Continuous_Fourier_series

# Continuous_Fourier_series

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Unformatted text preview: ct to the Fourier basis of sines and cosines. From the formulas above, with τ = −1, b0 For k ≥ 1, bk 0 1 = 2 2 = 2 1 dt = −1 0 1 2 t 2 cos 2πk −1 dt 0 = cos(πkt) dt −1 = ak = 1 sin(πkt) πk 2 2 t=0 =0 t=−1 0 sin 2πk −1 t 2 dt 6 0 = sin(πkt) dt −1 1 =− cos(πkt) πk t=0 t=−1 2 − πk , if k is odd 0, if k is even. = A diﬀerent method of ﬁnding the coeﬃcients is to notice that (a) Signal x is related to signal s of Example 1. If we set t0 = 1, A = 1, and T = 2 in Example 1 and shift s by t0 /2 to the left, we will obtain x: 0 x(t) = s(t + t2 ). (b) Complex exponential basis signals are related to the sine and cosine basis signals. Using the Fourier series coeﬃcients obtained in Example 1, we have the following representation of x(t) in the complex exponential Fourier basis: x(t) = s t + t0 2 ∞ = A sin πk k =−∞ ∞ = A sin πk k =−∞ ∞ = A sin πk k =−∞ ∞ = k =−∞ A sin πk t0 2 π kt0 T φk t + π kt0 T exp j 2πk (t + t0 /2) T π kt0 T exp j 2πk (t0 /2) T π kt0 T exp j πkt0 T exp j 2πkt T φk (t), which means that the coeﬃcients of x in the complex exponential Fourier basis are: A sin πk 1 sin = πk αk = j πkt0 π kt0 exp T T j πk πk exp . 2 2 (8) 7 But notice that the complex exponential basis functions are related to the sine and cosine bas...
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## This note was uploaded on 05/28/2012 for the course ECE 301 taught by Professor V."ragu"balakrishnan during the Fall '06 term at Purdue University-West Lafayette.

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