SolutionsVII - ECE 3250 HOMEWORK VII SOLUTIONS Fall 2009 1....

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Unformatted text preview: ECE 3250 HOMEWORK VII SOLUTIONS Fall 2009 1. (a) | a + b | 2 = ( a + b ) ( a + b ), so | a + b | 2 = a a + b b + a b + b a = | a | 2 + | b | 2 + 2Re { a b } . To get the last line, I used the fact that z + z = 2Re { z } for any complex number z . You can use the same argument to show that | a- b | 2 = | a | 2 + | b | 2- 2Re { a b } . Since | a- b | ≥ 0, we have 2Re { a b } ≤ | a | 2 + | b | 2 , so | a + b | 2 ≤ 2 | a | 2 + 2 | b | 2 . Now consider x ∈ l 2 . By definition, P ∞ n =-∞ | x ( n ) | 2 converges. If c o ∈ C , ∞ X n =-∞ | c o x ( n ) | 2 = | c o | ∞ X n =-∞ | x ( n ) | 2 also exists, so l 2 is closed under scalar multipication. As for addtion, if x 1 and x 2 are in l 2 , then ∞ X n =-∞ | x 1 ( n )+ x 2 ( n ) | 2 ≤ ∞ X n =-∞ (2 | x 1 ( n ) | 2 +2 | x 2 ( n ) | 2 ) = 2 ∞ X n =-∞ | x 1 ( n ) | 2 +2 ∞ X n =-∞ | x 2 ( n ) | 2 also exists, so l 2 is closed under addition as well. (b) This problem contained a slight typo, which I’ve since corrected. The series should feature y ( k ) rather than y ( k ). In any event, use the suggestion. ≤ ( | a | - | b | ) 2 = | a | 2 + | b | 2- 2 | a || b | , so | a || b | ≤ | a | 2 + | b | 2 2 . So if x and y are in l 2 , for each k ∈ Z we have | x ( k ) || y ( k ) | ≤ | x ( k ) | 2 2 + | y ( k ) | 2 2 , so ∞ X k =-∞ | x ( k ) || y ( k ) | ≤ ∞ X k =-∞ „ | x ( k ) | 2 2 + | y ( k ) | 2 2 « = k x k 2 2 + k y k 2 2 2 , so the series in the problem is absolutely summable and hence summable. It follows that if we set h x,y i = ∞ X k =-∞ x ( k ) y ( k ) , then we have a well defined candidate for an inner product. It’s easy to check that it has all the properties required of an inner product. 2. 1 2 (a) Let b H be the frequency response of the composite system. You can see in at least two ways that b H = b H 1 b H 2 . – If you put t 7→ e j Ω t into the composite system, then t 7→ b H (Ω) e j Ω t comes out. When you put t 7→ e j Ω t into the low-pass filter, t 7→ b H 1 (Ω) e j Ω t comes out. That signal enters the band-pass filter, whose output, by linearity, is t 7→ b H 1 (Ω) b H 2 (Ω) e j Ω t . Accordingly, for every Ω ∈ R , b H (Ω) e j Ω t = b H 1 (Ω) b H 2 (Ω) e j Ω t for all t ∈ R , so b H (Ω) = b H 1 (Ω) b H 2 (Ω) for all Ω ∈ R . – The impulse response of the composite system, as we saw in an earlier homework problem, is h = h 1 * h 2 , where h 1 and h 2 are the respective impulse responses of the low-pass and band-pass filters. By the convolution rule for Fourier transforms, b H = b H 1 b H 2 . The bottom line is that b H (Ω) = b H 1 ( ω ) b H 2 ( ω ) = 1 if 1737 π ≤ Ω ≤ 3100 π otherwise ....
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This note was uploaded on 03/26/2010 for the course ECE 3250 at Cornell University (Engineering School).

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SolutionsVII - ECE 3250 HOMEWORK VII SOLUTIONS Fall 2009 1....

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