Bio-Signals Tasks

Contributors

Youssef Ashraf

Mariam Ahmed

Main Points :

• The magnitude and phase characteristics of the 3-pole Butterworth low-pass filter
• Low Pass to Low pass Filters
• Low Pass to Band Pass Filters
• Low Pass to Band Elimination Filters

Task 1 Q1 :

 Use the MATLAB bode function to plot the magnitude and phase
 characteristics of the 3-pole Butterworth low-pass filter with unity gain and
 normalized frequency at ω = 1 rad/sec. Print your code, results and figures.

Solution :

Using the general rule at k=3 :

>> % it is a 3-pole Butterworth low-pass filter
>> % thus k=3 >> for the G(s) 
>> 
>> % Parameters : W = 1 rad/sec , b0=1 , a0=1 , a1=2 , a2=1 , a3=1 
>> % let us Create 2 vectors carrying the coefficents of  bn and an >> b & a Respectively 
>> b = [0 0 0 1]

b =

     0     0     0     1

>> a = [ 1 2 2 1 ]

a =

     1     2     2     1

>> % now send them to bode to start plotting magnitude and phase where W =1 rad/sec 
>> bode(b,a) 
>> % to add grid lines we will use Grid Command where grid on displays the major grid lines for the current axes
>> grid
>> 

Task 1 Q2 :

 Use the MATLAB buttap and lp2lp functions to find the transfer function
 of a third-order Butterworth low-pass filter with cutoff frequency fc =
 2KHz. Then, plot its frequency response. Print your code, results and figures.

Solution :

% it is a 3-pole Butterworth low-pass filter with cut off freq fc=2KHz 
[zeros , poles , k] = buttap(3);
% calcuate coeffecients of b(numerator) and a (denominator)
[b,a] = zp2tf ( zeros, poles , k);
% range the frequency and convert it to rad/sec(angular)
freq = 1000:30:10000;
w = 2*pi*freq;
% declare cutoff frequency and convert it to rad/sec(angular)
freqc = 2*1000 ;  % we multiplied by 1000 to convert kHZ to HZ
wc = 2*pi*freqc; 
% here calculate the new coefficents after low pass to low pass filter 
% let their names bnew , anew
% for explaination we are converting from analog values to angular domain 
[ bnew , anew ] = lp2lp(b , a, wc); 
% lets calculate The transfer function of the  new filtered  
Gofs = freqs(bnew , anew , w);
semilogx(w, abs(Gofs));      % plot it 
% add grid lines
grid;
hold on ;
title('Frequency respone plot for 3rd-order Butterworth low-pass filter with fc = 2KHz');
xlabel('Angular Frequency in :(rad/sec)');
ylabel('Magnitude of the Transfer Function in (dB) ');

Task 2 Q1 :

Use the MATLAB functions buttap and lp2bp to find the transfer function
of a 3-pole Butterworth analog band-pass filter with the pass band
frequency centered at f0 = 4 KHz, and bandwidth BW = 2 KHz. Then, plot
its frequency response. Print your code, results and figures.

Solution :

% it is a 3-pole Butterworth low-pass filter with pass band freq fo=4KHz , bandwidth = 2KHZ 
[zeros , poles , k] = buttap(3);
% calcuate coeffecients of b(numerator) and a (denominator) of the filter
[b,a] = zp2tf ( zeros, poles , k);
% declare band pass frequency f0=4KHZ and convert it to rad/sec(angular)
freq = 100:100:100000;
freqc = 4000 ;  
wc = 2*pi*freqc ;
% declare bandwidth frequency bwf= 2KHZ and convert it to rad/sec(angular)
bwf = 2000; 
wband = 2*pi*bwf ;  
% here calculate the new coefficents after low pass to band pass filter 
% let their names bnew , anew
% for explaination we are converting from analog values to angular domain 
[ bnew , anew ] = lp2bp(b , a, wc , wband ); 
% lets calculate the transfer function of the band-pass filter
w =2*pi*freq ; 
Gofs = freqs(bnew , anew , w);
semilogx(freq, abs(Gofs));      % plot freq response in hertz
% add grid lines
grid;
hold on;
title('Frequency respone plot for 3rd-order Butterworth band-pass filter with fo=4kHZ , Bw = 2KHZ');
xlabel('Frequency in :(HZ)');
ylabel('Magnitude of the Transfer Function in (dB)');

Task 2 Q2 :

Use the MATLAB functions buttap and lp2bs to find the transfer function of
a 3-pole Butterworth band-elimination (band-stop) filter with the stop band frequency
centered at f0 = 5 KHz, and bandwidth BW = 2 KHz. Then, plot its frequency response.
Print your code, results and figures.

Solution :

 % it is a 3-pole Butterworth with stop band frequency centered at f0 = 5 KHz , and bandwidth BW = 2 KHz

 [zeros , poles , k] = buttap(3);
 % calcuate coeffecients of b(numerator) and a (denominator)
 [b,a] = zp2tf ( zeros, poles , k); 
 % declare band pass frequency f0=5KHZ and convert it to rad/sec(angular)
 freq = 100:100:100000;
 freqc = 5000 ; 
 wc = 2*pi*freqc ; 
 % declare bandwidth frequency bwf= 2KHZ and convert it to rad/sec(angular)
 bwf = 2000; 
 wband = 2*pi*bwf ;  
 % here calculate the new coefficents after low pass to stop pass filter

 % let their names bnew , anew
 % for explaination we are converting from analog values to angular domain 
 [ bnew , anew ] = lp2bs(b , a, wc , wband ); 
 % lets calculate the transfer function of the stop-pass filter
 w =2*pi*freq ; 
 Gofs = freqs(bnew , anew , w);
 semilogx(freq, abs(Gofs));      % plot freq response in hertz
 % add grid lines
 grid;
 hold on;
 title('Frequency respone plot for 3rd-order Butterworth band-elimination filter with fo=5kHZ , Bw = 2KHZ');
 xlabel('Frequency in :(HZ)');
 ylabel('Magnitude of the Transfer Function in (dB)');