Answered: For the module and radar of problem 13…

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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Q14 For the module and radar of problem 13 (included below), what module power should be used to minimize the array aperture area? What would be the array area and number of elements?Prob 13 (An X-band (λ = 3 cm) T/R module uses a power amplifier that exhibits a power added efficiency on transmit of nominally 33% over a peak output power range from 0.1 W to 10 W at 20% transmit duty cycle. In the transmit mode the module also requires 250 mW of DC power to power the background control electronics. In the receive mode the module requires 500 mW of DC power, including background control electronics. A radar is required to have a power-aperture-gain product of 80 dBWm2. Assume nominally 4,000 elements per square meter as an element density to satisfy scan volume requirements. Tabulate and plot the DC prime power required by the array for peak module powers from 0.1 W to 10 W and examine the results.)

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Hints:
Namely, P = Pe x Ne and A = Ae x Ne where Ne is the number of elements and Pe is the power per
element and Ae is the area per element.
PAG = 80 dBWm2 = 10^8 and there are 4,000 elements in 1 m2. With the latter piece of info, we
can calculate Ae, the area per element.
G = 4 p A/12
With this information, we can now substitute for P and A and G and you'll end up with Pe and Ne as
unknowns. If you assume that Pe = 0.1 W, you can find Ne. Similarly, if increase Pe to 0.2W, 0.3W,
..., up to 10W, you can find the associated Ne and then solve the other part of the problem.

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