why wavelet doesn't show accurate results always?
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I have a signal that contains some modes in it. I've run a CWT transform but I haven't got my desired results. so I implement a test signal that contains freqs [0.1,0.3,0.6] Hz to see whats the problem with my codes. but the result was accurate. once I change the freqs to [0.1,0.4,0.6] or [0.1,0.5,0.6]Hz,The results wasn't accurate anymore on the plot. you can see the plot here, http://cubeupload.com/im/4msYBt.jpg.It might be for the damping effect of the 0.6Hz mode or ... I don't know actually :( I think I must fix why this happening first to get my actual signal modes. here's my codes.
t=linspace(0,30,300);
Fs=ceil(inv(t(2)-t(1))); % sampling freq
x=sin(2*pi*t*0.1).*(t<10)+sin(2*pi*t*0.3).*...
(t<30)+sin(2*pi*t*0.6).*(t<10).*exp(-t*.1); % my signal[0.1Hz,0.3Hz,0.6Hz]
wname = 'morl'; % define wavelet name
scales = 1:1:128; % scales range
coefs = cwt(x,scales,wname,'lvlabs'); % Get coefs of x
freq = scal2frq(scales,wname,1/Fs); %convert scales to freq range
surf(t,freq,abs(coefs));shading('interp'); % 3D surface plot
axis tight; xlabel('Seconds'); ylabel... % seting the axis 3D surface
('Pseudo-Frequency (Hz)'); % seting the axis 3D surface
axis([0 length(t)/Fs 0 1 0 max(coefs(:))*1.1]) % seting the axis 3D surface
figure;
sc=wscalogram('image',coefs,'scales',freq,'ydata',x); % get scalograme of x
xlabel('Time'); ylabel('Frequency of gen'); % set axis
hold on
abscof=abs(coefs)'; % |coefs|'
modI=max(abscof); % get max |coefs| coresponde to freqs of x
modI=modI/max(modI); % scale each clumn 0-1
figure;
plot(freq,modI) % Plot all modes. contain all
grid on
axis([0 1 0 max(modI)*1.1]) % seting the axis
xlabel('Pseudo-Frequency (Hz)'); ylabel('abs(coefs)'); % seting the axis
4 comentarios
Jan
el 4 de Ag. de 2013
Editada: Jan
el 4 de Ag. de 2013
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Respuesta aceptada
Wayne King
el 4 de Ag. de 2013
I'm not quite sure why you say this:
t=linspace(0,30,300);
Fs=ceil(inv(t(2)-t(1)));
x=sin(2*pi*t*0.1).*(t<10)+sin(2*pi*t*0.4).*...
(t<30)+sin(2*pi*t*0.6).*(t<10).*exp(-t*.1);
scales = 10:0.1:150;
wname = 'morl';
coefs = cwt(x,scales,wname);
freq = scal2frq(scales,wname,dt);
contour(t,freq,abs(coefs));shading('interp');
axis tight;
xlabel('Time'); ylabel('Pseudo-frequency (Hz)');
You have to keep in mind that the wavelet yields a bandpass analysis.
Más respuestas (1)
Wayne King
el 4 de Ag. de 2013
Editada: Wayne King
el 4 de Ag. de 2013
Wavelets have bandwidth, they are not like the complex exponentials of Fourier analysis. The larger the scale, the narrow the bandwidth of the analyzing wavelet in the Fourier domain.
Write down the expression for the wavelet transform in the Fourier domain, you'll see that it is a bandpass filtering of the signal with the bandwidth inversely proportional to scale. The actual bandwidth of course depends on the analyzing wavelet.
You are going to have difficulty with a signal where you want wavelet analysis to pick up the difference between 0.4 and 0.6 Hz.
Look at how well the wavelet analysis does when the separation between 0.4 and 0.6 is increased to 0.4 and 1 Hz and you actually include two periods of the 0.1 Hz wave.
t=linspace(0,30,300);
Fs=ceil(inv(t(2)-t(1)));
x=sin(2*pi*t*0.1).*(t<20)+sin(2*pi*t*0.4).*...
(t<30)+sin(2*pi*t*1).*(t<10).*exp(-t*.1);
scales = 5:0.5:150;
wname = 'morl';
coefs = cwt(x,scales,wname);
freq = scal2frq(scales,wname,dt);
contour(t,freq,abs(coefs));shading('interp');
axis tight; xlabel('Seconds'); ylabel...
('Pseudo-Frequency (Hz)');
set(gca,'ytick',[0.1 0.4 0.6 0.8 1 1.4])
But you can clearly see the effect of the bandwidth of the wavelet in the plot. Note how the oscillation at 0.4 Hz extends into the 0.6 region of frequency space. Further, note how the oscillation at 1 Hz extends further below and above 1 Hz than the effect at 0.4 and 0.1 Hz. That is because the bandwidth of the analyzing Morlet wavelet is larger at 1 Hz than at 0.4 or 0.1 Hz.
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