LEC3-PULSE & CONTINUOUS WAVE RDR

3/24/2011 LECTURES �

RADAR TECHNOLOGY. PROF. A.M.ALLAM

PULSE & CONTINUOUS WAVE RADAR

Speed radar

ft

ft ± fd


1. Pulse radar (PR)

Block diagram of pulse radar

Duplexer

Timer

ATRModulator HPMW osc. TR

Indic. Vid.Amp. IFAEnv.Det. MIX PRE.SEL LNRF.AMP

LO

Monostatic noncoherent radar

ANT



Timer: generates series of narrow pulses to start both the transmission and the indicator time base simultaneously.

Modulator : generates the pulse wave form with high amplitude to switch onthe HPMW amplifier.

Duplexer( ATR-TR):

TR: disconnect the RX during transmission to protect its input stage from high transmitted power and reconnect the RX after a short time called recovery time trecov hence:

Rmin = C (� + trecov )/ 2

ATR: disconnect the TX during reception and the power is delivered to RX .

Antenna : used for both TX&RX, the reflected energy from the target enters the RX through the same antenna.

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LNRF Amplifier: Decreases the over all noise of the RX.

Pre-selector filter: suppresses the image frequency.

Mixer: provides the difference between the carrier and local oscillator frequencies which is the IF frequency .It is selected at theresonance circuit at the mixer output and amplified by the IF amplifier.

Envelope detector: it demodulates or removes the pulse from the IFcarrier.

Video Amplifier: Increases the pulse amplitude to the level sufficient the output display.

Plane position indicator: The start of the time base center of the screen issynchronized with the start of transmission by means of servo system, hence the angle of the time base is directly measure the target bearing and the distance of the bright spot from the center indicates the range.



The wave forms of the transmitted signals

� Pulse

Trigger

Modulated signal

time

time

time

Bearing

TargetRange

RADAR

0o

BearingOrigin

Plane Position Indicator

North

Range



Radar range equation

rARR

tGtPrP24

124

1

πσ

π=

σπ

λ43)4(

22

R

GtPrP =

where Gt is the transmitter gain,Ar is the effective area of receiving antenna and � is the effective area of the target.

Assuming same transmitter and receiver antennas, A/G=�2/4�

The received power



Receiver noise: is originated in the receiver due to thermal noise of the agitation of the electrons as well as the noise enters thereceiver through the antenna :

Thermal noise = kTBwhere k is the Boltzmann’s constant = 1.38x10-23

B is the noise bandwidthT is the temperature in degrees Kelvin

The total noise N = kTBFwhere F is the noise figure, in some cases one uses the equivalent temperature Te , where Te = FT and the total noise is:

N = kTe B If the received power is denoted by S, the signal to noise ratio is :

Receiver noise

2 2

3 4(4 )tPGS

N R FKTBλ σ

π=

�


Other forms for radar equation

2

3 4(4 )t t rPGGS

N R FKTBλ σ

π=

2

3 2 2(4 )t t r

t r

PGGSN R R FKTB

λ σπ

=

1-If the antenna gains are different:

2-If the target ranges from antennas are different:

3-If the looses in trans., recep. and emw radiation are considered (L):

2 2

3 4(4 )tPGS

N R FKTBLλ σ

π=

4-the dB form:

LH side (S/N) = 10logS – 10logN

RH side 10logPt+ 20logG + 20log� +10log�- 40logR– 30log4� -10logFKTB-10logL

�


ExampleA radar with antenna gain 1591.88 is transmitting 1kw at f = 3000 MHz. The effective noise temperature FT = 800o K. The RX noise bandwidth is 1000 Hz and the loss factor is 10. Determine S/N at radar target ranges of 20, 40 and 100 nautical miles. Assume the radar crossection is 1000 ft2.

SolutionG=1591.87, �=0.1m, R=20nm=20x1.85km,

S/N = [(1000)(1591.87)2(0.1)21000x(30.5/100)2]/[(4�)3(20x1.85x1000)4

(800x1000x(1.38x10-23)(10)]= 5741.2877

(S/N)dB = 10 log(5741.2877)= 37.6 dB

At R = 40 nm (S/N)dB = 10 log(5741.2877 /16)= 10log(358.83)=25.55 dB

At R = 100 Km (S/N)dB = 10 log(5741.2877/625)= 10log(9.186)=9.63 dB

�


2 CW Radar (CWR)

• Continuous wave radar transmits continuously.

• Unmodulated CWR can not measure the rang but it measures the relative speed of the target using Doppler effect.

• It can used by the police to measure the speed of vehicle.

• FM CWR can be used to measure the altitude of the air vehicle. This system is called low power radar altimeter.

Speed radar

ft

ft ± fd

ft ft ± fd

��


The Doppler effect

• The change in frequency due to the relative motion between the radar and the target. The greater the speed , the greater the frequency:

fr = ft ± fd + for closing target, - for departing target

• This shift is due to the change in the round trip transit time which result in the change of the distances between wave front of the reflected wave ( this distance decreases for closing targets i.e., the frequency is increasing and this distance increases for departing targets i.e., the frequency is decreasing ) .

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-Motion toward

-Echo frequency increases

-Motion away

-Echo frequency decreases

Wave frontsDistances between wave fronts

Tx

Echo


• Since the wave traveled 2R, then the total phase:� = � t = 2�ft = (2�C/�t )(2R/C) = 4�R/�t

• The maximum range in CWR is hold when �= 2� (unambiguous range) , then:

Rmax = �t /2The change in phase is fd and

fd = (1/2�)(d�/dt)=(2/�t)(dR/dt) = (2ft /C)(dR/dt)

Hence fd = (2ft /C)(dR/dt)

• dR/dt is determined by both the relative velocity between the radar andthe target and the direction of motion:

dR/dt = vcos�

� is the angle between the direction of motion and LOSvcos� is the relative velocity along the LOS

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Target velocity v

Line of sight

Target velocity v�=0o, cos� = 1

�

�=90o, cos� = 0

�

Target velocity v�=45o, cos� = 0.7 vcos�

CWR

CWR

CWR

LECTURES ��


ExampleA police radar transmitting 10GHz detects a car moving directly toward the radar at velocity 30m/s.Calculate the Doppler frequency?

� = 0o

fd = (2ft /C)(dR/dt)= (2ft /C)(vcos�)= ( 2x1010 /3x108)(30x1)= 2x103= 2 kHz

ft

Speed radar

ft ± fd

3/24/2011 LECTURES ��


Filter

AMP Mixer

Indicator

CW transmitter

Antenna

Circulator

ft

ft ± fd

fd

ft

Block diagram of CWR

3/24/2011 LECTURES ��


Transmitter: generates a continuous unmodulated RF signal of frequency ft

Mixer:mixes the received with the original signal and generates2ft , 2ft ± fd (high comps.), fd (low freq.) and ft-ft (dc component of fixed target).

Filter: filters the unwanted comps. and leaves fd

The sign of Doppler frequency is lost. To recover the sign we add more mixer and we use two signals 90o

out of phase as shown in the following diagram.

3/24/2011 LECTURES ��


Filter

AMP Mixer

Indicator

CW transmitter

Antenna

Circulator

ft

ft ± fd

fd

ft

MixerAMPFilter

90o

Indicatorfd

The output of the two channels is fed to a synchronous

two phase motor at which the direction of motion indicates

the sign

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Measuring range in CWR

• To measure the range, the transmitted signals must include some sort of timing marks or modulation to be able to measure the round trip transit time. We present two methods:

(1)Multiple frequency MFCWR:Using f1 , f2 separated by �f for transmission:

f1Sum. Amp. Power Amp.

Amplifier

Amplifier

Mixer

Mixer

Mixer

Phase Det.

fd1

fd2

��

�f

Rx Ant.

Tx Ant.

f1

f2

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V1T = Asin(2�f1t + �1)V2T = Asin(2�f2t + �2)V1R = Bsin[2�(f1±fd)t - 2�f1td� 2�fdtd+ �1]V2R = Bsin[2�(f2±fd)t - 2�f2td� 2�fdtd+ �2]The mixers separate the received waveforms and extract the two Doppler frequencies ( consider same Doppler frequencies and �f << f1 )V1d = sin[±2�fdt - 4�f1R/C� 2�fdtd]V2d = sin[±2�fdt - 4�f2R/C� 2�fdtd]�� = 4 �(f2-f1)R/C = 4��f R/C

R = C �� /(4��f)

The max. (unambig. range) occurs when �� = 2�

Rmax = C/(2�f)



The transmitted signal is continuous but constantly ( linear ) changing in frequency. Then there will be a difference in the instantaneous frequency between the transmitted and received signals.

Modulator FMCWR

Mixer Rx Ant.

Tx Ant.

HP Filter

fb

The output frequency of the mixer is the difference which is called the beat frequency fb

(2) Frequency modulated FMCWR:

3/24/2011 LECTURES


Signal of FMCWR f

t

t

t

t

fo

fb

fmax

fmin

�f

Modulation period 1/fm

fb

Beat frequencytd=2R/C

• �f is the frequency deviation

• fb is the beat frequency

The rate of change of transmitted

frequency over one period (1/fm)

equals 4 �f fm, i.e., dft /dt = 4 �f fm ,

consequently the beat within td is

fb =( 2R/C) dft /dt

fb = 8R �f fm /C



ExampleAn aircraft is equipped with FMCWR altimeter of modulating frequency fm = 1 KHz and the frequency deviation of 0.6 MHz.(a) Find the relation between fb and the range?(b) Calculate the range if the beat frequency equals 160 KHz?

Solutionfb = (8 �f fm /C)R

= (8x0.6x106x103/3x108)R= 16R

R = fb /16 = 160x103/16= 10 Km.

�


PR(1) Single Antenna

(2) Gives range & altitude

(3) Range dead zone( Rmin)(4) High power (expensive)(5) Wide bandwidth

(Subject to jamming, low S/N)

(6) Physical range is determined by PW and PRF

CWR(1) In some times it needs two

antennas(2) FMCWR gives range and

altitude(3) No dead zone.(4) Low power(5) Narrow bandwidth (difficult to

be jammed but can be deceived, high S/N)

(6) Physical range is determined by �f and fm

Comparison between PR and CWR

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3 Coherent & noncoherent radar

- Non coherent radar• RF signal is generated by HP RF oscillators• The phase from one pulse to another is not

preserved.• Uses power oscillator tubes like Magnetron.• Can not be used in Doppler radar.

Magnetron

Modulator HP RF OSC.

Pulse Gen

DC power

Pulses

LP pulses

HP radar pulses

fo

Pulse width

P.R.P


TWT

Klystron

Coherent radar-• RF signal is generated by LP RF oscillators• The phase from one pulse to another is preserved.• Uses power amplifier tubes like Klystron and TWT.• Can be used in Doppler radar.

t

t

t

LP pulses

fo

Master OSC. Amplifier

Pulse Gen

LP RF HP radar pulses

LP pulses fo

Question: Compare between coherent and non coherent radars?



4 Millimeter wave radar

• MMWR exactly the same as the discussed radar with the exception of incorporating atmospheric losses.

• MMWR calculation can be made without atmospheric loss and then add the that losses to the result.

• It is mainly used for short range detection and tracking.



• The antenna gain is related to � and the effective aperture as:

G = (4�/�2)Aeff

i.e. the gain is inversely proportional the �, shorter � ( higher f ) gives higher gain for the same antenna size .

• The antenna beam width is related to � as:

BW = (70 � / D)o

i.e. shorter ( higher f ) gives lower BW , hence higher angular accuracy.

• Unfortunately the atmospheric attenuation increases with the frequency, so that we should select certain windows in mm band with low attenuations.

Selecting the frequency band of MMWR

�


The attenuation variation versus the frequency the following:

Atmospheric attenuations

1) 3- 10 GHz negligibleattenuation.

2) 35,94,140 and 240GHz (10-1mm) windows of lowattenuation. Usually we use 94 &95 GHz.

3) 1-10.6 m another windows used for laser radar (1.06&10.6m laser radar are used for accurate range finding with high power and good coherence characteristics).

0.01

0. 1

1

10

100

1000

3x1063x1053x1043x103300303

INFRAREDLASER

MM

VISIBLE

One

way

att.

dB

/Km

Frequency [GHz]

1 0. 110100 1 10100mm m

�

Atmospheric attenuation



Under adverse weather condition (consider rain and fog ) the atmospheric attenuation changes as:

1) There is attenuation beyond 3GHz.

2) Fog does not affect themm band but it hasadverse effect on laser radar (m).

mmm

0.01

0. 1

1

10

100

1000

3x1063x1053x1043x103300303

INFRARED

LASERMM

VISIBLEOne

way

att.

dB

/Km

Frequency [GHz]

1 0. 110100 1 10100

�

Rain and fog attenuation

FOG

RAIN

��


The following table summarize the atmospheric effect such that the frequency 35 GHz is not badly affected by the adverse weather conditions like the short � of 94 GHz.

0.0300.0300.3500.3503.5003.5000.0030.003

0.0100.0100.0600.0600.6000.6000.0070.007

Fog(g/mFog(g/m33))0.010(light)0.010(light)0.100(thick)0.100(thick)1.000(dense)1.000(dense)Snow(0Snow(0ooC)C)

0.170.170.950.953.003.007.407.40

0.070.070.240.241.001.004.004.00

Rain (mm/hr)Rain (mm/hr)0.25(drizzle)0.25(drizzle)1.00(light)1.00(light)4.00(moderate)4.00(moderate)16.0(heavy)16.0(heavy)

35.035.04.04.0

5.145.140.500.50

Cloud: RainCloud: RainDryDry

0.40.40.120.12Clear airClear air

94GHz94GHz35GHz35GHz94GHz94GHz35GHz35GHz

MMWR dB/kmMMWR dB/kmAtmospheric Atmospheric conditioncondition

MMWR dB/kmMMWR dB/kmAtmospheric Atmospheric conditioncondition

Table of atmospheric attenuation loss of MMWR

Questions: Explain the effect of atmosphere and adverse weather condition on the selection of the MMWR frequency?

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ExampleA MMWR radar has the following parameters:frequency f=94 GHz, power=3W, antenna gain=50000, the radar crossection =30m2, noise figure= 7dB, loss factor =10 dB, PRF = 10KHz, pulse width = 50ns. Calculate the following:a) the range resolution?b) the noise BW?c) range if S/N = 10dB?d) the atmospheric loss for moderate rain condition?

Solution:

• �=3x108/94x109=3.19x10-3m.

• L= invlog(10/10)=10.

• Noise factor F=invlog(7/10)=5.01

• Room temp = 293ok , 2 2

3 4(4 )tPGS

N R FKTBLλ σ

π=

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a) The range resolution is Rmin=C�/2 = 3x108x50x10-9/2= 7.5 m.b) The noise BW = 1/�= 1/(50x10-9)=2x107 Hz.c) From the radar equation we can get:

S/N=(3x500002x(3.19x10-3)2x30)/((4�)3x1.38x10-23x293 x5.01x2x107x10)= 2.85x1014/R4

for S/N = 10 db = invlog(1) = 10R = (2.85x1014/10)=2310.53 m= 2.31 km

d) The atmospheric loss for two way path = 2x2.31x 3db/km (moderate rain) =13.86db.

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LEC3-PULSE & CONTINUOUS WAVE RDR

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