chap4 sampling fda lecture

April 6, 2012 Digital Signal Processing 1

EEE & ECE Department

BITS-Pilani, Hyderabad campus

Sampling &

Reconstruction

Digital Signal Processing

April 6, 2012 3 Digital Signal Processing


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


............(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


Recovery of The Signal

The discrete time signal must pass through

an analog lowpass filter.


Recovery of The Signal


Aliasing

Digital Signal Processing 21


Critical Sampling


Under Sampling


Over Sampling


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 ?


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.



Butterworth

M4.2 (SK. Mitra)


M4.2 (SK. Mitra)


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.



(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


% 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


(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


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)

chap4 sampling fda lecture

Documents