# 1.7 The dft: frequency domain with a computer analysis

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This module will take the ideas of sampling CT signals further by examining how such operations can be performed in the frequency domain and by using a computer.

## Introduction

We just covered ideal (and non-ideal) (time) sampling of CT signals . This enabled DT signal processing solutions for CTapplications ( ):

Much of the theoretical analysis of such systems relied on frequency domain representations. How do we carry out thesefrequency domain analysis on the computer? Recall the following relationships: $x(n)\stackrel{\mathrm{DTFT}}{}X()$ $x(t)\stackrel{\mathrm{CTFT}}{}X()$ where  and  are continuous frequency variables.

## Sampling dtft

Consider the DTFT of a discrete-time (DT) signal $x(n)$ . Assume $x(n)$ is of finite duration $N$ ( i.e. , an $N$ -point signal).

$X()=\sum_{n=0}^{N-1} x(n)e^{-in}$
where $X()$ is the continuous function that is indexed by thereal-valued parameter $-\pi \le \le \pi$ . The other function, $x(n)$ , is a discrete function that is indexed by integers.

We want to work with $X()$ on a computer. Why not just sample $X()$ ?

$X(k)=X(\frac{2\pi }{N}k)=\sum_{n=0}^{N-1} x(n)e^{-i\times 2\pi \frac{k}{N}n}$
In we sampled at $=\frac{2\pi }{N}k$ where $k=\{0, 1, , N-1\}$ and $X(k)$ for $k=\{0, , N-1\}$ is called the Discrete Fourier Transform (DFT) of $x(n)$ .

The DTFT of the image in is written as follows:

$X()=\sum_{n=0}^{N-1} x(n)e^{-in}$
where  is any $2\pi$ -interval, for example $-\pi \le \le \pi$ .

where again we sampled at $=\frac{2\pi }{N}k$ where $k=\{0, 1, , M-1\}$ . For example, we take $M=10$ . In the following section we will discuss in more detail how we should choose $M$ , the number of samples in the $2\pi$ interval.

(This is precisely how we would plot $X()$ in Matlab.)

## Case 1

Given $N$ (length of $x(n)$ ), choose $(M, N)$ to obtain a dense sampling of the DTFT ( ):

## Case 2

Choose $M$ as small as possible (to minimize the amount of computation).

In general, we require $M\ge N$ in order to represent all information in $\forall n, n=\{0, , N-1\}\colon x(n)$ Let's concentrate on $M=N$ : $x(n)\stackrel{\mathrm{DFT}}{}X(k)$ for $n=\{0, , N-1\}$ and $k=\{0, , N-1\}$ $\mathrm{numbers}\mathrm{Nnumbers}$

## Discrete fourier transform (dft)

Define

$X(k)\equiv X(\frac{2\pi k}{N})$
where $N=\mathrm{length}(x(n))$ and $k=\{0, , N-1\}$ . In this case, $M=N$ .

## Dft

$X(k)=\sum_{n=0}^{N-1} x(n)e^{-i\times 2\pi \frac{k}{N}n}$

## Inverse dft (idft)

$x(n)=\frac{1}{N}\sum_{k=0}^{N-1} X(k)e^{i\times 2\pi \frac{k}{N}n}$

## Interpretation

Represent $x(n)$ in terms of a sum of $N$ complex sinusoids of amplitudes $X(k)$ and frequencies $\forall k, k\in \{0, , N-1\}\colon {}_{k}=\frac{2\pi k}{N}$

Fourier Series with fundamental frequency $\frac{2\pi }{N}$

## Remark 1

IDFT treats $x(n)$ as though it were $N$ -periodic.

$x(n)=\frac{1}{N}\sum_{k=0}^{N-1} X(k)e^{i\times 2\pi \frac{k}{N}n}$
where $n\in \{0, , N-1\}$

What about other values of $n$ ?

$x(n+N)=\mathrm{???}$

## Remark 2

Proof that the IDFT inverts the DFT for $n\in \{0, , N-1\}$

$\frac{1}{N}\sum_{k=0}^{N-1} X(k)e^{i\times 2\pi \frac{k}{N}n}=\frac{1}{N}\sum_{k=0}^{N-1} \sum_{m=0}^{N-1} x(m)e^{-i\times 2\pi \frac{k}{N}m}e^{i\times 2\pi \frac{k}{N}n}=\mathrm{???}$

## Computing dft

Given the following discrete-time signal ( ) with $N=4$ , we will compute the DFT using two different methods (the DFTFormula and Sample DTFT):

• DFT Formula
$X(k)=\sum_{n=0}^{N-1} x(n)e^{-i\times 2\pi \frac{k}{N}n}=1+e^{-i\times 2\pi \frac{k}{4}}+e^{-i\times 2\pi \frac{k}{4}\times 2}+e^{-i\times 2\pi \frac{k}{4}\times 3}=1+e^{-i\frac{\pi }{2}k}+e^{-i\pi k}+e^{-i\frac{3}{2}\pi k}$
Using the above equation, we can solve and get thefollowing results: $x(0)=4$ $x(1)=0$ $x(2)=0$ $x(3)=0$
• Sample DTFT. Using the same figure, , we will take the DTFT of the signal and get the following equations:
$X()=\sum_{n=0}^{3} e^{-in}=\frac{1-e^{-i\times 4}}{1-e^{-i}}=\mathrm{???}$
Our sample points will be: ${}_{k}=\frac{2\pi k}{4}=\frac{\pi }{2}k$ where $k=\{0, 1, 2, 3\}$ ( ).

## Periodicity of the dft

DFT $X(k)$ consists of samples of DTFT, so $X()$ , a $2\pi$ -periodic DTFT signal, can be converted to $X(k)$ , an $N$ -periodic DFT.

$X(k)=\sum_{n=0}^{N-1} x(n)e^{-i\times 2\pi \frac{k}{N}n}$
where $e^{-i\times 2\pi \frac{k}{N}n}$ is an $N$ -periodic basis function (See ).

Also, recall,

$x(n)=\frac{1}{N}\sum_{n=0}^{N-1} X(k)e^{i\times 2\pi \frac{k}{N}n}=\frac{1}{N}\sum_{n=0}^{N-1} X(k)e^{i\times 2\pi \frac{k}{N}(n+mN)}=\mathrm{???}$

## Illustration

When we deal with the DFT, we need to remember that, in effect, this treats the signal as an $N$ -periodic sequence.

## A sampling perspective

Think of sampling the continuous function $X()$ , as depicted in . $S()$ will represent the sampling function applied to $X()$ and is illustrated in as well. This will result in our discrete-time sequence, $X(k)$ .

Remember the multiplication in the frequency domain is equal to convolution in the time domain!

## Inverse dtft of s()

$\sum$ 2 k N
Given the above equation, we can take the DTFT and get thefollowing equation:
$N\sum$ n m N S n

Why does equal $S(n)$ ?

$S(n)$ is $N$ -periodic, so it has the following Fourier Series :

${c}_{k}=\frac{1}{N}\int_{-\left(\frac{N}{2}\right)}^{\frac{N}{2}} (n)e^{-i\times 2\pi \frac{k}{N}n}\,d n=\frac{1}{N}$
$S(n)=\sum$ 2 k N n
where the DTFT of the exponential in the above equation is equal to $(-\frac{2\pi k}{N})$ .

So, in the time-domain we have ( ):

## Connections

Combine signals in to get signals in .

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