Lect29Hertzian - ECE 3030 Electromagnetic Fields and Waves...

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1 ECE 3030 Electromagnetic Fields and Waves Fall 2009 Lecture 29 2009/11/4 Hertzian Dipole Antennas Gain Radiation Pattern 1 Swartz 08/11/4 Electromagnetic Fields and Waves – Fall 2009 Lecture 29: Instructor: Dr. Wesley E. Swartz Radiation Resistance x z d q ( t ) -q ( t ) I ( t ) Hertzian Dipole Antenna (1) A Hertzian dipole consists of two equal and opposite ± charge reservoirs located at a distance d from each other, as shown below. It is one of the simplest radiating elements for which analytical solutions for the fields can be obtained. x z d q ( t ) -q ( t ) I ( t ) Th t h i l t i ll t d d i id l 3 Swartz 08/11/4 Electromagnetic Fields and Waves – Fall 2009 Lecture 29: t I ( t ) q ( t ) -q ( t ) () dt t dq t I = The two charge reservoirs are electrically connected and a sinusoidal current I ( t ) flows between them. Consequently, the amount of charge in the reservoirs also changes sinusoidally: If the length of the dipole is much smaller than the wavelength, then Hertzian Dipole Antenna (2) Suppose we could write a current density for the Hertzian dipole. Then: The above integral represents the total “weight” or “strength” of the dipole x z d q ( t ) -q ( t ) I ( t ) r J d I z dz dy dx r J dv r J ˆ =  =  4 Swartz 08/11/4 Electromagnetic Fields and Waves – Fall 2009 Lecture 29: one may write: The above expression will give the same strength for the dipole, i.e.: ()()() r d I z z y x d I z r J 3 ˆ ˆ δ = = 3 3 m 1 r Amp I m d 2 m Amp r J Check units: d I z dv r J ˆ =  Hertzian Dipole Antenna (3) A Hertzian dipole is represented by an arrow whose direction indicates the positive direction of the current and the orientation of the dipole in space: x z d q 1 ( t ) q 2 ( t ) x z I ( t ) The length “ d ” of the Hertzian dipole is much smaller than the 5 Swartz 08/11/4 Electromagnetic Fields and Waves – Fall 2009 Lecture 29: Example : If then r d I z r J 3 ˆ = () = r J = I = d current density phasor Phasor for the current that flows in the dipole length of the dipole ( ) α ω + = t I t I o cos r e d I z r J j o 3 ˆ = wavelength of the radiation it emits. So with its current distribution represented by a delta function, we can move to its phasor form: Radiation Emitted by a Hertzian Dipole (1) To solve: use the superposition integral: r J r A k r A o μ = + 2 2 r k j o r k j o ' r r k j 3 o ' r r k j o e r 4 Id z ˆ e r 4 Id z ˆ r A ' r e ' r r 4 ) ' r ( Id z ˆ ' dv e ' r r 4 ' r J r A = = = =   π d 3 x z θ r d I z r J 3 ˆ = y φ with 6 Swartz 08/11/4 Electromagnetic Fields and Waves – Fall 2009 Lecture 29: In spherical coordinates the unit vector becomes , and [] r k j o e r Id r r A = 4 sin ˆ cos ˆ sin 1 1 4 ˆ o + = × = r k j e r Id k j r H r A r H r k j z ˆ sin ˆ cos ˆ r Find the H-field via:
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This note was uploaded on 11/26/2009 for the course ECE 3030 at Cornell University (Engineering School).

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Lect29Hertzian - ECE 3030 Electromagnetic Fields and Waves...

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