0.7 Lecture 8:the discrete time fourier transform (dtft)  (Page 3/4)

4/ Causal exponential

We have seen that

$x[n]={\alpha }^{n}u[n]\stackrel{Z}{\iff }\tilde{X}(z)=\frac{1}{1-{\mathrm{\alpha z}}^{-1}}\text{for}\mid z\mid >\mid \alpha \mid$

Provided |α|<1, we can evaluate the DTFT as

$\tilde{X}(\varphi )=\frac{1}{1-{\mathrm{\alpha e}}^{-\mathrm{j2}\text{πϕ}}}$

We have already examined these functions in connection with DT filters discussed in previous section.

A pole at 0<α<1 yields a DTFT with large spectral components at low frequencies. As |α| decreases, the unit sample response approaches a unit sample, the DTFT magnitude approaches 1 (0 dB) and the angle approaches 0.

A pole at −1<α<0 yields a DTFT with large spectral components at high frequencies. As |α| decreases, the unit sample response approaches a unit sample, the DTFT magnitude approaches 1 (0 dB) and the angle approaches 0.

IV. MODEL OF C/D CONVERTER

We now return to the model of the C/D converter and relate the DTFT of a sequence to the CTFT of a sampled time function.

1/ Representation of FT of CT impulse train and DT sample train

$\hat{x}(t)=\sum _{m}x(\text{mT})\delta (t-\text{mT})\stackrel{\text{CTFT}}{\iff }\hat{X}(f)=\sum _{m}x(\text{mT})e^{-\mathrm{j2\rho }\text{mfT}}$

$x[n]=\sum _{m}x(\text{mT})\delta [n-m]\stackrel{\text{DTFT}}{\iff }\hat{X}(\varphi )=\sum _{m}x(\text{mT})e^{-\mathrm{j2\rho m\varphi }}$

Note that X^(f) and X~(φ) are the same for

$\varphi =\text{fT}=\frac{f}{\frac{1}{T}}=\frac{\text{frequency}}{\text{sampling}\text{frquency}}$

2/ Completion of the model of a CD converter

3/ Model of C/D converter

4/ Model of D/C converter

Two-minute miniquiz problem

Problem 21-1 — DT filtering of a CT signal

The signal x(t) whose spectrum is band-limited to |f|<100 is sampled at the Nyquist rate and filtered by the DT filter shown. Find the spectrum of y(t), Y(f).

Solution

Sampling at the Nyquist rate implies that the sampling frequency is 2 × 100 = 200.

V. PERIODIC DISCRETE TIME SEQUENCES

For a periodic sequence of period N

x[n+N] = x[n]

This example shows a periodic sequence for which N = 4.

1/ Discrete time Fourier series (DTFS)

A periodic DT signal with period N can be expanded in a Fourier series of complex exponential sequences of the form

${\psi }_m[n]=e^{\mathrm{j2\pi }(\frac{m}{N})n}$

DT frequency

These exponentials have frequencies that are multiples of the fundamental frequency 1/N. Since

${\psi }_m+N^{[n]}=e^{\mathrm{j2\pi }(\frac{m+N}{N})n}=e^{\mathrm{j2\pi }(\frac{m}{N})n}={\psi }_m[n]$

there are only N distinct frequencies for a fundamental frequency of 1/N. As a consequence, only N frequencies are required to represent a periodic DT signal of period N.

The DT Fourier series can be expressed as

$x[n]=\sum _{m=0}^{N-1}\tilde{X}[m]e^{\mathrm{j2\pi }\frac{\text{mn}}{N}}$

$\tilde{X}[m]=\frac{1}{N}\sum _{n=0}^{N-1}x[n]e^{-\mathrm{j2\pi }\frac{\text{mn}}{N}}$

where

x[n] is periodic in n with period N,

and

is periodic in m with period N.

2/ DTFS — periodic unit sample train

The Fourier series of the periodic unit sample train is

$\tilde{S}[m]=\frac{1}{N}\sum _{n=0}^{N-1}\delta [n]e^{-\mathrm{j2\pi }\frac{\text{mn}}{N}}=\frac{1}{N}$

Therefore,

$s[n]=\sum _{m=-\infty }^{\infty }\delta [n-\text{mN}]=\frac{1}{N}\sum _{m=0}^{N-1}{e}^{-\mathrm{j2\pi }\frac{m}{N}}$

3/ DTFT of a periodic unit sample train

We have two expressions for a periodic unit sample train,

$s[n]=\sum _{m=-\infty }^{\infty }\delta [n-\text{mN}]=\frac{1}{N}\sum _{m=0}^{N-1}{e}^{-\mathrm{j2\pi }\frac{m}{N}}$

The DTFT of the first expression is

$\tilde{S}[\varphi ]=\sum _{m=-\infty }^{\infty }{e}^{-\mathrm{j2}\text{πϕ}\text{mN}}$

To obtain the DTFT of the second expression recall that

$e^{\mathrm{j2\pi }\frac{\text{mn}}{N}}\stackrel{F}{\iff }\sum _{k=-\infty }^{\infty }\delta (\varphi -\frac{m}{N}-k)$

so that

$\sum _{m=0}^{N-1}{e}^{\mathrm{j2\pi }\frac{\text{mn}}{N}}\stackrel{F}{\iff }\sum _{m=0}^{N-1}\sum _{k=-\infty }^{\infty }\delta (\varphi -\frac{m}{N}-k)=\sum _{m=-\infty }^{\infty }(\varphi -\frac{m}{N})$

The periodic unit sample train

$s[n]=\sum _{m=-\infty }^{\infty }\delta [n-\text{mN}]=\frac{1}{N}\sum _{m=0}^{N-1}{e}^{\mathrm{j2\pi }\frac{\text{mn}}{N}}$

has the Fourier transform

$\tilde{S}[\varphi ]=\sum _{m=-\infty }^{\infty }{e}^{-\mathrm{j2}\text{πϕ}\text{mM}}=\frac{1}{N}\sum _{m=-\infty }^{\infty }\delta (\varphi -\frac{m}{N})$

Therefore, the DT Fourier transform of a periodic unit sample train in time is a periodic unit impulse train in frequency.