How to find spectrum envelope from wav file
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How do I measure the energy change of the spectrum envelope 20 - 40 hz from from a wav file and plot it to graph? my recording is in 8000 sampling rate 16bits mono.
I would like to get something like this
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Thanks for great help, wish you all merry Christmas and happy new year :)
1 件のコメント
Image Analyst
2014 年 12 月 28 日
編集済み: Image Analyst
2014 年 12 月 28 日
Daemian, perhaps you overlooked this big hint given in Star's answer "I don’t have your signal, so I can’t test the filter I designed for you with it." HINT, HINT. What do you think you should do now?
回答 (2 件)
Star Strider
2014 年 12 月 26 日
If you have the Signal Processing Toolbox, such a bandpass filter is easy to design. First (to make things easier) convert your 16-bit signed integer signal (that I call ‘x’ here) to double:
xd = double(x); % Convert to ‘double’
Fs = 8000; % Sampling Frequency
Fn = Fs/2; % Nyquist Frequency
Fpb = [20 40]/Fn; % Passband
Fsb = [15 50]/Fn; % Stopband
[n,Wn] = buttord(Fpb, Fsb, 1, 10); % Filter Order, With Rp = 1, Rs = 10
[b,a] = butter(n,Wn); % Create Filter Transfer Function
[sos,g] = tf2sos(b,a); % Second-Order-Section Implementation
figure(1)
freqz(b,a) % Transfer Function Plot
figure(2) % Second-Order-Section Plot
freqz(sos)
yd = filtfilt(sos, g, xd); % Filter ‘xd’ To Get ‘y’
where ‘xd’ is your .wav signal and ‘yd’ is the filtered output. The filtfilt function will filter both channels of your signal at the same time (assuming your mono signal could have two channels, with the same information in both channels), so this code will work regardless of its having 1 or 2 channels. You can then change ‘yd’ to 16-bits with the int16 function if you want to.
Have fun with this!
Merry Christmas (belatedly) and happy New Year to you, too!
15 件のコメント
Youssef Khmou
2014 年 12 月 28 日
編集済み: Youssef Khmou
2014 年 12 月 28 日
Hilbert Transformation is used to obtain the envelope of signal, here is an example :
The envelope is decreasing exponential,
Fs=80;
F=10;
t=0:1/Fs:4-1/Fs;
x=exp(-t).*real(exp(j*2*pi*F*t));
figure; plot(t,x);
Y=abs(hilbert(x));
hold on;
plot(t,Y,'r');
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fx=fftshift(abs(fft(x))); fx=fx(floor(end/2:end));
fY=fftshift(abs(fft(Y))); fY=fY(floor(end/2:end));
figure; plot(fx); hold on
plot(fY,'r')
2 件のコメント
Image Analyst
2014 年 12 月 30 日
Cool - I didn't know hilbert() did that. Does it always get the envelope no matter what x looks like (like how fast it oscillates)? Why does it get a little squirrely around t=4?
Youssef Khmou
2014 年 12 月 30 日
編集済み: Youssef Khmou
2014 年 12 月 30 日
According to theory yes, oscillation in borders can be interpreted as Gibbs effect.
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