Envelope extraction of a time domain signal

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Lijie Zhao
Lijie Zhao 2018-1-24
回答： Mark Schwab 2018-1-26
I have a signal like this: t=[-10*10^(-9):0.0001*10^(-9):10*10^(-9)]; dt=394*10^(-15); betaL=1.2378*10^(-21); tao=15*10^(-12); i=0.5*exp(-(t./betaL.*dt).^2./(4*log(2))).*(1+cos(tao.*t./betaL)); I want to get its envelop so I tried Hilbert Transformation in this way: Y = hilbert(i); am = abs(Y); but it is not what I want, I want the envelope that goes through the peaks smoothly. Is there anybody know about this problem? Thank you very much!
CODE: t=[-10*10^(-9):0.0001*10^(-9):10*10^(-9)]; dt=394*10^(-15); betaL=1.2378*10^(-21); tao=15*10^(-12); i=0.5*exp(-(t./betaL.*dt).^2./(4*log(2))).*(1+cos(tao.*t./betaL)); Y = hilbert(i); am = abs(Y); plot(t,i,t,am)

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Mark Schwab
Mark Schwab 2018-1-26
If you are not concerned with the length of the envelope signal you can use the "findpeaks" function.
[pks, locs] = findpeaks(i); % Where pks is a vector with amplitude of peaks and locs is a vector of peak indeces
envelope = interp(pks,floor(length(i)/length(pks)); % Interpolate peaks to vector close in size to i
Please note that the interpolation function only works with integer multiples of length so this will not produce an envelope vector equal in size to the original signal. You can refer to the following MATLAB answers post for more information on this:
https://www.mathworks.com/matlabcentral/answers/338364-interpolate-using-interp-with-a-decimal-number
Also, I have included a link to the findpeaks function documentation for your convenience:
https://www.mathworks.com/help/signal/ref/findpeaks.html

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