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ee20-hw5-f08-sol

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

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EECS20N:StructureandInterpretationofSignalsandSystems Department of Electrical Engineering and Computer Sciences U NIVERSITY OF C ALIFORNIA B ERKELEY Problem Set 5 SOLUTIONS HW5.1 (a) We have for n = 0 , Y [0] = αY [ - 1] + X [0] = αA + X [0] n = 1 , Y [1] = αY [0] + X [1] = α 2 A + X [1] + αX [0] n = 2 , Y [2] = αY [1] + X [2] = α 3 A + X [2] + αX [1] + α 2 X [0] . Therefore: Y [ n ] = parenleftBig n summationdisplay k =0 α k X [ n - k ] parenrightBig + α 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 α k X 1 [ n - k ] parenrightBig + α n +1 A Y 2 [ n ] = H ( X 2 [ n ]) = parenleftBig n summationdisplay k =0 α k X 2 [ n - k ] parenrightBig + α n +1 A If: X 3 [ n ] = βX 1 [ n ] + γX 2 [ n ] Then: Y 3 [ n ] = H ( X 3 [ n ]) = parenleftBig n summationdisplay k =0 α k X 3 [ n - k ] parenrightBig + α n +1 A = β parenleftBig n summationdisplay k =0 α k X 1 [ n - k ] parenrightBig + γ parenleftBig n summationdisplay k =0 α k X 2 [ n - k ] parenrightBig + α n +1 A. Thus, you can verify that Y 3 [ n ] = βY 1 [ n ] + γY 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 0 ] , then we have: Y 2 [ n ] = H ( X 2 [ n ]) = parenleftBig n summationdisplay k =0 α k X 2 [ n - k ] parenrightBig + α n +1 A = parenleftBig n summationdisplay k =0 α k X 1 [ n - n 0 - k ] parenrightBig + α n +1 A. We can see that Y 2 [ n ] = Y 1 [ n - n 0 ] if and only if A = 0 . 1

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HW5.2 (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 δ [ n ] is equal to the input signal shifted by the same amount. Figure 1: Impulse response for 2.c HW5.3 (a) The frequency response A ( ω ) of the filter is given by A ( ω ) = summationdisplay n = -∞ a ( n ) e - iωn = a 0 + a 1 e - + . . . + a N e - iωN (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 δ ( n - i ) parenrightBig * parenleftBig M summationdisplay j =0 b j δ ( n - j ) parenrightBig = N summationdisplay i =0 M summationdisplay j =0 a i b j δ ( n - i ) * δ ( n - j ) = N summationdisplay i =0 M summationdisplay j =0 a i b j δ ( 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 δ ( 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 { 0 ,k - M } a i b k - i δ ( n - k ) = M + N summationdisplay k =0 c k δ ( n - k ) where c k = min { k,N } summationdisplay i =max { 0 ,k - M } a i b k - i . (ii) The frequency response C of the cascade interconnection is given by C ( ω ) = A ( ω ) B ( ω ) = ( a 0 + . . . + a N e - iωN )( b 0 + . . . + b M e - iωM ) = M + N summationdisplay k =0 c k e - iωk where c k is defined as before. (iii) Since the nonzero terms in C ( ω ) range from a 0 b 0 to a N b M e - ( M + N ) , the result of (a) tells us that the impulse response c ( n ) can only have at most M + N + 1 non-zero elements. Hence, C must be an FIR filter.

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

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