Notes 21 example Spring 2005

# Notes 21 example Spring 2005 - In figure b the antenna is...

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Notes 21 Spring 2005 Example : The antenna impedance of a half-wave dipole antenna also contains a reactive component, , in addition to the radiation resistance, . For this type of antenna . Including an ohmic antenna resistance of, say ant X rad R = 5 . 42 X ant 485 . 0 , the antenna’s impedance is + = 5 . 42 j 2 . 73 Z ant . For impedance matching puposes it is convenient to operate the antenna where its reactive component is zero. This can be achieved by making the antenna slightly shorter. An antenna length of λ = 485 . 0 d will result in a zero reactance and its resistive component will be about 73 . In figure a., below, we have a length half-wave dipole antenna transmitting with a power source that consists of a 12V ac supply in series with a λ 485 . 0 25 source resistance. We wish to determine the total power radiated from the antenna with and without an impedance matching network.
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Unformatted text preview: In figure b. the antenna is not matched to the line so a calculation for the current yields mA 122 73 25 12 R R V I rad s s s = + = + = The radiated power then is calculated as ( ) ( ) mW 547 73 mA 122 2 1 R I 2 1 P 2 rad 2 s = Ω = = In figure c. the antenna is matched to the line so the current calculation now yields mA 240 25 25 12 Z R V I in s s s = + = + = and the power is ( ) ( ) W 1 . 2 73 mA 240 2 1 P 2 = = This represents quite an improvement in efficiency simply because we matched the load to the line. Ω = 73 rad R V 12 s V = Ω = 25 R s λ = 485 . d Ω = 25 R s a. V 12 s V = s I Ω = 73 rad R b. Ω = 25 R s V 12 s V = s I Impedance matching network Ω = ∗ = 25 s Z in Z c....
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## This note was uploaded on 01/20/2012 for the course EE 4460 taught by Professor Czarnecki during the Fall '10 term at LSU.

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