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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: x⁢n ↔ DTFT X⁢ω x n ↔ DTFT X ω x⁢t ↔ CTFT X⁢Ω x t ↔ CTFT X Ω where ωω and ΩΩ are continuous frequency variables.

Sampling DTFT

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

X⁢ω=∑ n =0N−1x⁢n⁢e(−j)⁢ω⁢n X ω n N 1 0 x n ω n
(1)
where X⁢ω X ω is the continuous function that is indexed by the real-valued parameter −π≤ω≤π ω . The other function, x⁢n x n , is a discrete function that is indexed by integers.

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

X⁢k=X⁢2⁢πN⁢k=∑ n =0N−1x⁢n⁢e(−j)⁢2⁢π⁢kN⁢n X k X 2 N k n N 1 0 x n 2 k N n
(2)
In Equation 2 we sampled at ω=2⁢πN⁢k ω 2 N k where k=01…N−1 k 0 1 … N 1 and X⁢k X k for k=0…N−1 k 0 … N 1 is called the Discrete Fourier Transform (DFT) of x⁢n 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:

X⁢ω=∑ n =0N−1x⁢n⁢e(−j)⁢ω⁢n X ω n N 1 0 x n ω n
(3)
where ωω is any 2⁢π 2 -interval, for example −π≤ω≤π ω .

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

where again we sampled at ω=2⁢πN⁢k ω 2 N k where k=01…M−1 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 2⁢π 2 interval.

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

Choosing M

Case 1

Given NN (length of x⁢n x n ), choose M≫N ≫ 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 M≥N M N in order to represent all information in x⁢n  ,   n=0…N−1    n n 0 … N 1 x n Let's concentrate on M=N M N : x⁢n ↔ DFT X⁢k x n ↔ DFT X k for n=0…N−1 n 0 … N 1 and k=0…N−1 k 0 … N 1 numbers ↔ N  numbers numbers↔N  numbers

Discrete Fourier Transform (DFT)

Define

X⁢k≡X⁢2⁢π⁢kN X k X 2 k N
(4)
where N=length⁢x⁢n N length x n and k=0…N−1 k 0 … N 1 . In this case, M=N M N .

DFT

X⁢k=∑ n =0N−1x⁢n⁢e(−j)⁢2⁢π⁢kN⁢n X k n N 1 0 x n 2 k N n
(5)

Inverse DFT (IDFT)

x⁢n=1N⁢∑ k =0N−1X⁢k⁢ej⁢2⁢π⁢kN⁢n x n 1 N k N 1 0 X k 2 k N n
(6)

Interpretation

Represent x⁢n x n in terms of a sum of NN complex sinusoids of amplitudes X⁢k X k and frequencies ω k =2⁢π⁢kN  ,   k∈0…N−1    k k 0 … N 1 ω k 2 k N

Think:

Fourier Series with fundamental frequency 2⁢πN 2 N

Remark 1

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

x⁢n=1N⁢∑ k =0N−1X⁢k⁢ej⁢2⁢π⁢kN⁢n x n 1 N k N 1 0 X k 2 k N n
(7)
where n∈0…N−1 n 0 … N 1

Exercise 1

What about other values of nn?

Solution

x⁢n+N=??? x n N ???

Remark 2

Proof that the IDFT inverts the DFT for n∈0…N−1 n 0 … N 1

1N⁢∑ k =0N−1X⁢k⁢ej⁢2⁢π⁢kN⁢n=1N⁢∑ k =0N−1∑ m =0N−1x⁢m⁢e(−j)⁢2⁢π⁢kN⁢m⁢ej⁢2⁢π⁢kN⁢n=??? 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
    X⁢k=∑ n =0N−1x⁢n⁢e(−j)⁢2⁢π⁢kN⁢n=1+e(−j)⁢2⁢π⁢k4+e(−j)⁢2⁢π⁢k4⁢2+e(−j)⁢2⁢π⁢k4⁢3=1+e(−j)⁢π2⁢k+e(−j)⁢π⁢k+e(−j)⁢32⁢π⁢k 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: x⁢0=4 x 0 4 x⁢1=0 x 1 0 x⁢2=0 x 2 0 x⁢3=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:
    X⁢ω=∑n=03e(−j)⁢ω⁢n=1−e(−j)⁢4⁢ω1−e(−j)⁢ω=??? X ω n 0 3 ω n 1 4 ω 1 ω ???
    (10)
    Our sample points will be: ω k =2⁢π⁢k4=π2⁢k 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 X⁢k X k consists of samples of DTFT, so X⁢ω X ω , a 2⁢π 2 -periodic DTFT signal, can be converted to X⁢k X k , an NN-periodic DFT.

X⁢k=∑ n =0N−1x⁢n⁢e(−j)⁢2⁢π⁢kN⁢n X k n N 1 0 x n 2 k N n
(11)
where e(−j)⁢2⁢π⁢kN⁢n 2 k N n is an NN-periodic basis function (See Figure 7).

Figure 7
Figure 7 (sec8_fig7.png)

Also, recall,

x⁢n=1N⁢∑ n =0N−1X⁢k⁢ej⁢2⁢π⁢kN⁢n=1N⁢∑ n =0N−1X⁢k⁢ej⁢2⁢π⁢kN⁢(n+m⁢N)=??? 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 X⁢ω X ω , as depicted in Figure 9. S⁢ω S ω will represent the sampling function applied to X⁢ω X ω and is illustrated in Figure 9 as well. This will result in our discrete-time sequence, X⁢k 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 =−∞∞δ⁢ω−2⁢π⁢kN k δ ω 2 k N
(13)
Given the above equation, we can take the DTFT and get the following equation:
N⁢∑ m =−∞∞δ⁢n−m⁢N≡S⁢n N m δ n m N S n
(14)

Exercise 2

Why does Equation 14 equal S⁢n S n ?

Solution

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

c k =1N⁢∫−N2N2δ⁢n⁢e(−j)⁢2⁢π⁢kN⁢nd n =1N c k 1 N n N 2 N 2 δ n 2 k N n 1 N
(15)
S⁢n=∑ k =−∞∞e(−j)⁢2⁢π⁢kN⁢n S n k 2 k N n
(16)
where the DTFT of the exponential in the above equation is equal to δ⁢ω−2⁢π⁢kN 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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