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I like to graph the frequency of 440 Hz and analyse it

I use the following code:

figure(1)

sf = 1000; %sampling

f = 440;

a = 0.5;

t = 0:1/sf:1;

x = a*sin(2*pi*f*t);

sound(x,sf)

figure(1)

plot(t,x) %amplitude-time graph

axis([0 1/f min(x) max(x)])

figure(2) %Fourier analysis

n = length(t);

xhat = fft(x,n);

PSD = xhat.*conj(xhat)/n;

freq = 1/(dt*n)*(0:n);

L = 1:floor(n/2);

colormap jet(20)

plot(freq(L),PSD(L),'LineWidth',2.5)

First of: why is it, that the Frequency of amp-time Graph is jagged? If I increase the sampling frequency, why is, that the actual frequency changing?

Mathieu NOE
on 19 Jan 2021

hi - official answer this time

the PSD concept is valid only for random type signals , not sinus

mathematically speaking, a sinus has a PSD of infinite amplitude and zero bandwith, simply because a sine wave has its energy concentrated only at one discrete frequency and not spread accross a wider freq range

if you test your code with random signals, you should see that the peak amplitude stays the same whatever the sampling frequency. But this does not hold if you test now with a sine wave

for a sine wave , the psd concept must be replaced by power or rms or peak amplitude

in summary , a FFT analyser will operate this way :

it will first compute the ftt.*conj(fft) power spectrum

if you ask for "PSD" (knowing that you are looking at random type signals) , it will do : PSD = power / nfft (as you did)

if you ask for "RMS" (knowing that you are looking at sine / multi sine type signals) , it will do : RMS = sqrt(power) and there is no division by nfft

so I modified a bit your code to test the different scenarios :

sf = 2200; %sampling

f = 440;

a = 0.5;

t = 0:1/sf:1;

% x = a*sin(2*pi*f*t); % to test the PSD = 2*abs(xhat)/n

x = a*randn(size(t)); % to test the PSD = xhat.*conj(xhat)/n

sound(x,sf)

figure(1)

plot(t,x) %amplitude-time graph

axis([0 1/f min(x) max(x)])

figure(2) %Fourier analysis

n = sf; % so df = sf/n = 1 Hz

xhat = fft(x,n);

PSD = xhat.*conj(xhat)/n; % ok for random type signals

% PSD = 2*abs(xhat)/n; % ok for sine type signals

freq = 1/(1/sf*n)*(0:n);

L = 1:floor(n/2);

colormap jet(20)

plot(freq(L),PSD(L),'LineWidth',2.5)

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