2.6 The dft: frequency domain with a computer analysis

Fundamentals of signal processing Page 1 / 1

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: $x n \overset{DTFT}{} X$ $x t \overset{CTFT}{} X$ 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 n N 1 0 x n n

where

X

is the continuous function that is indexed by thereal-valued parameter

. 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 2 N k n N 1 0 x n 2 k N n

In we sampled at

2 N k

where

k 01 N 1

and

X k

for

k 0 N 1

is called the Discrete Fourier Transform (DFT) of

x n

Finite duration dt signal

The DTFT of the image in is written as follows:

X n N 1 0 x n n

where

is any

2

-interval, for example

Sample x()

where again we sampled at $2 N k$ where $k 01 M 1$ . 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 $2$ interval.

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

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Choosing m

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 $M N$ in order to represent all information in $n n 0 N 1 x n$ Let's concentrate on $M N$ : $x n \overset{DFT}{} X k$ for $n 0 N 1$ and $k 0 N 1$ $numbers Nnumbers$

Discrete fourier transform (dft)

Define

X k X 2 k N

where

N length x n

and

k 0 N 1

. In this case,

M N

Dft

X k n N 1 0 x n 2 k N n

Inverse dft (idft)

x n 1 N k N 1 0 X k 2 k N n

Interpretation

Represent $x n$ in terms of a sum of $N$ complex sinusoids of amplitudes $X k$ and frequencies $k k 0 N 1_{k} 2 k N$

Fourier Series with fundamental frequency

2 N

Remark 1

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

x n 1 N k N 1 0 X k 2 k N n

where

n 0 N 1

What about other values of $n$ ?

$x n N ???$

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Remark 2

Proof that the IDFT inverts the DFT for $n 0 N 1$

1 N k N 1 0 X k 2 k N n 1 N k N 1 0 m N 1 0 x m 2 k N m 2 k N n ???

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
$X k n N 1 0 x n 2 k N n 1 2 k 4 2 k 4 2 2 k 4 3 1 2 k k 32 k$
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:
$X n 03 n 141 ???$
Our sample points will be: $_{k} 2 k 4 2 k$ where $k 0123$ ( ).

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Periodicity of the dft

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

X k n N 1 0 x n 2 k N n

where

2 k N n

is an

N

-periodic basis function (See ).

Also, recall,

x n 1 N n N 1 0 X k 2 k N n 1 N n N 1 0 X k 2 k N n m N ???

Illustration

When we deal with the DFT, we need to remember that, in effect, this treats the signal as an

N

-periodic sequence.

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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()

k

∞ ∞ 2 k N

Given the above equation, we can take the DTFT and get thefollowing equation:

N m

∞ ∞ n m N S n

Why does equal $S n$ ?

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

c_{k} 1 N n N 2 N 2 n 2 k N n 1 N

S n k

∞ ∞ 2 k N n

where the DTFT of the exponential in the above equation is equal to

2 k N

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

Connections

Combine signals in to get signals in .

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Source: OpenStax, Fundamentals of signal processing. OpenStax CNX. Nov 26, 2012 Download for free at http://cnx.org/content/col10360/1.4

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