Continuous_Fourier_series

# K 1 matching these coecients with the coecients in eq

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Unformatted text preview: is functions as follows: φ0 (t) = 1 = c0 (t), 2πkt 2πkt For k ≥ 1, φk (t) = cos + j sin = ck (t) + jsk (t), T T 2πkt 2πkt + j sin − = ck (t) − jsk (t). φ−k (t) = cos − T T Therefore, ∞ x(t) = αk φk (t) k =−∞ −1 ∞ = α0 φ0 (t) + αk φk (t) + k =1 ∞ αk φk (t) k =−∞ [αk φk (t) + α−k φ−k (t)] = α0 φ0 (t) + k =1 ∞ [αk (ck (t) + jsk (t)) + α−k (ck (t) − jsk (t))] = α0 c0 (t) + k =1 ∞ = α0 c0 (t) + ∞ (αk + α−k )ck (t) + k =1 j (αk − α−k )sk (t). k =1 Matching these coeﬃcients with the coeﬃcients in Eq. (7), we get the following relationship between the exponential Fourier series coeﬃcients and the sine-cosine Fourier series coeﬃcients of any L2 (T ) signal: b0 = α0 , bk = αk + α−k for k = 1, 2, . . . , ak = j (αk − α−k ) for k = 1, 2, . . . . Using expression (8) we found for the coeﬃcients αk of the speciﬁc si...
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