I am analyzing a signal received by two antennas, where the second antenna is positioned farther than the first in the direction of wave propagation. I am encountering some confusion when it comes to the phase difference and gain between these two antennas. The wave equation of the signal is given by

$$ s(x, t) = \alpha \exp(i(\omega_0 t - kx)) = \alpha \exp(2\pi i(f_0 t - \frac{1}{\lambda}x)), $$

where \$f_0\$ (\$\omega_0\$) is the (angular) frequency of the wave.

Considering a scenario where the second antenna is at a distance \$\Delta x\$ farther from the transmitter than the first antenna (\$\Delta x = d \sin \theta\$ or \$\Delta x = d \cos \theta\$), the signal is received at the second antenna \$\frac{\Delta x}{c}\$ seconds later. This leads to the signal expression

$$ s(x + \Delta x, t + \frac{\Delta x}{c}) = \alpha \exp(2\pi i(f_0 (t + \frac{\Delta x}{c}) - \frac{1}{\lambda}(x + \Delta x))) = \alpha \exp(2\pi i(f_0 t - \frac{1}{\lambda}x)) = s(x, t), $$

due to the fact that \$\frac{f_0}{c} = \frac{1}{\lambda} \Rightarrow \frac{f_0}{c}\Delta x = \frac{1}{\lambda} \Delta x\$. It appears that the phase difference arising from the delay difference is canceled out by the phase difference due to the path length difference.

However, my confusion arises when considering Uniform Linear Array (ULA) antennas operating under the far-field assumption. These antennas are known to exhibit phase differences. Could someone elucidate where this phase difference is coming from?

Additionally, I am looking to understand the relation between the gains of the two antennas. If the first antenna has a gain of \$\beta\$, would the gain of the second antenna, which is farther than the first, be \$\exp(\frac{2\pi i}{\lambda}\Delta x)\$ or \$\exp(-\frac{2\pi i}{\lambda}\Delta x)\$ given that \$\Delta x > 0\$?

Any insights or clarifications on the phase and gain differences between the two antennas would be greatly appreciated.

  • \$\begingroup\$ The difference in arrival time of the signal between the two antennas can be expressed in time (ps, ns, etc), or in degrees (phase). The difference is that the former is just based on the geometry - path length difference between the two antennas - while the later also depends on the wavelength, or frequency of the signal. \$\endgroup\$
    – SteveSh
    Commented Jan 29 at 17:17

1 Answer 1


Let H(t-ax) be a function of time t and one dimensional place x. The function has actually a single variable t-ax. This makes it to present a propagating wave. Parameter a is the inverse of the propagation velocity.

To calculate right how changing the observation point affects you should change only x i.e. add to x the observation point placement difference Δx.
In the new observation point the function is H(t-a(x+Δx)) = H(t-a(Δx)-ax). That's like the original, but t is replaced by t-a(Δx). That means everything happens time interval a(Δx) later. That's the cause of phase differences implied by the distance differences.

You have for some obscure reason calculated separately also how long it takes to propagate distance Δx and shifted the time scale, too. It's no more the same wave observed in a different place. It has been started in the transmitter at a different moment of time and doesn't give a slightest hint how much phase lag moving the observation point further from the transmitter causes to the original wave.

The gain of an antenna is a property of the antenna. It normally depends on the operating frequency, but the gains of other antennas elsewhere and the distances to them do not affect as long as the other antennas are distant enough. If there's placed something in the near field, it effectively is no more the same antenna. Its properties can be recalculated by including the other items to the geometry, but it's essentially an analysis of a different antenna.


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