1. The fourier transform of the half cosine pulse as shown below is __________





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  • By: guest on 02 Jun 2017 12.56 am
    The given signal can be expressed as multiplication of x1(t) and x2(t) as shown below, where A = 2, T/2 = 0.25 ⇒ T = 0.5 ∴ x(t) = x1(t) x x2(t) ⇒ X(f) = X1(f) * X2(f) Now X1(f) = [δ(f - f0) + δ(f + f0)] where and X2(f) = T . sin c[fT] ⇒ X(f) = [δ(f - f0) + δ(f + f0)] *T. sinc(fT) = [sinc[T(f - f0)] + sinc [T(f + f0)]] Now, A = 2, T = 0.5 and ⇒ X(f) = 0.5[sin c(0.5(f - 1)) + sinc(0.5(f + 1))].
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