ee20-hw5-f08-sol - EECS 20N: Structure and Interpretation...

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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 time-shifted 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 time-invariant 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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ee20-hw5-f08-sol - EECS 20N: Structure and Interpretation...

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