short course on radar and electronic warfare kyle...
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Electronic Support Sensors
Short Course on Radar and Electronic Warfare
Kyle Davidson
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Classes of ES Sensors
β’ Radar Warning Receiver (RWR)
β’ Electronic Support Measures (ESM)
β’ Electronic Intelligence (ELINT)
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What are the challenges?
β’ Receiver sensitivity
β’ Probability of
Intercept (POI)
β’ Ability to discriminate an emitter within spatial and frequency coverage
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Wide Open Receiver
Instantaneous Frequency Measurement (IFM)
Amplitude envelope and related power estimation (A, PW, DOA, TOA)
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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
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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
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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)
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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€
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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
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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
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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
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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
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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
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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
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Pulses Received
β’ Number of pulses per unit time:π = ππππ πΉππ£π
β’ Where Nr is the number of radars
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Chance of Coincidence
ππ
ππ
ππ₯
ππ
ππ₯ + ππRadar Scan
ES Scan
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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
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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
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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‘
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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
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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
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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