Download presentation

Published byCheyanne Garrick Modified over 3 years ago

1
Symmetry and the DTFT If we know a few things about the symmetry properties of the DTFT, it can make life simpler. First, for a real-valued sequence x(n), the amplitude spectrum (magnitude of the DTFT) is an EVEN function of w. To show this first show that,

2
Symmetry and the DTFT But we’ve assumed that x(n) is a real sequence, so and the relationship is proven

3
**Symmetry and the DTFT The amplitude spectrum is**

If w is replaced by –w, th amplitude spectrum is unchanged. X(ejw) = X(e-jw), so the amplitude spectrum is an even function of w.

4
**Symmetry and the DTFT The phase function is**

For this to be an odd function of w, the imaginary part must be odd and the real part must be even. By euler’s formula,

5
Symmetry and the DTFT Since cos(wn) = cos(-wn), Re(X(ejw)) is an even function of w. Likewise, since sin(wn) = -sin(-wn), Im(X(ejw)) is an even function of w. Therefore, it is shown that the phase function is an odd function of w:

6
**Symmetry and the DTFT Suppose x(n) has even symmetry about the origin:**

Then Which means that X(ejw) is a real-valued function

7
**Periodicity and the DTFT**

The DTFT is a periodic function of w, with period = 2p So the frequency response function and all other DTFT’s are periodic. One period of a DTFT completely describes the function for all values of w, so it Is usually plotted only for the interval [-p, p]

8
**Periodicity and the DTFT**

If x(n) is a real-valued sequence, the amplitude spectrum is an even function of w, AND periodic. Therefore, it is completely described by the interval [0,p]. Since the phase function is an odd function, it is also completely described by the interval [0,p]. Since the phase function is an odd function, it is also completely described by the interval [0,p].

9
**DTFT of a Sinusoid Suppose x(n) is the complex exponential, ejfn**

This delta function is the Dirac (continuous time) delta function, so it has the sifting property. Taking the inverse DTFT,

10
DTFT of a Sinusoid So the complex exponential sequenc looks like this in the frequency domain: 2p 2p 2p -2p -2p+f -p f p 2p 2p+f

11
DTFT of a Sinusoid If x(n) is a sinusoid:

12
DTFT of a Sinusoid So the DTFT of a sinusoid is a pair of impulses int the frequency domain, at the positive and negative frequency of the sinusoid, then repeated every 2p radians per sample p p p p p p -2p-f -2p -2p+f -p -f f p 2p 2p+f 2p-f

13
**The DTFT and the Fourier Transform**

Suppose we have a continuous time signal xa(t), and its Fourier transform: We sample xa(t) at a sampling frequency fs, creating a discrete time signal (a sequence): This sequence has a spectrum (DTFT):

14
**The DTFT and the Fourier Transform**

I will state the following relationship now, without proof. The proof comes later: In other words, whatever the continuous time spectrum is, the spectrum of the discrete time signal obtained by sampling has the same shape, but it is periodic in frequency with period = fs

15
**The DTFT and the Fourier Transform**

If the continuous time signal has this spectrum Xa(f) -2fs -fs fs 2fs f Then the discrete time signal (sequence) has this spectrum: X(ej2pf) -2fs -fs fs 2fs f

16
**The DTFT and the Fourier Transform**

If the continuous time signal is bandlimited (has no significant spectral content above some frequency f=B), Xa(f) -2fs -fs -B B fs 2fs f Then the sequence obtained by sampling the continuous time signal looks like this in the frequency domain:

17
**The DTFT and the Fourier Transform**

X(ej2pf) -2fs -fs -B B fs 2fs f As long as fs > 2B, no problem. But if fs < 2B, it looks like this: X(ej2pf) -4fs -3fs -2fs -fs -B B fs 2fs 3fs 4fs f

18
**The DTFT and the Fourier Transform**

Note that the spectral replicas created by sampling overlap X(ej2pf) -4fs -3fs -2fs -fs -B B fs 2fs 3fs 4fs f This is called aliasing To avoid aliasing, the minimum sample frequency is given by the Nyquist criterion:

19
**The DTFT and the Fourier Transform**

And the minimum sampling frequency is called the Nyquist frequency or Nyquist rate. If the Nyquist criterion is met, the original continuous time signal xa(t) can be completely recovered from the sequence x(n) by an ideal digital to analog converter.

20
**The DTFT and the Fourier Transform**

To make sure the Nyquist criterion is met, the signal should pass through an antialiasing filter before it is sampled by the analog to digital converter (ADC): xa(t) Antiasing Filter ADC x(n) The antialising filter must be designed to cut off any spectral content above ½ the sampling frequency.

Similar presentations

OK

Fourier representation for discrete-time signals And Sampling Theorem

Fourier representation for discrete-time signals And Sampling Theorem

© 2018 SlidePlayer.com Inc.

All rights reserved.

To ensure the functioning of the site, we use **cookies**. We share information about your activities on the site with our partners and Google partners: social networks and companies engaged in advertising and web analytics. For more information, see the Privacy Policy and Google Privacy & Terms.
Your consent to our cookies if you continue to use this website.

Ads by Google

Ppt on unity in diversity images Ppt on condition based maintenance programs Ppt on neil bohr model of atom Ppt on object-oriented programming concepts in java Ppt on frank lloyd wright Ppt on multiplying decimals by powers of ten Ppt on summary writing Perspective view ppt on iphone Ppt on accounting standard 18 Download ppt on mutual fund