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I've been introduced to the concept of equivalent isotropically radiated power (EIRP) and so far I've used it to calculate RF power density at distance of 100m radiated by a directional antenna with a gain of 20dBi at 10W output power. The EIRP is 60dBm (1kW) so using the inverse square law 1kW/4pi100^2 got me ~7.9mW/m^2. I'm not sure I haven't made a mistake so please correct me if I'm wrong but under the assumption that I was in fact correct, a question arose.

If EIRP is supposed to represent a hypothetical isotropic radiator that would result in the same signal strength as produced by the directional antenna (at least in the area covered by the latter) then what happens if we were to measure power density close to the antenna? If we perform the same calculation as above but for 1m we end up with 1kW/4pi1^2 and that gives me ~79W/m^2. Now, ignoring near-field radiation and other circumstantial effects, why is the power density so high?

I feel like this is a good proof that I don't really understand gain but the argumentation that it's just the same power as from the isotropic source, just focused down to a beam doesn't cut it for me. I am assuming that if I input 1W into a 100% efficient isotropic antenna I get a total of 1W output power in all directions, now if I were to focus that to a point/small area I will get closer and closer to 1W but never above it. Where am I wrong?

Edited: I think it would be wise to clarify my misunderstanding now that I know the answer. The confusion in the above question stems from the fact that you can get a higher power density than input power. Of course, this doesn't violate any laws of physics as it's not the power that's higher but the area that gets smaller than m^2 so the density increases.

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    \$\begingroup\$ Ignore near field as that is not a point source and isotropic reference, EIRP includes gain beam width is define by half power threshold \$\endgroup\$ – Tony Stewart EE75 Aug 24 '20 at 3:01
  • \$\begingroup\$ High power density is always possible, but as Tony says it depends on the size of the near field region. For example, with visible light, you can imagine focusing a 10 W light bulb onto a tiny spot and achieving very high power density. But at 2.4 GHz, with a wavelength of 12cm, a 20 dBi antenna is at least half a metre across, so the beam is never smaller than this. Thus the power density at 1 m is only about 30 W/m2. \$\endgroup\$ – tomnexus Nov 30 '20 at 12:51
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When you have a "directional" antenna, it has a gain associated for every possible direction of arrival. When doing simple problems, a single value is quoted as the gain, and is usually the peak gain of the antenna and assumed that the target is in this direction, which is usually boresight.

In general, the gain of antenna is usually a function of spherical angles \$\theta\$ and \$\phi\$, \$G(\theta,\phi)\$. As mentioned before, an antenna's gain \$G_0\$ is usually quoted at boresight, or \$G(0,0) = G_0\$.

Remember that gain is how much better this antenna is at concentrating power in a particular direction when compared to an isotropic radiator, which radiates power equally in all directions.

The power density at a distance \$R\$ from a radiating antenna with transmit power \$P_t\$ and gain \$G\$ is given by

$$S_r = \frac{P_tG}{4{\pi}R^2}$$

Here, the direction is ambiguous. All we know is that from some generic direction, the antenna has a gain value of \$G\$.

In your case, your power density calculation is correct

$$S_r = \frac{P_tG}{4{\pi}R^2} = \frac{10(100)}{4{\pi}(100)^2} = 7.95 \space \frac{mW}{m^2}$$

Again, your value for EIRP is correct. If we want to get the same result with an isotropic radiator, then we are forced to set \$G = 1\$ (or 0 dB). Thus, we need a transmit power of 1 kW or 30 dB so that

$$S_r = \frac{P_t}{4{\pi}R^2} = \frac{1000}{4{\pi}(100)^2} = 7.95 \space \frac{mW}{m^2}$$

All this means is that for a low-gain antenna, we must compensate by increasing transmit power if we want to get the same power density at some desired distance.

As you already calculated, an isotropic radiator 1 m away with 1 kW of transmit power gives you

$$S_r = \frac{1000}{4{\pi}(1)^2} = 79.5 \space \frac{W}{m^2}$$

I think where your misunderstanding lies is in how the total transmit power is related to the power density. You transmit 1 kW, but that power is spread over an increasingly large spherical area. This is why as you get closer the power density increases, because the power has not spread out enough at a distance of 1 m versus 100 m. This is clearly seen by the expressions above where the denominator can be viewed as the surface of a sphere that has radius \$R\$.

When you introduce a directional antenna, the total transmitted power is distributed unequally in all directions since the antenna now focuses more power in a certain direction but not in others. Your goal is to deliver as much of the transmit power to a certain location at the cost of having to be accurately pointed. Using an antenna that is isotropic-like will spread power more evenly, but you must boost how much input power must be given to achieve the same result, everything else being equal.

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  • \$\begingroup\$ Overall I'm really happy with this answer but the last part doesn't sit well with me. The thing is, I don't "transmit 1kW", rather 10W but constrain it to a smaller portion of a sphere, then I say if I were to achieve similar power density from an isotropic source, it would have to output 1kW. It would make sense if I measured 80W/m^2 from a 1kW source, but if that 1kW is supposed to model the behavior of a directional 10W source then where is the missing power coming from? \$\endgroup\$ – dan dan Aug 24 '20 at 13:51
  • \$\begingroup\$ @dandan You are right, the real power you are transmitting is 10 W. The gain of an antenna will change how power is directed in a particular direction. So while your antenna uses a total of 10 W, the antenna will direct 20 dB (100x) more of that power in a particular direction when compared to an isotropic source. Take a lightbulb in a room that is emitting light and you stick a measurement device on a wall. Now put a reflecting surface by the lightbulb so that more of the light is directed at the device. Power hasn't changed, but is now being directed more towards the device: higher density! \$\endgroup\$ – Envidia Aug 24 '20 at 15:39
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I am assuming that if I input 1W into a 100% efficient isotropic antenna I get a total of 1W output power in all directions, now if I were to focus that to a point/small area I will get closer and closer to 1W but never above it. Where am I wrong?

With an isotropic antenna the total of 1 watt is spread out in all directions, so to collect the whole 1 watt the receiving antenna would have to completely surround the transmitting antenna. With a focused antenna it can be collected over a smaller area.

So while it is true that you can never receive more than 1 watt, you could (theoretically, with a narrow enough beam and large enough receiving antenna) collect that 1 watt at an infinite distance. That is how Voyager 1 is able to send a signal back to Earth from 22 billion kilometers away with only 22 watts of transmitter power.

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  • \$\begingroup\$ You've just rephrased what I said and attached a pointless trivia that can be easily deduced based on the aforementioned assumption. I'm sorry for being harsh but how does that help? \$\endgroup\$ – dan dan Aug 24 '20 at 3:20
  • \$\begingroup\$ You asked "the argumentation that it's just the same power as from the isotropic source, just focused down to a beam doesn't cut it for me... Where am I wrong?" But it is correct. So I rephrased what you said to (hopefully) reassure you that it is correct. Seems not though, so I might expand this answer when I get the time to do it. \$\endgroup\$ – Bruce Abbott Aug 24 '20 at 21:36
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If the energy is indeed focused, either by a dish parabolic or some yagi, then the main lobe will have high energy density, but as you model (or measure) slightly off bore-sight, the energy density will be much much lower.

In your example --- 79 watts/square meter --- the peak energy density will only exist for +-10 or +- 20 degrees around bore-sight, in both horizontal and in vertical axes.

At one meter distance, +- 10 degrees is only +- 15 centimeters; for both X and Y, the area is ( (2 * 15) * (2 * 15)) or 900 cm^2, compared to 10,000 cm^2 per square meter.

Where does this energy come from? from starving the other (360 - (2 * 10 )) or 340 degrees, in each axis.

Those other regions will be 0.1 watt/meter^2 or 0.01 watt/meter^2.

In general, the weaker the response of the off-bore-sight, the better. Especially for radar systems, that work with Range^4 issues, the side_lobes permit jammers to become a problem.

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