With on-off keying, you have a single carrier wave at a constant frequency transmitting its presence or absence. Now I understand that the crystal will not be perfect so you will have some minimal shift over time in the carrier frequency, but assuming the crystal was perfect you would have zero bandwidth because the frequency doesn't change. But datasheets say that the bandwidth is linked to the data rate somehow, what is that equation and what is using up all that extra spectrum if you don't change the frequency?
Electronic – Why do OOK transmissions have bandwidth
bandwidthRFsignal
Best Answer
The bandwidth is not due to carrier frequency drift. As a first approximation, the bandwidth is twice the frequency that modulates the carrier:
This may not be intuitive, but consider that the amplitude modulation (including OOK) is essentially a multiplication of carrier and modulation signals, and basic trigonometry tells us that
$$2 sinA sinB = cos(A - B) - cos(A + B)$$