Matlab R2015b snr output doesn't shows anything
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This following code is a simulation on FxLMS, developed by Agustinus Oey. I wanted to find the SNR of it. so, if i add snr command
SNR_1 = snr(Yd-e_cont,Yd);
at the end, it doesn't shows any output. what am i missing?
% Developed by Agustinus Oey <oeyaugust@gmail.com>
% Center of Noise and Vibration Control (NoViC)
% Department of Mechanical Engineering
% Korea Advanced Institute of Science and Technology (KAIST)
% Daejeon, South Korea
%--------------------------------------------------------------------------
% Set simulation duration (normalized)
clear
T=1000;
% We do not know P(z) and S(z) in reality. So we have to make dummy paths
Pw=[0.01 0.25 0.5 1 0.5 0.25 0.01];
Sw=Pw*0.25;
% Remember that the first task is to estimate S(z). So, we can generate a
% white noise signal,
x_iden=randn(1,T);
% send it to the actuator, and measure it at the sensor position,
y_iden=filter(Sw, 1, x_iden);
% Then, start the identification process
Shx=zeros(1,16); % the state of Sh(z)
Shw=zeros(1,16); % the weight of Sh(z)
e_iden=zeros(1,T); % data buffer for the identification error
% and apply least mean square algorithm
mu=0.1; % learning rate
for k=1:T, % discrete time k
Shx=[x_iden(k) Shx(1:15)]; % update the state
Shy=sum(Shx.*Shw); % calculate output of Sh(z)
e_iden(k)=y_iden(k)-Shy; % calculate error
Shw=Shw+mu*e_iden(k)*Shx; % adjust the weight
end
% Lets check the result
subplot(2,1,1)
plot([1:T], e_iden)
ylabel('Amplitude');
xlabel('Discrete time k');
legend('Identification error');
subplot(2,1,2)
stem(Sw)
hold on
stem(Shw, 'r*')
ylabel('Amplitude');
xlabel('Numbering of filter tap');
legend('Coefficients of S(z)', 'Coefficients of Sh(z)')
% The second task is the active control itself. Again, we need to simulate
% the actual condition. In practice, it should be an iterative process of
% 'measure', 'control', and 'adjust'; sample by sample. Now, let's generate
% the noise:
X=randn(1,T);
% and measure the arriving noise at the sensor position,
Yd=filter(Pw, 1, X);
% Initiate the system,
Cx=zeros(1,16); % the state of C(z)
Cw=zeros(1,16); % the weight of C(z)
Sx=zeros(size(Sw)); % the dummy state for the secondary path
e_cont=zeros(1,T); % data buffer for the control error
Xhx=zeros(1,16); % the state of the filtered x(k)
% and apply the FxLMS algorithm
mu=0.1; % learning rate
for k=1:T, % discrete time k
Cx=[X(k) Cx(1:15)]; % update the controller state
Cy=sum(Cx.*Cw); % calculate the controller output
Sx=[Cy Sx(1:length(Sx)-1)]; % propagate to secondary path
e_cont(k)=Yd(k)-sum(Sx.*Sw); % measure the residue (main path - secondary path)
Shx=[X(k) Shx(1:15)]; % update the state of Sh(z) %%previously Shx=[x_iden(k) Shx(1:15)];
Xhx=[sum(Shx.*Shw) Xhx(1:15)]; % calculate the filtered x(k),new Shx, old SSShw
Cw=Cw+mu*e_cont(k)*Xhx; % adjust the controller weight
end
% Report the result
figure
subplot(2,1,1)
plot([1:T], e_cont)
ylabel('Amplitude');
xlabel('Discrete time k');
legend('Noise residue')
subplot(2,1,2)
plot([1:T], Yd)
hold on
plot([1:T], Yd-e_cont, 'r:')
ylabel('Amplitude');
xlabel('Discrete time k');
legend('Noise signal', 'Control signal')
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回答(1 个)
Jerome Blair
2022-7-7
You have a semicolon after the command. That tells MATLAB to not show teh answere.
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