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There are 2 possible ways to calculated the signal sum: either by amplitude or by power. According to the paper http://authors.library.caltech.edu/824/1/HAJprocieee05.pdf, we should add the amplitudes together: if we have 2 synchronized signals with amplitude 1, their sum has an amplitude of 2 and a power of 4 ("if each element radiates P watts, the total power that will be seen at the receiver in the esired direction is m^2P watts). But I think that is against the low of energy conservation. Anyone has a good explanation?

Thanks in advance!

Bile

p.s. I understand in the paper above, what we calculate is not the total transmitted power but the power in a certain direction. So can this method be applied only in case of beamforming or can it be used also in a circuit?

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Energy is still conserved because by transmitting with two synchronized omnidirectional antennas, you make a directional antenna. If each antenna transmits \$1W\$, and there are two antennas, and the total power transmitted is \$2W\$. But, if your array has (for example) a gain of \$3dBd\$, then a receiver in the main lobe of that antenna will see effectively \$4W\$ (\$2W \cdot 3dBd\$). Energy is conserved because a receiving antenna not in the main lobe of the transmitting antenna will see less than \$2W\$.

This works because depending on the receiver's orientation to the transmitting antenna array, the radiation from each antenna in the transmitting array may arrive in phase, and thus add constructively, at the receiving antenna. Or, the radiation may be out of phase, and cancel.

Let's say you have two antennas, and you feed them \$180^\circ\$ out of phase. That is, the voltage at antenna A is always the opposite of the voltage at antenna B.

If these antennas occupy exactly the same space (somehow, as a thought experiment), then there will be no power. The exactly opposite fields of each antenna will mean that there is no field. There is no voltage, no current, and no power, as if you had done this:

schematic

simulate this circuit – Schematic created using CircuitLab

You can change the voltage all you like, but no current will flow. That is, the impedance of this antenna array is \$\infty\Omega\$: it looks like an open circuit.

However, if these antennas are placed half a wavelength apart, then an additional \$180^\circ\$ phase shift is introduced. Now, the voltages at one antenna are adding to the voltages of the other antenna, since they are in phase. Thus, in the directions aligned with the array, the voltage is twice what it was with just one antenna. Since power is proportional to the square of voltage, the apparent power is four times what it would be with just one antenna.

However, broadside to the array, one is equidistant to the antennas. Thus, the phase of each antenna is what it was at the transmitter: \$180^\circ\$ different, that is, exactly opposite. There is no voltage at all, and no power. That there isn't power here is how there can be a disproportionate amount of power in the direction in line with the array.

The antennas do interact, but this doesn't make anything impossible, or even really all that complicated. Don't let the fact that the other antenna is some distance away confuse you: this appears as a voltage across the feedpoint that is a function of the input signal. It will either add to or subtract from the voltage fed to the antenna, depending on the particular geometry of the array, and the result is just a change in feedpoint impedance.

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  • \$\begingroup\$ I think a bigger factor is that each antenna will receive some of the signal from the other. If the antennas are driven with voltage or current waveforms that are independent of whatever is coming in, then depending upon whether the signals are in phase or out of phase, this will increase or decrease the apparent load the antennas present to the driving circuit, and thus increase or decrease the amount of power taken from such circuitry. \$\endgroup\$ – supercat Jul 12 '13 at 15:03
  • \$\begingroup\$ I agree. Suppose there are 2 antennas at exactly the same position (just a thought experiment) by with a phase shift of pi/2. There would be no radiated power. Therefore the 2 antennas must feed back the power they have received from the driving circuit. \$\endgroup\$ – bile Jul 12 '13 at 16:47
  • \$\begingroup\$ @bile do you mean a phase shift of \$\pi\$ or \$180^\circ\$? \$\pi/2\$ is \$90^\circ\$. With a 180 degree phase shift, yes, there would be no radiated power, because the fields would exactly cancel. But it doesn't get transmitted, and fed back. You can't make any field with such an arrangement, so there's nothing to be transmitted. Fortunately, real phased arrays aren't like this, so they actually can transmit. \$\endgroup\$ – Phil Frost Jul 12 '13 at 17:39
  • \$\begingroup\$ @PhilFrost sorry I meant \$\pi\$... this example is just to show it's not like that if we feed power P in, we get power P out. The antenna feeds back power too. Unfortunately I am not familiar with circuits and can't do further quantitative analysis. \$\endgroup\$ – bile Jul 12 '13 at 19:40
  • \$\begingroup\$ @bile but you aren't feeding power P in. You are feeding 0 power in, because voltage is 0 at every point in space around your two antennas which exactly cancel each other out, by virtue of being in exactly the same space and having exactly opposite fields. No voltage means no power. It's no different than if you connect two identical batteries together, (+) to (+), and (-) to (-). One battery doesn't feed back into the other. There's no power. \$\endgroup\$ – Phil Frost Jul 12 '13 at 20:15
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You are making a simple mistake in your analysis. Consider this: -

enter image description here

  • In the top picture there is a single 1V battery and 1 ohm resistor.
  • The middle picture shows two batteries and two 1 ohm resistors in series. The total voltage is 2V.
  • The bottom picture is with the cross-link removed and both resistors joined to form a 2 ohm resistor.

The mistake you have made is assuming the bottom picture is with a single 1 ohm resistor. In that mistaken scenario the power will be 4W.

If the bottom mistaken scenario were reversed back to the middle scenario, the two resistors would become 0.5 ohms each and this is the same as having two circuits, each at 1V and feeding a 0.5 ohm resistor each. The power in each circuit would be 2W because the current is 2A.

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  • \$\begingroup\$ Thanks for that buddy. But my question is about the wireless communication: If 2 antenna elements transmit the same signal, what is the signal strength at a certain direction? I wanted to calculate the beamforming gain and encountered this problem. I didn't explain the background because I thought there won't be any misunderstandings. \$\endgroup\$ – bile Jul 12 '13 at 10:15
  • \$\begingroup\$ I'd suggest a different set of scenarios: 1V driving 1ohm (supply outputs one watt; load takes one watt); two additive series 1V supplies driving 1ohm (the power in the resistor is quadruple, because the power from each supply is doubled), two opposing 1V supplies driving 1ohm (power in the resistor is zero, because no supplies are doing any work), and a series string with two supplies in one direction and one the other driving 1 ohm (two supplies are outputting a watt, and one is receiving a watt). \$\endgroup\$ – supercat Jul 12 '13 at 21:45

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