A bistatic radar system shown in figure has following parameters: f = 5 GHz, G_dt = 34 dB, G_dr = 22 dB. To obtain a return power of 8p W the minimum necessary radiated power should be

Concept:

For the Bistatic radar system, Power received at the radar is given by:

\({P_r} = \frac{{{G_{dt}}{G_{dr}}{\lambda ^2}\sigma {P_{rod}}}}{{{{\left( {4\pi } \right)}^3}R_{tx}^2R_{rx}^2}}\)

Where,

P_r = received power

G_dt = Gain of Transmitting Antenna

G_dr = Gain of Receiving Antenna

λ = operating wavelength

σ = Area of the cross-section of the Target

R_tx = distance of Target from transmitting Antenna

R_rx = distance of the target from Receiving Antenna

Calculation:

R_tx = 4 km

\({R_{rx}} = \sqrt {{4^2} + {3^2}} = \sqrt {25} = 5\;km\)

f = 5 GHz

∴ \(\lambda = \frac{C}{f} = \frac{{3 \times {{10}^8}}}{{5 \times {{10}^9}}} = 0.06\;m\)

G_dt = 34 dB

10 log₁₀ (G_dt) = 34

∴ G_dt = 10^3.4 = 2511.86 ≈ 2512

G_dr = 22 dB

10 log₁₀ (G_dr) = 22

∴ G_dr = 158.5

P_r = 8 PW = 8 × 10^-12 W

Therefore

\({P_r} = \frac{{{G_{dt}}{G_{dr}}{\lambda ^2}\sigma {P_{rad}}}}{{{{\left( {4\pi } \right)}^3}R_{tx}^2R_{rx}^2}}\)

\(8 \times {10^{ - 12}} = \frac{{2512 \times 158.5 \times {{\left( {0.06} \right)}^2} \times 2.4 \times {P_{rad}}}}{{{{\left( {4 \times 3.14} \right)}^3} \times {{\left( 4 \right)}^2} \times {{\left( 5 \right)}^2} \times {{10}^6} \times {{10}^6}}}\)

\(8 \times {10^{ - 12}} = \frac{{3440.03 \times {P_{rad}}}}{{792554.08 \times {{10}^{12}}}}\)

∴ \({P_{rad}} = \frac{{6340432.7}}{{3440.03}}\)

= 1843.13

P_rad = 1.843 kWatt

Since there is no option is matching.

Now if we are considering the R_tx as the distance between the transmitter and receiver Antenna.

\({P_r} = \frac{{{G_{dt}}{G_{dr}}{\lambda ^2}\sigma {P_{rad}}}}{{{{\left( {4\pi } \right)}^3}R_{tx}^2R_{rx}^2}}\)

\(8 \times {10^{ - 12}} = \frac{{2512 \times 158.5 \times {{\left( {0.06} \right)}^2} \times 2.4 \times {P_{rad}}}}{{{{\left( {4 \times 3.14} \right)}^3} \times {{\left( 5 \right)}^2} \times {{\left( 3 \right)}^2} \times {{10}^6} \times {{10}^6}}}\)

P_rad = 1.038 kWatt