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TT The half-power beamwidths are equal to HPBW (az.) 2 [ 90°- sin (1/2)] = 2(90°-30°) = 120° HPBW (el.) = 2 [90°- sin' (12)] = 2(90°-30°) = 120° In a similar manner, it can be shown that for (b) U=sine sin' Do=5.09 = 7.07dB HPBW (el.) = 120°, HPBW(az.) = 90° (c) U= sine sin³ => Do = 6 = 7.78 dB HPBWCel.) = 120°, HPBWCA)= 74.93° (d) u = sine sing Do = 12/8=471 = 6.73 dB HPBWCel.) = 90°, HPBW Caz.)=120° 01-S (e) u sine sin³ Do = 6=778 dB, HPBW (az.) = HPBWCel) = 90°

TT The half-power beamwidths are equal to HPBW (az.) 2 [ 90°- sin (1/2)] = 2(90°-30°) = 120° HPBW (el.) = 2 [90°- sin' (12)] = 2(90°-30°) = 120° In a similar manner, it can be shown that for (b) U=sine sin' Do=5.09 = 7.07dB HPBW (el.) = 120°, HPBW(az.) = 90° (c) U= sine sin³ => Do = 6 = 7.78 dB HPBWCel.) = 120°, HPBWCA)= 74.93° (d) u = sine sing Do = 12/8=471 = 6.73 dB HPBWCel.) = 90°, HPBW Caz.)=120° 01-S (e) u sine sin³ Do = 6=778 dB, HPBW (az.) = HPBWCel) = 90°

Introductory Circuit Analysis (13th Edition)
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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I need  details for the steps azimuthal,elevation

TT The half-power beamwidths are equal to HPBW (az.) 2 [ 90°- sin (1/2)] = 2(90°-30°) = 120° HPBW (el.) = 2 [90°- sin' (12)] = 2(90°-30°) = 120° In a similar manner, it can be shown that for (b) U=sine sin' Do=5.09 = 7.07dB HPBW (el.) = 120°, HPBW(az.) = 90° (c) U= sine sin³ => Do = 6 = 7.78 dB HPBWCel.) = 120°, HPBWCA)= 74.93° (d) u = sine sing Do = 12/8=471 = 6.73 dB HPBWCel.) = 90°, HPBW Caz.)=120° 01-S (e) u sine sin³ Do = 6=778 dB, HPBW (az.) = HPBWCel) = 90°
Transcribed Image Text:TT The half-power beamwidths are equal to HPBW (az.) 2 [ 90°- sin (1/2)] = 2(90°-30°) = 120° HPBW (el.) = 2 [90°- sin' (12)] = 2(90°-30°) = 120° In a similar manner, it can be shown that for (b) U=sine sin' Do=5.09 = 7.07dB HPBW (el.) = 120°, HPBW(az.) = 90° (c) U= sine sin³ => Do = 6 = 7.78 dB HPBWCel.) = 120°, HPBWCA)= 74.93° (d) u = sine sing Do = 12/8=471 = 6.73 dB HPBWCel.) = 90°, HPBW Caz.)=120° 01-S (e) u sine sin³ Do = 6=778 dB, HPBW (az.) = HPBWCel) = 90°
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