ee20-hw3-s08-sol

# ee20-hw3-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 3 SOLUTIONS 0 3 1 2 1 1/2 1/4 1/8 -1 -2 Figure 1: f ( m ) 0 1 2 1 1 1 3 1 4 -2 -1 Figure 2: h ( m ) HW3.1 (a) Fig. 1 and Fig. 2 show the sketch for f ( m ) and h ( m ) respectively. 1

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(b) Fig. 3 and Fig. 4 show the sketch for f e ( m ) and f o ( m ) respectively. You cannot construct back f from f e only . Consider t ( m ) = f e ( m ) . Hence t ( m ) is an even function with t ( m ) = t ( m ) . Hence, t e ( m ) = t ( - m )+ t ( m ) 2 = 2 t ( m ) 2 = t ( m ) = f e ( m ) and t o ( m ) = 0 . Thus, we have found a signal t ( m ) for which t e ( m ) = f e ( m ) but t ( m ) negationslash = f ( m ) . Hence, while reconstructing from f e ( m ) , we could have guessed either of the two functions. Hence, we cannot reconstruct from f e ( m ) alone. Similarly, one can prove that f ( m ) cannot be reconstructed from f o ( m ) only . 0 3 1 2 1 1/ 16 1/4 1/8 -1 -2 -3 1/4 1/8 1/16 Figure 3: f e ( m ) 0 3 1 2 1/16 1/4 - 1/ 8 - 1 - 2 - 3 - 1/4 1/ 8 - 1/16 Figure 4: f o ( m ) 2
(c) Fig. 5 and Fig. 6 show the sketch for h u ( m ) and h d ( m ) respectively. 0 1 2 1 1 1 3 1 4 -2 -1 5 6 -4 -3 Figure 5: h u ( m ) 0 1 2 1 1 3 4 -2 -1 5 6 -4 -3 Figure 6: h d ( m ) 3

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(d) Fig. 7 and Fig. 8 show the sketch for g ( m ) = f ( m 1) and g ( m ) = f ( m + 1) respectively. 0 3 1 2 1 1/2 1/ 4 1/8 -1 4 Figure 7: f ( m 1) 0 1 2 1 1 /2 1/4 1/8 -1 -2 -3 Figure 8: f ( m + 1) (e) For v ( m ) = f ( m ) , we simply flip the graph of f ( m ) about the vertical access. Figure 9 plots v ( m ) . (f) To obtain w ( m ) = f ( n m ) , we first flip the graph of f ( m ) about the vertical access, and shift it to the right by n . Figures 10 and 11 plots w ( m ) for n = 1 and n = 1 respectively. (g) We simply multiply a flipped and shifted version of f ( m ) (which is the same as w ( m ) ) by h ( m ) . For n < 0 , w ( m ) = 0 for n 0 and does not overlap with h ( m ) , so p ( m ) is 0. 4
-5 -4 -3 -2 -1 0 1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 m v(m) Figure 9: Graph of v ( m ) = f ( m ) -5 -4 -3 -2 -1 0 1 2 0 0.2 0.4 0.6 0.8 1 m w(m) Figure 10: Graph of w ( m ) = f (1 m ) 5

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For 0 n 3 , w ( m ) overlaps with h ( m ) at n + 1 values, where the largest non-zero value is n . Figure ?? plots p ( m ) for n = 2 . For n > 3 , w ( m ) overlaps with h ( m ) for all values of h ( m ) . Figure 13 plots p ( m ) for n = 5 . (h) q ( m ) , which is the convolution of f ( m ) and h ( m ) , is obtained by adding all non-zero values of p ( m ) separately for each n , and plotting those values against the corresponding n . Figure 14 plots q ( m ) .
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