ee20-hw3-s08-sol

ee20-hw3-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 3 SOLUTIONS 3 1 2 1 1/2 1/4 1/8-1-2 Figure 1: f ( m ) 1 2 1 1 1 3 1 4-2-1 Figure 2: h ( m ) HW 3.1 (a) Fig. 1 and Fig. 2 show the sketch for f ( m ) and h ( m ) respectively. 1 (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 . 3 1 2 1 1/ 16 1/4 1/8-1-2-3 1/4 1/8 1/16 Figure 3: f e ( m ) 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. 1 2 1 1 1 3 1 4-2-1 5 6-4-3 Figure 5: h u ( m ) 1 2 1 1 3 4-2-1 5 6-4-3 Figure 6: h d ( m ) 3 (d) Fig. 7 and Fig. 8 show the sketch for g ( m ) = f ( m − 1) and g ( m ) = f ( m + 1) respectively. 3 1 2 1 1/2 1/ 4 1/8-1 4 Figure 7: f ( m − 1) 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 < , w ( m ) = 0 for n ≥ and does not overlap with h ( m ) , so p ( m ) is 0. 4-5-4-3-2-1 1 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 1 2 0.2 0.4 0.6 0.8 1 m w(m) Figure 10: Graph of w ( m ) = f (1 − m ) 5 For ≤ 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 ....
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This note was uploaded on 04/02/2008 for the course EECS 20n taught by Professor Babakayazifar during the Spring '08 term at Berkeley.

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ee20-hw3-s08-sol - EECS 20N: Structure and Interpretation...

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