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Published byBrett Holt Modified over 9 years ago
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Fundamentals of Digital Signal Processing
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Fourier Transform of continuous time signals with t in sec and F in Hz (1/sec). Examples:
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Discrete Time Fourier Transform of sampled signals with f the digital frequency (no dimensions). Example: since, using the Fourier Series,
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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
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Sampled Complex Exponential: no aliasing 1. No Aliasing digital frequency
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Sampled Complex Exponential: aliasing 2. Aliasing digital frequency
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Mapping between Analog and Digital Frequency
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Example Then: analog frequency FT: digital frequency DTFT: for
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Example Then: analog frequency FT: digital frequency DTFT: for
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Example Then: analog frequencies FT: digital frequencies DTFT
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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):
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Z-Transform Facts: Frequency Response of a filter:
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Digital Filters Ideal Low Pass Filter passband constant magnitude in passband… … and linear phase
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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.
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Non Ideal Ideal LPF The good news is that for the Ideal LPF
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Frequency Response of the Non Ideal LPF passstop transition region attenuation ripple LPF specified by: passband frequency passband ripple or stopband frequency stopband attenuation or
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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. attenuation ripple impulse response Linear Interpolation
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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
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Frequency response N=82 N=98
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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!!!
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Example: Low Pass Filter Passband f = 0.2 Stopband f = 0.25 with attenuation 40dB Choose order N=40 > 37 OK!!!
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General FIR Filter of arbitrary Frequency Response Weights for Error: Then apply: … and always check frequency response if it is what you expect!
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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
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not very good here!
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To improve it: 1.Increase order 2.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);
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