Report

ENE 428 Microwave Engineering Lecture 2 Uniform plane waves RS Propagation in lossless-charge free media • Attenuation constant = 0, conductivity = 0 • Propagation constant • Propagation velocity up 1 – for free space up = 3108 m/s (speed of light) – for non-magnetic lossless dielectric (r = 1), up RS c r Propagation in lossless-charge free media • intrinsic impedance • wavelength RS 2 Ex1 A 9.375 GHz uniform plane wave is propagating in polyethelene (r = 2.26). If the amplitude of the electric field intensity is 500 V/m and the material is assumed to be lossless, find a) phase constant = 295 rad/m b) wavelength in the polyethelene = 2.13 cm RS c) propagation velocity v = 2x108 m/s d) Intrinsic impedance = 250.77 e) Amplitude of the magnetic field intensity H = 1.99 A/m RS Propagation in dielectrics • Cause – finite conductivity – polarization loss ( = ’-j” ) • Assume homogeneous and isotropic medium H E j ( j ) E ' " H [( ) j ] E " Define RS eff ' " Propagation in dielectrics RS From j ( j ) and 2 2 ( j ) 2 Propagation in dielectrics We can derive RS ( 1 1) 2 2 ( 1 1) 2 and 2 1 1 j ( ) . Loss tangent • A standard measure of lossiness, used to classify a material as a good dielectric or a good conductor tan RS ' " eff ' Low loss material or a good dielectric (tan « 1) • If 1 or < 0.1 , consider the material ‘low loss’ , then 2 and RS (1 j ). 2 Low loss material or a good dielectric (tan « 1) • propagation velocity up 1 • wavelength RS 2 1 f High loss material or a good conductor (tan » 1) • In this case 1 or > 10, we can approximate 2 1 1) therefore and RS j 2 f j 45 e . High loss material or a good conductor (tan » 1) • depth of penetration or skin depth, is a distance where the field decreases to e-1 or 0.368 times of the initial field 1 f 1 • propagation velocity up • wavelength RS 2 2 1 m Ex2 Given a nonmagnetic material having r = 3.2 and = 1.510-4 S/m, at f = 30 MHz, find a) loss tangent tan = 0.03 b) attenuation constant = 0.016 Np/m RS c) phase constant = 1.12 rad/m d) intrinsic impedance = 210.74(1+j0.015) RS Ex3 Calculate the followings for the wave with the frequency f = 60 Hz propagating in a copper with the conductivity, = 5.8107 S/m: a) wavelength = 117.21 rad/m = 5.36 cm b) propagation velocity v = 3.22 m/s RS c) compare these answers with the same wave propagating in a free space = 1.26x10-6 rad/m = 5000 km v = 3x108 m/s RS Attenuation constant • Attenuation constant determines the penetration of the wave into a medium • Attenuation constant are different for different applications • The penetration depth or skin depth () is the distance z that causes E to reduce to E 0 e 1 z = 1 z = 1/ = RS Good conductor 1 1 f At high operation frequency, skin depth decreases A magnetic material is not suitable for signal carrier A high conductivity material has low skin depth RS Currents in conductor • To understand a concept of sheet resistance from R L R 1 L t w R sh eet A 1 L wt Rsheet () L w 1 t sheet resistance At high frequency, it will be adapted to skin effect resistance RS Currents in conductor E x E x0e z J x E x0e z Therefore the current that flows through the slab at t is I J x dS RS ; ds dydz Currents in conductor From I J x dS I ; ds dydz w E x0e z dydz z0 y0 w E x 0 e z I w E x 0 0 A. Jx or current density decreases as the slab gets thicker RS Currents in conductor For distance L in x-direction V E x0 L R V I E x0 L w E x 0 L R skin w w 1 L R is called skin resistance Rskin is called skin-effect resistance For finite thickness, t I w E x0e z z 0 y 0 RS R skin dydz w E x 0 (1 e 1 (1 e t / ) t ) Currents in conductor Current is confined within a skin depth of the coaxial cable RS Ex4 A steel pipe is constructed of a material for which r = 180 and = 4106 S/m. The two radii are 5 and 7 mm, and the length is 75 m. If the total current I(t) carried by the pipe is 8cost A, where = 1200 rad/s, find: a) The skin depth = 7.66x10-4 m b) RS The skin resistance c) The dc resistance RS The Poynting theorem and power transmission Poynting theorem ( E H ) d S J E dV Total power leaving Joule’s law the surface for instantaneous power dissipated per volume (dissipated by heat) E dV 2 t RS t 1 2 H dV 2 Rate of change of energy stored In the fields Instantaneous poynting vector S EH W /m 2 Example of Poynting theorem in DC case (E H ) d S J E dV E t 2 dV 1 H t 2 2 dV Rate of change of energy stored In the fields = 0 RS Example of Poynting theorem in DC case J From I a az 2 By using Ohm’s law, E J J E dV I a I 2 ( a ) I RS az 2 2 2 1 2 L a2 a 2 L 0 0 0 d d dz I R 2 Example of Poynting theorem in DC case Verify with (E H ) d S From Ampère’s circuital law, H dl I 2 a H I H RS I 2 a a Example of Poynting theorem in DC case S EH Total power I az a 2 S d S I 2 I a 2 RS 2 a 2 3 2 2 a 3 2 I 2 a a 2 2 a 2 3 a a d dz L d dz 0 I 0 I L 2 a 2 I R 2 W Uniform plane wave (UPW) power transmission • Time-averaged power density P a vg amount of power for lossless case, 1 2 R e( E H ) P P avg d S P avg 1 P avg 1 E x0 2 W/m2 E x0e j z ax E x0 2 RS 2 az W/m2 e j z ay Uniform plane wave (UPW) power transmission for lossy medium, we can write E ( z , t ) E x 0 e z cos( t z ) a x E E x0e The phasor is z e j z j e ax e intrinsic impedance for lossy medium H 1 a E RS 1 a z E x0e E x0 e z e z j z e j z j e e e j n j ax ay j n Homework Prob.6.19: In seawater, a propagating electric field is given by z 6 E ( z , t ) 20 e cos( 2 10 t z 0 . 5 ) a y V/m Assuming = 5 S/m, r’ = 72, r” = 0, find (a) and β and (b) the instantaneous form of H(z,t) z 6 cos( 10 t z / 6 ) a x A/m Prob. 6.26: In air, H ( z , t ) 12 e Determine the power passing through a 1.0 m2 surface that is normal to the direction of propagation. 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