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# hws - SOLUTIONS MANUAL for Extra Homework Problems from...

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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

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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 0 2 3 1 (b) y [ n ] = x [ n ] * h [ n ] n 2 0 1 -1 5 -2 (c) y [ n ] = x [ n ] * h [ n ] 0 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 0 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

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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 < 0 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. (c) e jωn + e j 2 ωn : y [ n ] = summationdisplay k = -∞ h [ k ] e ( n - k ) + summationdisplay k = -∞ h [ k ] e j 2 ω ( n - k ) = e jωn summationdisplay k = -∞ h [ k ] e - jωk + e j 2 ωn summationdisplay k = -∞ h [ k ] e - j 2 ωk = e jωn · H ( e ) + e j 2 ωn · H ( e j 2 ω ) Since the input cannot be extracted from the above expression, the sum of complex exponentials is NOT AN EIGENFUNCTION. (Although, separately the inputs are eigenfunctions. In general, complex exponential signals are always eigenfunctions of LTI systems.) (d) x [ n ] = 5 n : y [ n ] = summationdisplay k = -∞ h [ k ]5 ( n - k ) = 5 n summationdisplay k = -∞ h [ k ]5 - k YES, EIGENFUNCTION.

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hws - SOLUTIONS MANUAL for Extra Homework Problems from...

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