# ENE 428 Microwave Engineering

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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 for free space up = 3108 m/s (speed of light) for non-magnetic lossless dielectric (r = 1), RS

Propagation in lossless-charge free media
intrinsic impedance wavelength RS

Ex1 A 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 b) wavelength in the polyethelene  = 295 rad/m  = 2.13 cm RS

c) propagation velocity
d) Intrinsic impedance e) Amplitude of the magnetic field intensity v = 2x108 m/s  =  H = 1.99 A/m RS

Propagation in dielectrics
Cause finite conductivity polarization loss ( = ’-j” ) Assume homogeneous and isotropic medium Define RS

Propagation in dielectrics
From and RS

Propagation in dielectrics
We can derive and RS

Loss tangent A standard measure of lossiness, used to classify a material as a good dielectric or a good conductor RS

Low loss material or a good dielectric (tan « 1)
If or < 0.1 , consider the material ‘low loss’ , then and RS

Low loss material or a good dielectric (tan « 1)
propagation velocity wavelength RS

High loss material or a good conductor (tan » 1)
In this case or > 10, we can approximate therefore and RS

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 times of the initial field propagation velocity wavelength RS

Ex2 Given a nonmagnetic material having r = 3. 2 and  = 1
Ex2 Given a nonmagnetic material having r = 3.2 and  = 1.510-4 S/m, at f = 30 MHz, find a) loss tangent  b) attenuation constant  tan = 0.03  = Np/m RS

d) intrinsic impedance
c) phase constant  d) intrinsic impedance  = 1.12 rad/m  = (1+j0.015)  RS

Ex3 Calculate the followings for the wave with the frequency f = 60 Hz propagating in a copper with the conductivity,  = 5.8107 S/m: a) wavelength b) propagation velocity = rad/m  = 5.36 cm 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 to reduce to z = 1  z = 1/  =  RS

Good conductor 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
Rsheet () sheet resistance At high frequency, it will be adapted to skin effect resistance RS

Currents in conductor Therefore the current that flows through the slab at t   is RS

Currents in conductor From
Jx or current density decreases as the slab gets thicker RS

Currents in conductor For distance L in x-direction
R is called skin resistance Rskin is called skin-effect resistance For finite thickness, RS

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  = 4106 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 8cost A, where  = 1200 rad/s, find: The skin depth The skin resistance  = 7.66x10-4 m RS

c) The dc resistance RS

The Poynting theorem and power transmission
Total power leaving the surface Joule’s law for instantaneous power dissipated per volume (dissi- pated by heat) Rate of change of energy stored In the fields Instantaneous poynting vector RS

Example of Poynting theorem in DC case
Rate of change of energy stored In the fields = 0 RS

Example of Poynting theorem in DC case
From By using Ohm’s law, RS

Example of Poynting theorem in DC case
Verify with From Ampère’s circuital law, RS

Example of Poynting theorem in DC case
Total power W RS

Uniform plane wave (UPW) power transmission
Time-averaged power density W/m2 amount of power for lossless case, W/m2 RS

Uniform plane wave (UPW) power transmission
for lossy medium, we can write intrinsic impedance for lossy medium RS