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**Unformatted text preview: **SOLUTIONS MANUAL for Extra Homework Problems from Companion Website Discrete-Time Signal Processing, 3e by Alan V. Oppenheim and Ronald W. Schafer Prepared by B. A. Black, L. Lee, J. Rosenthal, M. Said, and G. Slabaugh Solutions – Chapter 2 Discrete-Time Signals and Systems 1 2.1. We use the graphical approach to compute the convolution: y [ n ] = x [ n ] * h [ n ] = ∞ summationdisplay k =-∞ x [ k ] h [ n- k ] (a) y [ n ] = x [ n ] * h [ n ] y [ n ] = δ [ n- 1] * h [ n ] = h [ n- 1] 1 n 2 2 3 1 (b) y [ n ] = x [ n ] * h [ n ] n 2 1-1 5-2 (c) y [ n ] = x [ n ] * h [ n ] 1 2 3 4 5 6 7 8 9 11 10 n 12 13 14 15 16 17 18 19 20 1 2 3 4 5 5 4 3 2 2 2 3 5 5 4 3 2 1 4 (d) y [ n ] = x [ n ] * h [ n ] n 1 2-1-2 5 3 4 6 1 3 3 2 1-1 1 2.2. The response of the system to a delayed step: y [ n ] = x [ n ] * h [ n ] = ∞ summationdisplay k =-∞ x [ k ] h [ n- k ] = ∞ summationdisplay k =-∞ u [ k- 4] h [ n- k ] 2 y [ n ] = ∞ summationdisplay k =4 h [ n- k ] Evaluating the above summation: For n < 4: y [ n ] = 0 For n = 4: y [ n ] = h [0] = 1 For n = 5: y [ n ] = h [1] + h [0] = 2 For n = 6: y [ n ] = h [2] + h [1] + h [0] = 3 For n = 7: y [ n ] = h [3] + h [2] + h [1] + h [0] = 4 For n = 8: y [ n ] = h [4] + h [3] + h [2] + h [1] + h [0] = 2 For n ≥ 9: y [ n ] = h [5] + h [4] + h [3] + h [2] + h [1] + h [0] = 0 2.3. The output is obtained from the convolution sum: y [ n ] = x [ n ] * h [ n ] = ∞ summationdisplay k =-∞ x [ k ] h [ n- k ] = ∞ summationdisplay k =-∞ x [ k ] u [ n- k ] The convolution may be broken into five regions over the range of n : y [ n ] = 0 , for n < y [ n ] = n summationdisplay k =0 a k = 1- a ( n +1) 1- a , for 0 ≤ n ≤ N 1 y [ n ] = N 1 summationdisplay k =0 a k = 1- a ( N 1 +1) 1- a , for N 1 < n < N 2 y [ n ] = N 1 summationdisplay k =0 a k + n summationdisplay k = N 2 a ( k- N 2 ) = 1- a ( N 1 +1) 1- a + 1- a ( n +1) 1- a = 2- a ( N 1 +1)- a ( n +1) 1- a , for N 2 ≤ n ≤ ( N 1 + N 2 ) y [ n ] = N 1 summationdisplay k =0 a k + N 1 + N 2 summationdisplay k = N 2 a ( k- N 2 ) = N 1 summationdisplay k =0 a k + summationdisplay m =0 N 1 a m 3 = 2 N 1 summationdisplay k =0 a k = 2 · parenleftbigg 1- a ( N 1 +1) 1- a parenrightbigg , for n > ( N 1 + N 2 ) 2.4. Recall that an eigenfunction of a system is an input signal which appears at the output of the system scaled by a complex constant. (a) x [ n ] = 5 n u [ n ]: y [ n ] = ∞ summationdisplay k =-∞ h [ k ] x [ n- k ] = ∞ summationdisplay k =-∞ h [ k ]5 ( n- k ) u [ n- k ] = 5 n n summationdisplay k =-∞ h [ k ]5- k Becuase the summation depends on n , x [ n ] is NOT AN EIGENFUNCTION. (b) x [ n ] = e j 2 ωn : y [ n ] = ∞ summationdisplay k =-∞ h [ k ] e j 2 ω ( n- k ) = e j 2 ωn ∞ summationdisplay k =-∞ h [ k ] e- j 2 ωk = e j 2 ωn · H ( e j 2 ω ) YES, EIGENFUNCTION....

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