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Firstly, I'm not a mathematician, I'm an engineer, so you can freely make fun of the question.

I have the following counter-intuitive behaviour in a sweep function. I have a sweep sine function (something like $$x(t) = \sin(\omega(t) \times t)$$, where $\omega(t)$ is a linear function of time, so it could be said that the function is $$x(t) = \sin(\omega t^2)$$ When I plot the function and take some instant in time and measure the frequency of the response (counting the time elapsed between two consecutive peaks of the wave) this do not give the supposed instant frequency of the wave, namely $f = \omega t$.

In fact, in my example, the instant frequency of the wave is approximately 2 times $\omega t$.

Of course, I feel bad about asking this trivial question here, but I could not find light anywhere.

Thanks !

GoodDeeds
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nodarkside
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1 Answers1

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Your function can be written in the form $$ x(t) = \sin(\phi(t)) $$ in this function, $\phi(t)$ tells you which "phase angle" (in radians) you hit at any particular time $t$.

Now, a question you should ask yourself is: "what exactly does frequency measure, anyway"? I would posit that the frequency for a non-constant $\phi(t)$ measures how quickly the phase angle changes. That is, the frequency at time $t$ is given by $\frac{d\phi}{dt}$.

So, if $\phi(t) = \omega t^2$, then the frequency at a time $t$ will be $\phi'(t) = 2\omega t$.

Of course, if $\phi(t) = \omega t$, then we get the expected constant frequency of $\phi'(t) = \omega$.

Ben Grossmann
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