6, 7,...,, X k = 0 3. (a) H(Ω) = 4 e jω 5. y[n] = cos ( π n) y[n] = ( 5 ( ) n 4 4 ( ) n ) u[n] 5 y(t) = (t 4 cos π ) 4

15 Answers to problems on the Final Examination, Fall 0. (a) H(Ω) = + cos(ω) y[n] =. Y (ω) = jω+3 3. y(t) = + 4 π cos(πt) (a) ω 0 = y(t) = 6 cos(t) 5. (a) y(t) = ( + e t e 3t ) u(t) y(t) = u(t)

16 Answers to problems on the Midterm Examination #, Fall 0. (a) TRUE TRUE (c) TRUE (d) TRUE (e) FALSE (f) TRUE. (a) x[n] n The plot for part is the same as for part (a). 3. x(t) = u(t + 3) r(t + 3) + 4r(t + ) u(t + ) r(t + ) + u(t ) u(t ) + 3r(t ) 3r(t 3) u(t 4) y[n] = 50 3 ((.06)n+ ) u[n] 5. (a) y(t) = (e t e t )u(t) y(t) = (e t e t )u(t) e 4 (e (t ) e (t ) )u(t )

17 Answers to problems on the Midterm Examination #, Fall 0. (a) x(t) = k= k 0 j ( cos kπ ( )) kπ e jkπt 3. (a) y(t) = k= ( ( ) ) kπ cos sin(kπt) kπ 3 e / e jω jω + (/4) e / π(jt (/4)) 3. (a) In this problem you derive that x(t) =. x(t) = 3 cos(6t + (π/4)) (a) v(t) = jtx(t) y(t) = jte jt x(t) 5. (a) We have Z(ω) = X L (ω) + X R (ω) + X L (ω ω ) + X L (ω + ω ) X R (ω ω ) X R (ω + ω ) where ω = 76000π. Z(ω) 3 30, 000π 30, 000π ω (c) y (t)

18 Answers to problems on the Midterm Examination #3, Fall 0. (a) Y (Ω) = X(Ω) + X( Ω)ejΩ Y (Ω) is even and purely real. Y (Ω) = 3/4 (5/4) cos(ω). (a) x[n] is periodic with period N = 4c (c) X = 4c and X 3 = 4c ( ) n u[n] y A (t) = 5 ( cos t + π ) 4 y B (t) = cos(3t π) 6 y(t) = cos(6t) cos(0t) + 35 cos(4t)

19 Answers to problems on the Final Examination, Fall 0. (a) y(t) = y 0 (t) y 0 (t ) = u(t) u(t ) u(t 4) + u(t 5) x (t) = x 0 (t+) x 0 (t ) so that y(t) = u(t+) u(t ) u(t 3)+u(t 5). y[n] = 4A cos ( π n) if x[n] = x [n], and y[n] = 0 if x[n] = x [n], so that we differentiate between the two possibilities based on whether or not y[n] = 0 or a multiple of cos ( π n). 3. y(t) = π cos(πt) y(t) = 4 ( e t e 4t) u(t) 4e ( e (t 3) e 4(t 3)) u(t 3) 5. y(t) = ( 5 e 3t + 3 e 5t ) u(t)

20 Answers to problems on Midterm Examination #, Spring 03. x(t) = r(t + ) + r(t) r(t ) + r(t ) + u(t ) r(t 3) + r(t 4) + u(t 4) u(t 6). x[n] n 3. (a) Causal, time-invariant Noncausal, time-invariant (c) Causal, time-invariant (d) Causal, time-varying 5. (a) 6. (a) y[n] = 9, n = 3, n = 7, n = 3 3, n = 4, n = 5 0, otherwise y(t) = 6 ( e t e 3t) u(t) y(t) = ( e (t ) e 3(t )) u(t ) ( e (t ) e 3(t )) u(t ) t, 0 t < y(t) = 4 t, t < 4 0, t < 0 or t 4 x(t) = p(t ) p(t 4) p(t 7)

21 Answers to problems on Midterm Examination #, Spring (a) TRUE x(t) = 6 + x e (t) = 3 + x o (t) = k= l= k= k 0 8 kπ sin ( 3kπ 4 ) e j kπ 4 t 4 cos ((l )πt) (l ) π ( ) k kπ sin(kπt) Y (ω) = ω (9 + ω ) The sketches for (a) and are as follows. (The curved portion of the plot in has formula + cos(πω/4).) 5. The sketch of Y (ω) is as follows.

22 Y (ω) 4π ω

23 Answers to problems on Midterm Examination #3, Spring 03. Y (Ω) =. (a) {X 0, X, X, X 3 } = {4, j, 0, j} c 0 =, c = j/, c = 0, c 3 = j/ (c) x[n] = sin(πn/) /8 ( + 4 e jω) ( y(t) = 8 4 cos 0.5t π ) ( ) sin t π 3 y(t) = 4 cos(5t) + cos(0t) cos(5t) y[n] = 3 4 ( )n

24 Answers to problems on the Final Examination, Spring 03. (a) y(t) = e t u(t) y(t) = e (t ) u(t ) e (t 4) u(t 4). y(t) = 3 cos(t) 3. X(ω) is purely real and with phase equal to 0 for all ω. The magnitude X(ω) is the same as X(ω), shown in the plot below. The output that results when p[n] is the input is shown below. 4 6y[n] n 8 Therefore, the smallest n0 with no overlap of pulse responses is (a) X(0) = 4, X(π/) =, X(π) = 0, X(3π/) = X0 = 4, X =, X = 0, X3 =

25 Answers to problems on Midterm Examination #, Fall 04. x(t) = r(t + ) r(t) u(t ) + r(t 4) r(t 5). 4 x(t) t 3. (a) y(t) = r(t) 4r(t ) + r(t 4) 4 y(t) 3 5 t 5. (a) { 4 ( ( y(t) = π + sin π t)), < t < 3 0, otherwise Ĥ(ω) = + jω y(t) = 3 ( + cos t 3π 4 ) 6. y(t) = u(t ) u(t 4)

26 Answers to problems on Midterm Examination #, Fall 04. (a) y(t) = 3 3t 3 5t 4t u(t) e 3e + e y(t) = 3 3(t ) 3 5(t ) 4(t ) u(t ) e 3e + e. y(t) = e t + e 4t, 3. t t y(t) = 3t + e u(t) 4 4 A sketch of the block diagram is as follows. 5. (a) No. There is a pole at s = 0. y(t) = tu(t)

27 Answers to problems on Midterm Examination #3, Fall 04.. (a) 3. x(t) = n= ( ( nπ cos nπ 3 X(ω) = ω y(t) = t )) sin(nπt) 4(ω + 3) ( (ω + 3) ), ω + 3 Y (ω) = 4(ω 3) ( (ω 3) ), ω 3 0, otherwise y(t) = (t 4 cos π ) 4

28 Answers to problems on the Final Examination, Fall 04. y(t) = ( e t )u(t) + ( e (t ) )u(t ) ( e (t ) )u(t ) ( e (t 3) )u(t 3). y(t) = ( 7 e t 3 ) e t u(t) 3. (a) 0 Ĥ(ω) = h(t)e jωt dt = e jωt dt = [ e j0ω ] 0 jω = jω ( e j0ω ) = jω (ej5ω e j5ω )e j5ω = j sin(5ω) jω e j5ω = sin(5ω) e j5ω ω y(t) = 0 ln() y[n] = 3 cos ( π 5 n )

29 Answers to problems on Midterm Examination #, Fall 05. x(t) = r(t + ) + r(t) + u(t ) r(t 4) + r(t 5). x(t) t 3. (a) Linear, time-invariant, causal, has memory Linear, time-invarient, non-causal, has memory y[n] n y(t) = 0, t < 0 t, 0 t < t + t, t <, t < 4 5 t, 4 t < 5 0, 5 t 0, t < ( y(t) = ) e (t ), t < 6 ( e 8 ), 6 t y(t) = [ e (t ) u(t ) e (t 3) u(t 3) ]

30 Answers to problems on Midterm Examination #, Fall 05. (a) T = (c) Even symmetry. (a) 3. x(t) = + X(ω) = y(t) = k= ( ) kπ kπ sin cos(kπt) ( + jω) e ( +jω) jω ω ( jt) e ( jt) 37 4 jt t ( x(t) = 8 cos 4t + π ) 6 (a) v(t) = x(t ) y(t) = x(t) (e t u(t)) 5. X(Ω) = 3/4 (5/4) cos(ω)

31 Answers to problems on Midterm Examination #, Fall 06. (a) TRUE FALSE (c) TRUE (d) TRUE (e) TRUE (f) TRUE. (a) x[n] n x[n] n 3. x(t) = u(t + 3) r(t + ) + 4r(t ) 3r(t 3) u(t 4) (a) 5. (a) y[n] = ( y[n] = ( ( ) n ) u[n ] ( ) n ) ( u[n ] y(t) = ( e t e 3t) u(t) y(t) = ( e t e 3t) u(t) ( ) ) n u[n 3]

32 Answers to problems on the Final Examination, Fall y(t) = + 9 ( 4 cos 4t π ) y(t) = 3 ( π cos n π ) 6 y(t) = π cos ( π t ) y(t) = 6(e t e t )u(t) 6(e (t 3) e (t 3) )u(t 3) y(t) = ( e t + e 4t) u(t)

33 Answers to problems on Midterm Examination #, Fall 06. (a) (c) c k is given by where ω 0 = π/t (d) x(t) is given by where ω 0 = π/t.. 3. (a) v(t) = t x(t) x(t) = 3 sin(5t) 5. (a) X(ω) = x(t)e jωt dt x(t) = X(ω)e jωt dω π c k = T x(t) = x(t) = k= T 0 x(t)e jkω 0t dt k= c k e jkω 0t ( ) ( ) 8 kπ kπ kπ sin cos 3 3 t V (ω) = j ω cos ( ) ( ω sin ω ) ω Y (ω) = 6 (jω + 4)(jω + ) y(t) = (e t e 4t )u(t)

34 Answers to problems on Midterm Examination #3, Fall Y (Ω) = e j4ω ( 3 e jω) x[n] = 3 4 cos ( π n ) ( y(t) = 6 + cos t 7π ) ( ) 3 6 sin t π y(t) = cos(πt) y[n] = ( π cos n 5π ) 6

35 Answers to problems on the Final Examination, Fall 06. (a) (c). 3. y(t) = 3 ( e 3t )u(t) y(t) = ( e 3(t ) )u(t ) ( e 3(t ) )u(t ) ( e 3(t 6) )u(t 6) y(t) = 3 y(t) = π sin(πt) (a) X(0) = 6, X(π/) = 0, X(π) =, X(3π/) = 0 X 0 = 6, X = 0, X =, X 3 = 0

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