ee20-hw4-s08-sol

ee20-hw4-s08-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 4 SOLUTIONS HW 4.1 (a) u [ n ] = k =0 d [ n- k ] . Then, using linearity and time-invariance, s [ n ] = summationdisplay k =0 h [ n- k ] . (b) d [ n ] = u [ n ]- u [ n- 1] . By LTI, h [ n ] = s [ n ]- s [ n- 1] . HW 4.2 (a) (i) From the convolution sum that flips and shifts f ( n ) , we have: y ( n ) = summationdisplay k x ( k ) f ( n- k ) = x (0) f ( n ) + x (1) f ( n- 1) = 1 4 ( n + 1) + 1 2 ( n ) + 1 4 ( n- 1) + 1 4 ( n ) + 1 2 ( n- 1) + 1 4 ( n- 2) = 1 4 ( n + 1) + 3 4 ( n ) + 3 4 ( n- 1) + 1 4 ( n- 2) (ii) From the convolution sum that flips and shifts x ( n ) , we have: y ( n ) = summationdisplay k f ( k ) x ( n- k ) = f (- 1) x ( n + 1) + f (0) x ( n ) + f (1) x ( n- 1) = 1 4 u ( n + 1) + 1 2 u ( n ) + 1 4 u ( n- 1) = 1 4 ( n + 1) + 3 4 ( n ) + u ( n- 1) (b) From the convolution sum that flips and shifts f ( n ) , we have: y ( n ) = x (0) f ( n ) + x (1) f ( n- 1) = f ( n )- f ( n- 1) = summationdisplay k =0 parenleftbigg 1 2 parenrightbigg k ( n- 4 k )- summationdisplay k =0 parenleftbigg 1 2 parenrightbigg k ( n- 1- 4 k ) = braceleftBigg ( 1 2 ) n 4 u ( n ) mod( n, 4) = 0- ( 1 2 ) n- 1 4 u ( n ) mod( n- 1 , 4) = 0 1 HW 4.3 (a) (i) Fig. 1 shows the sketch of x ( n ) = u ( n ) . From problem 2 , we know that we can write y ( n ) = ... + x (- 1) f ( n +1)+ x (0) f ( n )+ x (1) f ( n- 1)+ ... . Fig. 10 shows this summation and Fig. 2 shows the output for this summation....
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ee20-hw4-s08-sol - EECS 20N: Structure and Interpretation...

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