# Objective 3.9 — Electronic Warfare and Jamming
| LO# | Description |
|----------|----------|
| 3.9 | I can analyze jamming scenarios using the Friis equation, RADAR equation, and SNR. |
By the end of this reading you should be able to:
1. Classify a spectrum activity as **ES**, **EP**, or **EA**.
2. Compute the SNR of a link **under jamming** and judge it against the SNR the receiver requires.
3. Simplify the jamming SNR to its essential terms.
4. Compute the SNR of a **jammed radar** and find its **burn-through range**.
5. Explain **low probability of intercept** and how spread spectrum defeats a jammer.
## Electromagnetic Warfare Introduction
Electromagnetic warfare (EW) is defined as military action involving the
use of electromagnetic (EM) and directed energy to control the EM
spectrum or to attack the enemy.
The goal of EW is simple: keep the enemy from doing his job while making
sure we can still do ours. If the enemy needs their radios, we employ
jamming. If they jam us, then we must find a way to maintain
communications.
In our consideration of EW, we will not present any new equations, just
new ways of applying the ones you already know.
Nearly everything in EW has a countermeasure, and many countermeasures
have counter-countermeasures to work against them. It can quickly become
confusing without some way of categorizing EW. In accordance with Joint
Publication 3-85, dated 22 May 2020, EW includes three major
subdivisions:
- **Electromagnetic Attack (EA)** involves the use of EM energy,
directed energy, or antiradiation weapons to attack personnel,
facilities, or equipment with the intent of degrading, neutralizing,
or destroying enemy combat capability and is considered a form of
fires.
- **Electromagnetic Protection (EP)** involves actions taken to
protect personnel, facilities, and equipment from any effects of
friendly or enemy employment of the EM spectrum that degrade,
neutralize, or destroy friendly combat capability.
- **Electromagnetic Support (ES)** involves actions tasked by, or
under direct control of, an operational commander to search for,
intercept, identify, and locate or localize sources of intentional
and unintentional radiated EM energy for the purpose of immediate
threat recognition, targeting, planning, and conducting future
operations.
The topics discussed during this lesson will fall within one of these
three categories. Fortunately, the basic concepts of EW are fairly
simple, especially given the material you have already learned in this
course. However, as the nature of radio and radar propagation are
different in some aspects, we will consider two primary scenarios: radio
jamming, which is a type of *stand-off jamming*, and self-protection
radar jamming.
## Radio Jamming Techniques and Countermeasures
Radio jamming is a form of EA and is usually performed as *stand-off
jamming*. This means the jammer is separate from the transmitter and
receiver. Even though you may not immediately identify it with EA, most
people inherently understand jamming. For example, the home crowd at a
football game will attempt to jam the opposing team's quarterback by
making noise as he's calling the play. In ECE terms, we use the
signal-to-noise ratio (SNR) equation to describe the jamming:
$$SNR\ = \ \frac{P_{Signal}}{P_{Noise}}$$
In the football example, PSignal represents the volume of the
quarterback as he calls out his play to his teammates, and PNoise
represents the volume of the crowd. As the crowd gets louder, the SNR
becomes smaller, assuming the QB is already yelling as loud as possible.
At some point, the SNR becomes too small for the QB's team to hear him.
Now, we can translate this intuition to jamming a communications radio.
Figure 1 depicts a common jamming scenario. In this scenario, the
power received (and thus the PSignal in the SNR equation above) at the
receiving site is based on the Friis equation. Now, if you place a third
transmitter near the receiver, you can effectively reduce the SNR by
increasing the PNoise to a point where the receiver can no longer
receive the intended signal. If two transmitters with the same antenna,
power, and frequency characteristics both transmit a signal to a
receiver, the closer transmitter (the jammer) has a distinct advantage
in the Friis equation, since its range to the receiver is less than the
original transmitter.
Figure 1: A typical radio jamming scenario (stand-off jamming). Both
arrows end at the receiver: the range along the signal arrow is
$R_{txrx}$ and the range along the noise arrow is $R_{jam}$, both
defined below. The sketch is not to scale.
This example is similar to two people talking to you at the same time at
the same volume, but one is right next to your ear, while the other is
some distance away. Obviously, the person closer to your ear has the
advantage and can prevent you from hearing the desired transmitter
clearly.
For this same reason, a communications jammer on an aircraft gets more
effective as the aircraft gets closer to the receiver. Notice the jammer
must transmit on the same frequency as the transmitter that is to be
jammed, otherwise the signals will not couple.
Objective 3.8 put only thermal noise, $N = k\,T_{sys}\,BW$, under the
signal. A jammer adds the power the receiver collects from it, $J$, to
that noise. Writing $S$ for the signal power, the receiver works if and
only if
$$\frac{S}{N + J} \geq SNR_{\min}$$
where $SNR_{\min}$ is the SNR the receiver requires, the same number
with the jammer on or off. In every problem in this objective the
jamming is far stronger than the thermal noise. In Example Problem 1
below it is 21.62 nW, against the 11.04 fW of thermal noise in a 400 K,
2 MHz receiver: about two million times larger. So we drop $N$ and
treat the jamming as the noise: $SNR = S/J$.
*Vocabulary only, not assessed: EW references quote the
jamming-to-signal ratio $J/S$; jamming works when
$J/S > 1/SNR_{\min}$.*
A required SNR can be less than 1. It belongs to the receiver's
processing: a receiver that adds up many pulses, or de-spreads a
spread-spectrum signal (described later in this reading), can detect a
signal that arrives below the noise, so its $SNR_{\min}$ can be far
below 1. The receivers in Example Problems 1 and 3 are of that kind. A
receiver without that processing needs an $SNR_{\min}$ above 1.
To calculate the SNR, calculating the power of the signal and the power
of the noise independently -- using the Friis equation for both wireless
pathways -- is a perfectly valid approach. However, it can be tedious.
Fortunately, we can develop a simpler form of the SNR equation. We can
define the range from the transmitter to the receiver as Rtxrx, the
range from the jammer to the receiver as Rjam, and the power and gain
of the jammer as Pjam and Gjam, respectively. We then get:
$$SNR\ = \ \frac{P_{Signal}}{P_{Noise}} = \ \frac{\frac{P_{T}G_{T}G_{R}\lambda^{2}}{\left( 4\pi R_{txrx} \right)^{2}}}{\frac{P_{jam}G_{jam}G_{R}\lambda^{2}}{\left( 4\pi R_{jam} \right)^{2}}}$$
Since the frequencies are assumed to be the same, and since both the
transmitter and jammer are sending energy to the receiver, the GR
terms are also the same (we assume the receive antenna has the same gain
toward both), we can simplify to find the SNR equation for radio
jamming:
> **Key Concept**
>
> $SNR\ = \frac{P_{T}G_{T}R_{jam}^{2}}{P_{jam}G_{jam}R_{txrx}^{2}}\ $
The equation sheet writes this row as $SNR_{comm}$, with $R_{sig}$ for
our $R_{txrx}$, $P_{T,sig}$ and $G_{T,sig}$ for our $P_T$ and $G_T$, and
$P_{T,jam}$ and $G_{T,jam}$ for our $P_{jam}$ and $G_{jam}$. They are the
same quantities.
You may find this equation much more straightforward to use.
Furthermore, from this equation, it is easier to see how, as the jammer
gets closer to the receiver (i.e., Rjam decreases), the SNR also
decreases. This means our jamming becomes *more effective* the closer
our jammer is to the receiver.
Also, note the jamming platform can only control three variables in the
Friis equation:
- Amount of power in the jamming signal, Pjam
- Gain of our jamming transmitting antenna, Gjam
- Distance between our jammer and the object we are jamming, Rjam
:::{admonition} Key Concept
:class: key-concept
The jamming SNR is one Friis result divided by another. $G_R$, $\lambda^2$, and $(4\pi)^2$ cancel, leaving the equation sheet's $SNR_{comm} = P_{T,sig} G_{T,sig} R_{jam}^2 / (P_{T,jam} G_{T,jam} R_{sig}^2)$, where $R_{sig}$ is this reading's $R_{txrx}$. The cancellation assumes the receive antenna has the **same gain toward both** transmitters and that the jammer is **on the victim's frequency**, inside its bandwidth. Every gain is a linear ratio, and the two ranges only need the same unit.
:::
### Example Problem 1
A surface to air missile (SAM) site is receiving
position information about your aircraft from another location. You turn
on your jamming equipment to reduce the SAM's situational awareness. The
parameters for the radio transmitter and your jamming equipment are
indicated below. If the SAM site needs a minimum signal to noise ratio
of 0.06 to receive and process the intended communication signal, is
your jamming effective?
**Understand**: This is the scenario depicted in Figure 1, where
| SAM radio communications | Your jamming equipment |
|----------------------------|----------------------------|
| Power transmitted = 1 kW | Power transmitted = 2.0 kW |
| Transmit Antenna Gain = 3 | Transmit Antenna Gain = 20 |
| Receive Antenna Gain = 3 | Frequency = 750 MHz |
| Frequency = 750 MHz | Distance = 75 km |
| Distance = 80 km | |
We have two wireless channels -- one from the radio transmitter to the
SAM's receiver, and the other from your jammer to the SAM's receiver.
**Identify Key Information:**
- **Knowns:** We know the characteristics of the SAM transmitter and
receiver, as well as the specifications of the jamming system. We
also know the minimum SNR required by the SAM radio receiver.
- **Unknowns:** Power of the signal and power of the noise.
- **Assumptions:** The jammer is on the SAM radio's frequency, the
receive antenna has the same gain toward both transmitters, and the
jamming is far stronger than the receiver's thermal noise.
**Plan**: As both "communications links" are radio links, we can use the
Friis equation to determine the signal received at the SAM from the
intended source, as well as the noise received from the jammer. The
equation for SNR is straightforward:
$$SNR = \ \frac{P_{Signal}}{P_{Noise}}\ = \ \frac{P_{R - Radio}}{P_{R - Jamming}}$$
We can calculate Psignal and Pnoise separately, then calculate the
SNR.
**Solve:** First, we use the Friis equation to determine how much power
is received from the radio transmitter:
$$P_{R - Radio} = \ \frac{P_{T}G_{T}G_{R}\left( \frac{c}{f} \right)^{2}}{{(4\pi R)}^{2}}\ = \ \frac{(1000\ W)(3)(3)\left( \frac{3 \times 10^{8}\ m/s}{750 \times 10^{6}\ Hz} \right)^{2}}{(4\pi*80,000\ m)^{2}} = 1.425\ nW\ $$
Now, we can calculate the power received as a result of the jammer. In
essence, we use the jamming equipment as the transmitter and the SAM's
radio as the receiver. In this case, PT is the power transmitted by
the jammer and GT is the gain of the jammer. Since we are jamming the
SAM site, we use the SAM site antenna gain as the receiver gain, GR.
Finally, since the jammer is 75 km from the SAM site, our range is 75
km.
We can now calculate the amount of jamming power received by the SAM
site,
$$P_{R - Jamming} = \ \frac{P_{T}G_{T}G_{R}\left( \frac{c}{f} \right)^{2}}{{(4\pi R)}^{2}}\ = \ \frac{(2000\ W)(20)(3)\left( \frac{3 \times 10^{8}\ m/s}{750 \times 10^{6}\ Hz} \right)^{2}}{(4\pi*75,000\ m)^{2}} = 21.62\ nW\ $$
To calculate the SNR, we simply divide the power received from the radio
signal by the power received from the jamming signal:
$$SNR\ = \ \frac{P_{R - Radio}}{P_{R - Jamming}} = \ \frac{1.425\ nW}{21.62\ nW}\ = \ \ 0.0659$$
Therefore, our jamming is *not* effective, since we needed to get the
SNR below 0.06, but the actual SNR while jamming is 0.0659. Recall that
a higher SNR means a "better" communications link.
We can quickly validate this result by using the simplified SNR
equation:
$$SNR\ = \frac{P_{T}G_{T}R_{jam}^{2}}{P_{jam}G_{jam}R_{txrx}^{2}} = \ \frac{(1\ kW)(3)(75\ km)^{2}}{(2\ kW)(20)(80\ km)^{2}} = 0.0659$$
Both ranges went in as kilometers. They appear as a ratio, so they only
need to be in the same unit.
**Answer**: The jamming is not effective, since the actual SNR of 0.0659
is greater than the required SNR of 0.06.
### Example Problem 2
To increase the jamming effectiveness in the
previous problem, you fly closer to the SAM site. Is your jamming
effective when you are 50 km away from the site?
**Understand**: Distance is one of the three things jammers can control
to improve jamming effectiveness, so moving closer should help.
**Identify Key Information**:
- **Knowns:** In addition to the specifications of each system, we
also know the new range.
- **Unknowns:** The new SNR.
- **Assumptions:** None.
**Plan**: As we mentioned in the previous problem, the equation for
signal to noise ratio is
$$SNR = \ \frac{P_{R - Radio}}{P_{R - Jamming}}$$
Nothing has changed with the radio transmission, so
$$P_{R - Radio} = 1.425\ nW.\ $$
The only thing that has changed with the jamming is the distance.
**Solve**: Using the Friis equation:
$$P_{R - Jamming} = \ \frac{P_{T}G_{T}G_{R}\left( \frac{c}{f} \right)^{2}}{{(4\pi R)}^{2}}\ = \ \frac{(2000\ W)(20)(3)\left( \frac{3 \times 10^{8}\ m/s}{750 \times 10^{6}\ Hz} \right)^{2}}{(4\pi*50,000\ m)^{2}} = 48.63\ nW\ $$
Therefore, the new SNR is
$$SNR\ = \ \frac{P_{R - Radio}}{P_{R - Jamming}} = \ \frac{1.425\ nW}{48.63\ nW}\ = \ \ 0.02930$$
Again, we can validate this using the simplified SNR equation:
$$SNR\ = \frac{P_{T}G_{T}R_{jam}^{2}}{P_{jam}G_{jam}R_{txrx}^{2}} = \ \frac{(1\ kW)(3)(50\ km)^{2}}{(2\ kW)(20)(80\ km)^{2}} = 0.0293$$
**Answer**: At a distance of 50 km, the jamming is effective since the
new SNR is 0.0293 which is less than the required SNR of 0.06.
## Countermeasures for Communications Jamming
To prevent jamming, systems can implement one or more Low Probability of
Intercept (LPI) techniques. Three of the many LPI techniques used by the
military to protect their communications are chirp signals, frequency
hopping, and direct-sequence spread spectrum (DSSS). Strictly, LPI
means a signal is hard to *find*. These three also make it hard to
*jam*, because a jammer that cannot follow the signal has to spread its
power across the whole band. Transmitting no more power than the link
needs, and using a narrow antenna beam, are LPI techniques as well.
A *chirp signal's* carrier frequency starts at some point on the
frequency spectrum and sweeps up or down at a set rate. When it reaches
the end of its assigned spectrum, it jumps back to the beginning and
does it again. This is like changing the dial on an FM radio from one
end to another. As long as the transmitter and receiver are in sync and
move across the frequencies together, they are able to communicate. In
order to jam signals, the jammer must be at the same frequency as the
communications system. Therefore, by sweeping the transmission frequency
across a spectrum, chirp signals provide some jam resistance.
*Frequency hopping* entails the use of multiple carrier frequencies and
a pseudo-random noise (PRN) code to jump between various frequencies. Of
course, the transmitter and receiver must both jump to the same
frequencies at the same time. Therefore, like chirp signals,
transmitter-receiver synchronization is vital to successfully receive
the message. Frequency hopping is difficult to detect or jam since the
transmitter never stays on one frequency long enough to allow the jammer
to find the signal and effectively jam it. The quicker the frequency is
changed, the harder the signal is to detect or jam. As an example, the
U.S. military currently uses the SINgle Channel Ground and Airborne
Radio System (SINCGARS) as a primary combat net radio for voice and data
communications. SINCGARS includes an Electronic Counter-Countermeasures
(ECCM) frequency-hopping capability which hops frequencies at more than
100 changes per second across the 30 to 88 MHz frequency band. As this
is such a broad band, it provides significant counter-jamming protection
and ensures consistent, reliable battlefield communications.
Bluetooth is another example of a frequency hopping system. This system
operates in the 2.4000 -- 2.4835 GHz band (the same band as some
wireless routers, cordless phones, and microwave ovens), and the signal
hops 1,600 times per second to 79 different frequencies within that
band.
The third LPI technique is *direct-sequence spread spectrum (DSSS)*. The
digital signal to be transmitted is sampled at a much higher rate than
required (this is called *chipping*) and then multiplied by a PRN code
which makes the signal behave as a noise signal. The higher sampling
rate introduced by chipping causes the signal to occupy a much higher
bandwidth than the original signal, spreading out the transmitter power
making it very hard to detect. Additionally, spreading the power across
a large range of frequencies means the jammer has to transmit on a large
range of frequencies to jam effectively, which can be costly in terms of
equipment and power. The receiver applies the same PRN code, allowing it
to pull the signal out of the noisy received signal. That processing
is what lets such a receiver work with a required SNR below 1. This
received signal then passes through a low pass filter, which smooths out the
chips so the original bits sent by the transmitter can be recovered.
## Radar Jamming
Radar jamming is based on the same principles as communications jamming,
but is slightly different. With radar jamming, the radar returns are the
signal and the jamming is the noise. We will only consider a jammer
at the target's range. In *self-protection* jamming (also called
*self-screening*), the radar's target employs jamming to keep the radar
from effectively detecting it; an *escort* jammer flying with the
target works the same way. Because the jammer rides on the target or
flies with it, the echo and the jamming cross the same range $R$, and
the jammer sits in the radar's main beam. True *stand-off* radar
jamming, from well away from the target, needs two ranges and is
outside this course.
As with communications jamming, SNR is the key consideration:
$$SNR = \ \frac{P_{R - Radar}}{P_{R - Jamming}}$$
However, in this scenario, instead of using the Friis equation to
describe the power of the signal, we use the radar equation. Therefore,
the SNR becomes:
$$SNR = \ \frac{\frac{P_{RADAR}G_{RADAR}^{2}(RCS)\lambda^{2}}{{(4\pi)}^{3}R^{4}}}{\frac{P_{jam}G_{jam}G_{RADAR}\lambda^{2}}{{(4\pi)}^{2}R^{2}}}$$
With a little simplifying, we find the SNR equation for radar jamming.
> **Key Concept**
>
> $SNR = \ \frac{P_{RADAR}G_{RADAR}(RCS)}{P_{jam}G_{jam}4\pi R^{2}}$
The equation sheet writes this row as $SNR_{rad}$, with $P_T$ and
$G_T$ for our $P_{RADAR}$ and $G_{RADAR}$, and $P_{T,jam}$ and
$G_{T,jam}$ for our $P_{jam}$ and $G_{jam}$. Here $R$ must be in meters,
because the RCS is in square meters.
In the equation above, as R gets smaller, SNR gets *bigger*.
Therefore, as a jammer gets closer to a radar site, the jamming becomes
*less* effective, exactly the opposite of communications jamming. This
means at some point, the radar will *burn through* the jamming and now
detect the target. That point is the *burn-through range*: the range at
which the SNR reaches the minimum SNR the radar requires.
### Example Problem 3
A ground-based radar is searching for targets.
The radar transmits 3 kW, has an antenna gain of 2000, and requires an
SNR of at least 0.04 to detect a target. An aircraft with an RCS (σ) of
20 m² is trying to evade detection and has a jammer transmitting 100 W
with an antenna gain of 5. At what range does the jammer become
ineffective?
**Understand:** Since this is self-protection radar jamming, we know
there is a range at which the jamming becomes ineffective. That is, the
radar will *burn through* the jamming and be able to detect the target.
This is the range at which jamming will become ineffective.
**Identify Key Information:**
- **Knowns:** The transmission power, gain, and minimum required SNR
for the radar. We also know the transmission power and gain of the
jammer, and the RCS of the aircraft.
- **Unknowns:** The range at which jamming becomes ineffective.
- **Assumptions:** LOS is established and the radar has enough
power to see the targets at the given range.
**Plan:** We can use the SNR equation for radar jamming to solve for the
range at which the minimum SNR will occur. This is the range (between
the radar and jammer) at which jamming becomes ineffective.
**Solve:** The SNR equation for radar jamming is:
$$SNR = \ \frac{P_{RADAR}G_{RADAR}(RCS)}{P_{jam}G_{jam}4\pi R^{2}}$$
Solving for R, we have:
$$R = \ \sqrt{\frac{P_{RADAR}G_{RADAR}(RCS)}{P_{jam}G_{jam}4\pi(SNR)}}$$
Now, using the given parameters:
$$R = \ \sqrt{\frac{(3000\ W)(2000)(20\ m^{2})}{(100\ W)(5)(4\pi)(0.04)}} = 690.99\ m$$
**Answer:** The jamming becomes ineffective at 691 m.
Read that number against the geometry. An aircraft is never closer to a
radar than its own altitude, and even at 30,000 ft that is
$30{,}000 \times 0.305 = 9{,}150$ m. A burn-through range of 691 m is
never reached, so this radar never burns through: the jammer screens the
aircraft all the way in. A burn-through range this small is not a
mistake. It is what a jammer that works looks like.
:::{admonition} Key Concept
:class: key-concept
Against radar the jammer travels **one way** ($1/R^2$) while the echo travels **two** ($1/R^4$), so jamming wins at long range. The **burn-through range** is where the SNR reaches the radar's required $SNR_{\min}$: set the equation sheet's $SNR_{rad}$ equal to $SNR_{\min}$ and solve for $R$, in meters. Jamming does not move the radar equation's detection range. It adds a second, tighter limit, the burn-through range: a jammed radar detects only inside it.
:::
## Other Radar Countermeasures
Sometimes jamming may not be a possibility, but there are other things
we can do to protect against an enemy's radar. Consider the radar
equation again:
$$P_{R} = \ \frac{\mathbf{P}_{\mathbf{T}}G^{2}\mathbf{(RCS)}\lambda^{2}}{({4\pi)}^{3}*\mathbf{R}^{4}}$$
If our goal is to not be seen by the radar (which means reflecting the
smallest power possible back to the radar), there are three things we
can control:
- **R** -- distance from the radar site
- $\mathbf{RCS}$ -- radar cross section
- **PT** -- the amount of power transmitted by the radar
*Increasing distance* is fairly straightforward, meaning we reduce the
probability of detection by staying as far away from the site as
possible. This is one of the key roles the intelligence community plays
in electronic combat. If we know where the enemy radar sites are, we can
plan our routes to minimize exposure.
The modern military spent billions of dollars over the last few decades
to develop *stealthy* aircraft, meaning aircraft with very low RCS
values. The goal of these designs are to construct the angles of
aircraft so that energy is reflected away from the transmitting radar
and to coat the surface of the aircraft with highly radar-absorbent
materials. These are certainly effective strategies.
We typically use a single number to describe the RCS of any given
airplane, but doing so is not strictly accurate. The actual RCS depends
on the orientation of the airplane with regards to the radar system, as
well as other factors such as the load-out configuration (external fuel
tanks and weapons will increase RCS). Some aircraft, for instance, have
a lower RCS from the side than they do from the front. In such an
aircraft you can reduce your chances of being detected if you keep your
wing toward the radar site as much as possible.
One reason the US military consistently enjoys air superiority is we
combine these first two factors very well: using stealthy aircraft
flying carefully constructed flight plans around known threats.
In some situations, though, flying within the detection range of an
enemy radar is unavoidable. In this case, there are several potential
options.
The first two are related and are called *Suppression of Enemy Air
Defenses (SEAD)* and *Destruction of Enemy Air Defenses (DEAD)*. Both
are aimed at reducing the efficacy of enemy radar systems. SEAD can
include jamming enemy radar sites, or performing cyber attacks to take
them offline. In contrast, DEAD can include B-52 dropping bombs from
40,000 feet, a Special Forces team covertly taking out a SAM site from
the ground, or a submarine launching Tomahawk missiles. Although we
typically don't think of EW options as being kinetic, sometimes it can
be the best course of action.
Another more specialized kinetic option is the use of ***H**igh-speed
**A**nti-**R**adiation **M**issiles (HARMs)*, which can be carried by
aircraft such as the Navy's EA-18G Growler and the Air Force's F-16CM.
HARMs are designed to detect the
carrier signal of a radar transmission, locate its source, and destroy
the radar. When these missiles are in the air, the radar operator has
two choices: he can transmit and risk the missile destroying the radar
system or he can quit transmitting. Although HARMs can be thought of
primarily as DEAD, it can also serve as SEAD. In this case, it doesn't
actually matter if the radar is destroyed or not, the goal is to stop
the radar from transmitting long enough to accomplish the mission.
An interesting component to the use of HARMs is that it can be
psychological in nature. During the first Gulf War, for instance, some
American crews would simply broadcast "Magnum" over the radio whenever a
radar started transmitting. "Magnum" was the well-known code word for
launching HARMs. Usually, the radar operators would then cease
transmitting, not knowing there were no actual missiles inbound. Again,
regardless of whether the missiles were launched or not, the radars
stopped transmitting, leading to a significantly reduced probability of
detection.
Finally, we can tailor our flying tactics to reduce the probability of
radar detection. For example, recall the line of sight equation:
$$R = \ \sqrt{2\mathbf{h}_{\mathbf{RADAR}}} + \ \sqrt{2\mathbf{h}_{\mathbf{aircraft}}}$$
Since radar requires line of sight to detect a target, an attacking
aircraft can *fly lower* to avoid detection. Furthermore, the simplified
equation we use for line of sight is based on generally flat terrain. If
there are any hills or mountains, the aircraft can *terrain mask* by
working to keep terrain features between it and the radar.
The B-1 bomber was designed to do this very effectively. By flying very
low, and by virtue of a low nose-on RCS, it greatly limits the
opportunity for an enemy radar to detect it. Yet, even when detected by
a radar, the B-1 has another advantage. Before a SAM site can launch a
missile, several things need to happen. The site needs to see the
aircraft on radar, it needs to identify it as a threat, it needs to
target the missile, and then launch -- all of this takes time. However,
as it flies very fast, the B-1 can often cross through a SAM site's
effective engagement area before the SAM launches. Additionally, if
desired, the B-1 can jam the radar site to temporarily blind it, or it
can drop chaff to confuse the site as to the aircraft's actual location.
As discussed with the B-1, *chaff* can be used as a radar
countermeasure. *Chaff* is a bundle of metallic strips released into the
air. Each strip acts like a small antenna that absorbs the radar energy
and then retransmits it. The end result is that chaff creates a false
radar target that appears larger and brighter than the actual target,
thereby confusing the radar. This works particularly well for missiles
relying on radar to track their target, as chaff will distract the
targeting radar.
### Example Problem 4
An attacking force will be passing through an
area of radar coverage and desires to use chaff as a radar
countermeasure. If the radar operates at 2.5 GHz, how long should the
pieces of chaff be?
**Understand**: To be effective, the length of the chaff must depend on
the wavelength of the signal. In terms of what we have covered in this
class, think of chaff as a dipole antenna. In other words, the chaff
needs to be a *half-wavelength* to be effective.
**Identify Key Information**:
- **Knowns:** We know the frequency of the radar and the ratio of the
chaff length to the transmitted wavelength.
- **Unknowns:** The wavelength.
- **Assumptions:** None.
**Plan:** Since we want the chaff to be half of a wavelength, we can
simply calculate the wavelength of the frequency and divide by two.
**Solve**: The wavelength of the radar signal is
$$\lambda = \ \frac{c}{f}\ = \ \ \frac{3 \times 10^{8}\ m/s}{2.5 \times 10^{9}\ Hz}\ = 0.12\ m = 120\ mm.$$
Therefore, to be effective, the chaff needs to be half this length, or
60 mm long.
**Answer**: The chaff needs to be 60 mm long to be effective against a
2.5 GHz radar.
Another potential countermeasure is for aircraft to take advantage of
the limitation of a particular radar's range resolution. In this
scenario, aircraft can fly closer together than the range resolution and
mask their true numbers.
### Example Problem 5
A flight of four fighters wish to conceal their
numbers. If the search radar has a pulse width of 75 ns, how close must
the formation stay?
**Understand**: The radar resolution tells us the maximum separation of
the aircraft in order for them to appear as a single target to the
radar. This allows the formation to hide its true strength.
**Identify Key Information**:
- **Knowns:** We know the pulse width, *τ*, is 75 ns.
- **Unknowns:** The range resolution, $\Delta R$.
- **Assumptions:** LOS is established and the radar has enough
power to see the targets at the given range.
**Plan:** We can use the equation for radar resolution:
$$\Delta R = \ \frac{c\tau}{2}\ $$
where *τ* is the pulse width.
**Solve**: Solving for radar resolution.
$$\Delta R = \ \frac{c\tau}{2}\ = \ \frac{\left( 3 \times 10^{8}\ m/s \right)(75 \times 10^{- 9}\ s)}{2}\ = 11.25\ m\ $$
**Answer**: The aircraft's ranges from the radar must differ by less than
11.25 meters, for example by flying line abreast, for the formation to
appear as a single radar target.
Finally, many radar systems only track moving targets -- this is called
moving target indication (MTI). A previous reading explained how the
Doppler Effect may be used with a radar return to determine a target's
velocity. Consider again the Doppler equation:
$$f_{return} = \ f_{original}\left\lbrack 1 \pm \frac{2v\cos\theta}{c} \right\rbrack$$
If $\theta = 90^{\circ}$, then $\cos(90^{\circ}) = 0$, and we have:
$$f_{return} = \ f_{original}\left\lbrack 1 \pm \frac{2v(0)}{c} \right\rbrack = f_{original}$$
If a target is traveling perpendicular to the radar signal, then the
returned signal does not experience a change in frequency due to the
Doppler Effect. A radar measures only the part of the velocity that
lies along its line of sight, $v\cos\theta$, and with no shift that
part is zero:
$$v\cos\theta = \frac{c\left( f_{return} - f_{original} \right)}{2f_{original}} = \frac{c\left( f_{original} - f_{original} \right)}{2f_{original}} = 0$$
Seeing no motion toward or away from it, an MTI radar would determine
that the object is not a target of interest and would drop it from its
scope. One such radar was the AN/APY-7 mounted under the now-retired
E-8C JSTARS, whose ground moving target indicator (GMTI) mode tracked
moving vehicles and ignored everything that stood still.
Figure 2: A target moving perpendicular to the radar signal. The
figure's Θ is the angle θ in the Doppler equation.
Therefore, by knowing something about the orientation of the radar site,
an aircraft can fool certain radar systems into ignoring it. In reality,
an aircraft doesn't need to turn exactly 90° to disappear off of a
Doppler radar scope. Generally, a relative angle of about 80° to 100° is
sufficient.
::::{admonition} Type-along
:class: type-along
A threat radar transmits $P_T = 250$ kW through an antenna of gain $G_T = 2{,}000$ at 3 GHz. Its receiver has $T_{sys} = 400$ K and $BW = 2$ MHz, and it needs $SNR_{\min} = 2$. Your aircraft has an RCS of $10\ \text{m}^2$ and carries a self-protection jammer with $P_{T,jam} = 1$ W and $G_{T,jam} = 1$.
1. You are 50 km from the radar. What is $SNR_{rad}$? Does the radar detect you?
2. What is the burn-through range?
3. With your jammer off, how far out could this radar detect you?
4. A stronger pod has $P_{T,jam} = 20$ W and $G_{T,jam} = 5$. Find the new burn-through range. If you fly at 30,000 ft, can the radar ever burn through?
:::{admonition} Solution
:class: dropdown
**1.** With $R$ in meters, $SNR_{rad} = (250{,}000)(2000)(10)/[(1)(1)(4\pi)(50{,}000)^2] = 0.159$. That is below 2, so the radar does **not** detect you.
**2.** Set $SNR_{rad} = SNR_{\min}$ and solve for $R$: $R_{BT} = \sqrt{(250{,}000)(2000)(10)/[(1)(1)(4\pi)(2)]} = 14{,}100$ m $= 14.1$ km.
**3.** $P_{R,\min} = SNR_{\min}\,k\,T_{sys}\,BW = 2(1.38\times10^{-23})(400)(2\times10^6) = 22.08$ fW, and $\lambda = c/f = 0.1$ m. The radar range equation gives $R = [(250{,}000)(2000)^2(10)(0.1)^2/((4\pi)^3(22.08\times10^{-15}))]^{1/4} = 219$ km. A 1 W jammer pulls detection in from 219 km to 14.1 km.
**4.** The jammer's power times gain rose 100×, so the burn-through range falls by $\sqrt{100} = 10$: $R_{BT} = \sqrt{(250{,}000)(2000)(10)/[(20)(5)(4\pi)(2)]} = 1{,}410$ m $= 1.41$ km. At 30,000 ft you are never closer than $30{,}000 \times 0.305 = 9{,}150$ m $= 9.15$ km, which is more than 1.41 km, so the radar **never burns through**.
:::
::::
## Wrap-Up
**ES** senses, **EP** protects, **EA** attacks — and ES always comes first, because you cannot jam intelligently what you have not characterized. One rule decides every jamming problem: a receiver works when $S/(N + J) \ge SNR_{\min}$, and a jammer makes the thermal noise $N$ negligible. For a link that leaves the equation sheet's $SNR_{comm} = P_{T,sig} G_{T,sig} R_{jam}^2/(P_{T,jam} G_{T,jam} R_{sig}^2)$, so **geometry beats power**: both sides are fighting over $R^2$. For a radar jammed from the target's range it leaves $SNR_{rad}$, which grows as the target closes and reaches $SNR_{\min}$ at the **burn-through range**. **LPI** techniques deny the jammer the information it needs.
This objective uses everything in Block 3, and it is the last new material before **Quiz 10**. It also sets one of the limits in **Project 3**, the radar design lab: when the radar's band lies inside one of the bands the H-20's jammer covers, the H-20 adds a burn-through range $R_{BT}$, and each threat's detection range is the smallest of its noise-limited range, its line-of-sight range and, when jammed, $R_{BT}$.