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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...

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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.

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  • $\begingroup$ "Phase of a received signal is not an absolute property of a signal." So this answers the second part of my question, right? I will not see the same phase at the wavefront at all times? $\endgroup$ – fl0ta'' May 16 at 13:53
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    $\begingroup$ If you "follow" the wave along, as you described, then you'll see the same phase. What I was trying to say, was, you need a time reference. $\endgroup$ – auspicious99 May 16 at 13:55
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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.


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.

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  • $\begingroup$ Thanks! But I thought the propagating wavefront has the the phase of the signal and regardless of the traveled distance, the first received signal part has exactly this phase... $\endgroup$ – fl0ta'' May 12 at 19:33
  • $\begingroup$ From a fixed observation, the phase would vary continuously with time, right? So this is with reference to the two signal as received at exactly the same time, at any given time. At one instant when one is at 0.45 radians, the other would be 0.66 radians- for example. I hope that clears it up for you. $\endgroup$ – Dan Boschen May 12 at 19:38
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    $\begingroup$ That sounds right--- imagine waves hitting the beach, very similar. From crest to trough would be 180° of phase shift. At any given instant in time a different phase of the ocean wave will be in contact with the beach at the transition between land and water. For radio waves it is the change in electric field and magnetic field rather than height of the water. $\endgroup$ – Dan Boschen May 12 at 20:47
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    $\begingroup$ @fl0ta'' I updated my answer- but does the ocean wave analogy make sense to you? Resetting time is the same as resetting phase since phase is a relative measurement, so resetting $t=0$ at the receiver is non-sensical unless you are referencing the received signal to another signal in that same receiver. $\endgroup$ – Dan Boschen May 16 at 13:02
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    $\begingroup$ Yes that is correct. But phase is with reference to time so in order to compare the signals at the two receivers you must use the same time reference. The phase is a “pseudo-range”. $\endgroup$ – Dan Boschen May 16 at 13:59
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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.

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