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# HW9sol - ECEN 314 Signals and Systems Solutions to HW 9...

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ECEN 314: Signals and Systems Solutions to HW 9 1. (3.34) The frequency response of the system is H ( ) = Z -∞ ( e - 4 t u ( t ) + e 4 t u ( - t ) ) e - jωt dt = 0 Z -∞ e t (4 - ) dt + Z 0 e - t (4+ ) dt = 1 4 - + 1 4 + (a) Here T = 1 and ω o = 2 π and a k = 1 for all k . The FS coefficients of the output are: b k = a k H ( jkω o ) = 1 4 - j 2 + 1 4 + j 2 (b) Here T = 2 and ω o = π and a k = 0 , k even 1 , k odd The FS coefficients of the output are: b k = a k H ( jkω o ) = 0 , k even 1 4 - jkπ + 1 4+ jkπ , k odd (c) Here T = 1, ω o = 2 π and a k = 1 / 2 , k = 0 0 , k even, k 6 = 0 sin πk 2 πk , k odd The FS coefficients of the output are: b k = a k H ( jkω o ) = 1 / 4 , k = 0 0 , k even, k 6 = 0 sin πk 2 πk 1 4 - jk 2 π + 1 4+ jk 2 π k odd 1

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2. (3.35) The fourier series coefficients of y ( t ) are b k = H ( jkω o ) a k , where ω o is the fundamental fre- quency of x ( t ) and a k are the FS coefficients of x ( t ). Since y ( t ) = x ( t ), then b k = a k for all k . Noting that H ( ) = 0 for | ω | < 250, and since ω o = 2 π/ ( π/ 7) = 14, we can be sure that for | k | ≤ ¥ 250 14 ƒ = 17, b k = a k = 0.
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