chap4 sampling fda lecture
TRANSCRIPT
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April 6, 2012 Digital Signal Processing 1
EEE & ECE Department
BITS-Pilani, Hyderabad campus
Sampling &
Reconstruction
Digital Signal Processing
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April 6, 2012 Digital Signal Processing 2
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April 6, 2012 3 Digital Signal Processing
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April 6, 2012 4 Digital Signal Processing
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April 6, 2012 7 Digital Signal Processing
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April 6, 2012 Digital Signal Processing 9
nT)-(t(nT)g(t)p(t)g(t)gn
-n
aap
Since impulse is a periodic signal of period T, it
can be expressed as trigonometric Fourier series.
............2cos32cos22cos1T
1)()( ooo tttnTttp
n
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April 6, 2012 Digital Signal Processing 10
............(t)cos32g
(t)cos22g(t)cos2g(t)g
T
1)()()(
oa
oaoaa
t
tttptgtg ap
The FT of gp(t) is Gp(jω)
............)cos3(j2G
)cos2(j2G)cos(j2G)(jG
T
1)(
oa
oaoaa
t
ttjGp
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April 6, 2012 13 Digital Signal Processing
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Recovery of The Signal
The discrete time signal must pass through
an analog lowpass filter.
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Recovery of The Signal
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Aliasing
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Digital Signal Processing 21
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Critical Sampling
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Under Sampling
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Over Sampling
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Problem
(1) A continuous time signal xa(t) is composed of a linear combination of sinusoidal signals of frequencies 300 Hz, 500 Hz, 1.2 kHz, 2.15 kHz and 3.5 kHz. The signal xa(t) is sampled at a 2.0 kHz rate and the sampled sequence is passed through an ideal low pass filter with a cut-off frequency of 900 Hz, genearting a continuos time signal of ya(t)
What are the frequency components present in the output signal ?
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Filtering Using FDA tool
M4.2 (SK. Mitra)
Determine the lowest order of a lowpass Chebyshev Type I filter with a 0.25 dB passband frequency at 1.5 kHz and minimum attenuation 0f 25 dB at 6.0 kHz.
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Filtering Using FDA tool
Butterworth
M4.2 (SK. Mitra)
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M4.2 (SK. Mitra)
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M4.2 (SK. Mitra)
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M4.5 (SK. Mitra)
Determine the lowest order of a highpass Butterworth filter with a 0.5 dB passband frequency at 6.5 kHz and minimum attenuation 0f 40 dB at 1.5 kHz.
Filtering Using FDA tool
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(1) A Butterworth analog highpass filter is to be designed with the following specifications : Fp = 6.5 kHz and Fs = 1.5 kHz, peak passband ripple of 0.5 dB, and minimum stopband attenuation of 40 dB. What are the band edges and the order of the corresponding analog lowpass filter ? What is the order of the highpass filter ?
Problems
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% Designing of analog Butterworth High-pass filter
% Given specifications : wp = 6500 Hz, ws = 1500 Hz
% Given Specifications : alphap =0.5 dB, alphas = 40 dB
% Initially design analog prototype butterworth lowpass filter
% Then design highpass filter using frequency transformation
% Prototype Lowpass filter specifications : wp = 1,
% ws = wp(cap)/ws(cap)= 2*pi*6500 / 2*pi*1500 = 4.333
[n,wn]=buttord(1,4.333,0.5,40,'s')
[num,den]=butter(n,wn,'s')
[num1,den1]=lp2hp(num,den,2*pi*6500)
tf(num1,den1)
%Directly designing High-pass filters
[n2,wn2]=buttord(2*pi*6500,2*pi*1500,0.5,40,'s')
[num2,den2]=butter(n2,wn2,'high','s')
tf(num2,den2)
Matlab-coding
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(2) A Butterworth analog highpass filter is to be designed with the following specifications : Fp = 4 kHz and Fs = 1 kHz, peak passband ripple of 0.1 dB, and minimum stopband attenuation of 40 dB. What are the band edges and the order of the corresponding analog lowpass filter ? What is the order of the highpass filter ?
Problems
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April 6, 2012 Digital Signal Processing 77
Matlab-coding
% Designing of analog Butterworth High-pass filter
% Given specifications : wp = 4000 Hz, ws = 1000 Hz
% Given Specifications : alphap =0.1 dB, alphas = 40 dB
% Initially design analog prototype butterworth lowpass filter
% Then design highpass filter using frequency transformation
% Prototype Lowpass filter specifications : wp = 1
% ws = wp(cap)/ws(cap)= 2*pi*4000 / 2*pi*1000 = 4
[n,wn]=buttord(1,4,0.1,40,'s')
[num,den]=butter(n,wn,'s')
tf(num,den)
[num1,den1]=lp2hp(num,den,2*pi*4000)
tf(num1,den1)
%Directly designing High-pass filters
[n2,wn2]=buttord(2*pi*4000,2*pi*1000,0.1,40,'s')
[num2,den2]=butter(n2,wn2,'high','s')
tf(num2,den2)