Solutions of Chapter 3 Part 1/2

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Page 1 of 7 Solutions of Chapter 3 Part 1/ Problem 3.1-1 Find the energy of the signals depicted in Figs.P3.1-1. Figure 1: Fig3.

1 Page 1 of 7 Solutions of Chapter 3 Part 1/ Problem Find the energy of the signals depicted in Figs.P Figure 1: Fig3.1-1 (a) E x n x[n] (b) E x n x[n] (c) E x n x[n] (d) E x n x[n]

2 Page of 7 Problem Show that a power of a signal De j(π/n0)n is D. Hence, show that a power of a signal x[n] N 0 1 r0 D re jr(π/n0)n is P x N 0 1 r0 D r. Use the fact that { e j(r m)πk/n 0 N0, r m; 0, otherwise. k0 (1) Let x[n] De j(π/n 0)n D(cos(π/N 0 )n + j sin(π/n 0 )n), we see that x[n] is periodic with N 0 its period. For a complex number z Z Z, so we have De j(π/n 0)n (De j(π/n 0)n )(D e j(π/n 0)n ) D. hus, its power is given by P x 1 N 0 n0 De j(π/n0)n 1 N 0 D D. n0 () Clearly, the signal x[n] is periodic, its period is N 0. hus, the power can be written as Since r0 P x 1 N 0 n0 D r e jr(π/n 0)n x[n] 1 N 0 n0 r0 r0 D r e jr(π/n 0)n. D r e jr(π/n 0)n N 0 1 Dme jm(π/n0)n, m0 the above equation can be written as follows by interchanging the order of summation. [ ] P x 1 N0 N 0 D r Dm 1 e j(r m)(π/n 0)n. r0 m0 n0 he summation within square brackets is N 0 when r m and 0 otherwise. herefore, we have P x r0 D r Dr r0 D r. Problem 3.-1 If the energy of a signal x[n] is E x, then find the energy of the following: (a) x[ n] (b) x[n m] (c) x[m n] (d) Kx[n] (m integer and K constant) (a) he energy of x[ n] is given by E 1 n x[ n].

3 Page 3 of 7 Let m n, we have E 1 (b) he energy of x[n m] is given by E m n (c) he energy of x[m n] is given by E 3 n (d) he energy of Kx[n] is given by E 4 x[m] x[n m] x[m n] n r Kx[n] K m r x[r] n x[m] E x. x[r] E x. r x[r] E x. x[n] K E x. Problem 3.-3 and 3.-4 For the signal depicted in Fig. P3.1-1 (a) and (c), sketch the following signals: (a) x[ n] (b) x[n + 6] (c) x[n 6] (d) x[3n] (e) x[n/3] (f) x[3 n] Figure : Fig3.-4

4 Page 4 of 7 Problem Describe each of the signals in Fig.P3.1-1 by a signal expression valid for all n. here are many different ways of viewing x[n]. Here is just one possible expression. (a) x[n] x 1 [n] + x [n] (n + 3)(u[n + 3] u[n]) + ( n + 3)(u[n] u[n 4]) Other possible solutions: x[n] (n + 3)(u[n + 3] u[n]) + ( n + 3)(u[n] u[n 3]) x[n] (n + 3)(u[n + ] u[n]) + ( n + 3)(u[n] u[n 3]) x[n] (n + 3)(u[n + 3] u[n 1]) + ( n + 3)(u[n 1] u[n 4]) (b) x[n] n(u[n] u[n 4]) + ( n + 6)(u[n 4] u[n 7]) (c) x[n] n(u[n + 3] u[n 4]) (d) x[n] n(u[n + ] u[n]) + n(u[n] u[n 3]) Problem A moving average is used to detect a trend of a rapidly fluctuating variable such as the stock market average. A variable may fluctuate (up and down) daily, masking its long-term (secular) trend. We can discern the long-term trend by smoothing or averaging the past N values of the variable. For the stock market average, we may consider a 5-day moving average y[n] to be the mean of the past 5 days market closing values x[n],x[n 1],,x[n 4]. (a) Write the difference equation relating y[n] to the input x[n]. (b) Use time-delay elements to realize the 5-day moving-average filter. (a) y[n] 1 5 (x[n] + x[n 1] + x[n ] + x[n 3] + x[n 4]) (b) Let D represent unit delay, the realization is as follows. X[n] 1/5 D D D D Y[n] Figure 3: Fig3.4-3

5 Page 5 of 7 Problem Approximate the following second-order differential equation with a difference equation. d y dt + a dy 1 dt + a 0y(t) x(t) Let x[n] and y[n] represent the samples seconds apart of the signals x(t) and y(t), respectively. i.e. x[n] x(n ), y[n] y(n ) Suppose is small enough so that the assumption 0 may be made. hen, we have y(t) y[n] dy y[n] y[n 1] dt d y dt y[n] y[n 1] y[n 1] y[n ] y[n] y[n 1] + y[n ] Substituting the above expressions into the given differential equation, we can obtain the following approximated difference equation where A 1 y[n] + A y[n 1] + A 3 y[n ] Bx[n] A a 1 + a 0 A ( + a 1 ) A 3 1 B

Problem 3.4-8 A linear, time-invariant system produces output y1 [n] in response to input x1 [n], as shown in Fig.P3.4-8. Determine and sketch the output y [n] that results when input x [n] is applied to the same system.

6 Problem A linear, time-invariant system produces output y1 [n] in response to input x1 [n], as shown in Fig.P Determine and sketch the output y [n] that results when input x [n] is applied to the same system. Figure 4: Fig3.4-8 he output y1 [n] can be expressed as y1 [n] δ [n] + δ [n 1] + δ [n ]. he input x [n] can be expressed as x [n] x1 [n 1] x1 [n ]. Since the system is LI, from the property of superposition, we have ě y [n] y1 [n 1] y1 [n ] δ [n 1] + 3δ [n ] 4δ [n 4] he output is sketched as follows. 4 y[n] n Figure 5: Fig3.4-8a Page 6 of 7

7 Page 7 of 7 Problem A system is described by y[n] 1 k (a) Explain what this system does. (b) Is the system BIBO stable? Justify your answer. (c) Is the system linear? Justify your answer. (d) Is the system memoryless? Justify your answer. (e) Is the system causal? Justify your answer. (f) Is the system time invariant? Justify your answer. x[k](δ[n k] + δ[n + k]) In the summation, n is a constant, the signal x[k] is sampled at k n and k n. hus, the system can be rewriten as y[n] 1 (x[n]δ[n k] + x[ n]δ[n + k]) 1 (x[n] + x[ n]) (a) Each signal can be expressed as a sum of an even component and an odd component as he system extracts the even portion of the input. x[n] 1 (x[n] + x[ n]) + 1 (x[n] x[ n]). (b) he system is BIBO stable. If the input is bounded, then the output is necessarily bounded. Assume input x[n] is bounded as x[n] M x <, then its reverse x[ n] is also bounded and x[ n] M x <. hus, y[n] 1 (x[n] + x[ n]) 1 (x[n] + x[ n]) M x <. (c) Yes. he system is linear. Let y 1 [n] 1 (x 1[n] + x 1 [ n]), y [n] 1 (x [n] + x [ n]) Applying x[n] a 1 x 1 [n] + a x [n] to the system yields y[n] 1 (x[n] + x[ n]) 1 (a 1x 1 [n] + a x [n] + (a 1 x 1 [ n] + a x [ n])) a 1 y 1 [n] + a y [n]. (d) No, he system is not memoryless since the output does not only depend on its current input. For example, at time n, the output y[] 1 (x[] + x[ ]) depends on a past value of input x[ ]. (e) No, he system is not causal. For example, at time n 1, the output y[ 1] 1 (x[ 1] + x[1]) depends on a future value of input x[1]. (f) No, he system is not time invariant. For example, let the input be x 1 [n] u[n+10] u[n 11]. Since the input is already even, the output equals to the input as y 1 [n] x 1 [n] u[n + 10] u[n 11]. Shifting by a non-zero integer N, then the signal x [n] x 1 [n N] is not even; the output y [n] y 1 [n N] x 1 [n N]. herefore, the system is time varying.

New Mexico State University Klipsch School of Electrical Engineering. EE312 - Signals and Systems I Spring 2018 Exam #1

New Mexico State University Klipsch School of Electrical Engineering EE312 - Signals and Systems I Spring 2018 Exam #1 Name: Prob. 1 Prob. 2 Prob. 3 Prob. 4 Total / 30 points / 20 points / 25 points /