How to apply Hilbert Huang transfer method on the signal Sc(t):

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Jan Ali
Jan Ali el 27 de Oct. de 2020
Comentada: Jan Ali el 16 de En. de 2021
How to run Hilbert spectrum of the following function(signal):

Respuestas (1)

Ayush Bansal
Ayush Bansal el 30 de Oct. de 2020
Editada: Ayush Bansal el 30 de Oct. de 2020
Create a t vector varying from 0 to 0.3s with sampling rate fs
t=[0:1/fs:0.3];
[value,idx1]=min(abs(t-0.1)); which will give index for ta
Similarly find index for tb and tc.
Create a signal vector of size t;
Signal=zeros(length(t));
Then define signal for 0-ta,ta-tb and tb-tc.
Use emd(x) to find intrinsic mode function.
Use hht(imf) to find Hilbert spectrum where imf is intrinsic mode function.
  2 comentarios
Jan Ali
Jan Ali el 19 de Nov. de 2020
Editada: Jan Ali el 19 de Nov. de 2020
Thanks a lot Ayush, as I am new with Matlab would you please write the script for the function given above?
Jan Ali
Jan Ali el 16 de En. de 2021
Hi Ayush,
This pice of code: [value,idx1]=min(abs(t-0.1)); is difficult for me to realize.
Also I don't know how to generate a signal in this style. However, I have created my own signal but I don't get the result as expected:
fs=2000;
t = 0:0.0001:0.3; % defining time axis
ut1=t>=0; % unitsetp1 takes 1 only when t is greater than equal to 0
ut2=t>=0.1; % unitsetp2 takes 1 only when t is greater than equal to 0.1
ut3=t>=0.18;% unitsetp3 takes 1 only when t is greater than equal to 0.18
ut4=t>=0.3; % unitsetp4 takes 1 only when t is greater than equal to 0.3
t1=ut1-ut2; % t1 takes 1 only when 0<=t<0.1
t2=ut2-ut3; % t2 takes 1 only when 0.1<=t<0.18
t3=ut3-ut4; % t2 takes 1 only when 0.18<=t<0.3
s1=sin(2*pi*50*t)+0.2*sin(2*pi*250*t);
s2=sin(2*pi*50*t)+0.2*sin(2*pi*250*t)+sin(2*pi*500*t).*exp(-30*t-0.1);
% s2=sin(2*pi*50*t)+0.2*sin(2*pi*250*t)+sin(2*pi*500*t).*exp(-30*t2-0.1)+0.7*rand(size(t2));
s3=sin(2*pi*50*t)+0.2*sin(2*pi*250*t);
s = s1.*t1 + s2.*t2 + s3.*t3;
plot(t,s);
I would highly appreciate if you could help me where I am wrong and how to acheive the true result.
Thanks in advance,

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