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Applied Electromagnetic Waves

by: Karolann Wiegand

Applied Electromagnetic Waves ECE 3317

Marketplace > University of Houston > Electrical Engineering > ECE 3317 > Applied Electromagnetic Waves
Karolann Wiegand
GPA 3.51

David Jackson

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David Jackson
Class Notes
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This 18 page Class Notes was uploaded by Karolann Wiegand on Saturday September 19, 2015. The Class Notes belongs to ECE 3317 at University of Houston taught by David Jackson in Fall. Since its upload, it has received 38 views. For similar materials see /class/208291/ece-3317-university-of-houston in Electrical Engineering at University of Houston.


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Date Created: 09/19/15
ECE 3317 Prof David R Jackson Notes 1 6 Name in Good d Dehote 5 d SKih depth I heh We have Exz Ex0 6 ij kk39 jkquotz1 j Hence Ex Z ExO 6 25 25 1 Hz 66 cm 10 Hz 21 cm Example copper 100 Hz 66 mm 7 1 kHz 21 mm 1 2 10 24W mm 10 kHz 66 mm 039 58gtlt1O7 Sm 100 kHz 21 mm 1 MHZ 66 pm 10 MHZ 201 mm 100 MHz 66 pm 1 GHz 21 pm 10 GHz 066 pm 100 GHz 021 pm The same penetration principle holds for curved conductors as long as the radius of curvature is large compared with the skin depth E E 0 e jz396 e ZI IE lm PEC coax Penetration into conductor Regions of strong currents Surface Impedance x I Z Equivalent surface current I IJXZdS Ay fJxZdZ actual current S I JsxAy surface current model Hence Define the surface impedance ExZ Ex0 e jkz sx J Jowazdz I aExO 617 de 00 aExO e 1 dz 0 Hence Jsx GExo JkJk 0Ex0 ku aExO 1 k 1 aSExO We then have ZS h 2 1 0396 J sx Define surface resistance and surface reactance We then have 1 Hz 26gtlt1O397 Q 10 Hz 83gtlt10397 9 Example copper 100 Hz 26gtlt10396Q 7 1 kHz 83gtlt10396 Q 2 10 24mm Hm 10 kHz 26gtlt10395Q 039 58gtlt107 Sm 100 kHz 83gtlt10395Q 1 MHZ 26gtlt10394 Q 10 MHZ 83gtlt10394 Q 100 MHZ 00026 9 1 GHZ 00083 9 10 GHZ 0026 9 100 GHz 0083Q Find the highfrequency resistance and inductance for a solid wire Note The current mainly flows on the outside surface of the wire I 2 27m Jsz Z R jX impedance Z V Ezo ZXiLj E20 I 27zaJSZ 27rd JSZ l 1 Hence Z 2 ZS ZS ljRS1J 275a 0395 Rena 0395 2039 Equivalent circuit Example 0 mm Assume 1 5 cm 7 f10 GHz 10 um 63957 9 01 mm 0657 9 2 76 1 mm 00657 9 5 Z a 2090gtlt10 m 10 mm 000657 g R Z 2 E R3 000825 9 10 mm 27 Q 01 mm 0027 Q 1 mm 27gtlt10394 Q 10 mm 27gtlt10396Q We use the surface resistance concept to calculate the resistance per unit length of coax Resistance per unit length


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