Homework Solution #3.pdf

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Department
Electrical & Computer Engineering
Course
ENEE 322
Professor
Steven Marcus
Semester
Spring

Description
Homework #3 2.3 Solution 1: 1 n▯2 Since the output is the convolution between the inp2t xu[n▯2]( ) and the impulse response h[n] = u[n + 2]. y[n] = x[n] ▯ h[n] = ( )n▯2u[n ▯ 2] ▯ u[n + 2] 2 X1 1 k▯2 = (2) u[k ▯ 2]u[n ▯ k + 2] k=▯1 By replacing k with m + 2 in the summation, we obtain 1 X 1 m y[n] = ( ) u[m]u[n ▯ m] m=▯1 2 Note that u[m] and u[n ▯ m] implies that m ▯ 0 and m ▯ n respectively. n X 1m y[n] = ( ) m=0 2 = 2[1 ▯ ( )+1]u[n] 2 Solution 2: Note that 1 n x[n] = (2) u[n] ▯ ▯[n ▯ 2] h[n] = u[n] ▯ ▯[n + 2] Thus, the output y[n] is y[n] = x[n] ▯ h[n] 1 = ( ) u[n] ▯ ▯[n ▯ 2] ▯ u[n] ▯ ▯[n + 2] 2 1 n = ( ) u[n] ▯ ▯[n ▯ 2] ▯ ▯[n + 2] ▯ u[n] 2 1 n = ( ) u[n] ▯ u[n] 2 Xn 1 = ( )k k=0 2 1 n+1 = 2[1 ▯ 2) ]u[n] 1 8 x[n]*h[n] 6 4 2 0 0 5 10 15 20 25 Figure 1: Output y[n] (Problem 2.4) 2.4 Solution 1 ▯ 1; 3 ▯ n ▯ 8 x[n] = 0; otherwise = u[n ▯ 3] ▯ u[n ▯ 9] ▯ 1; 4 ▯ n ▯ 15 h[n] = 0; otherwise = u[n ▯ 4] ▯ n[n ▯ 16] Thus, the output is y[n] = x[n] ▯ h[n] = (u[n ▯ 3] ▯ u[n ▯ 9]) ▯ (u[n ▯ 4] ▯ u[n ▯ 16]) = u[n ▯ 3] ▯ u[n ▯ 4] ▯ u[n ▯ 3] ▯ u[n ▯ 16] ▯u[n ▯ 9] ▯ u[n ▯ 4] + u[n ▯ 9] ▯ u[n ▯ 16] = u[n] ▯ u[n] ▯ (▯[n ▯ 7] ▯ ▯[n ▯ 13] ▯ ▯[n ▯ 19] + ▯[n ▯ 25]) Xn Xn Xn Xn = 1 ▯ 1 ▯ 1 + 1 k=7 k=13 k=19 k=25 8 < n ▯ 6; 7 ▯ n < 13 6; 13 ▯ n < 19 = : 24 ▯ n;19 ▯ n < 25 0; otherwise The output y[n] is shown in Figure 1. Solution 2: We can directly calculate the convolution of x[n] and h[n], which is shown in Figure 2. 2 1. Filp x[m], then we obtain x[▯m]. 2. Shift x[▯m] by n, then we obtain x[n▯m]. When n increases to 7, x[n▯m] and h[n] overlap. And the overlapping area continues to increase when n increases. 3. When n 2 [12;18], the overlapping area does not change, which means the convolution value is constant at [12;18]: 4. When n increases to 19, the overlapping area begins to decrease. 5. There is no overlap when n > 25. Therefore, the output 8 > n ▯ 6; 7 ▯ n < 13 < 6; 13 ▯ n < 19 y[n] = : 24 ▯ n;19 ▯ n < 25 0; otherwise 2.7 (a) Given that x[n] = ▯[n ▯ 1] We see that X y[n]= x[k]g[n ▯ 2k] k=▯1 X = ▯[k ▯ 1]g[n ▯ 2k] k=▯1 = g[n ▯ 2] = u[n ▯ 2] ▯ u[n ▯ 6] (b) Given that x[n] = ▯[n ▯ 2] We see that 1 X y[n]= x[k]g[n ▯ 2k] k=▯1 1 X = ▯[k ▯ 2]g[n ▯ 2k] k=▯1 = g[n ▯ 4] = u[n ▯ 4] ▯ u[n ▯ 8] 3 2 2 x[­m] x[7­m] 1.8 1.8 h[m] h[m] 1.6 1.6 1.4 1.4 1.2 1.2 1 1 0.8 0.8 0.6 0.6 0.4 0.4 0.2 0.2 0 0 ­10 ­5 0 5 10 15 20 ­10 ­5 0 5 10 15 20 Step 1 Step 2 2 2 x[13­m] 1.8 x[19­m] 1.8 h[m] h[m] 1.6 1.6 1.4 1.4 1.2 1.2 1 1 0.8 0.8 0.6 0.6 0.4 0.4 0.2 0.2 0 0 ­10 ­5 0 5 10 15 20 ­10 ­5 0 5 10 15 20 Step 3 Step 4 2 1.8 x[24­m] h[m] 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 ­10 ­5 0 5 10 15 20 Step 5 Figure 2: 2.8 4 (c) The input to the system in part (b) is the same as the input in part (a) shifted by 1 to the right. If S is time invariant then the system output obtained in part (b) has to be the same as the output in part (a) shifted by 1 to the right. Clearly, this is not the case. Therefore, the system is not LTI. (d) If x[n] = u[n], then 1 X y[n] = x[k]g[n ▯ 2k] k=▯1 1 X = g[n ▯ 2k]
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