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 of2 kW (peak power). If it is desired to detect a target with a cross section of 12 m², and the minimumdetectable signal is Pmin=-90 dBm, what is the maximum range of the radar?Ans:8114 m. Palsed vadar systemf = 10GHz5 = 12m² Phến -90dBmorR4&Radar equation Pr = Pt G² 2²0(47)³ R4PminG: 1028/102 2= 10€so R4.G= 28 dBPEYZPrPtG²²0Pr (47)³=DJ= 631- 90/10mW3x1081010PE = 2 kW=) R = 8114 m= 10-12= 0.0 3 мW2000 x (631) ² x 12 x (0.03)²(47)³x10-12or 8.1 km= 12m²

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

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

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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Question

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.

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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