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DATE: 25-27/06/07 ENE 429 Antenna and Transmission lines Theory Lecture 2 Uniform plane waves Review Wave equations 2 E E 2 E 2 t t 2 H H 2 H 2 t t Time-Harmonics equations 2 E s 2 E s 0 2 H s 2 H s 0 where j ( j ) Time-harmonic wave equations or where 2 H s 2 H s 0 j ( j ) This term is called propagation constant or we can write = +j where = attenuation constant (Np/m) = phase constant (rad/m) Transverse ElectroMagnetic wave (TEM) http://www.edumedia.fr/a185_l2-transverseelectromagnetic-wave.html Solutions of Helmholtz equations The instantaneous forms of the solutions E E0 e z cos(t z )a x E0e z cos(t z )a x H H 0 e z cos(t z )a y H 0e z cos(t z )a y The phasor forms of the solutions E s E0 e z e j z a x E0e z e j z a x incident wave reflected wave H s H 0 e z e j z a y H 0e z e j z a y 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, 1 E E is the distance z that causes to reduce to 0 e z = 1 z = 1/ = 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 Currents in conductor To understand a concept of sheet resistance from L 1 L R A wt 1 L R Rsheet () L t w w Rsheet 1 t sheet resistance At high frequency, it will be adapted to skin effect resistance Currents in conductor Ex Ex 0e z J x Ex 0e z Therefore the current that flows through the slab at t is I J x dS ; ds dydz Currents in conductor From I J x dS ; ds dydz w I Ex 0e z dydz z 0 y 0 w Ex 0e z I w Ex 0 0 A. Jx or current density decreases as the slab gets thicker Currents in conductor For distance L in x-direction V Ex 0 L R Ex 0 L V 1 L L Rskin I w Ex 0 w w R is called skin resistance Rskin is called skin-effect resistance For finite thickness, t w I Ex 0e z dydz w Ex 0 (1 e t ) z 0 y 0 Rskin 1 t / (1 e ) Currents in conductor Current is confined within a skin depth of the coaxial cable Ex 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 b) The skin resistance c) The dc resistance The Poynting theorem and power transmission Poynting theorem 1 2 1 2 ( E H ) d S J E dV E dV H dV t 2 t 2 Total power leaving Joule’s law the surface for instantaneous power dissipated per volume (dissipated by heat) Rate of change of energy stored In the fields Instantaneous poynting vector S EH W/m 2 Example of Poynting theorem in DC case 1 2 1 2 ( E H ) d S J E dV E dV H dV t 2 t 2 Rate of change of energy stored In the fields = 0 Example of Poynting theorem in DC case From I J az 2 a By using Ohm’s law, J I E az 2 a a 2 L I2 d d dz 2 2 ( a ) 0 0 0 1 L 2 I I R 2 a 2 Example of Poynting theorem in DC case Verify with E H d S From Ampère’s circuital law, H dl I 2 aH I H I 2 a a Example of Poynting theorem in DC case I I I 2 S E H 2 az a 2 3 a 2 a a 2 a 2 I Total power S d S 2 3 a d dz 2 a I 2 a 2 L I 2 L 2 d dz I R 2 3 2 2 a 0 a 0 W Uniform plane wave (UPW) power transmission Time-averaged power density Pavg 1 Re( E H ) W/m2 2 P P avg d S for lossless case, P avg 1 E e j z a x Ex 0 e j z a y 2 x0 1 Ex20 P avg az W 2 amount of power Uniform plane wave (UPW) power transmission for lossy medium, we can write E Ex 0e z e j z e j a x intrinsic impedance for lossy medium H 1 a E 1 e j a z Ex 0e z e j z e j a x Ex 0 e z e j z e j e jn a y n Uniform plane wave (UPW) power transmission from Pavg 1 Re( E H ) 2 1 Ex20 2 z j Re e e az 2 2 E 1 x 0 2 z e cos a z 2 W/m2 Question: Have you ever wondered why aluminum foil is not allowed in the microwave oven? Polarization UPW is characterized by its propagation direction and frequency. Its attenuation and phase are determined by medium’s parameters. Polarization determines the orientation of the electric field in a fixed spatial plane orthogonal to the direction of the propagation. Linear polarization Consider E in free space, E ( z, t ) E0 cos(t z )a x At plane z = 0, a tip of E field traces straight line segment called “linearly polarized wave” Linear polarization A pair of linearly polarized wave also produces linear polarization E ( z, t ) Ex 0 cos(t z )a x E y 0 cos(t z )a y At z = 0 plane E (0, t ) Ex 0 cos(t )a x E y 0 cos(t )a y At t = 0, both linearly polarized waves Have their maximum values E (0, 0) Ex 0 a x Ex 0 a y t E (0, ) 0 4 More generalized linear polarization More generalized of two linearly poloraized waves, E ( z, t ) Ex 0 cos(t z x )a x E y 0 cos(t z y )a y Linear polarization occurs when two linearly polarized waves are in phase out of phase y x 0 y x 180 Elliptically polarized wave Super position of two linearly polarized waves that y x 0 or 180 If x = 0 and y = 45, we have E (0, t ) Ex 0 cos(t )a x Ey 0 cos(t )a y 4 Circularly polarized wave occurs when Exo and Eyo are equal and y x 90 Right hand circularly polarized (RHCP) wave y x 90 E (0, t ) Ex 0 cos(t )a x Ey 0 cos(t )a y 2 Left hand circularly polarized (LHCP) wave y x 90 E (0, t ) Ex 0 cos(t )a x Ey 0 cos(t )a y 2 Circularly polarized wave Phasor forms: from E ( z 0) Ex 0 e jx a x E y 0e for RHCP, E ( z 0) Ex 0 (a x ja y ) for LHCP, E ( z 0) Ex 0 (a x ja y ) j y ay Note: There are also RHEP and LHEP Ex Given E( z, t ) 3cos(t z 30 )a x 8cos(t z 90 )a y ,determine the polarization of this wave Ex The electric field of a uniform plane wave in free space is given by E s 100(a z ja x )e j 50 y , determine a) f b) The magnetic field intensity H s c) S d) Describe the polarization of the wave