View
249
Download
5
Category
Preview:
Citation preview
Electronic Support Sensors
Short Course on Radar and Electronic Warfare
Kyle Davidson
Classes of ES Sensors
β’ Radar Warning Receiver (RWR)
β’ Electronic Support Measures (ESM)
β’ Electronic Intelligence (ELINT)
What are the challenges?
β’ Receiver sensitivity
β’ Probability of
Intercept (POI)
β’ Ability to discriminate an emitter within spatial and frequency coverage
Wide Open Receiver
Instantaneous Frequency Measurement (IFM)
Amplitude envelope and related power estimation (A, PW, DOA, TOA)
Crystal Detector
β’ Crystal/square law detector allows detection of the entire RF bandwidth, but eliminates any phase or frequency information
β’ Performs the following measurements:β Pulse amplitude
β Pulse width
β Direction of arrival (DOA)
β Time of arrival
β Antenna scan type
β Antenna scan period
ESM Architectures
ESM Receiver
β’ Typically super heterodyne
β’ Often uses a wideband IF switching amongst channels
β’ Compromise between POI and resolution
ELINT Receiver
β’ Focus is on a single signal at a time
β’ Selectable IF with a sample and hold receiver
β’ High gain antenna
POI vs Receiver Architecture
1. Omni directional UWB antenna, with a wideband receiver covering the entire RF bandwidth (100 % POI)
2. WB antenna with a narrow band receiver swept quickly across the RF spectrum (β 10 % POI)
3. Rotating high gain wideband antenna, with narrowband, selective receiver (β 2 % POI)
Who sees the other first
β’ Range Advantage Factor (RAF)
π π΄πΉ =π ππ π
= 1 + π
β’ π π β‘ RWR detection range
β’ π π β‘ radar detection range
β’ Closing velocityπ£π = π£π + π£π
β’ Warning Timeπ π = π π + π£πΆππ€
Understanding RAF
β’ Received Power
ππ =ππ‘πΊπ‘
4ππΏπ‘π 2
π
4ππ 2πΊππ
2
4ππΏπ
ππ =ππ‘πΊπ‘
4ππΏπ‘ππ 2
πΊππ2
4ππΏπ
β’ Assume a ππ· = 90% or ππΉπ΄ = 10β6
β SNR0 β₯ 13 dB
β’ For a good DOA assessment , SNR0 β₯ 18 dB
Understanding RAF
β’ Required radar receiver powerππ0 = ππ Γ SNR0 = πππ΅ππΉπ SNR0
β’ π΅π = equivalent noise bandwidthπ΅π = π΅π/πΊπ
β’ π΅π‘ = transmitter bandwidth
β’ πΊπ = processing gain
β’ Required EW receiver powerππ0 = ππ Γ SNR0/πΊππ
Range Equations
π π =4 ππ‘πΊπ‘πΊππ
2ππΊπ
4π 3πΏπ‘πΏπ πππ΅π‘πΉπ πππ 0
π π =ππ‘πΊπ‘ππΊππ
2ππΊππ
4π 2πΏπ‘πΏπ πππ΅π‘πΉπ πππ 0
β’ Note the radar antenna main lobe may not always be pointed at the target
πΊπ‘π = πΊπ‘/ππΏπΏ
Range Advance Factor
π ππ π
=πππ‘π
2πΊπ‘π2πΊπ
2πΊππ2π΅π‘
4ππΊπ‘2πΊππ΅π
2π
4
π =πΉπ
πΉπ2πΏπ
2
1
πππππ 0= 83 dBm/MHz
β’ Assumes the following β Monostatic
β Transmitter and receiver losses are equal
β πππ 0 = 13 ππ΅ = 20
β πΉπΈ β 5πΉπ
Case 1: Prior Gen EW & Radar
β’ Radar characteristicsππ‘ = 100 ππ = 80 ππ΅π;πΊπ‘ = πΊπ = 35 ππ΅π; π΅π‘= 1ππ»π§ = 0 ππ΅ππ»π§; π = 0.1; πΊπ = 13 ππ΅; πΉπ = 3 ππ΅; πΏπ‘ = πΏπ = 2 ππ΅; π πΆπ = 5 π2 = 7 ππ΅π2
β’ RWR characteristicsπΊπ = πΊππ = 0 ππ΅; π΅π£ = 20 ππ»π§;π΅π πΉ = 16 πΊπ»π§; πΉπ= 10 ππ΅; πΏπ = 2ππ΅;π΅π = 2π΅π£π΅π πΉ
1/2 = 800 ππ»π§
β’ With the above parameters π π = 188.4 ππ and π π = 3548 km in the main lobe and 64 km in the side lobe
Case 2: LPI Radar
β’ Radar characteristicsππ‘ = 100π = 80 ππ΅π;πΊπ‘ = πΊπ = 35 ππ΅π; π΅π‘= 500 ππ»π§; π = 0.03; πΊπ = 30 ππ΅; πΉπ = 3 ππ΅; πΏπ‘= πΏπ = 2 ππ΅; π πΆπ = 1000 π2
β’ RWR characteristicsπΊπ = πΊππ = 0 ππ΅; π΅π£ = 20 ππ»π§;π΅π πΉ = 16 πΊπ»π§; πΉπ= 10 ππ΅; πΏπ = 2ππ΅;π΅π = 2π΅π£π΅π πΉ
1/2 = 800 ππ»π§
β’ With the above parameters π π = 40 ππ and π π =35.5 km in the main lobe and 0.63 km in the side lobe
Pulses Received
β’ Number of pulses per unit time:π = ππππ πΉππ£π
β’ Where Nr is the number of radars
Chance of Coincidence
ππ
ππ
ππ₯
ππ
ππ₯ + ππRadar Scan
ES Scan
Probability of Coincidence
πππ + ππ₯ = πππβ’ Need to find the least common multiple of
m and n to find the coincident period
β’ For a common window duration
πΆ = π/ππ where ππ = πππ ππ , ππ
Coincidence Metrics
β’ The case of different window durations can be deduced by setting ππ = πΌπ and ππ = π½π
β’ It can be shown that the coincidence fraction for different window widths is
πΆ ππ , ππ =πΌπ½
βπ=ππππππππ
β’ The mean period between coincidences is
ππ =ππππππ + ππ
β’ The average duration of the coincidence is
ππ =ππππππ + ππ
Probability of Intercept
β’ Assuming the probability of coincidence, π π‘ , is independent from observation to observation
β’ The probability of a coincidence at time π‘ +Ξπ‘ is increased with respect to π π‘ by the ratio of the increment in time, Ξπ‘, to the mean period between coincidences ππ
π π‘ + Ξπ‘ β π π‘ =Ξπ‘
ππ1 β π π‘ Ξπ‘
Probability of Intercept
β’ Taking the derivative this givesππ π‘
ππ‘=Ξπ‘
ππ1 β π π‘ Ξπ‘
β’ This can then be solved for the probability of coincidence
π π‘ = 1 β 1 β πΆ ππ‘/ππ
Example of POI
β’ An EW sensor with a wide beam antenna and a fast stepped scan frequency receive able to cover 1 GHz of bandwidth in 100 MHz search steps, at a 10 ms duration per step.
β’ The targeted radar is a surveillance radar with a 2 degree beam width, rotating at 10 rpm
Example of POI (continued)
ππ = 10 ππ and ππ =1000ππ»π§
100ππ»π§10 ππ = 100 ππ
ππ = 6 π and ππ =2β
360β6 π = 33.3 ππ
The coincidence fraction is then:
πΆ ππ , ππ =ππππππππ
=0.033 Γ 0.01
0.1 Γ 6= 0.55 Γ 10β3
Mean period between coincidences ππ = 0.1 Γ 6 / 0.033 + 0.01 = 13.86 π ec
Average duration of coincidences ππ = 33.3 Γ 10 Γ 10β3 / 33.3 + 10 = 7.7 ππ
Time of InterceptπππΌ = 2.3ππ = 31.9 π ππ
Recommended