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Unformatted text preview: (c) No. If we let y ( t ) = t R-∞ x ( τ ) dτ with x ( t ) = u ( t )-u ( t-1). Then x ( t ) = 0 for t > 1, but y ( t ) = 1 for t > 1. Problem 1.43 (a) We have x ( t ) S → y ( t ) Since S is time-invariant, x ( t-T ) S → y ( t-T ) Now if x ( t ) is periodic with period T, x ( t ) = x ( t-T ). Therefore, we may we conclude that y ( t ) = y ( t-T ). This implies that y ( t ) is periodic with period T . A similar argument can be made in discrete-time. (b) Consider the system y ( t ) = ∞ X k =-∞ x 2 ( t-2 k ) The above system is time-invariant. Consider the input x ( t ) = u ( t )-u ( t-1). Then it is easy to see that while the input is not periodic, the output is periodic. 2...
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- Spring '08
- Greatest common divisor, 2k, dτ, discrete-time