Steps involved in performing FIR filter operation

Step I: Let’s enter the pass band frequency (fp) and stop band frequency (fq).

Step II: And then get the sampling frequency (fs), length of window (n).

Step III: Next calculate cut off frequency.

Step IV: Use the boxcar, hamming, Blackman Commands for designing window.

Step V: And then design filter by using above parameters.

Step VI: To find frequency response of the filter by using matlab command freqz.

Step VII: Then plot the magnitude response and phase response of the filter.

Program:

clc;

clear all;

close all;

n=20;

fp=300;

fq=200;

fs=1000;

fn=2*fp/fs;

window=blackman(n+1);

b=fir1(n,fn,’high’,window);

[H W]=freqz(b,1,128);

subplot(2,1,1);

plot(W/pi,abs(H));

title(‘mag res of lpf’);

ylabel(‘gain in db——–>’);

xlabel(‘normalized frequency——>’);

subplot(2,1,2);

plot(W/pi,angle(H));

title(‘phase res of lpf’);

ylabel(‘angle——–>’);

xlabel(‘normalized frequency——>’);

Steps involved in performing IIR filter operation

Step I: Now enter the pass band ripple (rp) and stop band ripple (rs).

Step II: while the pass band frequency (wp) and stop band frequency (ws).

Step III: Then Get the sampling frequency (fs).

Step IV: Then calculate the normalized pass band frequency, and normalized stop band frequency w1 and w2 respectively.

Step V : Make using the following function for calculating the Butterworth filter order [n,wn]=buttord(w1,w2,rp,rs ) and Chebyshev filter order [n,wn]=cheb1ord(w1,w2,rp,rs)

 Step VI: Let’s design the nth order digital high pass Chebyshev or Butterworth filter using the following commands. Butterworth filter [b,a]=butter (n, wn,’high’) Chebyshev filter [b,a]=cheby1 (n, 0.5, wn,’high’)

Step VII: In order to find the digital frequency response of the filter by using ‘freqz ( )’ function

Step VIII: Then calculate the magnitude for the frequency response in decibels (dB) mag=20*log10 (abs (H))

Step IX: Then plot the magnitude response [magnitude in dB Vs normalized frequency]

Step X: Next calculate the phase response using angle (H).

Step XI: Then plot the phase response [phase in radians Vs normalized frequency (Hz)].

Program:

clc;

clear all;

close all;

disp(‘enter the IIR filter design specifications’);

rp=input(‘enter the passband ripple’);

rs=input(‘enter the stopband ripple’);

wp=input(‘enter the passband freq’);

ws=input(‘enter the stopband freq’);

fs=input(‘enter the sampling freq’);

w1=2*wp/fs;w2=2*ws/fs;

[n,wn]=buttord(w1,w2,rp,rs,’s’);

disp(‘Frequency response of IIR HPF is:’);

[b,a]=butter(n,wn,’high’,’s’);

w=0:.01:pi;

[h,om]=freqs(b,a,w);

m=20*log10(abs(h));

an=angle(h);

figure, subplot(2,1,1);plot(om/pi,m);

title(‘magnitude response of IIR filter is:’);

xlabel(‘(a) Normalized freq. –>’);

ylabel(‘Gain in dB–>’);

subplot(2,1,2);plot(om/pi,an);

title(‘phase response of IIR filter is:’);

xlabel(‘(b) Normalized freq. –>’);

ylabel(‘Phase in radians–>’);

The DTMF stands for “Dual Tone Multi-Frequency”, and maybe a method of representing digits with tone frequencies, so as to transmit them over an analog communications network, for instance, a telephone line. In telephone networks, DTMF signals are wont to encode dial trains and other information. Dual-tone Multi-Frequency (DTMF) signaling is widely used for voice communications control and is mostly used worldwide in modern telephony to configure switchboards and to dial numbers. It is also utilized in systems like voice mail, electronic message, and telephone banking. The DTMF signal is considered as the sum of two sinusoids – or tones – with frequencies taken from two mutually exclusive groups. Where these frequencies are chosen to stop any harmonics from being incorrectly detected by the receiver as another DTMF frequency. while each pair of tones consist of one set frequency of the low group (697 Hz, 770 Hz, 852 Hz, 941 Hz) and another set frequency of the high group ( 1336 Hz, 1477Hz,1209 Hz) and represents a unique symbol.

Program:

clc;

clear all;

close all;

t = -2:0.05:2;

x=input(‘enter the input number’);

fr1=697;

fr2=770;

fr3=852;

fr4=941;

fc1=1209;

fc2=1336;

fc3=1477;

fc4=1633;

y0 = sin(2*pi*fr4*t) + sin(2*pi*fc2*t); % 0

y1 = sin(2*pi*fr1*t) + sin(2*pi*fc1*t); % 1

y2 = sin(2*pi*fr1*t) + sin(2*pi*fc2*t); % 2

y3 = sin(2*pi*fr1*t) + sin(2*pi*fc3*t); % 3

y4 = sin(2*pi*fr2*t) + sin(2*pi*fc1*t); % 4

y5 = sin(2*pi*fr2*t) + sin(2*pi*fc2*t); % 5

y6 = sin(2*pi*fr2*t) + sin(2*pi*fc3*t); % 6

y7 = sin(2*pi*fr3*t) + sin(2*pi*fc1*t); % 7

y8 = sin(2*pi*fr3*t) + sin(2*pi*fc2*t); % 8

y9 = sin(2*pi*fr3*t) + sin(2*pi*fc3*t); % 9

y_start = sin(2*pi*fr4*t) + sin(2*pi*fc1*t); % *

y_canc = sin(2*pi*fr4*t) + sin(2*pi*fc3*t); % #

if (x==1)

 plot(t,y1)

 xlabel(‘time(t)’)

 ylabel(‘amplitude’)

elseif (x==2)

 plot(t,y2)

 xlabel(‘time(t)’)

 ylabel(‘amplitude’)

elseif (x==3)

 plot(t,y3)

 xlabel(‘time(t)’)

 ylabel(‘amplitude’)

elseif (x==4)

 plot(t,y4)

 xlabel(‘time(t)’)

 ylabel(‘amplitude’)

elseif (x==5)

 plot(t,y5)

 xlabel(‘time(t)’)

 ylabel(‘amplitude’)

elseif (x==6)

 plot(t,y6)

 xlabel(‘time(t)’)

 ylabel(‘amplitude’)

elseif (x==7)

 plot(t,y7)

 xlabel(‘time(t)’)

 ylabel(‘amplitude’)

elseif (x==8)

 plot(t,y8)

 xlabel(‘time(t)’)

 ylabel(‘amplitude’)

elseif (x==9)

 plot(t,y9)

 xlabel(‘time(t)’)

 ylabel(‘amplitude’)

elseif (x==0)

 plot(t,y0)

 xlabel(‘time(t)’)

 ylabel(‘amplitude’)

elseif (x==11)

 plot(t,y_start)

 xlabel(‘time(t)’)

  ylabel(‘amplitude’)

 elseif (x==12)

 plot(t,y_canc)

 xlabel(‘time(t)’)

 ylabel(‘amplitude’)

else

 disp(‘enter the correct input’)

 end