Question

1. Design a 10th-order lowpass FIR filter using the window method (fir1) to cut frequencies

above 30Hz in an application where the sampling frequency is 125 Hz.

2. Plot the filter coefficients that define the filter (stem).

3. Plot the frequency response of the FIR filter designed (freqz)

4. Design a 100th-order lowpass FIR filter using the window method (fir1) to cut frequencies

above 30Hz in an application where the sampling frequency is 125 Hz. Plot the filter

coefficients that defining the filter (stem). Plot the frequency response of the FIR filter

designed (freqz). Compare the frequency response of the 10th-order with the 100th-order

filter.

5. Design a 10th-order highpass FIR filter using the window method (fir1) to cut frequencies

below 30Hz in an application where the sampling frequency is 125 Hz.

6. Plot the filter coefficients that define the filter (stem).

7. Plot the frequency response of the FIR filter designed (freqz)

8. Design a 100th-order highpass FIR filter using the window method (fir1) to cut

frequencies below 15Hz in an application where the sampling frequency is 125 Hz. Plot

the filter coefficients that define the filter (stem). Plot the frequency response of the

FIR filter designed (freqz). Compare the frequency response of the 10th-order with the

100th-order filter.

9. Design a 100th-order bandpass FIR filter using the window method (fir1) to cut

frequencies below 15Hz and above 30 Hz in an application where the sampling frequency

is 125 Hz. Plot the filter coefficients that defining the filter (stem). Plot the frequency

response of the FIR filter designed (freqz).

0. Design a 100th-order bandstop FIR filter using the window method (fir1) to cut

frequencies between 15Hz and 30 Hz in an application where the sampling frequency

is 125 Hz. Plot the filter coefficients that defining the filter (stem). Plot the frequency

response of the FIR filter designed (freqz).

11. Design a 100th-order bandpass FIR filter using the window method (fir2) to cut

frequencies below 15Hz and above 30 Hz in an application where the sampling frequency

is 125 Hz. Plot the filter coefficients that defining the filter (stem). Plot the frequency

response of the FIR filter designed (freqz).

Answer 1

10th order LPF FIR

clc;close all;clear all;

N=10;fc=30;Fs=125;

n=0:1:N;

wc=2*fc/Fs;

h=fir1(N,wc);

w=0:pi/1000:pi;

H=freqz(h,1,w);

figure;

subplot(221)

stem(n,h);grid;

xlabel('n');ylabel('h(n)')

title('FIR filter coeffiecients for N=10 ')

subplot(222)

plot(w/pi,20*log10(abs(H)));grid;

xlabel('w/pi');ylabel('|H(exp(jw)|')

title('Magnitude response')

subplot(223)

plot(w/pi,angle(H));grid;

xlabel('w/pi');ylabel('<H(exp(jw)>')

title('Phase response')

__________________________________________________________________________

100th order LPF FIR

N=100;fc=30;Fs=125;

n=0:1:N;

wc=2*fc/Fs;

h=fir1(N,wc);

w=0:pi/1000:pi;

H=freqz(h,1,w);

figure;

subplot(221)

stem(n,h);grid;

xlabel('n');ylabel('h(n)')

title('FIR filter coeffiecients for N=100 ')

subplot(222)

plot(w/pi,20*log10(abs(H)));grid;

xlabel('w/pi');ylabel('|H(exp(jw)|')

title('Magnitude response')

subplot(223)

plot(w/pi,angle(H));grid;

xlabel('w/pi');ylabel('<H(exp(jw)>')

title('Phase response')

____________________________________________________________________________

clc;close all;clear all;

N=10;fc=15;Fs=125;

n=0:1:N;

wc=2*fc/Fs;

h=fir1(N,wc,'high');

w=0:pi/1000:pi;

H=freqz(h,1,w);

figure;

subplot(221)

stem(n,h);grid;

xlabel('n');ylabel('h(n)')

title('FIR filter coeffiecients for N=10 ')

subplot(222)

plot(w/pi,20*log10(abs(H)));grid;

xlabel('w/pi');ylabel('|H(exp(jw)|')

title('HPF-Magnitude response')

subplot(223)

plot(w/pi,angle(H));grid;

xlabel('w/pi');ylabel('<H(exp(jw)>')

title('Phase response')

N=100;fc=30;Fs=125;

n=0:1:N;

wc=2*fc/Fs;

h=fir1(N,wc,'high');

w=0:pi/1000:pi;

H=freqz(h,1,w);

figure;

subplot(221)

stem(n,h);grid;

xlabel('n');ylabel('h(n)')

title('FIR filter coeffiecients for N=100 ')

subplot(222)

plot(w/pi,20*log10(abs(H)));grid;

xlabel('w/pi');ylabel('|H(exp(jw)|')

title('HPF-Magnitude response')

subplot(223)

plot(w/pi,angle(H));grid;

xlabel('w/pi');ylabel('<H(exp(jw)>')

title('Phase response')

clc;close all;clear all;

%band pass filter

N=100;fc1=15;fc2=30;Fs=125;

n=0:1:N;

wc=[2*fc1/Fs 2*fc2/Fs] ;

h=fir1(N,wc,'bandpass');

w=0:pi/1000:pi;

H=freqz(h,1,w);

figure;

subplot(221)

stem(n,h);grid;

xlabel('n');ylabel('h(n)')

title('FIR filter coefficients for N=100 ')

subplot(222)

plot(w/pi,20*log10(abs(H)));grid;

xlabel('w/pi');ylabel('|H(exp(jw)|')

title('BPF-Magnitude response')

subplot(223)

plot(w/pi,angle(H));grid;

xlabel('w/pi');ylabel('<H(exp(jw)>')

title('Phase response')

%Band stop filter

N=100;fc1=15;fc2=30;Fs=125;

wc=[2*fc1/Fs 2*fc2/Fs] ;

n=0:1:N;

h=fir1(N,wc,'stop');

w=0:pi/1000:pi;

H=freqz(h,1,w);

figure;

subplot(221)

stem(n,h);grid;

xlabel('n');ylabel('h(n)')

title('FIR filter coefficients for N=100 ')

subplot(222)

plot(w/pi,20*log10(abs(H)));grid;

xlabel('w/pi');ylabel('|H(exp(jw)|')

title('BSF-Magnitude response')

subplot(223)

plot(w/pi,angle(H));grid;

xlabel('w/pi');ylabel('<H(exp(jw)>')

title('Phase response')