Continuous_Fourier_series

# 4 t cos 2 k i t 2 t sin 2k eq 5 1 2 t t t t t t sin

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Unformatted text preview: (α − β ) + sin(α + β )) 2 1 (cos(α − β ) + cos(α + β )) cos α cos β = 2 (4) sin α sin β = (5) (6) Now we compute the inner products, keeping in mind that sk (t) is deﬁned for k ≥ 1 while ck (t) is deﬁned for k ≥ 0: τ +T sk , si = sin 2πk τ 1 2 Eq. (4) = τ +T cos 2π (k − i) τ T , 2 τ +T = sin 2πk τ Eq. (5) = 1 2 t T t T t T sin 2π (k − i) τ = cos 2πk τ Eq. (6) = = cos 2πi τ +T τ +T ck , ci sin 2πi dt − cos 2π (k + i) t T dt k=i 0, k = i. = sk , ci t T t T t T t T dt + sin 2π (k + i) cos 2πi t T t 1 τ +T cos 2π (k − i) 2 T τ T, k = i = 0 T , k=i=0 2 0, k = i. t T d t = 0. dt + cos 2π (k + i) t T dt We are now ready to derive formulas for the coeﬃcients a1 , a2 . . . and b0 , b1 , b2 . . . of the expansion of a CT T -periodic signal s ∈ L2 (T ): ∞ s(t) = b0 + ak sk (t) + k =1 b0 = s, c 0 c0 , c0 ∞ bk ck (t). k =1 (7) 5 x(t) 1 ttt ttt −2 −1 0 1 2 t Figure 2. Signal x(t) of Example 2. = bk = = ak = = 1 τ +T s(t) dt Tτ s, c k ck , ck 2 τ +T t s(t) cos 2πk Tτ T s, s k sk , sk 2 τ +T t s(t) sin 2πk Tτ T dt, k = 1, 2 , . . . dt, k = 1, 2, . . . Example 2. Suppose that the period is T = 2, and let signal x be deﬁned by: x(t) = 1, −1 ≤ t &lt; 0 0, 0 ≤ t &lt; 1, as illustrated in Fig. 2. Let us compute the Fourier series coeﬃcients with respe...
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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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