# Frequency Swept sine wave — chirp

I am experiencing what I think is really simple confusion.

Take $y(t) = \sin(2 \cdot \pi \cdot t \cdot\omega(t))$ and $\omega(t) = a \cdot t+b$ for $t \in [0,p)$ and let $\omega(t)$ have a periodic extension with period $p$. The values $a,b,p$ are parameters

$\omega(t)$ looks like the blue function here

with $\omega$ as such, $y(t)$ should be two chirps -- a sine wave whose frequency sweeps. My intuition is that the two chirps should be identical, but they're not. The second chirp is higher in frequency. The argument to $\sin$ , the quantity $2 \cdot \pi \cdot t \cdot \omega(t)$ looks like the blue function here:

Two piecewise quadratics. The second quadratic should just be a shifted version of the first one, which is what is depicted in red. I can't argue with the math, but there is something very simple wrong with my intuition. The function $\omega$ should modulate the "instantaneous frequency" of the sinewave.

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If you want the instantaneous frequency to be $\omega(t)$, the argument to $\sin$ should be $2\pi\int_0^t\omega(\tau)d\tau$. This is equal to your $2\pi t\omega$ when $\omega$ is constant, but not otherwise. – Rahul Jan 13 '11 at 5:34
@Rahul Thank you! My intuition was broken and I should have remembered this from my signals book. – Gus Jan 13 '11 at 5:43

For a chirp signal of the form $\sin(2\pi\phi(t))$, the instantaneous frequency $\omega(t)$ is the time derivative of the instantaneous phase $\phi(t)$. So for given $\omega(t)$, your signal should be $\sin\left(2\pi\int_0^t\omega(\tau)d\tau\right)$.
For what it's worth, the incorrect expression $\sin(2\pi t\omega(t))$ gives a signal with instantaneous frequency $\omega(t) + t\omega'(t)$. In your example, $\omega'(t)$ is positive almost everywhere, which is why the signal seems to be too high in frequency.