Why does altering path length in radio transmission alter the phase of the received signal?

Question

I do have a very naive yet bothering question: Assume I send a sinusoid signal

\begin{equation} A*cos(2\pi ft+ \phi_0) \end{equation}

over a radio channel. Assume there is only a LOS-path for each receiver antenna and no interference or other fading effects.

Next, I receive the signal at two different locations with different distances from the receiver. Will the phase of the received signals be different for the two sites?

In order to deepen my understanding:

Let's assume I record my received signal and I set $t=0$ for each receiver individually when the signal arrives at the respective receiver. Will the phase of the received signal at each receiver be different for different distances? Obviously, for one observation instant (time on reference clock equal for all receivers) the phases vary at the different sites. However, what I mean is: Does the signal phase vary for the situation I explained above (individual receiver clocks)? This question boils down to the understanding of the physical reality of an EM wave propagating. Assume I follow the wave and I 'look' at the wavefront at different times; will I see the same phase at all times? Or does the 'front' of the signal wave oscillate in time and space? I really hope that does not sound too stupid; it's really a question that bothers me...

Answer 1

Next, I receive the signal at two different locations with different distances from the receiver. Will the phase of the received signals be different for the two sites?

It depends on the time reference used at the receiver. If it is perceived as being phase shifted, it needs to be with respect to a particular time reference. Phase of a received signal is not an absolute property of a signal.

If the signal is a simple one as in your example, the receiver cannot separate the two paths, so it may time synchronize to the first arriving path (so when the first path starts arriving, that becomes $t=0$.) Then, the phase of the signal on the second arriving path, would be considered with respect to the time reference from the signal on the first arriving path. Then, as explained in the other answers, you'll need an exact integer multiple of the wavelength, to not perceive the signal on the second arriving path as being phase shifted.

Now, there are real, practical receivers where the 2 paths can be distinguished. For example, in CDMA systems, because of the PN codes applied to the signal, the signal arriving at the 2 paths can be distinguished. Then, they can be separated and not have to share a time reference. Hence, in the so-called Rake Receiver for CDMA systems, each "finger" of the Rake Receiver can be time-synced to a different path, and each be received coherently, and then combined.

Answer 2

Yes indeed it will be (although we could argue that it may not be necessarily for all distances since the phase is cyclical!).

In free space the signal propagates at the speed of light, therefore this sets the wavelength in distance based on the frequency transmitted according to:

$$\lambda = c/f$$

Where $c$ is the speed of light in meters/second (or other units of choice), and $f$ is the frequency of the signal in units of Hz if time is in units of seconds. A fixed tone at a given frequency with have a length in phase of $2\pi$ over the distance of one wavelength.

If the channel has a dielectric constant higher than free space given by $\epsilon_r$, the wavelength will be decreased by $1/\sqrt{\epsilon_r}$.

For example the wavelength of a 1 GHz signal in free space (or air which has a dielectric constant that is very close to free space) is approximately $3e8/1e9 = 0.3$ meters, while over a cable with a polyethylene dielectric ($\epsilon_r = 2.3$) would be $3e8/(1e9\sqrt{2.3}) = 0.198$ meters.

For this case of a 1 GHz signal in free space given a wavelength of $0.3 m$ (corresponding to the phase over one wavelength of $2\pi$ radians), if you receive the direct signal with no multi-path at two different locations with a $1$ cm path difference between the two, the expected phase difference will be $0.21$ radians.

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As a follow up to the OP's question; if you set $t=0$ upon arrival of the wavefront, then the phase everywhere at the new $t=0$ will be equal to the phase when the signal was initiated at the transmitter. Although this is somewhat of a meaningless result: it is equivalent to measurement the distance of two objects and resetting the distance to 0 once you determine the objects length.

Answer 3

Phase measurement depends on a reference time or point.

If your time reference or clock sync is the same speed-of-light signal, then the phase of a signal with reference to itself is zero.

If your clock or time reference is broadcast perpendicular to your Signal, then the phase could be different if the distance between the two points isn’t an exact integer multiple of the wavelength.