FFT of sound wave frequency graph

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Leon Ellis
Leon Ellis 2021-10-31
评论： Star Strider 2021-11-1
Good day, I am plotting the sound wave in the frequincy domain and applying a low pass filter in the time and frequincy domain. My problem is the last graph i get in the frequincy domain subplot(3,2,6). I want it to look similar to subplot(3,2,2) in the sense that it appears as the outline of the graph. (movmax does not solve this).
After this my job is to compare the audio file's frequincy domains and determine if a spesific word i.e "one" or "five" has been said by the user. Any hints and function that could come in handy?
My code is:
[CompareWord, Fs]=audioread("C:\Users\leone\OneDrive\Desktop\Year 2\Semester 2\EERI 222\Practical1\Sounds\WordL1_5.wav"); %%.mat file is attached.
Fs=8000;
Ts=1/Fs;
dt=(0:length(CompareWord)-1)*Ts;
nfft=length(CompareWord);
nfft2=2.^nextpow2(nfft);
ff=fft(CompareWord,nfft2);
ff=ff(1:nfft2/2);
ffm=movmax(ff,50);
xfft=Fs*(0:nfft2/2-1)/nfft2;
cut_off=1.2e3/Fs/2;
order=32;
h=fir1(order,cut_off);
fh=fft(h,nfft2);
fh=fh(1:nfft2/2);
mul=fh.*ff;
con=conv(CompareWord,h);
subplot(3,2,1);
plot(dt,CompareWord);
subplot(3,2,2);
plot(xfft,abs(ffm/max(ffm))); %#ok<ADPROPLC>
hold on;
%pks=findpeaks(abs(ffm));
%%Gets center x-coordinates of local maximum values.
TF2=islocalmax(abs(ffm),'FlatSelection',"center");
x=1:length(xfft);
hold on;
plot(x,abs(ffm)/max(abs(ffm)),x(TF2),abs(ffm(TF2)/max(ffm)),'r*');
hold off;
subplot(3,2,3);
stem(h);
subplot(3,2,4);
plot(abs(fh/max(fh))); %#ok<ADPROPLC>
subplot(3,2,5);
plot(con);
subplot(3,2,6);
plot(abs(mul));

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采纳的回答

Star Strider
Star Strider 2021-10-31
I have no idea what the desired result actually is, even when I run the code and plot it.
See if the envelope function will produce the desired result. It has additional options and arguments.
.
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Leon Ellis
Leon Ellis 2021-11-1
So instead of mul=fh.*ff; I used conv(fh,ff);
Star Strider
Star Strider 2021-11-1
Multiplying them in the frequency domain is equivalent to convolving them in the time domain, and convolving them in the frequency domain is equivalent to multiplying them in the time domain, if I remember correctly.
I am not certain what the desired result is here.
.

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