RADAR (Radio Detection and Ranging) RADAR EQUATIONS no appreciable directivity and the power density of its Ap gives,

Also, having Rmax in terms of

radiation at the radar station is
- A system that measures distance (and usually Pt Ap At
P
direction) to an object by determining the amount of PD   1
time required by an electro-magnetic energy to travel 4 R 2 (4 R 2 ) 2  P A A2  2 4
and return from the object. Rmax   t t3 p 
 (4 ) P '
Radar basic types fall into two broads categories: The reradiated power intercepted by the receiving  min 
antenna will be
A. CW (continuous wave) radar–CW or continuous
wave, Pt Ap At Ac Where
radars are so called because the transmitters used in P  PD Ae 
these The power density at the target will be (4 R 2 ) 2 Rmax  maximum range
systems transmit almost continuously.
Where Pmin  minimum received power
One of the simplest forms of CW radar is so Pt Ap
called
PD  Pt  peak pulse power of the transmitter
doppler radar, wherein any reflecting object moving
4 R 2 P power received by the radar antenna
toward or away from the transmitter will return echo where PD  power density at the radar At  radar cross-section of the target
signals slightly different in the frequency from the PD  power density
station Ap  antenna power gain
transmitted frequency due to doppler shift.
Ap  antenna power gain Ae  effective capture area of the   operating wavelength
The other most familiar type of CW radar is
frequency modulated, or FM radar. This is type of Pt  peak pulsed power of the receiving antenna
EXAMPLE
radar is differ from the first, because the frequency Calculate the maximum range of an X-band radar
modulation signal transmitted from an fm system transmitter For the radar antenna, the effective capture area is
which operates at 10 GHz with peak pulse power of
permits range measurement to be made on both R  distance of target from 750 kW, if its minimum receivable power is 1013 W,
moving and stationary targets.
the radar Ap  2 the effective capture area of its antenna is 5 m and
Ae  the radar cross-sectional area of the target is 25 m.
B. Pulse radar – as the name implies, the 4
transmitted energy from a pulse radars system is in Since the target has an effective radar cross-section,
the form of short burst of RF energy. the power impinging on the target will be SOLUTION
Considering the same antenna is used for both
reception and transmission. So, 1
Pulse radar system is divided into two general Pt A p At  P A A 4
groups: (1) primary pulse radar and (2) secondary P  PD At  Pt At Ae A 4 Rmax   t 2e t 
pulse radar. 4R 2 P  e2  4 Pmin 
(4 R )2 2
 1
Primary pulse radar where
- is the name given to the simple echo system
P power received by the target  4
wherein the target reflects the transmitted energy Pt At Ae2  
from the system and the time lapse between At  radar cross-section of the target P  (750k )(25)(5)
2

transmission and reception is measured and 4 R 4  2 Rmax  
8 2
converted to range information. One of the limitations
 4  3  10  (1013 ) 
is that it does not supply positive identification of Solving for the maximum range (the received power   10G  
targets. is equal to the minimum received power of the    
receiver) gives,
Secondary pulse radar Rmax  802.365 km
- radar used to overcome identification problem 1
and it was developed to supplement primary radar  P A A2  4
information. Rmax   t 2t e  EXAMPLE:
 4 Pmin  A radar transmitter has a power of 25 kW and
operates at a frequency of 11.5 GHz. The radar pulse
The target will cause this power to be reflected reflects from a target 20 km away with a radar cross-
isotropically, since the target is considered as having
page 1
section of 10 m². Calculate the received signal power cT EXAMPLE: Pave
if the antenna gain is 20 dB. R A pulse radar emits pulses with a duration of 5 µs dc 
2 and a repetition rate of 1500 Hz. Calculate the Ppk
SOLUTION: where maximum and minimum range for this radar. where
R distance (range) to the target
SOLUTION: dc  duty cycle
Pt A  At
2
p
2
c speed of light
Pave 
P For the maximum unambiguous range average power
(4 )3 R 4 T round trip time taken for the radar
c
signal to return Rmax  Ppk  peak pulse power
First, solve for the wavelength 2f
EXAMPLE: EXAMPLE

c 3  108
f
Calculate the range of a target when the between the Rmax  A radar transmitter has a peak pulse power of 500
transmission and reception of the signal is found to 2(150) kW, a PRR of 1000 pps and a pulse duration of 0.85
be 167 µs. µs. Determine:
3 108 Rmax  100 km a) the maximum unambiguous range

11.5G SOLUTION: b) the duty cycle
cT and for the minimum range c) the average transmitted power
  0.0261 m R
2
cW SOLUTION
and express the antenna gain as a power ratio (3  108 )(167  ) Rmin  a)
R
2 2 cT
20dB = 10 log Ap Rmax 
R  25.05 km (3 108 )(5 ) 2
Rmin 
 20  MAXIMUM UNAMBIGUOUS RANGE 2  1 
Ap  anti log   (3  108 )  
 10  Rmin  750 m  1000 
cT Rmax 
Ap  100 Rmax  2
2 Rmax  150 km
Also, DUTY CYCLE
for the received power
c b)
Rmax 
2 2 2f dc = W x PRR
25k (100) (0.0261) (10)
P where dc = (0.85µ)(1000) x 100
(4 )3 (20k ) 4 dc = 0.085%
Rmax  maximum unambiguous range
P  5.358 10 W 15
c  speed of light c)
T  pulse period Pave
P  5.358 fW f  pulse repetition rate dc 
Ppk
MINIMUM USABLE RANGE W Pave  dc  Ppk
PULSE RADAR DC 
PRT
cW also Pave  (0.00085)(500k )
Rmin  DC  W  PRR Pave  425W
2 where
where DC = duty cycle
Rmin  minimum usable range W = pulse width (duration)
CW DOPPLER RADAR
PRT = pulse repetition time
c speed of light PRR = pulse repetition rate
W pulse duration 2VR cos 
Similarly, the ratio between the average power and
fD 

peak power may be used in expressing in terms of
the duty cycle,
where
page 2
 f D  frequency change between the transmitted 
c
and reflected signal dc  W ( PRR )or f
VR  relative velocity between the radar set and Pave 3  108
dc  
the target Ppk 5GHz
 angle between target direction and radar set
dc  1µ(900)   0.06 m
  0° if the radar is in line with the direction of
the target dc  900 106

EXAMPLE 45 1(0.06)(3k )
900 s  Vb 
A police radar operates at a frequency of 1.024 GHz Ppk 2
in direct line with a car. The reflected signal from the Vb  90 m/s
45
car is shifted 275 Hz in frequency. Determine the Ppk 
speed of the car in miles per hour. 900 
Ppk  50 kW
SOLUTION
fD
VR  MTI RADAR
2
but,
- a type of radar that employs pulsed radar
c technique and Doppler shift to minimize clutter
 effects and detect only targets that are moving.
f
3 108
 BLIND SPEED
1.024G
- The relative velocity that results in a Doppler
 0.293 m frequency equal to the PRF or some multiple of it.

(275)(0.293) n PRF
VR  Vb 
2 2
VR  40.283 m/s where

Converting to mi/hr Vb  blind speed

n = any integer (1,2,…)
λ = operating wavelength
m  1km  1mi   3600s 
VR  40.283     PRF = pulse repetition frequency
s  1000m  1.609km   1hr 
EXAMPLE
VR  90.13 mi/hr Determine the lowest blind speed for MTI radar
operating at 5 GHz when the PRF is 3 kHz.

EXAMPLE
(ECE Board Problem, November 2002) What is the SOLUTION
peak power of a radar having a pulse width of 1µs
and PRR of 900 Hz? The average power is 45 W? n PRF
Vb 
2

SOLUTION But the wavelength is

page 3