How did they get G= 10 and Pmin= 10 ?

Introductory Circuit Analysis (13th Edition)
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How did they get G= 10 and Pmin= 10 ?
Exercise 17.
A pulse radar operating at 10 GHz has an antenna with a gain of 28 dB, and a transmitter power of
2 kW (peak power). If it is desired to detect a target with a cross section of 12 m², and the minimum
detectable signal is Pmin=-90 dBm, what is the maximum range of the radar?
Ans:
8114 m.
Transcribed Image Text:Exercise 17. A pulse radar operating at 10 GHz has an antenna with a gain of 28 dB, and a transmitter power of 2 kW (peak power). If it is desired to detect a target with a cross section of 12 m², and the minimum detectable signal is Pmin=-90 dBm, what is the maximum range of the radar? Ans: 8114 m.
Palsed vadar system
f = 10GHz
5 = 12m² Phến -90dBm
or
R4
&
Radar equation Pr = Pt G² 2²0
(47)³ R4
Pmin
G: 1028/10
2 2
= 10
€
so R4.
G= 28 dB
PE
YZ
Pr
Pt
G²²0
Pr (47)³
=
DJ
= 631
- 90/10
mW
3x108
1010
PE = 2 kW
=) R = 8114 m
= 10-12
= 0.0 3 м
W
2000 x (631) ² x 12 x (0.03)²
(47)³x10-12
or 8.1 km
= 12m²
Transcribed Image Text:Palsed vadar system f = 10GHz 5 = 12m² Phến -90dBm or R4 & Radar equation Pr = Pt G² 2²0 (47)³ R4 Pmin G: 1028/10 2 2 = 10 € so R4. G= 28 dB PE YZ Pr Pt G²²0 Pr (47)³ = DJ = 631 - 90/10 mW 3x108 1010 PE = 2 kW =) R = 8114 m = 10-12 = 0.0 3 м W 2000 x (631) ² x 12 x (0.03)² (47)³x10-12 or 8.1 km = 12m²
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