# What is the power increase required to double the range of RADAR?

Studying for a Pivate Pilot's Licence (PPL) and I've hit something in the study material that doesn't make sense to me. Typical of the PPL study material, it's quite shallow, so the full explanation may have been omitted.

To double the range of a radio wave, you need 4x the power.

I understand this, because of the inverse square law.

To double the range of RADAR you need 16x the power.

This is where I'm stuck.

If a radar system has a range of 3km, the wave must travel 6km, so the received return has 1/36 the power of the transmission.

Now if I want to double the range of the system, to 6km, the wave must now travel 12km. Without power increase, the return signal has 1/144 the original transmission. To get back to 1/36, I need to multiply by 4. - Twice the range, 4x the power.

Are my PPL notes wrong, or am I misunderstanding?

• Radar systems work by reflecting a radio wave off an object, a 16x boost in transmitted power results in a 2x greater limit in signal range because there is a double path (to object, and returned, both inverse square law), so that would give you the 4x factor, and another 4x because the returning radio wave is diffused. Feb 27 '16 at 2:31

Here's what you are thinking: if the radar station and the target are 1km away, then the distance there-and-back is 2km. If the distance doubles to 2km, then the distance there-and-back is 4km, just twice what it was. So the there-and-back distance also doubles, so you should need only a 4x increase in power, right?

If the radar target were a flat reflector aimed in just the right direction, this would be true. A target can also employ a corner reflector, which works like a mirror but without the aiming requirement. $$P_\text{received} \propto {P_\text{transmitted} \over 2d^2}$$

That $\propto$ symbol means "proportional to". The power received is also going to depend on the gain of the antenna, the size of the target (more specifically, the solid angle subtended by the target), etc. But if all the factors (except distance) are held constant, then the relationship of power received to power transmitted will obey this proportionality.

A flat target, like a bathroom mirror, forms a virtual image behind it. So the radar station "sees itself" behind the target, twice the distance away. Since in radar the receiver and transmitter are the same station, we can also think of this as the virtual image transmitting to the real station.

But most targets aren't mirrors. The power they receive from the radar station isn't reflected back towards where it came from (like a mirror). Aircraft aren't shaped like mirrors. Since the wavelength of radar is many orders of magnitude greater than light, diffraction will also make the target less like an ideal mirror. Given all these variables, it's a reasonable assumption that the power intercepted by the aircraft will be scattered randomly in all directions. That is, it's a diffuse reflector. $$P_\text{received} \propto {P_\text{transmitted} \over d^2 \cdot d^2}$$

Here's one way to think of it:

Imagine the radar target isn't passively reflecting, but instead is a transmitter in itself.

If the target doubles its distance to the station, the target must now "transmit" with 4 times the power.

But targets aren't transmitters: they are passive reflectors of the power they intercept from the station. So to get 4 times the power at the target which is now twice as far away, you need to multiply the power by 4 again. 4*4 gives you 16.

Another example: a mirror can reflect nearly all of the power it receives from a laser pointer back at the source. A white sheet of paper receives the same power from the laser pointer, but that power is diffusely reflected everywhere, and so the power received back at the source is much less.

If it's still not making sense, think about how the reflection off the target affects the divergence of the beam. We can also do better than a flat reflector: we can use a parabolic reflector, with the radar station at the focal point: $$P_\text{received} \propto P_\text{transmitted}$$

As long as the antenna is in the focal point, a parabolic reflector will reverse the divergence of the beam. Thus, it will make a real image of the antenna right on top of the actual antenna. All of the power transmitted (in the direction of the reflector) is received, and there's no distance term at all!

• Firstly, excellent answer, exactly what I was hoping for. But suffer some ignorance a little further. Why doesn't your argument apply to a perfect reflector? I.e. a flat reflector 3km away receives a single 1/9th the original, the reflection must travel the same distance and so arriving at the origin again will be 1/9th as intense again. So 1/9 * 1/9 = 1/81 = 1/(3^4).
– Matt
Feb 25 '16 at 20:12
• Both my reasoning above and in my original question make sense to me, but arrive at different answers, hence my confusion. Your answer of perfect vs imperfect reflection makes sense, but isn't that a separate factor and not one that can be quantified so easily with a function such as square root? It would vary from object to object. e.g. why wouldn't it apply to the perfect reflector? Thanks :)
– Matt
Feb 25 '16 at 20:15
• @Matt Here's one thing you've missed: "a flat reflector 3km away receives a single 1/9th the original" is false. You can tell because the dimension of length (unit of km) disappeared. What you're being taught is a rule of proportionality; it is true that the power 3 km away from a transmitter is 1/9 of the power 1 km away. But you don't know what that power exactly is, and one of the factors that affects it in the radar case is the characteristics of the target — just like in a voice radio link the antennas on each end affect things. Feb 25 '16 at 20:28
• @Matt I added a few more examples. Make sense now? Feb 25 '16 at 21:03
• A corner reflector is the same as a flat reflector without an aiming requirement, right? Feb 25 '16 at 21:08

• Hello and welcome to ham.stackexchange.com! Jul 19 at 13:34

In addition to Phil Frost's answer. I believe what's missing is a coefficient sigma, which represents the amount of signal returned to the transceiver. Let's assume some units (m and m-squared if you like). Pictured below is a signal that at 1 unit from transmitter, is spread over 1 unit squared of area. At 3 units from the transmitter, covers 9 units squared of area. This is the basic inverse square law. (the intensity is spread over this area - hence the inverse part).

Now let's assume sigma is 1 unit square (the amount of signal returned - not necessarily equivalent to surface area of the target as it is also shape dependent - see Phil Frost's answer). The red illustrates a simple application of the same inverse square law and shows that the signal picked up at the point of transmission again will be proportional to 1/r^4.

The mistake I made in my original post was assuming that the entire signal is reflected. This was implicit in assuming that the return signal is equivalent to simply extending the signal over the same distance again. This would only be the case if sigma was proportional to r-squared. So if, at r = 3, sigma was 9 units square, then it would be equivalent to continuing the original signal over double the distance. (sigma sits on the top the intensity equation, so it it's proportional to r^2, the remaining r in the denominator is r^2).

To put it another way, if only a single unit square is returned, this unit square is projected onto 9 units square back at r=1 (shown in red). So the intensity is spread of 9 units square of area. If the signal continues to r=5 (same extra distance covered) that single unit square projects onto less than 9 units squared (9 units squared needs to project onto 25, so each can only project onto 25/9 units squared) so the signal here is more intense. (as expected - proportional to 1/r^2 not 1/r^4. Hope that makes some sense, anyone have any corrections?

• Did you mean, "At 3 units from the transmitter, covers 9 units squared of area." Mar 9 '16 at 22:38
• "The mistake I made in my original post was assuming that the entire signal is reflected." I'm not sure if that's all of the mistake, or at least, if that phrasing entirely captures it. It's not just how much of the signal is reflected, but how it's reflected. Think of a laser pointer hitting a mirror, versus a white sheet of paper. In each case, the entire signal is reflected, but the power received back at the laser pointer is very different. Mar 9 '16 at 22:44
• It's also the case that if you could focus the radar beam so tightly that all the radar's energy hit the target (ignoring for the moment that it would be impossible), then the energy received will still be $\propto 1/r^4$. Mar 9 '16 at 22:45
• Yes I meant 9,fixed that. I agree with your second point, I don't know if sigma accounts for that or if another variable accounts for the reflective nature of the material. For your third point, I'm not sure I follow. Wouldn't the at the target have lost no energy in that case, and only dissipate on the return, so 1/r²?
– Matt
Mar 11 '16 at 6:04
• "Wouldn't the at the target have lost no energy in that case" Yes, I concede you are right there. But also consider, this would mean the transmitter would need to focus the beam to an increasingly narrow beam as the target moves away, which is equivalent to increasing the antenna gain, so it's not an entirely apples-to-apples comparison. Mar 11 '16 at 15:02

To (roughly) double the range of radar, you not only need to double the range of the transmitted signal, but double the range of the reflected signal, in order to produce the same received signal level. A 4X power increase for the returning reflected signal on top of a 4x power increase for the initial transmitted signal thus requires 16X the transmit power for roughly the same received reflected power at double the range.