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?

  • \$\begingroup\$ 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. \$\endgroup\$ – Zulu Apr 21 '15 at 22:59
  • \$\begingroup\$ 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. \$\endgroup\$ – cuddlyable3 Apr 21 '15 at 23:18
  • \$\begingroup\$ Yes I meant RX. Do active mixers implement a phase-lock circuit? The simple rx circuits I have seen have no PLL. \$\endgroup\$ – crocboy Apr 21 '15 at 23:21
  • \$\begingroup\$ 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. \$\endgroup\$ – Zulu Apr 21 '15 at 23:36
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    \$\begingroup\$ Here is a schematic: mightydevices.com/wp-content/uploads/2014/04/receiver.png \$\endgroup\$ – 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:

  • Use a phase-recovery circuit in the receiver, such as a PLL. This adds a bit to the system's complexity and cost. This is the most common solution these days for all but the simplest receivers.
  • For a very cheap system, let the user adjust the antenna positions to improve the reception.

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.


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