Question

P7. Use the Fourier transform of the signal (t) and Parseval's theorem to: duw that「.sinc.(ka)dz-t; determine sinc(t - 2)dt

Answer 1

Given x (t) = 1/2 pi sin c (t/2) we know Fourier transform pair of tau sin c (tau t/2) rightarrow_IFT^FT 2 pi rect (w/tau) in our case tau = 1 1/2 pi sin (t/2) rightarrow^FT rect (w) Parseval's theorem integral_-infinity^infinity Vertical-bar x (t) Vertical-bar ^2 dt = 1/2 pi integral_-infinity^infinity Vertical-bar x (w) Vertical-bar ^2 dw \r\n (a) integral_-infinity^infinity sin c^2 (k x) dx = pi/k Let x (t) = sin c (k t) on comparing with tau sin c (tau/2 t) k = tau/2 rightarrow tau = 2 k x (t) = 1/2 k (tau sin (tau/2 t) x (w) = 1/2 k (2 pi rect (w/2 k)) x (w) = pi/k rect (w/2 k) integral_-infinity^infinity x (t)^2 dt = 1/2 pi integral_-infinity^infinity Vertical-bar x (w) Vertical-bar ^2 dw = 1/2 pi integral_-k^k (pi/k)^2 dw = 1/2 pi [(pi/k)^2 w]_-k^k \r\n = 1/2 pi (pi/k)^2 [k - (-k)] = pi^2/2 pi k^2 [2 k] integral_-infinity^infinity x (t)^2 dt = pi/k Since x (t) = sin c (k t)