0
\$\begingroup\$

My question is regarding RF measurement systems which use antennas signal arrival time as principle of operation e.g. GPS. It is known that impedance of well-matched antenna immersed in lossy dielectric is changed. Like in this case of an antenna covered by snow. Thus the complex input impedance Z of the antenna depends on the surrounding environment. From the circuit theory we know a relation between Z and a phase shift associated to it as: $$ \Theta=atan(X/R) $$ where $$ Z=R+jX $$

So a radar altimeter on airplane passing through a dense cloud would give false altitude reading?

In the literature I see the topic of antenna equivalent circuits vs. lossy dielectric environment is well covered, however no one considered what does it mean from the time domain point of view. Is my reasoning, that antenna intrinsically adds a phase shift, expressed by formula given above, correct?

\$\endgroup\$
  • \$\begingroup\$ Just how much of a fraction of a micro second do you consider significant to creating an error that is noteworthy? \$\endgroup\$ – Andy aka Dec 2 '14 at 13:44
  • \$\begingroup\$ What do you mean by \$\Im\$ and \$\Re\$? \$\endgroup\$ – Phil Frost Dec 2 '14 at 13:56
  • \$\begingroup\$ @PhilFrost - I think those refer to the imaginary and real parts of Z. \$\endgroup\$ – Reinstate Monica Dec 2 '14 at 14:02
  • \$\begingroup\$ I used GPS for illustration purpose only. I am interested in general, if this phenomenon occurs or not. I now design an underwater sensing device for sea application, and 5ns delay, caused by sediments makes a great difference fo me. \$\endgroup\$ – Kris Dec 2 '14 at 14:05
  • \$\begingroup\$ @Kris perhaps you can explain where you got that equation, or remove it if it's not relevant. \$\endgroup\$ – Phil Frost Dec 2 '14 at 14:06
1
\$\begingroup\$

I don't think you understand how GPS works. Each satellite transmits a pseudo-random sequence of bits, and the receiver compares the difference in time of arrival between them. By comparing the relative delays of three or more satellites, combined with knowledge of the satellite's position at that instant (which is also transmitted), the receiver can triangulate its position.

Because the receiver is comparing relative times, any delay that is equal to all the signals is irrelevant.

It would not make sense for a GPS receiver to rely on absolute time of arrival because that would require each receiver to contain an extremely accurate atomic clock for reference.

It also would not make sense for a GPS receiver to rely on phase. All you need to do to alter the phase of the received signal is rotate the antenna. Does rotating your GPS receiver change your reported position? I hope not!

I'd further point out that the formula you reference (\$\theta=\tan^{-1}(X/R)\$) does not mean the antenna is introducing a phase shift. It's simply the phase angle of the impedance of the antenna.

complex impedance plane

A phase angle of \$\theta\$ means that if you put apply a sinusoidal voltage to that impedance, the phase of the current will lag that of voltage by \$\theta\$. This may or may not introduce a phase shift to the received signal: it depends on what you mean by "shift" and what you are using as a reference, the impedance of the receiver, and whether you are considering the current or the voltage as the signal.

That, and between the receiver and antenna are a bunch of complex impedances and transmission lines that each will introduce phase shifts of their own, so the impedance of the antenna isn't making any new problem.

\$\endgroup\$
  • \$\begingroup\$ Thank you Phil for the answer. So in principle the received signal many contain several delay factors but due the architecture of the GPS they compensate each other. \$\endgroup\$ – Kris Dec 3 '14 at 11:13
  • \$\begingroup\$ Sorry for confusion with the formula, I made a typo. Now it is correct \$\endgroup\$ – Kris Dec 3 '14 at 11:14
  • \$\begingroup\$ By comparing the relative delays of signals received at the antenna, you also get rid of another factor: the length of the cable between antenna and receiver. In most cases, this variable length would be more significant than changes in characteristic impedance. You can have an antenna on a building roof, with the receiver in a rack several metres away / down. \$\endgroup\$ – Alan Campbell Dec 3 '14 at 12:06
  • \$\begingroup\$ Actually the really good gps ones do use the carrier frequency phase to achieve higher accuracy. L1 wavelenght is ~20cm so given a reasonable phase detection accuracy you can get into millimeters. But this all is jeopardized by the fact that athmospheric instabilities are changing each signal differently, so this is only really useful for differential gps (and is being used that way in surveying for quite a while.) \$\endgroup\$ – PlasmaHH Dec 3 '14 at 12:57
  • \$\begingroup\$ @PlasmaHH even so, they are comparing the phase of the satellite signal with the phase of another signal received from a transmitter at a precisely known terrestrial location. Hence the "differential". Anything that delays all signals equally is still irrelevant. \$\endgroup\$ – Phil Frost Dec 3 '14 at 13:01
0
\$\begingroup\$

Yes, the phenomenon is real, but usually of no significance. For GPS in particular, all of the received signals are delayed by the same amount, so it has no effect on the position calculation. It's effectively the same thing as adding a few inches of antenna cable.

If you're designing a system in which tiny absolute delays are significant, then you need to find a way to measure them and compensate for them.

\$\endgroup\$
  • \$\begingroup\$ Thank you Dave. That's the answer I was looking for. Do you know how is this effect compensated in real systems? \$\endgroup\$ – Kris Dec 3 '14 at 11:23
  • \$\begingroup\$ If the extra delay is relatively stable over time, you can periodically calibrate it out. Otherwise, you need to find a way to turn an absolute measurement into a differential one, perhaps by adding a reference signal generator to the system. \$\endgroup\$ – Dave Tweed Dec 3 '14 at 12:57

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.