How to remove awgn from and ecg signal?

조회 수: 3 (최근 30일)
v shikhashastri
v shikhashastri 2021년 5월 12일
댓글: Star Strider 2021년 5월 14일
I'am doing this project on removing both PLI and AWGN in the ecg. I have downloaded the ecg code for generating the signal from matlab. I tried using Chebyshev2 and the Wiener filter to filter out the two noises. But that makes it even worse. Please to help me .I'll provide my code :
Fs = 1000;
Ts = 1/Fs;
x=0:Ts:10-Ts;
li=30/72;
a_pwav=0.25;
d_pwav=0.09;
t_pwav=0.16;
a_qwav=0.025;
d_qwav=0.066;
t_qwav=0.166;
a_qrswav=1.6;
d_qrswav=0.11;
a_swav=0.25;
d_swav=0.066;
t_swav=0.09;
a_twav=0.35;
d_twav=0.142;
t_twav=0.2;
a_uwav=0.035;
d_uwav=0.0476;
t_uwav=0.433;
pwav=p_wav(x,a_pwav,d_pwav,t_pwav,li);
%qwav output
qwav=q_wav(x,a_qwav,d_qwav,t_qwav,li);
%qrswav output
qrswav=qrs_wav(x,a_qrswav,d_qrswav,li);
%swav output
swav=s_wav(x,a_swav,d_swav,t_swav,li);
%twav output
twav=t_wav(x,a_twav,d_twav,t_twav,li);
%uwav output
uwav=u_wav(x,a_uwav,d_uwav,t_uwav,li);
%ecg output
ecg=pwav+qrswav+twav+swav+qwav+uwav;
figure(1)
plot(x,ecg);
xlabel('Time[s]')
ylabel('Amplitude')
title('ECG')
%pli
fNoise = 50; % Frequency [Hz]
aNoise = 1; % Amplitude
noise = aNoise*sin(2*pi.*x.*fNoise);
pliecg=ecg + noise;
%adding awgn
SNR=0;
noisy = awgn(pliecg,SNR,'measured')
figure(2)
plot(x,noisy)
xlabel('Time[s]')
ylabel('Amplitude')
title('Noisy')
function [pwav]=p_wav(x,a_pwav,d_pwav,t_pwav,li)
l=li;
a=a_pwav;
x=x+t_pwav;
b=(2*l)/d_pwav;
n=100;
p1=1/l;
p2=0;
for i = 1:n
harmonic1=(((sin((pi/(2*b))*(b-(2*i))))/(b-(2*i))+(sin((pi/(2*b))*(b+(2*i))))/(b+(2*i)))*(2/pi))*cos((i*pi*x)/l);
p2=p2+harmonic1;
end
pwav1=p1+p2;
pwav=a*pwav1;
end
function [qrswav]=qrs_wav(x,a_qrswav,d_qrswav,li)
l=li;
a=a_qrswav;
b=(2*l)/d_qrswav;
n=100;
qrs1=(a/(2*b))*(2-b);
qrs2=0;
for i = 1:n
harm=(((2*b*a)/(i*i*pi*pi))*(1-cos((i*pi)/b)))*cos((i*pi*x)/l);
qrs2=qrs2+harm;
end
qrswav=qrs1+qrs2;
end
function [qwav]=q_wav(x,a_qwav,d_qwav,t_qwav,li)
l=li;
x=x+t_qwav;
a=a_qwav;
b=(2*l)/d_qwav;
n=100;
q1=(a/(2*b))*(2-b);
q2=0;
for i = 1:n
harm5=(((2*b*a)/(i*i*pi*pi))*(1-cos((i*pi)/b)))*cos((i*pi*x)/l);
q2=q2+harm5;
end
qwav=-1*(q1+q2);
end
function [swav]=s_wav(x,a_swav,d_swav,t_swav,li)
l=li;
x=x-t_swav;
a=a_swav;
b=(2*l)/d_swav;
n=100;
s1=(a/(2*b))*(2-b);
s2=0;
for i = 1:n
harm3=(((2*b*a)/(i*i*pi*pi))*(1-cos((i*pi)/b)))*cos((i*pi*x)/l);
s2=s2+harm3;
end
swav=-1*(s1+s2);
end
function [twav]=t_wav(x,a_twav,d_twav,t_twav,li)
l=li;
a=a_twav;
x=x-t_twav-0.045;
b=(2*l)/d_twav;
n=100;
t1=1/l;
t2=0;
for i = 1:n
harm2=(((sin((pi/(2*b))*(b-(2*i))))/(b-(2*i))+(sin((pi/(2*b))*(b+(2*i))))/(b+(2*i)))*(2/pi))*cos((i*pi*x)/l);
t2=t2+harm2;
end
twav1=t1+t2;
twav=a*twav1;
end
function [uwav]=u_wav(x,a_uwav,d_uwav,t_uwav,li)
l=li;
a=a_uwav
x=x-t_uwav;
b=(2*l)/d_uwav;
n=100;
u1=1/l
u2=0
for i = 1:n
harm4=(((sin((pi/(2*b))*(b-(2*i))))/(b-(2*i))+(sin((pi/(2*b))*(b+(2*i))))/(b+(2*i)))*(2/pi))*cos((i*pi*x)/l);
u2=u2+harm4;
end
uwav1=u1+u2;
uwav=a*uwav1;
end
  댓글 수: 16
v shikhashastri
v shikhashastri 2021년 5월 14일
Yes yes... Sure.Thank you.
Star Strider
Star Strider 2021년 5월 14일
My pleasure!

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