Statement:

If a finite energy signal x(t) contains no frequency. higher than “w” Hz( i.e. it is a band limited signal ) then it is completely determined by specifying its values at the instants of time which are spaced (1/2w) secs apart.

If a finite energy signal x(t) contains no frequency components higher than “w” Hz then it may be completely recovered from its samples which are spaced (1/2w) secs apart

Proof:

Let x(t) – signal with finite energy and infinite duration

X(t) – bandlimited signal

$∴Ts=\frac{1}{f_s}$ = sampling period

And $fs = \frac{1}{T_s}$ = sampling rate

S (t) – train of impulses

Spacing between adjacent unit impulses – Ts

∴ Frequency of sampling function is equal to the sampling frequency. fs

Sampled signal – xs(t)

$∴s(t)=s(t+2Ts)+S(t+Ts)+S(t)+S(t-Ts)+S(t-2Ts)+⋯.. \\ ∴S(t)=∑_{n=-∞}^∞ S(t-nTs)$

$⇒xS(t)→$ obtained by x(t) and S(t)

$∴xS(t)=x(t)×S(t)=x(nTs)×S(t) \\ ∴xS(t)=∑_{n=-∞}^∞x(nTs).S(t-nTs)$

Obtain Fourier transform:

$X(f)=fo ∑_{n=-∞}^∞S(f-nfo)$

We have similar pulse train as sampling frequency S(t)

$∴S(f)=fs ∑_{n=-∞}^∞S(f-nfs)$

∴ fo is been replaced by fs

The sampled sig in time domain is x(t) and S(t)

i.e. $xS(t) = x(t) \times S(t)$

Taking Fourier on both the sides,

$XS(f) = X(f) * S(f)$

Therefore $Xs(F) = X(f) * [fs ∑_{n=-∞}^∞S(f-nfs)]$

Where ‘*’ denotes convolution

From the properties of delta function, we find that convolution of $x(f)$ and $S(f- n fs)$ is equal to $x(f-n fs)$

F.T of sampled sig Xs(f) will be,

$Xs(t) = fs ∑_{n=-∞}^∞X(f-n fs)$