# Application of the Fourier Transform to the Sampling

In analog systems, signals are processed in their entirety. However, in modern digital systems, only samples of signals are required for processing. The sampling can be done by using a train of pulses or impulses. We will use impulse sampling here.
Fig. 1: (a) Continuous (analog) signal to be sampled, (b) train of impulses, (c) sampled (digital) signal.
Consider the continuous signal $g(t)$ shown in Fig. 1(a). This can be multiplied by a train of impulses $\delta\left(t-n T_{s}\right)$ shown in Fig. $1(\mathrm{~b})$, where $T_{s}$ is the sampling interval and $f_{s}=1 / T_{s}$ is the sampling frequency or the sampling rate. The sampled signal $g_{s}(t)$ is therefore
$$g_{s}(t)=g(t) \sum_{n=-\infty}^{\infty} \delta\left(t-n T_{s}\right)=\sum_{n=-\infty}^{\infty} g\left(n T_{s}\right) \delta\left(t-n T_{s}\right)$$
The Fourier transform of this is
$$G_{s}(\omega)=\sum_{n=-\infty}^{\infty} g\left(n T_{s}\right) \mathcal{F}\left[\delta\left(t-n T_{s}\right)\right]=\sum_{n=-\infty}^{\infty} g\left(n T_{s}\right) e^{-j n \omega T_{s}} \tag{1}$$
It can be shown that
$$\sum_{n=-\infty}^{\infty} g\left(n T_{s}\right) e^{-j n \omega T_{s}}=\frac{1}{T_{s}} \sum_{n=-\infty}^{\infty} G\left(\omega+n \omega_{s}\right)$$
where $\omega_{s}=2 \pi / T_{s}$. Thus, Eq. (1) becomes
$$G_{s}(\omega)=\frac{1}{T_{s}} \sum_{n=-\infty}^{\infty} G\left(\omega+n \omega_{s}\right)$$
This shows that the Fourier transform $G_{s}(\omega)$ of the sampled signal is a sum of translates of the Fourier transform of the original signal at a rate of $1 / T_{s}$.In order to ensure optimum recovery of the original signal, what must be the sampling interval? This fundamental question in sampling is answered by an equivalent part of the sampling theorem:
A band-limited signal, with no frequency component higher than W hertz, may be completely recovered from its samples taken at a frequency at least twice as high as 2W samples per second.
In other words, for a signal with bandwidth $W$ hertz, there is no loss of information or overlapping if the sampling frequency is at least twice the highest frequency in the modulating signal. Thus, $$\frac{1}{T_{s}}=f_{s} \geq 2 W$$ The sampling frequency $f_{s}=2 W$ is known as the Nyquist frequency or rate, and $1 / f_{s}$ is the Nyquist interval.
Example 1: A telephone signal with a cutoff frequency of $5 \mathrm{kHz}$ is sampled at a rate 60 percent higher than the minimum allowed rate. Find the sampling rate.
Solution: The minimum sample rate is the Nyquist rate $=2 \mathrm{~W}=2 \times 5=$ $10 \mathrm{kHz}$. Hence, $$f_{s}=1.60 \times 2 W=16 \mathrm{kHz}$$