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113_1_midterm 2010 solutions

113_1_midterm 2010 solutions - EE113 Digital Signal...

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EE113: Digital Signal Processing Prof. Mihaela van der Schaar Spring 2010 Name: Student ID: MIDTERM SOLUTIONS 1. Problem 1 A system is described by the block diagram shown in the figure below with x ( n ) denoting the input sequence, y ( n ) denoting the output sequence, w ( n ) = e 0 n , and v ( n ) = e - 0 n . h ( n ) is known to be a linear, time-invariant, stable, and causal system. (a) ( 10 PTS ) Is the system linear? The system is linear . y ( n ) = [ x ( n ) w ( n ) * h ( n )] v ( n ) = £ x ( n ) e 0 n * h ( n ) / e - 0 n = " X k = -∞ h ( k ) x ( n - k ) e 0 ( n - k ) # e - 0 n = " X k = -∞ h ( k ) x ( n - k ) e 0 n e - 0 k # e - 0 n = e 0 n e - 0 n " X k = -∞ h ( k ) x ( n - k ) e - 0 k # = X k = -∞ h ( k ) x ( n - k ) e - 0 k where the last line follows from the second-to-last because e 0 n e - 0 n = 1. We define h 1 ( n ) = h ( n ) e - 0 n . Since y ( n ) = x ( n ) * h 1 ( n ) the overall system S is basically a convolution, which we know is linear (see the derivation of the convolution sum in chapter 5 in the section titled ’The Convolution Sum’). To verify this, let y 1 ( n ) = x 1 ( n ) * h 1 ( n ) = S [ x 1 ( n )] and y 2 ( n ) = x 2 ( n ) * h 1 ( n ) = S [ x 2 ( n )]. Then, ay 1 ( n ) + by 2 ( n ) = [ ax 1 ( n ) * h 1 ( n ) + bx 2 ( n ) * h 1 ( n )] = [ ax 1 ( n ) + bx 2 ( n )] * h 1 ( n ) = S [ ax 1 ( n ) + bx 2 ( n )] .
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