hi very valuable group members. I wish you all a good day. I ask you to help Hilbert Huan transformation point. I have two different nonlinear signals. I obtained imfs for these signals. I need the matlab code for the hilbert huang transform of these
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EXAMPLE SİGNAL 1 CODE
>> nsamples=10000;
fs=1e3;
time=(1:nsamples-1)/fs;
signal=0.5*time+sin(pi*time)+sin(2*pi*time)+sin(6*pi*time);%
>> plot(signal)
>>
EXAMPLE SİGNAL 1 RESULT
EXAMPLE SİGNAL 1 İMF CODE
>> clear all
close all
nsamples=10000;
fs=1e3;
time=(1:nsamples-1)/fs;
signal=0.5*time+sin(pi*time)+sin(2*pi*time)+sin(6*pi*time);%
threshouldSifting=0.1;
monotone=0;
indexIMF=0;
signalmem=signal;
while (not(monotone))
h=signal;
SD=1;
while(SD>threshouldSifting)
%% upper
[pksMax,locsMax] = findpeaks(h);
pksMax=[h(1), pksMax, h(end)];
locsMax=[1, locsMax, length(h)];
upperSpline=spline(time(locsMax),pksMax);
spmax=ppval(upperSpline,time);
%% lower
[pksMin,locsMin] = findpeaks(-1*h);
pksMin=-1*pksMin;
pksMin=[h(1), pksMin, h(end)];
locsMin=[1, locsMin, length(h)];
lowerSpline=spline(time(locsMin),pksMin);
spmin=ppval(lowerSpline,time);
%% evaluate the mean
mspline=(spmax+spmin)/2;
hprec=h;
h=h-mspline;
SD=sum((hprec-h).^2)/sum(hprec.^2); %criterio di stop proposto da by Huang et al. (1998)
end
indexIMF=indexIMF+1;
IMF(indexIMF,:)=h;
signal=signal-h;
monotone=all(diff(signal)>0)||all(diff(signal)<0);
end
subplot(indexIMF+2,1,1), plot(time,signalmem),title('Segnale di ingresso'),xlabel('Tempo [s]');
for index=1:indexIMF
subplot(indexIMF+2,1,index+1), plot(time,IMF(index,:)),title(['IMF_',num2str(index)]),xlabel('Tempo [s]');
end
subplot(indexIMF+2,1,indexIMF+2), plot(time,signal),title('Residuo'),xlabel('Tempo [s]');
>>
>>
EXAMPLE SİGNAL 1 IMF RESULT
HELP TOPİC
WHAT I WANT : HİLBERT HUANG TRANSFORM OF THİS SİGNAL AND RESULT GRAPHS. SO; FREQUENCY , TİME AND ENERGY RESULTS.
MATLAM CODE AND GRAPHS ????
………………………………………………………………………………………………………..
EXAMPLE SİGNAL 2 CODE
>> Fs= 10000;
Ts=2/Fs;
L=150;
t=(0:L-1);
x=36*sin(2*pi*50*t)+sin(2*pi*120*t);
plot(t,x);
>>
EXAMPLE 2 SİGNAL RESULT
EXAMPLE 2 SİGNAL İMF CODES
>> clear all
close all
nsamples=10000;
fs=1e3;
time=(1:nsamples-1)/fs;
signal=36*sin(2*pi*50*time)+sin(2*pi*120*time);
%
threshouldSifting=0.1;
monotone=0;
indexIMF=0;
signalmem=signal;
while (not(monotone))
h=signal;
SD=1;
while(SD>threshouldSifting)
%% upper
[pksMax,locsMax] = findpeaks(h);
pksMax=[h(1), pksMax, h(end)];
locsMax=[1, locsMax, length(h)];
upperSpline=spline(time(locsMax),pksMax);
spmax=ppval(upperSpline,time);
%% lower
[pksMin,locsMin] = findpeaks(-1*h);
pksMin=-1*pksMin;
pksMin=[h(1), pksMin, h(end)];
locsMin=[1, locsMin, length(h)];
lowerSpline=spline(time(locsMin),pksMin);
spmin=ppval(lowerSpline,time);
%% evaluate the mean
mspline=(spmax+spmin)/2;
hprec=h;
h=h-mspline;
SD=sum((hprec-h).^2)/sum(hprec.^2); %criterio di stop proposto da by Huang et al. (1998)
end
indexIMF=indexIMF+1;
IMF(indexIMF,:)=h;
signal=signal-h;
monotone=all(diff(signal)>0)||all(diff(signal)<0);
end
subplot(indexIMF+2,1,1), plot(time,signalmem),title('Segnale di ingresso'),xlabel('Tempo [s]');
for index=1:indexIMF
subplot(indexIMF+2,1,index+1), plot(time,IMF(index,:)),title(['IMF_',num2str(index)]),xlabel('Tempo [s]');
end
subplot(indexIMF+2,1,indexIMF+2), plot(time,signal),title('Residuo'),xlabel('Tempo [s]');
>>
>>
EXAMPLE 2 İMF RESULTS
HELP TOPİC
WHAT I WANT : HİLBERT HUANG TRANSFORM OF THİS SİGNAL AND RESULT GRAPHS. SO; FREQUENCY , TİME AND ENERGY RESULTS.
MATLAM CODE AND GRAPHS ????
………………………………………………………………………..
SAME THİS ARTİCLE APPLİCATİON
MAİL : sukruozdemir37@gmail.com
27 commentaires
Walter Roberson
le 8 Nov 2019
The imf (intrinsic mode functions) is what you would get as output of HHT, so you would appear to have already done the work.
Is there are reason you did not use edm and hht ?
Réponses (1)
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