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. 

It seems my comment has answered the question, so I suppose I should turn it into an actual answer.
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.
