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Unformatted text preview: EECS 20N: Structure and Interpretation of Signals and Systems Department of Electrical Engineering and Computer Sciences UNIVERSITY OF CALIFORNIA BERKELEY Problem Set 5 SOLUTIONS HW 5.1 (a) We have for n = 0 , Y [0] = Y [ 1] + X [0] = A + X [0] n = 1 , Y [1] = Y [0] + X [1] = 2 A + X [1] + X [0] n = 2 , Y [2] = Y [1] + X [2] = 3 A + X [2] + X [1] + 2 X [0] . Therefore: Y [ n ] = parenleftBig n summationdisplay k =0 k X [ n k ] parenrightBig + n +1 A. Let H stands for the system response to an input signal, then Y 1 [ n ] = H ( X 1 [ n ]) = parenleftBig n summationdisplay k =0 k X 1 [ n k ] parenrightBig + n +1 A Y 2 [ n ] = H ( X 2 [ n ]) = parenleftBig n summationdisplay k =0 k X 2 [ n k ] parenrightBig + n +1 A If: X 3 [ n ] = X 1 [ n ] + X 2 [ n ] Then: Y 3 [ n ] = H ( X 3 [ n ]) = parenleftBig n summationdisplay k =0 k X 3 [ n k ] parenrightBig + n +1 A = parenleftBig n summationdisplay k =0 k X 1 [ n k ] parenrightBig + parenleftBig n summationdisplay k =0 k X 2 [ n k ] parenrightBig + n +1 A. Thus, you can verify that Y 3 [ n ] = Y 1 [ n ] + Y 2 [ n ] if and only if A = 0 . (b) We define X 2 a timeshifted version of X 1 such that X 2 = X 1 [ n n ] , then we have: Y 2 [ n ] = H ( X 2 [ n ]) = parenleftBig n summationdisplay k =0 k X 2 [ n k ] parenrightBig + n +1 A = parenleftBig n summationdisplay k =0 k X 1 [ n n k ] parenrightBig + n +1 A. We can see that Y 2 [ n ] = Y 1 [ n n ] if and only if A = 0 . 1 HW 5.2 (a) Since the input signal is zero prior to time t = 4 while the output is not, therefore the system can not be causal. (b) The system is not memoryless because it is not causal. (c) If the system is timeinvariant it means that the impulse response of the system is the current output shifted four samples to the left. This is because [ n ] is equal to the input signal shifted by the same amount. Figure 1: Impulse response for 2.c HW 5.3 (a) The frequency response A ( ) of the filter is given by A ( ) = summationdisplay n = a ( n ) e in = a + a 1 e i + ... + a N e iN (b) (i) The impulse response c of the cascade interconnection is given by c ( n ) = a ( n ) * b ( n ) = parenleftBig N summationdisplay i =0 a i ( n i ) parenrightBig * parenleftBig M summationdisplay j =0 b j ( n j ) parenrightBig = N summationdisplay i =0 M summationdisplay j =0 a i b j ( n i ) * ( n j ) = N summationdisplay i =0 M summationdisplay j =0 a i b j ( n ( i + j ) ) We define k = i + j , thus we get c ( n ) = N summationdisplay i =0 i + M summationdisplay k = i a i b k i ( n k ) 2 You can verify that after changing the summation order we get c ( n ) = M + N summationdisplay k =0 min { k,N } summationdisplay i =max { ,k M } a i b k i ( n k ) = M + N summationdisplay k =0 c k ( n k ) where c k = min { k,N } summationdisplay i =max { ,k M } a i b k i ....
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 Spring '08
 BabakAyazifar
 Computer Science, Electrical Engineering

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