0.7 Lecture 8:the discrete time fourier transform (dtft)

Extends notion of the frequency response of a DT system to the frequency content of a DT signal. The development of the FFT algorithm to compute the DT Fourier series efficiently has revolutionized signal processing.

Lecture #8:

THE DISCRETE TIME FOURIER TRANSFORM (DTFT)

Motivation:

Extends notion of the frequency response of a DT system to the frequency content of a DT signal.

Basis of much of digital signal processing.

The development of the FFT algorithm to compute the DT Fourier series efficiently has revolutionized signal processing.

Outline:

DT processing of CT signals

The discrete time Fourier transform (DTFT)

Transform properties

Transform pairs

DT processing of CT signals revisited

Periodic DT signals

Summary of frequency domain representations of signals

The computation of Fourier transforms — the DFT and the FFT algorithm

Conclusion.

I. DT PROCESSING OF CT SIGNALS

DT filtering of CT signals can be modeled as a cascade of signal transformations. The C/D converter transforms a CT signal to a DT signal, the D/C converter transforms a DT signal to a CT signal.

1/ Model of C/D converter

The C/D converter has two components —a CT sampler and an I/S converter that transforms a CT impulse train to a DT sample train. The CT sampler is modeled as an impulse modulator with so that. Hence,

$s\left(t\right)={\sum }_{m}\delta \left(t-\text{mT}\right)$

$\stackrel{ˆ}{x}\left(t\right)=x\left(t\right)×s\left(t\right)$

$x\left(t\right)=\sum _{m}x\left(\text{mT}\right)\delta \left(t-\text{mT}\right)$

$x\left[n\right]=\sum x\left(\text{mT}\right)\delta \left[n-m\right]$

2/ Motivation for the DTFT

The relation among the x(t), s(t), (t), and x[n] is shown below as is the relation among the X(f), S(f), and (f).

How should we represent the Fourier transform of x[n]?

How are the Fourier transforms of (t) and of x[n] related?

II. THE DISCRETE TIME FOURIER TRANSFORM (DTFT)

1/ Notation for CT and DT transforms

2/ Relation of Z transform and Fourier transform of a DT signal

The Z transform pair is defined as

$\stackrel{˜}{X}\left(z\right)=\sum _{n=-\infty }^{\infty }{x\left[n\right]z}^{-n}\stackrel{z}{⇔}x\left[n\right]=\frac{1}{2\pi j}\int \stackrel{˜}{X}\left(z\right){z}^{n-1}\text{dz}$

If the ROC of (z) includes the unit circle then the Z transform can be evaluated on the unit circle,

$\stackrel{˜}{X}\left({e}^{j\Omega }\right)=\sum _{n=-\infty }^{\infty }x\left[n\right]{e}^{-j\Omega n}\stackrel{z}{⇔}x\left[n\right]=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }\stackrel{˜}{X}\left({e}^{j\Omega }\right){e}^{j\Omega n}d\Omega$

Finally, we let Ω = 2πφ and define the DTFT as

3/ Properties of the DTFT

a/ Periodicity

$\stackrel{˜}{X}\left(\varphi \right)=\sum _{n=-\infty }^{\infty }x\left[n\right]{e}^{-\mathrm{j2}\text{πϕ}n}$

Note that

$\stackrel{˜}{X}\left(\varphi +1\right)=\sum _{n=-\infty }^{\infty }x\left[n\right]{e}^{-\mathrm{j2\pi }\left(\varphi +1\right)n}$

$=\sum _{n=-\infty }^{\infty }x\left[n\right]{e}^{-\mathrm{j2\pi n}}{e}^{-\mathrm{j2}\text{πϕ}n}$

$=\stackrel{˜}{X}\left(\varphi \right)$

Therefore, (φ) is periodic with period equal to 1.

b/ Symmetry

$\stackrel{˜}{X}\left(\varphi \right)=\sum _{n=-\infty }^{\infty }x\left[n\right]{e}^{-\mathrm{j2}\text{πϕ}n}$

Note that

$\stackrel{˜}{X}\left(\varphi \right)=\sum _{n=-\infty }^{\infty }\left({x}_{e}\left[n\right]+{x}_{0}\left[n\right]\right)\left(\text{cos}\left[2\text{πϕ}n\right]-j\text{sin}\left[2\text{πϕ}n\right]\right),$

$\stackrel{˜}{X}\left(\varphi \right)=\sum _{n=-\infty }^{\infty }{x}_{e}\left[n\right]\left(\text{cos}\left[2\text{πϕ}n\right]-j\sum _{n=-\infty }^{\infty }{x}_{0}\left[n\right]\text{sin}\left[2\text{πϕ}n\right],$

$\stackrel{˜}{X}\left(\varphi \right)={\stackrel{˜}{X}}_{r}\left(\varphi \right)+j{\stackrel{˜}{X}}_{i}\left(\varphi \right)$

where

${\stackrel{˜}{X}}_{r}\left(\varphi \right)=\sum _{n=-\infty }^{\infty }{x}_{e}\left[n\right]\text{cos}\left[2\text{πϕ}n\right]$

${\stackrel{˜}{X}}_{i}\left(\varphi \right)=-\sum _{n=-\infty }^{\infty }{x}_{0}\left[n\right]\text{sin}\left[2\text{πϕ}n\right]$

We can infer symmetry properties of the DTFT of a real time function x[n].

${\stackrel{˜}{X}}_{r}\left(\varphi \right)=\sum _{n=-\infty }^{\infty }{x}_{e}\left[n\right]\text{cos}\left[2\text{πϕ}n\right],\text{even function of}\varphi$

${\stackrel{˜}{X}}_{i}\left(\varphi \right)=-\sum _{n=-\infty }^{\infty }{x}_{0}\left[n\right]\text{sin}\left[2\text{πϕ}n\right],\text{odd function of}\varphi$

$\mid \stackrel{˜}{X}\left(\varphi \right)\mid =\sqrt{{\stackrel{˜}{X}}_{r}^{2}\left(\varphi \right)+{\stackrel{˜}{X}}_{i}^{2}\left(\varphi \right)},\text{even function of}\varphi$

Therefore, if

x[n] (φ)

x[n] $\stackrel{˜}{X}$ (φ)

Real and even function of n Real and even function of φ

Real and odd function of n Imaginary and odd function of φ

The angle can be computed as follows,

$\begin{array}{}{\text{tan}}^{-1}\left(\frac{{\stackrel{˜}{X}}_{i}\left(\varphi \right)}{{\stackrel{˜}{X}}_{r}\left(\varphi \right)}\right)\text{}\text{for}\text{{}\stackrel{˜}{X}\\ \\ \pi +{\text{tan}}^{-1}\left(\frac{{\stackrel{˜}{X}}_{i}\left(\varphi \right)}{{\stackrel{˜}{X}}_{r}\left(\varphi \right)}\right)\text{}\text{for}\text{{}\stackrel{˜}{X}\\ \\ \\ \angle \stackrel{˜}{X}\left(\varphi \right)=\begin{array}{}\left\{r\left(\varphi \right)>0\end{array}\\ \end{array}$

But, since ±n2π can always be added to the angle, and since ${\text{tan}}^{-1}\left(\frac{{\stackrel{˜}{X}}_{i}\left(\varphi \right)}{{\stackrel{˜}{X}}_{r}\left(\varphi \right)}\right)$ is an odd function of φ,

Questions & Answers

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Source:  OpenStax, Signals and systems. OpenStax CNX. Jul 29, 2009 Download for free at http://cnx.org/content/col10803/1.1
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