Some Stuff About RADARs, pt. 1 – Range, Power And Doppler

Life proceeds as it does. Working in a large corporate has its advantages, no doubt. But the technology industry has always been a risky one, no doubt. The shifting trends tend to exist in contradiction to the long haul hardware projects. Every small glitch in the economy is a trigger to downsizing steps, be it big or small. In my case? The (almost) complete elimination of the RF\Analog design group in my workplace.

So as a kick-start for whatever may come in the near future, I am starting to recollect some of my areas of expertise, in the form of a series (maybe multiple series?) of posts. This may be just for fun, for attention, or just a nice summary for myself to start preparing for whatever interviews may come. In this one, I am going to recollect a bit about some of the RADAR basics. I have a small fantasy of trying to actually build a working RADAR system, but do yourselves a favor, don’t hold me to it. On top of the amount of work required, building a well-isolated transmitter-receiver pair (required for any RADAR configuration) in home conditions next to impossible.

So what shall we start with today? Why not talk about how to even measure distance using propagating fields. I’m going to discuss EM fields, naturally, however the principles for acoustic distance measuring systems should be pretty similar. I’m not sure that the special relativity principles hold, though.

Propagation Time

So to kick it off, let’s talk about propagation time. A pulse of known length is transmitted from an Antenna with (hopefully) known properties. As demonstrated here, the pulse moves towards the target in a constant speed, which is the speed of light. The time that takes the pulse to return is measured, hence determining the distance by calculating

R = \frac{\Delta_T}{2c_0}

where c_0 is the speed of light in vacuum. It’s close enough to the speed of light in air, don’t worry. How this time is actually calculated? Not in this article.

Power And Range

Here’s the thing, though. This is not how field propagation works. The field source here is an Antenna. Generally speaking, RADAR Antennas are designed to radiate most of the power in a given direction. The proper terminology is called the Gain in that direction. This doesn’t mean that the Antenna gives actual power amplification, rather focuses the power in that direction. The two illustrations below show the difference between an Isotropic (Omni-Directional) Antenna and a Directional Antenna. All of you Antenna Engineers reading this now, ranting about me not explaining the difference properly between Gain, Directivity and whatever Realized Gain is, leave it be, please… This isn’t the main issue here.

So we have the transmitted power P_t, the transmitter (Tx) Antenna gain G_t. The pulse reflects from the target and returns to the receiver Antenna, that has its own Gain, G_r. Spoiler: I’m going to build the RADAR range equation now.

As shown in the Isotropic case, the fields dissipates in a perfect sphere. This assumes an isotropic medium, but we are in regular ole’ open space. Meaning that at any given point, starting from the point source, the power will decay by a factor of 1/4\pi R^2, where 4\pi R^2 is the sphere area. A more appropriate term, if you like, would be that the power spreads across the sphere.

Let’s build an expression for the power density at the target,

\frac{P_{target} [Watt]}{A [m^2]} = P_t G_t \frac{1}{4 \pi R_{Tx \rightarrow Target}^2}.

In order for us to determine what is the power at the receiver Antenna, we first need to resolve the unit crisis. A quantity, given in area units, namely m^2 is needed. Without proof at the moment, we will call this the RADAR cross section (RCS) of the target. It is commonly denoted by \sigma. This property is frequency dependent, as the wavelength varies with frequency. Now the power density at the receiver Antenna (Rx) can be determined by

\frac{P_{Rx} [Watt]}{A [m^2]} = P_t G_t \frac{\sigma}{4 \pi R_{Tx \rightarrow Target}^2} \frac{1}{4 \pi R_{Target \rightarrow Rx}^2}.

Again, a quantity is required to convert the units back to Watts. This quantity is well known in Antenna theory as an effective aperture, commonly denoted by A_e. Now we have a version of the RADAR equation. In the case where the Tx and Rx are in the same place, commonly known as the monostatic case, the equation takes the form of

P_{Rx} = P_t G_t A_e \frac{\sigma}{(4 \pi)^2 R^4}.

The relation between the effective aperture A_e and receiver Gain is

G_r = \frac{4\pi}{\lambda^2}A_e.

One more thing remains. Can a RADAR determine the speed of the target? Well, yes. A simple solution would be to send two pulses and measure the distance difference between them. Something like

v = \left|{R_2 - R_1}\right|/\Delta_T.

What if I told you it can be done with one? Let’s see how.

Doppler Effect

For much desired simplicity, let’s assume that the target is advancing in a single direction, \hat{x}. Namely, the time dependent range is given by R(t) = R + vt at any given measured range. Let’s assume that the transmitted field takes the form of

E_{inc} = E_0 f\left( { t - \frac{R(t)}{c_0} } \right) ,

where f(t + x/c_0)|_{x = R} is the modulated transmitted signal along the path. If you didn’t catch this correctly, worry not. I’ll show more practical examples in later posts. The reflected field takes the form of

E_{ref} = E_0 \Gamma \tilde{f}\left( { t + \frac{R(t)}{c_0} } \right) ,

where \tilde{f}(t + x/c_0)|_{x = R} is the backwards propagating signal and \Gamma is the reflection coefficient of the target, that has everything to do with the RCS quantity. To make it easier for us to determine the field\signal back at the receiver, the substitution \tau = t + R(t)/c_0 is suggested. At the point of interaction between the target the forward and backward propagating signal are equal, namely

\tilde{f}\left( { u } \right) = f\left( { \tau - 2\frac{R(t)}{c_0} } \right).

A bit of algebra yields

\tau - 2\frac{R(t)}{c_0} = \alpha \left({ \tau - \frac{R}{c_0} }\right) - \frac{R}{c_0},

where

\alpha = \frac{1 - \frac{v}{c_0}}{1 + \frac{v}{c_0}},

is the so called Doppler scale factor.

Almost done, breath deep! Now let us back-propagate the signal to the receiver. In this highly simplified case, at x = 0. Substituting back \tau = t + x/c_0 obtains

\tilde{f}\left( {t} \right)|_{x=0}  = f\left[ {\alpha \left({ t - \frac{R}{c_0} }\right) - \frac{R}{c_0}} \right] |_{x=0}.

It is finally time to start discussing the transmitted signal. It is composed of a low frequency signal, or base-band (BB) and modulated to a higher frequency, \omega_0. Hence it can be presented as the product

f(t) = s_{BB}(t)\cos (\omega_0 t).

For better understanding of what the Doppler scale factor actually does, let’s approximate it until the 1st order:

\alpha = 1 - 2\frac{v}{c_0} + \mathcal{O}\left( {\frac{v^2}{c^2_0}} \right).

Let’s look at the cosine part, or modulation part, of the signal:

\cos \left[ {\omega_0 \left( {\alpha t - \alpha\frac{R}{c_0} - \frac{R}{c_0}} \right)} \right] \approx \cos \left[ {(\omega + \omega_d)\left({t - \frac{R}{c_0} }\right) -\frac{R}{c_0} } \right] .

Ok, so we can see that the frequency shifts by the amount \omega_d = -2v/c_0, which is the so called Doppler shift. Wicked! This is that Doppler effect that (certain) Dad’s like to tell their kids every other time an ambulance drives by.

What about the BB signal? Well, a common assumption to make would be that due to the lower frequency it is bounded in, the Doppler shift is negligible for it. Namely, the received signal (and I am not discussing its amplitude right now) can be written as

f_{rec}(t) = s_{BB}(t)\cos \left[ {(\omega + \omega_d)\left({t - \frac{R}{c_0} }\right) -\frac{R}{c_0} } \right].

Let’s see if this approximation has any merit. For example, let’s take a RADAR that modulates its BB signal to the area of 10_{GHz}. This is a reasonable assumption, as there are many RADAR frequency bands in that area. A target moving in Mach 1, \approx 343_{m/sec} will shift the frequency by a factor of \approx 2.3\cdot 10^{-6}. Namely, this will reflect as a \approx 22_{KHz} shift (not perturbation!!) the BB signal and may allow detecting the speed of the target.

That’s enough for this post. In the next one, I’ll deal with how the received signal is processed and how to shape it. ‘Till next time!

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