This is an RF-based question. My professor told me that demodulating a TX signal that is 90 degrees out of phase with the receiver oscillator is impossible; there will be no baseband signal. He said that the TX and RX signal need to be in-phase in order for demodulation to work correctly. He also said that the way around this is quadrature demodulation. However, I have seen many RX circuits that use a single oscillator and a single mixer. How do they solve the problem of TX/RX phase synchronization?
Demodulation with a phase difference
demodulationRF
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Best Answer
Let's say you want to transmit the message signal \$m(t)\$ and you upconvert it by multiplying it by a cosine, so that the actual signal transmitted is \$s(t)=m(t)\cos(2\pi f_ct)\$ . This is called AM DSB-SC (double side-band suppressed carrier). If you multiply \$s(t)\$ with \$\cos(2\pi f_ct)\$ and low-pass filter the result, you get \$m(t)\$.
However, if you multiply \$s(t)\$ with \$\cos(2\pi f_ct+\phi)\$, then the result after the low-pass filter is \$m(t)\cos(\phi)\$. If \$\phi\$ is close to \$\pi/2\$ or \$3\pi/2\$, the received signal is close to zero. There are several solutions, along these two lines:
A third solution is to transmit \$s(t)=(A+m(t))\cos(2\pi f_ct)\$, where \$A\$ is large enough to make \$A+m(t)\$ be always positive. This is called AM DSB-LC (double side-band large carrier). Essentially, you're transmitting the carrier along with your signal, so the receiver can figure out exactly what phase you used. An envelope detector is a circuit that recovers \$m(t)\$ in this case. The downside is that a lot of power goes to transmit the carrier instead of the message.
Quadrature transmission is a way to profit from this fact by transmitting two messages with the same carrier frequency; one message is transmitted with a carrier that is \$\pi/2\$ radians out of phase with the other. Using similar carriers in the receiver, both messages can be recovered. Of course, in this case a very good phase estimation is essential.