Given the following specifications, determine using the monostatic radar equation based entity required. Appropriate graphical data-sheets...

   

 

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Given the following specifications, determine using the monostatic radar equation based entity required. Appropriate graphical data-sheets are supplied and assume ko 1.3806488 x 1023 m² kg s² K¹ as for standard temperature, T = 290⁰K. Radar band: Operating frequency: Operating wavelength (WL): Peak transmitted power: Average transmitted power Pav Pulse width of transmitted pulse: T Pulse-width of the matched receiver: w = T Pulse repetition frequency (PRF): f, IEEE Standard f 2 Pr Duty cycle: PAV/PT = Duty cycle RCS: Probability of detection: Probability of false alarm: 0 Pa Pra (L through W) Hz meter watts watts second second pulse/second PRF XT meter² H Number of pulses integrated: Antenna gain (Monostatic system): Noise figure (NF) of the receiver: Receiver bandwidth Total system loss; RCS fluctuation model: Monostatic radar equation max = P,GA,onE, (n) S (k TBW)F, (S/N), = min P,GA,σ (4n) (k TBW)F(S/N), L (4n) x (k TBW)F (S/N) D I,(n): = n E,(n): (S/N), = n (S/N), - Modified: G=GT=GR Fn BW = 1/t Ls Swerling 1, 2, 3 Rmax Maximum range in meters Ae: Effective antenna aperture = (2²/47)G (G in ratio) Ei(n): Integration efficiency=(S/N)/(S/N)n = [(SNR with n = 1 pulse)/(SNR with n pulses)] to obtain a given probability of detection dB NF in dB Hz dB SW (Use Graph U.A supplied) [(SNR with n = 1 pulse] given pa and Pra) + Integration improvement factor under fluctuating target scenario (Use Graph U.B supplied for t [(SNR with n= 1 pulse] PGA o (47)×(Smin) Additional SNR needed for fluctuating moc (GRAPH U.C) Compensation for system loss Data Band S for λ NF 6 dB Example Problem with Solution Hints/Steps Pay fr PT STEP 1: Solution Hints ... 10° λ== 0.01 m W 3 GHz 100 kW Data (Continued) L, 6 dB X 10-13 STEP 1: Solution Hints ... 3.98 6 dB 3 (ratio) STEP 2: Calculate... Ac 0.8 m SW 3 (S/N) 14.6 dB For pa 0.95 and Pia 10 Use Graph U.B STEP 2: Calculate... Additional dB loss due to Swerling model fluctuations 5.6 dB from Graph U.C STEP 4: Calculate... Smin ? 14.75 76.5 KM Rmax ? 41.3 nmi T 0.5 μS 0.5 us 0.5 ×10 s 0 Pa Pra 1 m² 0.95 10 1 m² BW 1/t 2×106 Hz (S/N)-Modified 26.2 dB n 10 -8 0.95 10 10 using Graph U.A (D). (n = 10). (S/N) 9 W=T = 46.32 (ratio) 0.5 μs [(S/N)1] 1, (n = 10) G 50 dB 105 Group # 5: Data & Calculation of Parameter(s) Needed (?) Band f PT Pav fr σ Pd Pfa K 22 GHz 15 KW T ? 400 0.5 pulse/sus 0.3 W G T 0.90 10 10 0.5 n 33 us dB Given the following specifications, determine using the monostatic radar equation based entity required. Appropriate graphical data-sheets are supplied and assume ko 1.3806488 x 1023 m² kg s² K¹ as for standard temperature, T = 290⁰K. Radar band: Operating frequency: Operating wavelength (WL): Peak transmitted power: Average transmitted power Pav Pulse width of transmitted pulse: T Pulse-width of the matched receiver: w = T Pulse repetition frequency (PRF): f, IEEE Standard f 2 Pr Duty cycle: PAV/PT = Duty cycle RCS: Probability of detection: Probability of false alarm: 0 Pa Pra (L through W) Hz meter watts watts second second pulse/second PRF XT meter² H Number of pulses integrated: Antenna gain (Monostatic system): Noise figure (NF) of the receiver: Receiver bandwidth Total system loss; RCS fluctuation model: Monostatic radar equation max = P,GA,onE, (n) S (k TBW)F, (S/N), = min P,GA,σ (4n) (k TBW)F(S/N), L (4n) x (k TBW)F (S/N) D I,(n): = n E,(n): (S/N), = n (S/N), - Modified: G=GT=GR Fn BW = 1/t Ls Swerling 1, 2, 3 Rmax Maximum range in meters Ae: Effective antenna aperture = (2²/47)G (G in ratio) Ei(n): Integration efficiency=(S/N)/(S/N)n = [(SNR with n = 1 pulse)/(SNR with n pulses)] to obtain a given probability of detection dB NF in dB Hz dB SW (Use Graph U.A supplied) [(SNR with n = 1 pulse] given pa and Pra) + Integration improvement factor under fluctuating target scenario (Use Graph U.B supplied for t [(SNR with n= 1 pulse] PGA o (47)×(Smin) Additional SNR needed for fluctuating moc (GRAPH U.C) Compensation for system loss Data Band S for λ NF 6 dB Example Problem with Solution Hints/Steps Pay fr PT STEP 1: Solution Hints ... 10° λ== 0.01 m W 3 GHz 100 kW Data (Continued) L, 6 dB X 10-13 STEP 1: Solution Hints ... 3.98 6 dB 3 (ratio) STEP 2: Calculate... Ac 0.8 m SW 3 (S/N) 14.6 dB For pa 0.95 and Pia 10 Use Graph U.B STEP 2: Calculate... Additional dB loss due to Swerling model fluctuations 5.6 dB from Graph U.C STEP 4: Calculate... Smin ? 14.75 76.5 KM Rmax ? 41.3 nmi T 0.5 μS 0.5 us 0.5 ×10 s 0 Pa Pra 1 m² 0.95 10 1 m² BW 1/t 2×106 Hz (S/N)-Modified 26.2 dB n 10 -8 0.95 10 10 using Graph U.A (D). (n = 10). (S/N) 9 W=T = 46.32 (ratio) 0.5 μs [(S/N)1] 1, (n = 10) G 50 dB 105 Group # 5: Data & Calculation of Parameter(s) Needed (?) Band f PT Pav fr σ Pd Pfa K 22 GHz 15 KW T ? 400 0.5 pulse/sus 0.3 W G T 0.90 10 10 0.5 n 33 us dB

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To determine the entity required using the monostatic radar equation we need to use the provided dat... View the full answer