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Unformatted text preview: Solved Example 19.1 We have x ( t ) = s ( t ) + s ( t ), and thus X k = S k + S k Since s ( t ) is realvalued, S k = S * k and thus also X k = 2 < e { S k } The spectrum is real and even, as is x ( t ). For y ( t ), we have y ( t ) = x ( t ) + x ( t T / 2) and thus the two sets of Fourier series coefficients—both evaluated with respect to the fundamental period T of x ( t )—are related by Y k = S k + e jk Ω T / 2 · S k = (1 + e jkπ ) S k = (1 + ( 1) k ) S k = ‰ 2 S k , k even; , k odd. The fact that Y k = 0 for all odd k is not surprising, since the fundamental period of y ( t ) is T / 2. Thus y ( t ) is a sum of sinusoids of frequency k (2Ω ) = (2 k )Ω . If we were to express y ( t ) as a Fourier series with respect to its true fundamental period T / 2, then the k th coefficient of the series would be given by 2 S 2 k (note the doubling in the subscript). Note: No change in time scale is involved between s ( t ) and y ( t ), i.e., y ( t ) 6 = s (2 t )....
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This note was uploaded on 10/18/2011 for the course ENEE 241 taught by Professor Staff during the Spring '08 term at Maryland.
 Spring '08
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