The purpose of this project is to design the filter using MATLAB to filter this signal below

The task to complete this project is:

1. Generate the signal above using Matlab.

2. Plot the DFT of this signal.

3. Design a digital Butterworth LPF to remove all frequencies except 1000 Hz.

4. Design a digital Chebyshev HPF filter to maintain only the 5000Hz frequency.

5. Design a BPF of your choice to remove all frequencies except 3000Hz.

From the signal given, the maximum frequency is 5000Hz

In order to take the sample of the signal and to avoid aliasing, the sample frequency selected is

Fs ≥ 2FN = 15000Hz

The sample time will be 1/Fs = 0.0067ms

The sample for 0 ≤ t ≤ 10ms is N = 10ms/Ts = 150 samples

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1. Generate x(t) in MATLAB

a) MATLAB code

b) Output waveform

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Compute figure 2

Compute figure 3

Figure 2 : Signal Corrupted with Zero-Mean Random Noise of x(t)

Figure 1 : Single-Sided Amplitude Spectrum of x(t)

Figure 1: signal x(t)

2) Plot the DFT of x(t)

Basically, to compute the DFT of a time signal, x(t), we need to transform the signal into discrete

– time signal , x[n]. After that, the x[n] will be transform into Fourier transform which is X(w) before

transform to DFT of X(k). The X(k) value was compute with the length of N-1 which N is the number of

sample from X(w). For this signal, L = N-1 = 149.

a) MATLAB code

b) Output waveform

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Show value of X[k] from k=0 to 149

Figure 4 Magnitude of X[k] Figure 5 Phase of X[k]

c) Value of X[k]

k

value

11121314151617181920

3.4894 - 1.1269i 0.3217 - 0.0968i 0.1710 - 0.0469i 0.1175 - 0.0286i 0.0902 - 0.0188i0.0735 - 0.0125i 0.0623 - 0.0079i 0.0542 - 0.0042i 0.0480 - 0.0012i 0.0431 + 0.0015i

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k value21222324252627282930..

0.0390 + 0.0039i 0.0356 + 0.0062i 0.0326 + 0.0084i 0.0299 + 0.0106i 0.0274 + 0.0129i0.0250 + 0.0152i 0.0227 + 0.0176i 0.0204 + 0.0203i 0.0180 + 0.0232i 0.0154 + 0.0264i

…………..

k value0123456789

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1.0e+002 -0.0293 -0.0295 + 0.0016i - 0.0301 + 0.0032i -0.0313 + 0.0049i -0.0330 + 0.0067i-0.0353 + 0.0086i -0.0385 + 0.0107i -0.0427 + 0.0132i -0.0485 + 0.0160i -0.0565 + 0.0196i

k value141142143144145146147148149

-0.0565 - 0.0196i -0.0485 - 0.0160i -0.0427 - 0.0132i -0.0385 - 0.0107i -0.0353 - 0.0086i-0.0330 - 0.0067i -0.0313 - 0.0049i -0.0301 - 0.0032i -0.0295 - 0.0016i

FILTER DESIGNING APPROACH

To design the filter, there is several basic information that needs to be focus. The first

step to design a filter is to determine the filter’s type whether it is ‘lowpass’, ‘highpass’ or

‘bandpass’. After that, we need to find the cutoff frequency of the filter by referring to its ideal

design. In this MATLAB, the function that was provided such as buttord(for butterworth) and

cheb1ord(for chebyshev type I) will find the value of order,n and cut off frequency, Wn for the

filter with the given passband frequency, stopband frequency and sample frequency. After the

value of n an Wn has been obtained, the function butter(for butterworth) and cheby1(fo

chebyshev type I) will create the filter equation based on the information found in previous

step. The design may consist of analog domain filter or digital domain filter. To validate that the

filter designed is correct, the original signal, x(t) will be convolved with filter signal, h[n] to

obtain the output signal, y[n]. The magnitude spectrum of the output signal will validate

whether the filter has function as it’s needed or not.

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3) Design a digital Butterworth LPF to remove all frequencies except 1000 Hz.

To design this filter, we need to determine the cutoff frequency which is 1000Hz

So the pass band will be, Wp = 1000/7500 and the stopband is Ws = 2000/7500

The filter contain less than 3dB ripple in the pass band and 60dB of attenuation in the stopband

From the code generated, the order, n and the new cutoff frequency, Wn is;

n = 10 and Wn = 0.1398

a) MATLAB code

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close all;clear all;f1=1000;f2=3000;f3=5000;fs=15000;t=0:1/fs:.01;

%original signalx1=5*cos(2*pi*f1*t); x2=2*cos(2*pi*f2*t); x3=.5*cos(2*pi*f3*t); Xw=x1+x2+x3;

%butterworth lowpass FilterWp = 1000/7500; %Passband corner frequencyWs = 2000/7500; %Stopband corner frequency Rp = 3; %Passband ripple, in decibelsRs = 60; %Stopband attenuation, in decibels

%Returns the lowest order n of the butterworth filter,%and the corresponding cutoff frequencies, Wn[n wn]=buttord(Wp,Ws,Rp,Rs)[b a]=butter(n,wn,'low')[z,p,k]= butter(n,wn,'low');figure(1); plot([b a]); title('filter signal');

%Converts a discrete-time zero-pole-gain representation of a given digital%filter to an equivalent second-order section representation.[sos,g]=zp2sos(z,p,k);

%Discrete-time, second-order section, direct-form II filterHd = dfilt.df2sos(sos,g); y=filter(Hd,Xw); %Passing signal through filter

%plotting%magnitude and phase responsefvtool(Hd);title('Butterworth lowpass Filter')fvtool(y);title('Frequency domain Filtered Y');

%Time Domainsubplot(511);plot(t,x1,'r');hold on;stem(t,x1);title('Original X1=5cos(2*pi*1000*t)');subplot(512);plot(t,x2,'r');hold on;stem(t,x2);title('Original X2=2cos(2*pi*3000*t)');subplot(513);plot(t,x3,'r');hold on;stem(t,x3);title('Original X3=0.5cos(2*pi*5000*t)');subplot(514);plot(t,Xw,'r');hold on;stem(t,Xw);title('X1+X2+X3=5cos(2*pi*1000*t)+2cos(2*pi*3000*t)+0.5cos(2*pi*5000*t)');subplot(515);plot(t,y,'r');hold on;stem(t,y);title('Filtered Y signal');

Determine order, n and the new cutoff frequency Wn

Show filter window tool

b) Output waveform

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Figure 6 : magnitude response of filter

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Figure 7 : phase response of filter

Figure 8 : signal of the filter

c) Filter validation

Figure 9: Shows that the filtered signal is almost the same with X1 signal with 1000Hz and magnitude of 5 which means that X2 and X3 is being filtered.

Alternative method:

To validate the filter that have been designed, the input signal, x(t) was convolved with the filter, h[n] using the MATLAB code as below;

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ts=0:1:100;x=5*cos(2*pi*f1*ts)+2*cos(2*pi*f2*ts)+0.5*cos(2*pi*f3*ts);res=2*pi/360;w=0:res:2*pi-res;lpf=[b a]

out=conv(x,lpf);figure(1);plot(out)

out_mag=abs(out);figure(2);plot(w(1:75).*15000/(2*pi),out_mag(1:75))

[b a] Is the filter equation

The output waveform after the signal has been filtered out is;

From figure 11, it shows that the frequency at 1000Hz is remain and proved that this filter is valid.

4) Design a digital Chebyshev HPF filter to maintain only the 5000Hz frequency.

In order to design the HPF, the bandpass and stoppass band was first determined, to get the frequency

at 5000Hz only

The band pass frequency, Wp = 5000/7500

Stop band frequency, Ws = 4000/7500

The Rp, ripple at pass band is 3dB and Rs, ripple at stop band is 40dB

The order, n and the new cutoff frequency is:

n = 6 and Wn = 0.6667

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Figure 10: output signal after filtered Figure 11: magnitude spectrum of the filtered signal

Cheby1 indicate that 1st order chebyshev is used

a) MATLAB code

b) Output waveform

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close all;clear all;f1=1000;f2=3000;f3=5000;fs=15000;t=0:1/fs:.01;

%original signalx1=5*cos(2*pi*f1*t); x2=2*cos(2*pi*f2*t); x3=.5*cos(2*pi*f3*t); Xw=x1+x2+x3;

%Chebyshev Type 1 highpass FilterWp = 5000/7500; %Passband corner frequencyWs = 4500/7500; %Stopband corner frequency Rp = 3; %Passband ripple, in decibelsRs = 40; %Stopband attenuation, in decibels

%Returns the lowest order n of the Chebyshev type I filter,%and the corresponding cutoff frequencies, Wn[n,Wn] = cheb1ord(Wp,Ws,Rp,Rs); [b,a] = cheby1(n,Rp,Wn,'high');[z,p,k]= cheby1(n,Rp,Wn,'high');

%Converts a discrete-time zero-pole-gain representation of a given digital%filter to an equivalent second-order section representation.[sos,g]=zp2sos(z,p,k);

%Discrete-time, second-order section, direct-form II filterHd = dfilt.df2sos(sos,g); y=filter(Hd,Xw); %Passing signal through filter

%Plotting%Frequency Domainfvtool(Hd);title('Chebyshev Type I Highpass Filter')fvtool(x1);title('Frequency domain X1');fvtool(x2);title('Frequency domain X2');fvtool(x3);title('Frequency domain X3');fvtool(Xw);title('Frequency domain X1+X2+X3');fvtool(y);title('Frequency domain Filtered Y');

%Time Domainsubplot(511);plot(t,x1,'r');hold on;stem(t,x1);title('Original X1=5*cos(2*pi*1000*t)');subplot(512);plot(t,x2,'r');hold on;stem(t,x2);title('Original X2=2*cos(2*pi*3000*t)');subplot(513);plot(t,x3,'r');hold on;stem(t,x3);title('Original X3=0.5*cos(2*pi*5000*t)');subplot(514);plot(t,Xw,'r');hold on;stem(t,Xw);title('X1+X2+X3=5cos(2*pi*1000*t)+2cos(2*pi*3000*t)+0.5cos(2*pi*5000*t)');subplot(515);plot(t,y,'r');hold on;stem(t,y);title('Filtered Y signal');

Figure 12: magnitude response

c) Filter validation:

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Figure 13: phase response

Figure 14: the magnitude and phase response of the filtered signal.

Figure 15: the filtered signal is almost the same with X3 signal with 5000Hz and magnitude of 0.5 which means that X1 and X2 is being filtered.

5) Design a BPF of butterworth filter to remove all frequencies except 3000Hz

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To design this filter so that only 3000Hz frequency remains, the passband needed is from 3000Hz to

3500Hz and the stop band is 500Hz different before and after the pass band. From the MATLAB code

generates, the order, n and new cutoff frequency, Wn is;

n = 9 and Wn = [0.3975 0.4693]

The Rp, ripple at the pass band is 3dB and the ripple at stop band is 80dB

a) MATLAB code

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close all;clear all;f1=1000;f2=3000;f3=5000;fs=15000;t=0:1/fs:.01; %original signalx1=5*cos(2*pi*f1*t); x2=2*cos(2*pi*f2*t); x3=.5*cos(2*pi*f3*t); Xw=x1+x2+x3; %butterworth Bandpass FilterWp = [3000 3500]/7500; %Passband corner frequencyWs = [2500 4000]/7500; %Stopband corner frequency Rp = 3; %Passband ripple, in decibelsRs = 80; %Stopband attenuation, in decibels %Returns the lowest order n of the butterworth filter,%and the corresponding cutoff frequencies, Wn[n wn]=buttord(Wp,Ws,Rp,Rs)[b a]=butter(n,wn)[z,p,k]= butter(n,wn);figure(1); plot([b a]); title('filter signal'); %Converts a discrete-time zero-pole-gain representation of a given digital%filter to an equivalent second-order section representation.[sos,g]=zp2sos(z,p,k); %Discrete-time, second-order section, direct-form II filterHd = dfilt.df2sos(sos,g); y=filter(Hd,Xw); %Passing signal through filter %plotting%magnitude and phase responsefvtool(Hd);title('Butterworth bandpass Filter')fvtool(y);title('Frequency domain Filtered Y'); %Time Domainsubplot(511);plot(t,x1,'r');hold on;stem(t,x1);title('Original X1=5cos(2*pi*1000*t)');subplot(512);plot(t,x2,'r');hold on;stem(t,x2);title('Original X2=2cos(2*pi*3000*t)');subplot(513);plot(t,x3,'r');hold on;stem(t,x3);title('Original X3=0.5cos(2*pi*5000*t)');subplot(514);plot(t,Xw,'r');hold on;stem(t,Xw);title('X1+X2+X3=5cos(2*pi*1000*t)+2cos(2*pi*3000*t)+0.5cos(2*pi*5000*t)');subplot(515);plot(t,y,'r');hold on;stem(t,y);title('Filtered Y signal');

b) Output waveform

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Figure 16: magnitude response

Figure 17: phase response

c) Filter validation

Figure 19: the filtered signal is almost the same with X2 signal with 3000Hz and magnitude of 2 which means that X1 and X3 is being filtered.

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Figure 18: filter signal

Alternative method:

To validate the filter designed, the same approach is use by using the same MATLAB code as below;

The output waveform for the signal that has been filtered out is;

From figure 21, it shows that the frequency at 3000Hz remains and other frequency were eliminated

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Figure 20: output signal after filtered Figure 21: magnitude spectrum of the signal

ts=0:1:100;x=5*cos(2*pi*1000/15000*ts)+2*cos(2*pi*3000/15000*ts)+0.5*cos(2*pi*5000/15000*ts);res=2*pi/120;w=0:res:2*pi-res;

bpf=[b a]

out=conv(x,bpf);figure(1);plot(out)

out_mag=abs(out);figure(2);plot(w(1:75).*15000/(2*pi),out_mag(1:75))