ee20-hw4-s08-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 4 SOLUTIONS HW4.1 (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] . HW4.2 (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 δ ( n + 1) + 1 2 δ ( n ) + 1 4 δ ( n - 1) + 1 4 δ ( n ) + 1 2 δ ( n - 1) + 1 4 δ ( n - 2) = 1 4 δ ( n + 1) + 3 4 δ ( n ) + 3 4 δ ( n - 1) + 1 4 δ ( 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 δ ( n + 1) + 3 4 δ ( 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 δ ( n - 4 k ) - summationdisplay k =0 parenleftbigg 1 2 parenrightbigg k δ ( 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
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HW4.3 (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. 0 3 1 2 1 -1 -2 1 1 1 x(n) = u(n) Figure 1: x ( n ) 0 3 1 2 1 -1 -2 y(n) Figure 2: y ( n ) 2
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(ii) x ( n ) is shown in Fig. 3. Working out as done in (i), we get y ( n ) as in Fig. 4. 0 3 1 2 1 -1 x(n) 1 1 1 4 Figure 3: x ( n ) 0 3 1 2 1 -1 y(n) -1 4 Figure 4: y ( n ) 3
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(iii) x ( n ) is shown in Fig. 5. Working out as done in (i), we get y ( n ) as in Fig. 6. 0
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