# Demodulation with a phase difference

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?

• Do you mean to say, "I have seen many RX circuits that use a single oscillator and single mixer"? Frankly, the TX circuit doesn't care what the phase is; the responsibility of phase locking falls on the RX circuit. – Zulu Apr 21 '15 at 22:59
• True. Also, "the TX and RX signal need to be in-phase" implies that the TX (transmitter) and RX (receiver) must be exactly a whole number of wavelengths apart. The question needs rewording so it addresses the subject of quadrature demodulating a received signal using a phase-locked local oscillator. – cuddlyable3 Apr 21 '15 at 23:18
• Yes I meant RX. Do active mixers implement a phase-lock circuit? The simple rx circuits I have seen have no PLL. – crocboy Apr 21 '15 at 23:21
• Can you provide an example RX circuit that you're thinking of? We can make speculations, but it would be best to see what you're seeing. – Zulu Apr 21 '15 at 23:36
• Here is a schematic: mightydevices.com/wp-content/uploads/2014/04/receiver.png – crocboy Apr 22 '15 at 1:40

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.