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How to calculate SNR of an exponential sine sweep?

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I have an exponential sine sweep and I want to calculate its SNR. How can this be done
  2 Commenti
Dimitris Kalogiros
Dimitris Kalogiros il 26 Lug 2018
Hi Trishita.
I think you have to give some more info about your problem.
TRISHITA BANERJEE
TRISHITA BANERJEE il 26 Lug 2018
%Testing using both interleaved and overlapped signal
% close all clear all; clc
fs = 44100;
e=3; % no of interleaved system
K=3; % no of harmonic impulse response
f_start = 55;%553
f_end = fs/2;%24100; %Possible values 20k and 24k
T =3;
A = 0.04;%0.025; for 0.6m%0.6;%for 1m distance%0.7;%for1.5m
L1 =9350; % for -70 db
L2 =2250; % for -70 db
W1=2*pi*f_start/fs;
W2=2*pi*f_end/fs;
tau_2=15300;
T_1 =(((e-1)*L1)+L2).*log(W2/W1)./log(2);
tau_k= T_1/log(W2/W1).*log(K);
siglen_MESM = T_1;
x_m = A*sin(W1*siglen_MESM/log(W2/W1).*(exp(((0:siglen_MESM)./siglen_MESM)*log(W2/W1))-1)); %Multiple Sine Sweep
% t_i = ceil(L1*fs); % t_o = ceil((T_1/log(W2/W1)).*log(K)*fs); % t_i = ceil(L1); % t_o = ceil(L1+ tau_k ); imax=5; t = delay (imax, L1 ,L2, tau_k, tau_2); t_max=max(t);
x = zeros(length(x_m)+t_max+fs,imax);
for n = 1:imax
x(:,n) = [zeros(t(n)+fs,1);x_m';zeros(t_max -t(n),1)];
end
x1 = x(:,1);
x2 = x(:,2);
x3 = x(:,3);
x4 = x(:,4);
x5 = x(:,5);
y_m = x1+x2+x3+x4+x5;% Simulated Measurement
y_m=y_m';
%Testing using both interleaved and overlapped signal
% close all clear all; clc
fs = 44100;
e=3; % no of interleaved system
K=3; % no of harmonic impulse response
f_start = 55;%553
f_end = fs/2;%24100; %Possible values 20k and 24k
T =3;
A = 0.04;%0.025; for 0.6m%0.6;%for 1m distance%0.7;%for1.5m
L1 =9350; % for -70 db
L2 =2250; % for -70 db
W1=2*pi*f_start/fs;
W2=2*pi*f_end/fs;
tau_2=15300;
T_1 =(((e-1)*L1)+L2).*log(W2/W1)./log(2);
tau_k= T_1/log(W2/W1).*log(K);
siglen_MESM = T_1;
x_m = A*sin(W1*siglen_MESM/log(W2/W1).*(exp(((0:siglen_MESM)./siglen_MESM)*log(W2/W1))-1)); %Multiple Sine Sweep
% t_i = ceil(L1*fs); % t_o = ceil((T_1/log(W2/W1)).*log(K)*fs); % t_i = ceil(L1); % t_o = ceil(L1+ tau_k ); imax=5; t = delay (imax, L1 ,L2, tau_k, tau_2); t_max=max(t);
x = zeros(length(x_m)+t_max+fs,imax);
for n = 1:imax
x(:,n) = [zeros(t(n)+fs,1);x_m';zeros(t_max -t(n),1)];
end
x1 = x(:,1);
x2 = x(:,2);
x3 = x(:,3);
x4 = x(:,4);
x5 = x(:,5);
y_m = x1+x2+x3+x4+x5;% Simulated Measurement
y_m=y_m';
xinv = fliplr(x_m) .* (W2/W1).^(-(0:siglen_MESM)/siglen_MESM); %Inverse sweep
f_t = f_start*exp(((0:siglen_MESM)./siglen_MESM)*log(W2/W1));
[h1, h2, h1_nl, h2_nl] = farina_deconvolution2(x_m,y_m,xinv,17.6584,1 );
figure subplot(2,1,1); spectrogram(h1,1024,1000,1024,fs,'yaxis'); title('Linear Impulse Response H1');ylabel('h1 \rightarrow dB') subplot(2,1,2); spectrogram(h2,1024,1000,1024,fs,'yaxis'); title('Linear Impulse Response H2');ylabel('h2 \rightarrow dB')
Here x_m is my desired signal...And when I calculate the h1 that is the impulse response of my signal i usually take the peaks above -70db.So I can consider anything above -70db is original signal and less than -70db is noise.. So how can I approach to find SNR of this system

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