html5-img
1 / 26

Fundamentals of Digital Signal Processing

Fundamentals of Digital Signal Processing. Fourier Transform of continuous time signals. with t in sec and F in Hz (1/sec). Examples:. Discrete Time Fourier Transform of sampled signals. with f the digital frequency (no dimensions). Example:. since, using the Fourier Series,.

Télécharger la présentation

Fundamentals of Digital Signal Processing

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Fundamentals of Digital Signal Processing

  2. Fourier Transform of continuous time signals with t in sec and F in Hz (1/sec). Examples:

  3. Discrete Time Fourier Transform of sampled signals with f the digital frequency (no dimensions). Example: since, using the Fourier Series,

  4. Property of DTFT • f is the digital frequency and has no dimensions • is periodic with period f = 1. • we only define it on one period

  5. Sampled Complex Exponential: no aliasing 1. No Aliasing digital frequency

  6. Sampled Complex Exponential: aliasing 2. Aliasing digital frequency

  7. Mapping between Analog and Digital Frequency

  8. Example • Then: • analog frequency • FT: • digital frequency • DTFT: for

  9. Example • Then: • analog frequency • FT: • digital frequency • DTFT: for

  10. Example • Then: • analog frequencies • FT: • digital frequencies • DTFT

  11. Linear Time Invariant (LTI) Systems and z-Transform If the system is LTI we compute the output with the convolution: If the impulse response has a finite duration, the system is called FIR (Finite Impulse Response):

  12. Z-Transform Facts: Frequency Response of a filter:

  13. Digital Filters Ideal Low Pass Filter constant magnitude in passband… … and linear phase passband

  14. Impulse Response of Ideal LPF Assume zero phase shift, This has Infinite Impulse Response, non recursive and it is non-causal. Therefore it cannot be realized.

  15. Non Ideal Ideal LPF The good news is that for the Ideal LPF

  16. Frequency Response of the Non Ideal LPF ripple attenuation stop pass stop transition region • LPF specified by: • passband frequency • passband ripple or • stopband frequency • stopband attenuation or

  17. Best Design tool for FIR Filters: the Equiripple algorithm (or Remez). It minimizes the maximum error between the frequency responses of the ideal and actual filter. ripple attenuation impulse response Linear Interpolation

  18. The total impulse response length N+1 depends on: • transition region • attenuation in the stopband Example: we want Passband: 3kHz Stopband: 3.5kHz Attenuation: 60dB Sampling Freq: 15 kHz Then: from the specs We determine the order the filter

  19. Frequency response N=82 N=98

  20. Example: Low Pass Filter Passband f = 0.2 Stopband f = 0.25 with attenuation 40dB Choose order N=40/(22*(0.25-0.20))=37 Almost 40dB!!!

  21. Example: Low Pass Filter Passband f = 0.2 Stopband f = 0.25 with attenuation 40dB Choose order N=40 > 37 OK!!!

  22. General FIR Filter of arbitrary Frequency Response Weights for Error: Then apply: … and always check frequency response if it is what you expect!

  23. Example: for fp=0:0.01:0.2; % vector of passband frequencies fs=[0.25,0.5]; % stopband frequencies M=[1./sinc(fp), 0, 0]; % desired magnitudes Df=0.25-0.2; % transition region N=ceil(A/(22*Df)); % first guess of order h=firpm(N, [ fp, fs]/0.5,M); % impulse response

  24. not very good here!

  25. To improve it: • Increase order • Add weights w=[1*ones(1,length(fp)/2), 0.2*ones(1, length(fs)/2)]; h=firpm(N, [fp, fs]/0.5,M,w);

More Related