Design of a low pass FIR filter to remove the white noise that has corrupted the message received

MR=load('MessageReceived.mat'); testData=MR.UNKNOWNSIGNAL(); Msig=abs(testData); t=1:length(Msig); figure(1) plot(Msig) grid on title('Corrupted Received Message y(n)') %Designing the Low Pass filter for FIR Mb=fir1(42,fc,chebwin(43,30) figure(2) freqz(Mb,1,Mb;length(

DIGITAL SIGNAL PROCESSING
NEE3208 SIGNAL PROCESSING
ASSIGNMENT 3
STUDENT NAME
STUDENT ID NUMBER
INSTRUCTOR (PROFESSOR/TUTOR)
DATE OF SUBMISSION

TASK
Design of a low pass FIR filter to remove the white noise that has corrupted the message
received. The FIR filter is of the form,
y ( n )= ∑
i=0
M−1
h ( i ) x ( n−i )
Part 1
Designing the kernel filter h(n) by defining its impulse response.
The non-recursive finite impulse response is given as,
y ( n )=h ( 0 ) x ( n ) +h ( 1 ) x ( n−1 )+ …+h ( N ) x ( n−N )
N−order of the filter
For a finite duration, the impulse response is given as,
To obtain the ideal low pass filter,
  c
sin1
( ) sinc
2
c
c
cj n c
d
n
h n Ae d A A n
n
 

  


  


 
    
 


For windowing,
hw ( n ) =hd ( n ) w ( n )
¿ 1
2 π Hd ( ω )∗W ( ω )
The impulse response of the FIR filter, following the Hamming window, is, therefore, given as,
h [n]= sin ( ωc ( n−L ) )
π ( n−L ) w [ n ]
Part 2
Evaluate the transfer function using DFT. Magnitude of the transfer functions.
Hd ( ω )=DTFT {hd ( n ) }= ∑
n=−∞
+ ∞
hd ( n ) e− jωn
Part iii
Apply the filter in the frequency domain to X(k). Obtaining the inverse DFT to get filtered signal
and determining the cut-off frequency
hd ( n )=IDTFT { Hd ( ω ) }
¿ 1
2 π 
−ωc
+ωc
A e− jωn dω
¿ A ( sin ( ωc n )
πn )
¿ A ( ωc
π )sinc ( ωc
π n )
Part iv
Applying the filter in the time domain to obtain the filtered signal y(n)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Normalized Frequency ( rad/sample)
-2000
-1500
-1000
-500
0
Phase (degrees)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Normalized Frequency ( rad/sample)
-100
-50
0
Magnitude (dB)
0 5 10 15
105
0
5000
10000
15000
Original Signal
0 5 10 15 20 25 30 35 40 45
Time(s)
-0.2
0
0.2
0.4
0.6 LowPass Filtered Signal

0 10 20 30 40 50 60 70 80 90
-5
0
5
10
15
20
25
30
35
40
45 Impulse response of the Message After Filtering
Discussion
Filters are used to remove disturbances; in this case, the message signal is corrupted by white
noise during transmission. At the receiver end, the message received is distorted by the noise and
a low pass filter FIR is designed to remove the white noise. Let the low pass filter have
ω p=0.4 π , ωs=0.6 π
Attenuation=50 dB(stopband )
The transition band,
ωs−ωp=0.2 π
Using the Hamming window,

8 π
M , M =40
The window is given as,
ω [ n ] = {0.54−0.46 cos ( 2 nπ
M ) 0≤ n ≤ M
0 otherwise }
h [ n ] = [ 0.54−0.46 cos ( 2 nπ
M ) ] [ 0.5 sinc n−20
2 ] 0 ≤ n ≤ M
The MATLAB implementation,
%% ASSIGNMENT 3
%VICTORIAL UNIVERSITY-MELBOURNE AUSTRALIA
%NEE3208 SIGNAL PROCESSING
%introduction
clear
close all
format short
clc
%% Initializing several parameters
N=10; %filter order- length of the test input signal
fs=36e2; %sample frequency
fc=0.48; %Cut off frequency
Ap=0.01; %pass band ripple
Ast=80; %stop band attenuation
% loading the data.
MR=load('ReceivedMessage.mat');
testData=MR.UNKNOWNSIGNAL();
Msig=abs(testData);
t=1:length(Msig);
figure(1)
plot(Msig)
grid on
title('Corrupted Received Message y(n)')
%Designing the Low Pass filter for FIR
Mb=fir1(42,fc,chebwin(43,30));
Outr=filter(Mb,1,length(Mb));
figure(1)
subplot(2,1,1)

plot(t,Msig,'b-*');
title('Original Signal');
grid on
subplot(2,1,2)
plot(Mb,'r-*')
grid on
title('LowPass Filtered Signal')
xlabel('Time(s)')
%Using the window size
figure(2)
freqz(Mb,1);
figure(3)
te=1:length(Mb);
hy=conv(Mb,te);
plot(hy)
grid on
title('Impulse response of the Message After Filtering')