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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 DTFT X ω x t CTFT X Ω where ω and Ω 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 ω n N 1 0 x n ω n
where X ω is the continuous function that is indexed by thereal-valued parameter ω . The other function, x n , is a discrete function that is indexed by integers.

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

X k X 2 N k n N 1 0 x n 2 k N n
In we sampled at ω 2 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 .

Finite duration dt signal

The DTFT of the image in is written as follows:

X ω n N 1 0 x n ω n
where ω is any 2 -interval, for example ω .

Sample x(Ω)

where again we sampled at ω 2 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 interval.

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

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 DFT X k for n 0 N 1 and k 0 N 1 numbers N  numbers

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 ω 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 ???

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 3 2 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 ω n 0 3 ω n 1 4 ω 1 ω ???
    Our sample points will be: ω k 2 k 4 2 k where k 0 1 2 3 ( ).

Periodicity of the dft

DFT X k consists of samples of DTFT, so X ω , 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.

A sampling perspective

Think of sampling the continuous function X ω , as depicted in . S ω will represent the sampling function applied to X ω 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(Ω)

k δ ω 2 k N
Given the above equation, we can take the DTFT and get thefollowing equation:
N m δ 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 δ 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 δ ω 2 k N .

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

Connections

Combine signals in to get signals in .

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Source:  OpenStax, Ece 454 and ece 554 supplemental reading. OpenStax CNX. Apr 02, 2012 Download for free at http://cnx.org/content/col11416/1.1
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