# How much power does a receiver antenna obtain from a radio station?

I have this problem where a transmitting antenna radiated uniformly in all directions at a radius of r. The station broadcasts at 10 kilowatts. How much power is obtained by a receiver antenna 20km away? The antenna of the receiver is 50cm^2. I am unsure of how to set up an equation for this problem, I know the distance and power, but how would I set this in a equation to find the power obtained by a receiver?

• You must start here: en.wikipedia.org/wiki/Poynting%27s_theorem Commented Sep 28, 2017 at 19:15
• hint: google free space path loss. Commented Sep 28, 2017 at 19:15
• Probably you need to start with right equation before "setting it up". What does your textbook say? Commented Sep 28, 2017 at 19:17
• 1. find the area of a sphere 40km in diameter. 2. find the proportion of your antenna's area to that spherical area. 3. multiply the proportion by 10kw. hint: it's a small number. Commented Sep 28, 2017 at 21:22

If the antenna radiates in all directions (isotropic radiator), you can consider it as the center of a sphere, where the radius is the distance to the RX. If there isn't attenuation on the path, the power remains the same, then divide by the sphere area and you got power/area relation.

Knowing antenna's area, you can calculate the power received.

• Would I leave the power as kilowatts and distance in km? Commented Sep 28, 2017 at 19:29
• @Kytex Checkout the question!!! In what units is the response expected? Commented Sep 28, 2017 at 19:38
• Doesn't say which is why I asked, I will just leave it as is then. Oh and after getting the relation, would I multiply by the antenna area? Commented Sep 28, 2017 at 19:39
• @Kytex The SI units are [W/m²], for power density and, of course, [W] for power. Convert units first, and make the calculations to get the answer in [W]. Commented Sep 28, 2017 at 19:40

Here is a practical example. I have a cluster of FM broadcast antennas 5.25 miles away from my home, or 8400 meters, seven or eight masts. Their total transmission power is about 640 kW, all in the FM range 96 – 100 MHz.

In first approximation, let’s assume an uniform emission into a sphere with radius 8400 meters. The surface area of a sphere is S = 4*pi*R^2, = 9e+8 (=886233600) m2, and the RF emission will be spread out evenly. Again, assuming no losses or absorbers on the way, the power density at my location should be about 640000/88623360 = 7.22e-4 W/m2, or about 700uW/m2. This is assuming the spherical space.

For commercial broadcast antennas they use an array of vertical dipoles, so they won’t emit up or down, but mostly into horizontal plane. The antenna gain will be at least 2 or more, I assume 3x. So the field is likely about 0.7mW x 3 = 2mW/m2 at my location.

If a receiver loop is, say, ~ 3cm x 3cm = ~10 cm2, it covers 0.001m2. So the 10 cm2 loop will get 2uW.

Now a practical question, is it much, or not really? For example, how much voltage can be registered by an oscilloscope from this loop? This can be a tricky part. One way is to assume that the wire loop gets the power from free space which is having an impedance of 300 Ohms. Then the open loop will produce about 25 mV RMS:

P= V^2/R, so V^2 = 2e-6*300 = 6e-4; so the sqrt() gives V = 25 mV.

Surprisingly, this is close to what I can see on all my scopes, for example:

This is very annoying when dealing with low-voltage signals, the interference is everywhere. Note to retired engineers and serious hobbyists: when shopping for a home to live, pay attention to broadcast towers around.

I have this problem where a transmitting antenna radiated uniformly in all directions at a radius of r. The station broadcasts at 10 kilowatts. How much power is obtained by a reciever antenna 20km away? The antenna of the reciever is 50cm^2.

If the transmitter emits uniformly in all directions then all the 10 kW transmitted is passing notionally through a surface area of $4\pi \times20,000^2$ square metres. That's the surface area of a sphere of radius 20 km.

This means that your receiver only gets a tiny fraction because it has an area of 0.25 square metres.

So, 10,000 watts x 0.25/($4\pi \times20,000^2$) = 487 nW.

The power per square metre is clearly 4 times bigger at ~2 $\mu W/m^2$ and, given that the impedance of free space is 377 ohms you can calculate the local E field at the receiver as being 27.4 mV/metre using P = V^2/R. Not a bad sized signal really. H field is 377 times lower at 72.6 uA/m.