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The DFT: Frequency Domain with a Computer Analysis

Module by: Robert Nowak. E-mail the author

Summary: 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 CT applications (Figure 1):

Figure 1
Figure 1 (sec8_fig1.png)

Much of the theoretical analysis of such systems relied on frequency domain representations. How do we carry out these frequency domain analysis on the computer? Recall the following relationships: xn DTFT Xω x n DTFT X ω xt CTFT XΩ x t CTFT X Ω where ωω and ΩΩ are continuous frequency variables.

Sampling DTFT

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

Xω= n =0N1xne(j)ωn X ω n N 1 0 x n ω n
(1)
where Xω X ω is the continuous function that is indexed by the real-valued parameter πωπ ω . The other function, xn x n , is a discrete function that is indexed by integers.

We want to work with Xω X ω on a computer. Why not just sample Xω X ω ?

Xk=X2πNk= n =0N1xne(j)2πkNn X k X 2 N k n N 1 0 x n 2 k N n
(2)
In Equation 2 we sampled at ω=2πNk ω 2 N k where k=01N1 k 0 1 N 1 and Xk X k for k=0N1 k 0 N 1 is called the Discrete Fourier Transform (DFT) of xn x n .

Example 1

Figure 2
Finite Duration DT Signal
Finite Duration DT Signal (sec8_fig2.png)

The DTFT of the image in Figure 2 is written as follows:

Xω= n =0N1xne(j)ωn X ω n N 1 0 x n ω n
(3)
where ωω is any 2π 2 -interval, for example πωπ ω .

Figure 3
Sample X(ω)
Sample X(ω) (sec8_fig3.png)

where again we sampled at ω=2πNk ω 2 N k where k=01M1 k 0 1 M 1 . For example, we take M=10 M 10 . In the following section we will discuss in more detail how we should choose MM, the number of samples in the 2π 2 interval.

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

Choosing M

Case 1

Given NN (length of xn x n ), choose MN M N to obtain a dense sampling of the DTFT (Figure 4):

Figure 4
Figure 4 (sec8_fig4.png)
Case 2

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

In general, we require MN M N in order to represent all information in xn  ,   n=0N1    n n 0 N 1 x n Let's concentrate on M=N M N : xn DFT Xk x n DFT X k for n=0N1 n 0 N 1 and k=0N1 k 0 N 1 numbers N  numbers numbersN  numbers

Discrete Fourier Transform (DFT)

Define

XkX2πkN X k X 2 k N
(4)
where N=lengthxn N length x n and k=0N1 k 0 N 1 . In this case, M=N M N .

DFT

Xk= n =0N1xne(j)2πkNn X k n N 1 0 x n 2 k N n
(5)

Inverse DFT (IDFT)

xn=1N k =0N1Xkej2πkNn x n 1 N k N 1 0 X k 2 k N n
(6)

Interpretation

Represent xn x n in terms of a sum of NN complex sinusoids of amplitudes Xk X k and frequencies ω k =2πkN  ,   k0N1    k k 0 N 1 ω k 2 k N

Think:

Fourier Series with fundamental frequency 2πN 2 N

Remark 1

IDFT treats xn x n as though it were NN-periodic.

xn=1N k =0N1Xkej2πkNn x n 1 N k N 1 0 X k 2 k N n
(7)
where n0N1 n 0 N 1

Exercise 1

What about other values of nn?

Solution

xn+N=??? x n N ???

Remark 2

Proof that the IDFT inverts the DFT for n0N1 n 0 N 1

1N k =0N1Xkej2πkNn=1N k =0N1 m =0N1xme(j)2πkNmej2πkNn=??? 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 ???
(8)

Example 2: Computing DFT

Given the following discrete-time signal (Figure 5) with N=4 N 4 , we will compute the DFT using two different methods (the DFT Formula and Sample DTFT):

Figure 5
Figure 5 (sec8_fig5.png)

  1. DFT Formula
    Xk= n =0N1xne(j)2πkNn=1+e(j)2πk4+e(j)2πk42+e(j)2πk43=1+e(j)π2k+e(j)πk+e(j)32πk 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 3 2 k
    (9)
    Using the above equation, we can solve and get the following results: x0=4 x 0 4 x1=0 x 1 0 x2=0 x 2 0 x3=0 x 3 0
  2. Sample DTFT. Using the same figure, Figure 5, we will take the DTFT of the signal and get the following equations:
    Xω=n=03e(j)ωn=1e(j)4ω1e(j)ω=??? X ω n 0 3 ω n 1 4 ω 1 ω ???
    (10)
    Our sample points will be: ω k =2πk4=π2k ω k 2 k 4 2 k where k=0123 k 0 1 2 3 (Figure 6).

Figure 6
Figure 6 (sec8_fig6.png)

Periodicity of the DFT

DFT Xk X k consists of samples of DTFT, so Xω X ω , a 2π 2 -periodic DTFT signal, can be converted to Xk X k , an NN-periodic DFT.

Xk= n =0N1xne(j)2πkNn X k n N 1 0 x n 2 k N n
(11)
where e(j)2πkNn 2 k N n is an NN-periodic basis function (See Figure 7).

Figure 7
Figure 7 (sec8_fig7.png)

Also, recall,

xn=1N n =0N1Xkej2πkNn=1N n =0N1Xkej2πkN(n+mN)=??? 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 ???
(12)

Example 3: Illustration

Figure 8
Figure 8 (sec8_fig8.png)

note:

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

A Sampling Perspective

Think of sampling the continuous function Xω X ω , as depicted in Figure 9. Sω S ω will represent the sampling function applied to Xω X ω and is illustrated in Figure 9 as well. This will result in our discrete-time sequence, Xk X k .

Figure 9
Figure 9 (sec8_fig9.png)

Recall:

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

Inverse DTFT of S(ω)

k =δω2πkN k δ ω 2 k N
(13)
Given the above equation, we can take the DTFT and get the following equation:
N m =δnmNSn N m δ n m N S n
(14)

Exercise 2

Why does Equation 14 equal Sn S n ?

Solution

Sn S n is NN-periodic, so it has the following Fourier Series:

c k =1NN2N2δne(j)2πkNnd n =1N c k 1 N n N 2 N 2 δ n 2 k N n 1 N
(15)
Sn= k =e(j)2πkNn S n k 2 k N n
(16)
where the DTFT of the exponential in the above equation is equal to δω2πkN δ ω 2 k N .

So, in the time-domain we have (Figure 10):

Figure 10
Figure 10 (sec8_fig10.png)

Connections

Figure 11
Figure 11 (sec8_fig11.png)

Combine signals in Figure 11 to get signals in Figure 12.

Figure 12
Figure 12 (sec8_fig11b.png)

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