Chirp: Why do you need to integrate the frequency function?

Question

From the Wikipedia page on chirp, if I provide some time-dependent frequency function $f(t)$, this has to be integrated if I want to use it as the argument to, say, a sinusoid. Suppose $f(t) = f_0+kt$; then

$x(t) = \sin\left(2\pi\int_0^t f(\tau)d\tau\right) = \sin\left[2\pi\left(f_0t+\frac{k}{2}t^2+\phi_0\right)\right]$

Why is it not good enough to claim that $x(t) = \sin(2\pi f(t)t) \equiv \sin(\omega(t)t)$?

Answer 1

If you just said $\sin(2\pi f(t)t)$, the instant frequency would be $\frac{d}{dt}\bigl[f(t)t\bigr]$ which is (by the product rule) $f(t) + f'(t)t$. The first of these terms is what you want; but the second contributes an unwanted error which can be as large as your desired frequency variation (or even larger) if $f'(t)$ is large.

For example, if you set $f(t) = \frac{1}{4t}$ you get $\sin(2\pi f(t)t)=\sin(\frac{\pi}{2})=1$, so instead of a chirp with falling frequency you get a constant, i.e. frequency 0 everywhere. The error has completely swallowed the intended frequency history.

Answer 2

The instantaneous radian frequency of a signal $x(t) = \sin(y(t))$ is defined as the derivative of the phase function $y(t)$ and the instantaneous frequency is smaller by a factor of $2\pi$. So if you want the instantaneous frequency to have value $f_0$ at $t=0$ and to increase as a linear function $kt$ with $t$, then the phase function $y(t)$ needs to be the solution to the differential equation: $$\frac{\mathrm d}{\mathrm dt}y(t) = 2\pi(f_0 + kt), ~ t \geq 0.$$ It follows that for $t \geq 0$, $$y(t) = 2\pi \int_0^t f_0 + k\tau\,\mathrm d\tau + y(0)$$ where $y(0)$ is denoted by $\phi_0$ in your question. Choosing $y(t) = 2\pi(f_0 + kt)t$ does not quite give you what you wanted since the instantaneous frequency increases at the rate of $2kt$ instead of $kt$, and the value of the phase at $t = 0$ is also restricted to be $0$ which might not hold in an actual situation.