Received power from multiple transmitters

Question

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.

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?

Answer 1

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.

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.

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.

Answer 2

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:

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.

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