Wave with oscillating frequency

AI Thread Summary
A user seeks to generate sine waves with a slowly oscillating frequency but encounters unexpected behavior in their function. They initially used f(t) = sin(2πt[5 + cos(2πt/10)]) but found the maxima and minima too close together. Another participant explains that the issue lies in misunderstanding frequency modulation and suggests a corrected approach. The correct function involves integrating the instantaneous frequency to derive the phase, leading to f(t) = sin(φ) where φ is defined as 10πt + 10sin(πt/5). The user confirms that this solution meets their needs.
Nikratio
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Hello,

In order to test-drive a data analysis program, I am looking for a function that generates sine waves with slowly oscillating frequency, i.e. the distance between the maxima should be slowly changing.

I thought that I could simply achieve this by using a function of the form
<br /> f(t) = \sin\Bigl( 2 \pi t \bigl[ 5 + \cos( 2 \pi t / 10 ) \bigr] \Bigl)<br />
and expected to get something that oscillates with a frequency that slowly changes between 4 and 6 over when t goes from 0 to 10.

However, the function behaves in a very different way. It reaches a maximum of f(t)~0.5 at t~3 and has to minima very close to each other at t~4.2.

Can someone help me to understand (a) why the above function doesn't behave the way I expect it to and (b) what function I could use to generate the desired data?


Nikolaus
 
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Hi Nik, you've fallen for a really common mistake with regard to frequency modulation.

In general the instantaneous radian frequency (w=2 pi f) is the time rate of change of phase (phi), this of course is not the same thing as "phase" divided by "t" except for the special case where the phase is a linear function of "t", that is where \phi = 2 \pi f t[/tex].<br /> <br /> So what you really want is,<br /> <br /> \frac{d\phi}{d t} = 2 \pi ( 5 + \cos(\pi t /5) )<br /> <br /> Which upon integration gives,<br /> <br /> \phi = 10 \pi t + 10 \sin(\pi t /5)<br /> <br /> Try using f(t)=sin(phi), with phi as given above and I think you&#039;ll find it does what you&#039;re looking for.
 
Yes, that was it! Thank you very much!
 
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