# Lesson 39 -- Radio Detection And Ranging (RADAR) 2
## Learning Outcomes
1. Explain the RADAR parameters of range resolution, unambiguous
range, pulse width, pulse repetition interval (PRI), and pulse
repetition frequency (PRF) and how each parameter influences RADAR
performance.
2. Explain the concept of a Doppler shift.
3. Apply the Doppler equation to a given scenario to determine target
velocity or returned frequency.
## Radio Detection And Ranging (RADAR) 2
### Introduction
Now that we've introduced the basic concepts of RADAR, let's look at
some more advanced topics. Up to this point, we've assumed most RADARs
use the same antenna for both transmitting and receiving. We then
calculated how much power would actually return for some operational
scenarios. However, there's a key concept that complicates the scenario
a bit. Consider a RADAR that transmits 1 kW of power at a frequency of
300 MHz using an antenna with a gain of 150 towards an aircraft with a
RADAR cross section (RCS) of 100 $m^2$, which is 86 km away. How much
power will the RADAR receive and what will be the signal to noise ratio
(SNR) at the RADAR's antenna? Solving for the received power, using the
RADAR equation, gives:
$$P_{R} = \ \frac{P_{T}G^{2}\sigma\left( \frac{c}{f} \right)^{2}}{({4\pi)}^{3}R^{4}} = \ \frac{(1000\ W)(150)^{2}(100\ m^{2})\left( \frac{3 \times 10^{8}\ m/s}{300 \times 10^{6}\ Hz} \right)^{2}}{({4\pi)}^{3}{(86,000\ m)}^{4}} = 20.73\ fW$$
Now for many ultra-sensitive receivers, 20.73 fW is an acceptable power.
However, what if the RADAR is still transmitting its 1 kW of power at
the same time? If we treat this transmitted power as noise, we can
calculate a signal to noise ratio of:
$$SNR = \ \frac{P_{signal}}{P_{noise}}\ = \ \ \frac{20.73\ fW}{1\ kW} = 20.73 \times 10^{- 18}$$
This number is unacceptably small -- no receiver will be able to
distinguish the returned signal from the "noise" of the transmitter.
There are two distinct ways to solve this problem: use separate antennas
for transmitting and receiving (this is called *bistatic*) or send
pulses of energy, instead of a continuous stream, with a *monostatic*
RADAR.
While there are many bistatic RADARs in operation today, this
essentially doubles the required hardware and increases the complexity
of the processing algorithms. Some weapon systems simply cannot
accommodate the necessary hardware. Therefore, the second solution,
using a *pulsed RADAR*, is the most common. In the rest of this lesson,
we will discuss some of the ramifications of pulsed RADAR systems.
### Pulsed RADAR Parameters
A monostatic RADAR works by periodically sending a high-frequency pulse
(typically in the UHF to EHF frequency band) for a short period of time,
waiting for a return, sending out the next pulse, waiting for a return,
and so on. The Figure 1 shows the various parameters that determine how
this process works:
- **Pulse width or pulse length (τ, sec):** Amount of time the RADAR
transmits a pulse
- **Pulse Repetition Frequency (PRF, Hz):** The number of pulses
transmitted per second.
- **Pulse Repetition Interval (PRI, sec):** The duration of the
transmit and wait cycle. Related to the PRF by:
Key Concept — Pulse Repetition Interval
$$PRI = \frac{1}{PRF}$$
*Fig. 1. Pulsed RADAR parameters.*
Now, clearly, these parameters impact the performance of a RADAR. For example, a shorter pulse width leads to a smaller average power, which reduces the RADAR's maximum detection distance. However, making the pulse width too long presents a challenge with resolving objects that are too close together. Therefore, as with all engineering, we must carefully balance the trade-offs with our specific requirements.
### Maximum Unambiguous Range
The maximum unambiguous range for a RADAR is the maximum range at which a RADAR can both detect two objects and differentiate between them. This range is different than the maximum ranges calculated last lesson from the LOS and RADAR equations. Those equations tell us at what distance the RADAR will see anything. However, the concept of pulsed RADAR has now introduced a problem -- we have multiple pulses traveling through space at one time. As such, a RADAR's PRF (PRI) determines its *maximum* *unambiguous* *range*.
An example will help clarify the situation. Consider a RADAR system with
a PRF of 10 kHz, which results in a PRI of 100 μs. If the first pulse is
transmitted at *t* = 0 s, the pulse travels through the air until it
strikes an aircraft at a given time (in this case, *t* = 30 μs). We can
calculate the range of the aircraft by multiplying the time by the speed
of light ($30\ \mu s \times 3 \times 10^{8}\ m/s = 9\ km$). The signal
then reflects off the aircraft and travels back to the RADAR, a process
that takes another 30 μs. This means the pulse from this aircraft --
which is 9 km away from the RADAR -- will return 60 μs into the 100 μs
pulse interval. In this case, the RADAR signal can make its entire
two-way trip and get back to the antenna before the next RADAR pulse is
transmitted.
*Fig. 2. A RADAR pulse transmitted and received with no range ambiguity.*
Our range ambiguity appears when we receive echoes from farther away.
Consider the case where the aircraft is 18 km away from the RADAR. In
this situation, the first RADAR pulse doesn't reach the aircraft until
$t = 60\ \mu s$. That means the RADAR won't receive the return signal
until 120 μs. Therefore, the echo from the aircraft arrives at the RADAR
20 μs after the second pulse has been sent. How does the RADAR know
whether this echo came from the first pulse (from an object 18 km away)
or the second one (from an object only 3 km away)? In fact, a simple
RADAR can't distinguish between the two cases, because it will always
assume any received energy comes from the most recent pulse.
*Fig. 3. A RADAR pulse transmitted and received with range ambiguity.*
The maximum unambiguous range, $R_{unamb}$, is calculated from the PRI
as:
Key Concept — Maximum Unambiguous Range
$$R_{unamb} = \frac{c(PRI)}{2}$$
In other words, a RADAR's PRI setting directly determines the
unambiguous range where:
- If a RADAR target is beyond the maximum unambiguous range, then the
RADAR will not know exactly where the target is located; its
position will be ambiguous.
- If the RADAR target is inside (or right at) the maximum unambiguous
range, then the RADAR will know exactly where the target is located;
its position will be unambiguous.
In the previous example, the RADAR's maximum unambiguous range was 15
km, so the return from the aircraft at 18 km would actually be
interpreted as being only 3 km away. To help prevent this from
happening, many ground RADAR sites set their maximum unambiguous range
to be equal to their $R_{max}$(determined by LOS or by the minimum
detectable power, $P_{R,min}$) to prevent ambiguities.
In real RADAR systems, there are ways around range ambiguities. Modern
RADARs frequently stagger the interval between pulses to allow the
correct range to be determined. With a staggered PRF, a \'packet\' of
pulses is transmitted, and each pulse is on a slightly different
interval from the last pulse (or viewed a different way -- delayed
variable amounts from the reference trigger). At the end of the packet,
the timing returns to its original value, in sync with the trigger, and
the process repeats.
Example Problem 1
A ground RADAR station's antenna is located 15 ft
AGL and is looking for targets below 500 ft AGL. What PRF should be used
to prevent range ambiguities?
**Understand:** In this case, we want to set our maximum unambiguous
range to our maximum LOS range, so we can be absolutely sure we know
where any target in our detection range is. To do this, we will adjust
the RADAR's PRI (and therefore its PRF).
**Identify Key Information:**
- **Knowns:** We know the height of our RADAR tower and the maximum
height of any expected targets.
- **Unknowns:** The LOS range, PRI, and PRF.
- **Assumptions:** None.
**Plan:** First we will calculate the maximum LOS range, based on the
highest expected target (500 ft AGL). Then, we will use that to
calculate the longest possible PRI.
**Solve:** We need to calculate the antenna's maximum LOS range first:
$$r_{LOS,max\ } = \ \sqrt{2h_{RADAR}}\ + \ \sqrt{2h_{target}}\ = \ \sqrt{2*15\ ft} + \sqrt{2*500\ ft}\ = \ 37.1\ miles = 59.73\ km$$
We will need to set our PRI such that the maximum unambiguous range is
equal to the LOS range.
$$PRI = \ \frac{2R_{unamb}\ }{c} = \frac{2(59.73\ km)\ }{3\ x\ 10^{8}\ m/sec} = 398.2\ \mu s$$
We can now calculate the PRF:
$$PRF = \frac{1}{PRI} = \frac{1}{398.2\ \mu s} = 2.51\ kHz$$
**Answer:** The ground RADAR station should be set to a PRF of 2.51 kHz.
### RADAR Range Resolution
In addition to being able to unambiguously identify a target's location,
a RADAR's ability to resolve two objects separated by a given distance
(∆*R*) is called the RADAR's *range resolution.* Range resolution is
limited by the duration of each pulse, known as the pulse width (*τ*). A
longer pulse means there's more energy in the received signal, which
increases the RADAR's range. However, a wider pulse also causes less
resolution in range information.
Consider a RADAR with PRF of 10 kHz and PRI of 100 μs, which sends out a
10 μs-long pulse. The transmitted wave strikes two separate aircraft --
the first at *t* = 25 μs (7.5 km) and the second at *t* = 40 μs (12 km).
The RADAR's receiver will then receive two distinct returns, the first
at 50 μs and the second at 80 μs, meaning the RADAR will have no trouble
resolving the aircraft as unique.
*Fig. 4. A RADAR resolving two aircraft separated by more than the minimum range resolution.*
However, if the separation between the two aircraft reduces to 1.2 km,
we will still receive two returns, but they are so close they overlap
and can't be distinguished. For example, the pulse hits the first
aircraft at *t* = 25 μs (7.5 km) and the second aircraft at *t* = 29 μs
(8.7 km). The returns arrive at the receiver at *t* = 50 μs and *t* = 58
μs, respectively. Because the pulses are now overlapping, the RADAR will
only see one continuous pulse. A RADAR operator may suspect that there
is more than one aircraft out there, but there is no way to determine
how many there are and how far apart they are from each other.
*Fig. 5. A RADAR unable to resolve two aircraft because they are separated by less than the minimum range resolution.*
A RADAR's pulse width determines the range resolution, ∆*R*:
Key Concept — Range Resolution
$$\Delta R = \frac{c(\tau)}{2}$$
In other words, a RADAR's pulse width setting directly determines ∆*R*
where:
- If a RADAR's targets are within ∆R of each other, the RADAR will not
know exactly how many targets exist or exactly where the targets
are.
- If the RADAR's targets are separated by more than ∆R, the RADAR will
be able to resolve them as individual entities.
In the example above, the RADAR's resolution is 1.5 km, so when the
distance between the aircraft closed to 1.2 km, the planes were no
longer seen as being separate.
Example Problem 2
Two inbound aircraft have a nose-tail separation
of 20 m. If a tracking RADAR has a pulse width of *τ* = 1 μs, will it be
able to tell if there is more than one airplane?
**Understand:** The pulse width of the RADAR determines the RADAR
resolution.
**Identify Key Information:**
- **Knowns:** We know the pulse width of the RADAR and the separation
of the aircraft.
- **Unknowns:** The range resolution for this pulse width.
- **Assumptions:** None.
**Plan:** Calculate the range resolution of the RADAR and determine if
the aircraft are separated by more than this distance.
**Solve:** In the given problem, *τ* = 1 μs. We can use this to solve
for the RADAR resolution,
$$\Delta R = \ \frac{c\tau}{2}\ = \ \frac{\left( 3 \times 10^{8}\ m/s \right)(1 \times 10^{- 6}\ s)}{2}\ = 150\ m$$
Therefore, with a pulse width of 1 μs, the RADAR cannot distinguish
between objects with less than a 150-m separation. Since our planes are
20 m apart, they will be seen as a single object.
**Answer:** No, the RADAR will not be able to tell that there is more
than one airplane, because its RADAR resolution, Δ*R*, is 150 m.
### Target Speed Determination
RADARs are well known for their ability to calculate the relative
velocity of targets. To do so, they measure the frequency shift in a
returned signal relative to the known frequency of the transmitted
signal. It may seem counterintuitive that a system could receive a
different frequency than it transmitted. What could cause the frequency
of the RADAR return to differ from the transmitter frequency? This
phenomenon is known as the *Doppler shift*. For example, a passing siren
has a higher pitched tone as it approaches than it does after it passes
by and moves away. The sound waves are traveling at a fixed speed, but
the waves are compressed as the car approaches, creating an increase in
frequency. As the car passes by, the waves stretch so that the peaks
spread apart. This causes a decrease in the frequency. For a simple
system such as that shown in Figure 6, the Doppler shift only occurs on
the one-way trip from the source to the observer. However, in a RADAR,
the Doppler shift is experienced twice: once as the wave strikes the
target, since, in the plane's frame of reference, the RADAR is moving
and the plane is stationary; and again as the wave returns to the RADAR,
as now, in the RADAR's frame of reference, the plane is moving and the
RADAR is stationary.
*Fig. 6. Doppler Effect as experienced by a target moving towards the observer.*
For a stationary target, a RADAR return a pulse would be
$V_{m}\cos{(360{^\circ}f_{o}t)}$ where $f_o$ is the RADAR's transmitter
frequency. However, an approaching target with a closing speed of *v*
(m/s) will return a pulse of
$$V_{m}\cos\left\lbrack 360{^\circ}\ f_{o}\left( 1 + \frac{2v}{c} \right)t \right\rbrack$$
A retreating target will present a negative Doppler shift, leading to:
$$A\cos\left\lbrack 360{^\circ}\ f_{o}\left( 1 - \frac{2v}{c} \right)t \right\rbrack$$
If the target is not traveling straight toward the RADAR, but rather at
an angle θ, the frequency for closing targets will be:
Key Concept — Doppler Shift (RADAR)
**Closing target:**
$$f = f_{o}\left( 1 + \frac{2v\cos\theta}{c} \right)$$
**Retreating target:**
$$f = f_{o}\left( 1 - \frac{2v\cos\theta}{c} \right)$$
where θ is the angle between the target's velocity vector and the line from the target to the RADAR.
By determining the difference between the transmitted frequency and the
frequency of the received RADAR return, the RADAR system can determine
the relative speed with respect to the RADAR.
Example Problem 3
A supersonic aircraft is approaching a RADAR
system at a speed of 400 m/s with an approach angle of 30°. If the RADAR
uses a frequency of 1 GHz, at what frequency will the signal return?
**Understand**: The Doppler shift will help us determine the received
frequency. Note that, for a RADAR, the Doppler shift is a two-way shift.
Also, since the airplane is moving towards the RADAR, the frequency will
increase. If the airplane were moving away, the frequency would
decrease.
**Identify Key Information**:
- **Knowns:** We know the velocity of the incoming aircraft and the
angle of its velocity vector with respect to the RADAR. We also know
the transmitting frequency of the RADAR system.
- **Unknowns:** The expected frequency of the RADAR return.
- **Assumptions:** None.
**Plan:** We can use the RADAR Doppler equation to calculate the
expected return frequency.
**Solve**: Using the RADAR Doppler equation,
$$f = \ f_{o}\left\lbrack 1 + \ \frac{2v\cos\theta}{c} \right\rbrack = (1\ GHz)\left\lbrack 1 + \ \frac{2(400\ m/s)\cos(30{^\circ})}{3 \times 10^{8}\ m/s} \right\rbrack = 1.000002309\ GHz$$
**Answer**: The return signal will be at a frequency of 1.000002309
GHz.
Notice the use of significant figures here. Since the real information
is in how much the signal changes (and the change in frequency is very
small), we need to keep enough digits to capture that change.
Example Problem 4
An airplane is flying overhead with an approach
angle of 60°. If a RADAR signal departs at 300 MHz and returns at
299.999875 MHz, how fast is the airplane travelling?
**Understand**: We can use Doppler shift to determine the speed of
objects.
**Identify Key Information**:
- **Knowns:** We know the transmitted and received frequencies and the
angle of the velocity vector with respect to the RADAR.
- **Unknowns:** The aircraft's velocity.
- **Assumptions:** None.
**Plan**: First, we will solve the Doppler equation for velocity. Then,
we will use the given values to determine the velocity.
**Solve:** Solving for velocity, our original equation (for a closing
target) becomes:
$$v = \frac{c(f - \ f_{o})}{{2f}_{o}\cos\theta}$$
We can now use this equation to solve for v:
$$v = \ \frac{c(f - \ f_{o})}{2f_{o}\cos\theta} = \ \frac{(3 \times 10^{8}\ m/s)(299.999875\ MHz - \ 300\ MHz)}{2(300\ MHz)\cos{60{^\circ}}} = \ - 125\ m/s$$
Notice that the velocity is a *negative* 125 m/s, which means the
target is *moving away* from the RADAR.
**Answer**: The airplane is moving away from the RADAR with a velocity
of 125 m/s.
## Key Takeaways
- **Pulsed RADAR.** A monostatic RADAR uses the same antenna for transmitting and receiving by sending short pulses and listening for echoes in between, avoiding the otherwise catastrophic SNR problem of transmitting and receiving simultaneously.
- **Pulse width ($\tau$).** The pulse width sets how long the RADAR transmits each pulse; a longer pulse carries more energy (improving detection range) but degrades range resolution, requiring engineers to balance the two needs.
- **Pulse Repetition Frequency and Interval.** PRF is the number of pulses per second and PRI = 1/PRF is the time between the start of successive pulses; both directly determine the RADAR's maximum unambiguous range.
- **Maximum unambiguous range.** $R_\text{unamb} = c \cdot PRI / 2$; targets beyond this range produce echoes that arrive after the next pulse is transmitted, causing the RADAR to misidentify their range.
- **Range resolution ($\Delta R$).** $\Delta R = c\tau / 2$; two targets separated by less than $\Delta R$ cannot be distinguished as separate objects, so a shorter pulse width is required to resolve closely spaced targets.
- **Doppler shift.** A moving target changes the frequency of the returned RADAR signal by $f = f_0[1 \pm 2v\cos\theta / c]$; a closing target increases the frequency while a retreating target decreases it, allowing the RADAR to calculate target velocity.
- **Perpendicular motion and Doppler blind spots.** A target moving exactly perpendicular to the RADAR beam ($\theta = 90°$) produces zero Doppler shift and appears stationary, which is exploited tactically to evade moving-target-indicator RADARs.