FFT of sound wave frequency graph

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));

Answer 1

[Star Strider]

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]

Thanks, will have a look

[Star Strider]

My pleasure!

.

[Leon Ellis]

Ok so I updated the code to work noww, and the problem is, my output after the low pass filter consists of MANY lines for some reason. It's I want to produce the same as the figure top right (Ignore the red dots btw, I know they are wrong). the envelope works but it doesn't erase all the lines that cause the graph. I think I'll start looking at it tomorrow again. I'll keep you whoever else is interested updated. Thanks again

[Star Strider]

My pleasure.

The lines are because ther are several peaks in the signal, and envelope outlines all of them, since I thought that was the desired result.

[Leon Ellis]

So instead of mul=fh.*ff; I used conv(fh,ff);

[Star Strider]

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.

.