Sinewave with log frequency - Shows the wrong frequency content

I'm struggling with a rather basic problem. I create a sinewave with a continuously increasing frequency:

NPTS=500000;

A=(log(50)-log(7))/450;

time = linspace(0,450,NPTS)';

freq = 7*exp(A*time); % The frequency varies from 7 to 50 Hz

Y = sin(2*pi*freq.*time);

Then, when I do an FFT vs. Time of the Y trace, I find that the highest frequency in the signal is close to 150 Hz.

What am I missing here?

Answer 1

[Mathieu NOE]

hello

you have to compute the angle increment vs time then take the sine of that angle (or, in other words, do the integral of the time varying frequency and then take the sine of that angle )

NPTS=500000;
t_final = 450;
A=(log(50)-log(7))/t_final;
time = linspace(0,t_final,NPTS)'; 
dt = mean(diff(time));
fs = 1/dt

fs = 1.1111e+03

freq = 7*exp(A*time);       % The frequency varies from 7 to 50 Hz
freq(1)

ans = 7

freq(end)

ans = 50

ang = cumsum(2*pi*freq)*dt; %  integral of the time varying frequency
Y = sin(ang);

% spectrogram plot
nfft = 1000;
overlap = 0.75; % 75% overlap
spectrogram(Y,hanning(nfft),round(overlap*nfft),nfft,fs,'yaxis')
ylim([0 100])

[Djamil Boulahbal]

That's an interesting calculation. Thank you very much for sharing. I still do not understand WHY my initial implementation fails to provide the correct result.

[Mathieu NOE]

hello again

yes this is the usual remark when someone tries to generate variable frequency signals (sweeps)

always remember you are dealing with variable frequency and what you want at the end is the sine of an angle (not a frequency)

if your frequency is steady , you can write Y = sin(2*pi*freq.*time);

but if you have a variable frequency, start with the equation that describes how this frequency varies , multiply by 2*pi to convert to pulsation and then time integrate to get the angle

[Djamil Boulahbal]

Ok, thank you very much for the explanation.