0
\$\begingroup\$

Assuming I have two transmitting systems, feed by identical signals of the same power, \$P_{TX}\$, which are also in-phase. They are both placed at the same distance to the receiver, and therefore has the same path loss.

enter image description here

What is the power, \$P_{RX}\$, at the receiver? \$2P_{TX}\$ or \$4P_{TX}\$?

Some quick calculations assuming:
\$P_{TX} = +30dBm\ (1\ W)\$,

and

\$FSPL = 30dB\$

Would mean that the received power from each system would be \$0dBm\ (1mW)\$ and at a quick glance would indicate that \$P_{RX} = +3dBm\ (2mW)\$ in total.

But, if we sum up the voltages(assuming \$50 \Omega\$ system), the voltage from one transmitter would be:

\$V = \sqrt{1\ mW\cdot 50 \Omega\ } = 224\ mV\$ And as both signals are in-phase, the total voltage would add up to:

\$V_{RX} = 2\cdot 224\ mV = 448\ mV\$,

which equals a power of: \$ P_{RX} = \frac{448\ mV^2}{50 \Omega} = 4\ mW\$ and therefore quadrupling the power.

But which is it? A fourfold increase in power does not seem right to me. What am I missing?

\$\endgroup\$

2 Answers 2

2
\$\begingroup\$

The correct answer is 4x the power. Let me give an example to answer your comment (on Tesla23's great answer) on whether an antenna pattern changes with the number of antennas.

Assume the antennas in your image are separated by distance d and that both antennas are isotropic radiators so that radiated field for each INDIVIDUAL antenna is \$ E_{iso} = \dfrac{e^{-jkr}}{4\pi |r|} \$ where \$r \$ is a vector pointing from the antenna to any location we wish to know the fields at. In your case, the fields at the receiving antenna are of primary interest.enter image description here

The image above is the scenario you described, but now I have added two vectors, one showing the path of fields from TX1 to the RX \$ r_1 \$ and one showing the path from TX2 to the RX \$ r_2 \$. Furthermore, assume that each antenna is excited with power \$ P_{TX} \$. The field associated with this power will be \$ E_{TX} \propto \sqrt{P_{TX}} \$. Now then, the fields at the receiver due to both antennas is \$ E_{RX} = E_{TX1}+E_{TX2} = E_{TX}\dfrac{e^{-jkr_1}}{4\pi |r_1|}+E_{TX}\dfrac{e^{-jkr_2}}{4\pi |r_2|} \$

Since in the given situation \$r_1 = r_2 \$ then the equation is simplified as

\$ E_{RX} = 2 E_{TX}\dfrac{e^{-jkr_1}}{4\pi |r_1|} \$

This is to say that the received fields at the receiver are double what they would have been for one transmitter. Since \$ E_{TX} \propto \sqrt{P_{TX}} \$, the power received is quadrupled due to the doubling of the field intensity. You are right to wonder where this power comes from, and that is what I will answer next.

An ideal isotropic antenna radiates fields in all directions equally. So no matter where I place my receive antenna, the expected gain would be constant for a SINGLE isotropic element. But now consider the case where there are two. Assume now for simplicity that the separation between antennas is \$ d = \lambda/2 \$. Now also assume that I placed an antenna to the left of TX1, as shown below.

enter image description here

Now the fields at the RX are again given by

\$ E_{RX} = E_{TX1}+E_{TX2} = E_{TX}\dfrac{e^{-jkr_1}}{4\pi |r_1|}+E_{TX}\dfrac{e^{-jkr_2}}{4\pi |r_2|} \$

In the case above \$ r_2 = r_1 + d \$. if I assume that \$ |r_1| \approx |r_2| \$ which is equivalent to \$ |r_1| >> d \$ then the total fields at the receiver are given by

\$ E_{RX} = E_{TX1}+E_{TX2} = E_{TX}\dfrac{e^{-jkr_1}}{4\pi |r_1|}+E_{TX}\dfrac{e^{-jk(r_1+d)}}{4\pi |r_1|} \$

Which is equivalent to

\$ E_{RX} = E_{TX}\dfrac{e^{-jk(r_1)}}{4\pi |r_1|}(1+e^{-jk(d)}) \$

Using the definition of k

\$ E_{RX} = E_{TX}\dfrac{e^{-jk(r_1)}}{4\pi |r_1|}(1-1) = 0 \$

So the fields for this RX location would be 0. This is the essence of where the power is coming from. The fields add up constructively in some locations (RX location 1) and destructively in some locations (RX location 2). So if I integrated along the entire polar sphere, the total power would still be the same, so power is still conserved, but because of these field interactions at a specific location, the power can change drastically (as demonstrated).

Antenna Theory by Constantine Balanis has a whole chapter dedicated to this subject. Phased arrays are a very interesting field I would encourage you to look into. Finally, one disclaimer I should make about this simple analysis is that I assumed the power radiated by each antenna did not change by their proximity to one another. Generally, this is not true, and antennas will "couple." This coupling changes how each antenna radiates its power and adds another complexity to antenna analysis.

\$\endgroup\$
1
  • \$\begingroup\$ Thanks a lot for your answer. This really helped me understand what is going on, and your examples are very clear. Antenna theory has never been my strong side, but I will definitely look for a copy of that book. \$\endgroup\$
    – Linkyyy
    Mar 5, 2023 at 7:21
5
\$\begingroup\$

It's complicated.

As you have drawn it, if the transmitting antennas don't interfere with each other, then as they are in phase, if they arrive in phase at the receiver they will add coherently and so the receiver will get twice the amplitude, hence 4x the power.

This 'extra' power simply comes from the the array pattern of the two antennas, radiating more power in the direction where the signals add in phase and less in the directions where they tend to cancel. Think of it as some form of focusing the energy. If the transmitting antennas are many wavelengths apart then you will get many peaks and nulls. Look up phased arrays if you are interested.

Life gets more complicated when the two transmitting antennas get close, in this case they interact significantly and this simple analysis fails.

Edit: some patterns -

if the two transmitting antennas are half a wavelength apart, you get a max on boresight and a null on axis. So the 'extra' power you get at 90deg is at the expense of power at 0 and 180:

enter image description here

if they are one wavelength apart, you get a max on boresight and another max on axis and a null in between. So the 'extra' power you get at 90deg and 180 deg is at the expense of power at the null in between.

enter image description here

images from: www.ece.rutgers.edu/~orfanidi/ewa/ch22.pdf which contains the theory.

\$\endgroup\$
10
  • \$\begingroup\$ Thanks for your answer! I am not sure i understand the part about antenna pattern, and why you say that they radiate more power in some direction - The antenna pattern(gain) doesn't change depending on how many antennas there is? There was another answer which apparently he removed again, talking about conservation of energy. In that regard, it doesn't seem to be right that the power received is quadrupled. Can you comment on that? \$\endgroup\$
    – Linkyyy
    Mar 4, 2023 at 21:42
  • \$\begingroup\$ @Linkyyy The part where you specify that the signals are in phase means that the gain of the overall array is increased. If the signals were incoherent you'd get the more expected summing in power, but coherent signals add in amplitude (when in phase), thus 2^2 times the power. \$\endgroup\$ Mar 4, 2023 at 23:14
  • \$\begingroup\$ @user1850479 and Tesla23: This seems to be the case when the antennas placed in the near-field of each other, right? What if they are spaced far apart ( d >> λ)? Their fields wouldnt interfere with each other, and the radiation pattern is not affected then? \$\endgroup\$
    – Linkyyy
    Mar 4, 2023 at 23:35
  • \$\begingroup\$ @Linkyyy No, not related to near field. This is just linearity, it holds as long as the signal is coherent. \$\endgroup\$ Mar 4, 2023 at 23:40
  • \$\begingroup\$ Im sorry if this was not clear in the question, but im not talking about phased array or beamforming where the antenna elements are spaced closely to each other. But just 2 different systems spaced far apart, with the same distance to the receiver. They are still coherent and synchronized in phase. \$\endgroup\$
    – Linkyyy
    Mar 4, 2023 at 23:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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