Continuous_Fourier_series

In other words these signals are pairwise orthogonal

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Unformatted text preview: 2, . . . forms an orthogonal basis for L2 (T ). In other words, these signals are pairwise orthogonal (as shown below), and we can represent any T -periodic CT signal s ∈ L2 (T ) as a linear combination of these complex exponentials: ∞ ∞ ak exp ak φk (t) = s(t) = k =−∞ k =−∞ j 2πkt T . (1) 2 The first “=” sign in Eq. (1) needs careful interpretation: unlike the finiteduration DT case, the equality here is not pointwise. Instead, the equality is understood in the following sense: M s− ak φk → 0 as N → −∞ and M → ∞. k =N Nevertheless, the coefficient formula previously derived for orthogonal representations in CN , is still valid: s, φ k ak = . (2) φk , φk The inner product of φk and φi is: τ +T φk , φi = = = j 2πit j 2πkt exp − T T τ τ +T j 2π (k − i)t exp dt T τ T if k = i 0 if k = i exp dt The fact that φk , φi = 0 shows that our vectors are indeed pairwise orthogonal.1 Substituting φk , φk = T back into Eq. (2), we get: 1 s, φ k = ak = T T τ +T s(t) exp − τ j 2πkt T dt. (3) Example 1. Let T , t0 , and A be three positive real numbers such that T > t0 > 0. Consider the following periodic signal: s(t) = A if |t| ≤ 0 if |t| ≤ t0 2 t0 , 2 periodically extended with period T , as shown in Fig. 1. Using Eq. (3) with τ = −T /2, its Fourier series coefficients are: a0 = 1 1 T t0 /2...
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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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