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ENE 428 Microwave Engineering
Lecture 2 Uniform plane waves RS
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Propagation in lossless-charge free media
Attenuation constant = 0, conductivity = 0 Propagation constant Propagation velocity for free space up = 3108 m/s (speed of light) for non-magnetic lossless dielectric (r = 1), RS
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Propagation in lossless-charge free media
intrinsic impedance wavelength RS
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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 RS
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c) propagation velocity
d) Intrinsic impedance e) Amplitude of the magnetic field intensity RS
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Propagation in dielectrics
Cause finite conductivity polarization loss ( = ’-j” ) Assume homogeneous and isotropic medium RS
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Propagation in dielectrics
Define From and RS
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Propagation in dielectrics
We can derive and RS
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Loss tangent A standard measure of lossiness, used to classify a material as a good dielectric or a good conductor RS
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Low loss material or a good dielectric (tan « 1)
If or < 0.1 , consider the material ‘low loss’ , then and RS
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Low loss material or a good dielectric (tan « 1)
propagation velocity wavelength RS
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High loss material or a good conductor (tan » 1)
In this case or > 10, we can approximate therefore and RS
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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
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Ex2 Given a nonmagnetic material having r = 3. 2 and = 1
Ex2 Given a nonmagnetic material having r = 3.2 and = 1.510-4 S/m, at f = 3 MHz, find a) loss tangent b) attenuation constant RS
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d) intrinsic impedance
c) phase constant d) intrinsic impedance RS
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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 b) propagation velocity RS
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c) compare these answers with the same wave propagating in a free space
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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
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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
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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
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Currents in conductor Therefore the current that flows through the slab at t is RS
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Currents in conductor From
Jx or current density decreases as the slab gets thicker RS
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Currents in conductor For distance L in x-direction
R is called skin resistance Rskin is called skin-effect resistance For finite thickness, RS
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Currents in conductor Current is confined within a skin depth of the coaxial cable RS
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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: The skin depth The skin resistance RS
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c) The dc resistance RS
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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
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Example of Poynting theorem in DC case
Rate of change of energy stored In the fields = 0 RS
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Example of Poynting theorem in DC case
From By using Ohm’s law, RS
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Example of Poynting theorem in DC case
Verify with From Ampère’s circuital law, RS
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Example of Poynting theorem in DC case
Total power W RS
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Uniform plane wave (UPW) power transmission
Time-averaged power density W/m2 amount of power for lossless case, W RS
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Uniform plane wave (UPW) power transmission
for lossy medium, we can write intrinsic impedance for lossy medium RS
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