EECS 206 Solutions to Midterm Exam 2 July 12, Problems 1 to 10 are multiple-choice. Each has 7 points. No partial credit will be given.

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1 EECS 06 Solutions to Midterm Exam July, 00 Instructions: Answer on this questionnaire Print your name Sign the pledge below Closed book and notes One 8 / x sheet of paper allowed Calculators allowed Read the questions carefully. Problems to 0 are multiple-choice. Each has 7 points. No partial credit will be given. In Problems and, partial credit will be given. You must show your derivations/calculations to get full credit. Name: PLEDGE: I have neither given nor received any aid on this exam, nor have I concealed any violations of the Honor Code. SIGNATURE: DO NOT TURN THIS PAGE OVER UNTIL TOLD TO DO SO! Good Luck!

2 Mark your answers for Problems ( (0 ( (a (b (c (d (e ( (a (b (c (d (e (3 (a (b (c (d (e ( (a (b (c (d (e (5 (a (b (c (d (e (6 (a (b (c (d (e (7 (a (b (c (d (e (8 (a (b (c (d (e (9 (a (b (c (d (e (0 (a (b (c (d (e Subtotal ( ( Total

3 ( The fundamental period of the following discrete-time signal is (a 8 (b (c 8 (d (e None of the above Answer: (e cos( π n cos(3π 7 n The signal consists of two sinusoids. The component signals are periodic: Let N and N denote their respective periods (they ust be positive integers. π N = πm, 3π 7 N = πl, where m and l are positive integers. This implies that Since N and N are positive integers, N = 8m, N = l/3. N = 8, 6,..., N =, 8,.... Therefore, common multiples of 8 and will be periods of the given signal: 56,,.... ( A real discrete-time signal x[n] is periodic with the fundamental period 8 and some of its 8-point DFT coefficients are given below. X 0 = 0, X =, X = j, X 3 = + j. Then X 6 is (a 0 (b (c + j (d j (e Not enough information is given Answer: (c A real signal has the complex conjugate property: X N k = X k. Then X 6 = X 8 = X = + j. (3 A real discrete-time signal x[n] is periodic with the fundamental period n 3

4 Let X k be the DFT coefficients of the 8-point DFT of x[n]. Then the value of (a (b (c (d (e None of the above Answer: (b Parseval s theorem states that Note that for the given signal Therefore, By evaluation, we get x[n] = n=0 7 x[n] = n=0 X 0 = X k. k=0 7 x[n] = 0. k=0 7 X k = X 0 + }{{} 0 k=0 7 x[n] =. n=0 7 X k. k= 7 X k equals k= ( The signal 0 cos ( π5t + π, when sampled at f s = 0 samples/sec, will alias to (a 0 cos ( π5t π (b 0 cos ( π5t + π (c 0 cos ( π5t π (d 0 cos ( π5t + π (e None of the above Answer: (b Aliases of 0 cos ( π5t + π 0 cos ( π(5 + f s lt + π or 0 cos ( π(f s m 5t π, where l and m are integers. Among these candidates the answer is the one with frequency less than half the sampling frequency, f s / = 0. Therefore, from the following candidates, l =, 0,,... = f = 5, 5, 5,..., m =, 3,,... = f = 5, 35, 55,.... the frequncy that is less than f s / = 0 is obtained when l = and the corresponding signal is 0 cos ( π5t + π.

5 (5 The ideal interpolation of cos( π n cos(7π n 0. using the sampling period T s = /f s = 0.0 will produce (a cos( π 00t cos( 7π 00t 0. (b cos( π 00t + 0. cos( 7π 00t + 0. (c cos( π 00t + 0. (d cos( π 00t 0. (e None of the above Answer: (c We note that if 0 ˆω < π. Since the second term of the given signal the given signal is, in effect, cos(ˆωn + φ = cos(ˆωf s t + φ cos( 7π n 0. = cos((7π πn 0. = cos( π n 0. = cos(π n + 0., Therefore, the ideal reconstruction will be cos( π n cos(7π n 0. = cos(π n cos( π n + 0. = cos(π 00t (6 The system whose input/output relationship is given by is y[n] = n x[n] x[n] y[n] n (a (b (c (d (e linear and time-invariant nonlinear and time-invariant linear and time-varying noninear and time-varying Not enough information is given Answer: (c The system is linear and time-varying. 5

6 (i Linearity T {αx [n] + βx [n]} = n ( αx [n] + βx [n] = αnx [n] + βn x [n] = αt {x [n]} + βt {x [n]}. (ii Time-varying (7 The condition for the system to be linear and time-invariant is (a ABC 0. (b A = B = 0 and C 0. (c A 0 and B = C = 0. (d C = 0. (e A = C = 0. Answer: (e T {x[n n 0 ]} = n x[n n 0 ] (n n 0 x[n n 0 ] = y[n n 0 ]. y[n] = Ax [n] + Bx[n 3] + C A must be zero, because otherwise the system will create a cross term, making it nonliner. Also C must be zero, because a nonzero constant term violate linearity. (8 The convolution of the following two signals x[n] = ( nu[n] and y[n] = u[n] x[n] y[n] n n is given by, for n 0, (a ( n (b + ( n (c ( n+ (d + ( n+ (e None of the above Answer: (a 6

7 x[n] y[n] = = = n k=0 x[k]y[n k] ( k u[k] }{{} =,k 0 ( k n 0 = (/n+ / (9 An LTI filter is described by the following difference equation: Which of the following statements is incorrect? u[n k] }{{} =,n k = ( n. y[n] = x[n + ] + x[n] + x[n ]. (a Its order is 3. (b It is noncausal. (c Its impulse response h[n] is given by h[n] = δ[n + ] + δ[n] + δ[n ]. (d y[n] is periodic whenever x[n] is periodic. (e y[n] is sinusoidal whenever x[n] is sinusoidal. Answer: (a The order is. (0 The overall impulse response h[n] of the following two cascaded LTI filters x[n] Unit Delay + h [n] = δ[n] + δ[n ] +... y[n] is (a δ[n] δ[n ]. (b δ[n] + δ[n ]. (c δ[n] δ[n ]. (d (δ[n] (δ[n ]. (e None of the above. Answer: (c (Also (d is acceptable. 7

8 Two LI filters are cascaded. So, h[n] = h [n] h [n] = (δ[n] δ[n ] (δ[n] + δ[n ] = δ[n] δ[n ]. Choice (d happens to give the same answer: in general δ[n k] = (δ[n k] m, for m =,,.... ( (5 points The following continuous-time signal is to be sampled and reconstructed. x(t = cos ( π0t + π ( π ( + cos π0t + + cos π60t. Let the sampling frequency f s = 00 samples/sec and x[n] = x(nt s = x(n/f s. (a Plot the double-sided spectrum of x(t as a functon of f Hz. x(t = cos ( π0t + π ( π ( + cos π0t + + cos π60t = e j π e jπ0t + e j π e jπ0t + e j π e jπ0t + e j π e jπ0t + e jπ60t + e jπ60t. e jπ/ e jπ/ e jπ/ e jπ/ f (b Determine whether x(t can be perfectly reconstructed from x[n] using the ideal interpolation. Justify your answer. No, because aliasing occurs due to the fact that f s = 00 < f max = 60 = 0. (c Find and simplify the reconstruction x(t from x[n] using the ideal interpolation. When sampled at f s = 00 (T s = 0.0, we get x[n] = cos ( π0 0.0n + π ( π ( + cos π0 0.0n + + cos π60 0.0n = cos ( 0.πn + π ( π ( + cos 0.8πn + + cos.πn }{{} = cos ( 0.8πn = cos ( 0.πn + π ( π ( + cos 0.8πn + + cos 0.8πn } {{} combine these = cos ( 0.πn + π +.7 cos ( 0.8πn Note that each of frequencies in the last expression is less than π. reconstruct x(t = cos ( 0.π00t + π +.7 cos ( 0.8π00t = cos ( π0t + π +.7 cos ( π0t This reconstructed waveform differs from x(t due to undersampling and hence aliasing. Then the ideal interpolation will 8

9 ( (0 points Answer for the LTI filter whose input/output relationship is given by y[n] = 3x[n] + x[n ] + x[n ]. (a Find and plot the impulse response h[n]. h[n] = 3δ[n] + δ[n ] + δ[n ] n (b Draw the block diagram implementation in the direct form. x[n] Unit Delay Unit Delay y[n] (c Find the response of the filter to input x[n] = e j( π n 0. and simplify it. y[n] = h[k]x[n k] = h[k]e j0. e j π (n k = e j0. e j π n h[k]e j π k = e j0. e j π n( 3 + e j π + e j π = e j0. e j π n 5.033e j = 5.033e j( π n

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