Frequency domain view of the relationship between a signal and a sampling of that signal

A slecture by ECE student Botao Chen

Partly based on the ECE438 Fall 2014 lecture material of Prof. Mireille Boutin.

## Outline

1. Introduction
2. Derivation
3. Example
4. Conclusion

## Introduction

In this slecture I will discuss about the relations between the original signal $X(f)$ (the CTFT of $x(t)$ ), sampling continuous time signal $X_s(f)$ (the CTFT of $x_s(t)$ ) and sampling discrete time signal $X_d(\omega)$ (the DTFT of $x_d[n]$ ) in frequency domain and give a specific example showing the relations.

## Derivation

The first thing which need to be clarified is that there two different types of sampling signal: $x_s(t)$ and $x_d[n]$. $x_s(t)$ is created by multiplying a impulse train $P_T(t)$ with the original signal $x(t)$ and actually $x_s(t)$ is $comb_T(x(t))$ where T is the sampling period. However the $x_d[n]$ is $x(nT)$ where T is the sampling period.

Now we first concentrate on the relationship between $X(f)$ and $X_s(f)$.

We know that $x_s(t) = x(t) \times P_T(t)$, we can derive the relationship between $x_s(t)$ and $x(t)$ in the following way:

\begin{align} F(comb_T(x(t)) &= F(x(t) \times P_T(t))\\ &= X(f)*F(P_T(t))\\ &= X(f)*\frac{1}{T}\sum_{n = -\infty}^\infty \delta(f-\frac{n}{T})\\ &= \frac{1}{T}X(f)*P_\frac{1}{T}(f)\\ &= \frac{1}{T}rep_\frac{1}{T}X(f)\\ \end{align}

Show this relationship in graph below:

## Derivation

Then we are going to find the relation between $X_s(f)$ and $X_d(\omega)$

We know another way to express CTFT of $x_s(t)$:

\begin{align} X_s(f) &= F(\sum_{n = -\infty}^\infty x(nT)\delta(t-nT))\\ &= \sum_{n = -\infty}^\infty x(nT)F(\delta(t-nT))\\ &= \sum_{n = -\infty}^\infty x(nT)e^{-j2\pi fnT}\\ \end{align}

compare it with DTFT of $x_d[n]$:

\begin{align} X_d(\omega) &= \sum_{n = -\infty}^\infty x_d[n]e^{-j\omega n}\\ &= \sum_{n = -\infty}^\infty x(nT)e^{-j\omega n}\\ \end{align}

we can find that:

\begin{align} X_d(2\pi Tf) &= X_s(f)\\ \end{align}

if $f = \frac{1}{T}$

we have that:

\begin{align} X_d(2\pi ) &= X_s(\frac{1}{T})\\ \end{align}

from this equation, we can know the relationship between $X_s(f)$ and $X_d(\omega)$ and the relationship is showed in graph as below:

## conclusion

So the relationship between $X(f)$ and $X_s(f)$ is that $X_s(f)$ is a a rep of $X(f)$ in frequency domain with period of $\frac{1}{T}$ and magnitude scaled by $\frac{1}{T}$. the relationship between $X(f)$ and $X_d(\omega)$ is that $X_d(\omega)$ is also a a rep of $X(f)$ in frequency domain with period $2\pi$ and magnitude is also scaled by $\frac{1}{T}$, but the frequency is scaled by $2\pi T$