1
\$\begingroup\$

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?

\$\endgroup\$
  • \$\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
  • 1
    \$\begingroup\$ Here is a schematic: mightydevices.com/wp-content/uploads/2014/04/receiver.png \$\endgroup\$ – crocboy Apr 22 '15 at 1:40
1
\$\begingroup\$

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.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.