The out put x,(t), which is the sampled version of x(t), in time domain can be written as x;(t) = x(t)8(t) Using the sifiting property of the impulse function, x,(t) can be rewritten as 00 x:(t) = ,...

Extracted text: The out put x,(t), which is the sampled version of x(t), in time domain can be written as x;(t) = x(t)8(t) Using the sifiting property of the impulse function, x,(t) can be rewritten as 00 x:(t) = , x(t)ô(t – nT5) = ), x(nT;)8(t – nT;) %3D n=ー0 n=-00 Using the frequency convolution properity of the Fourier transform, the time domain product x(t)8(t) can be transformed to frequency domain convolution X (f)Xs(f) 1 xs(f) : T.2 Xf-nf,) | n=ー00 Example: Suppose that an analog signal is given as x(t) 8kHz. Sketch the spectrum for the original signal and the sampled signal from 0 to 20kHz. 5cos (2n x 103t) and is sampled at the rate %3D I need the solution step by step and clear line please