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# ENE 428 Microwave Engineering - PowerPoint PPT Presentation

ENE 428 Microwave Engineering. Lecture 2 Uniform plane waves. Propagation in lossless-charge free media. Attenuation constant  = 0, conductivity  = 0 Propagation constant Propagation velocity for free space u p = 310 8 m/s (speed of light)

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ENE 428Microwave Engineering

Lecture 2 Uniform plane waves

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• Attenuation constant  = 0, conductivity  = 0

• Propagation constant

• Propagation velocity

• for non-magnetic lossless dielectric (r = 1),

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• intrinsic impedance

• wavelength

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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

b) wavelength in the polyethelene

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d) Intrinsic impedance

e) Amplitude of the magnetic field intensity

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• Cause

• finite conductivity

• polarization loss ( = ’-j” )

• Assume homogeneous and isotropic medium

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Define

From

and

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We can derive

and

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

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• If or < 0.1 , consider the material ‘low loss’, then

and

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• propagation velocity

• wavelength

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High loss material or a good conductor (tan » 1)

• In this case or > 10, we can approximate

therefore

and

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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-1or 0.368 times of the initial field

• propagation velocity

• wavelength

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Ex2 Given a nonmagnetic material having r= 3.2 and  = 1.510-4 S/m, at f = 3 MHz, find

a) loss tangent 

b) attenuation constant 

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d)intrinsic impedance

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Ex3 Calculate the followings for the wave with the frequency f = 60 Hz propagating in a copper with the conductivity,  = 5.8107 S/m:

a) wavelength

b) propagation velocity

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Attenuation constant free space

• 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/  = 

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Good conductor free space

• At high operation frequency, skin depth decreases

• A magnetic material is not suitable for signal carrier

• A high conductivity material has low skin depth

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Currents in conductor free space

• To understand a concept of sheet resistance

from

Rsheet()

sheet resistance

At high frequency, it will be adapted to skin effect resistance

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Currents in conductor free space

Therefore the current that flows through the slab at t   is

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Currents in conductor free space

From

Jxor current density decreases as the slab gets thicker

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Currents in conductor free space

For distance L in x-direction

Ris called skin resistance

Rskinis called skin-effect resistance

For finite thickness,

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Currents in conductor free space

Current is confined within a skin depth of the coaxial cable

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Ex free space 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

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c) free spaceThe dc resistance

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Poynting theorem

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

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Example of Poynting theorem in DC case free space

Rate of change of energy stored

In the fields = 0

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Example of Poynting theorem in DC case free space

From

By using Ohm’s law,

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Example of Poynting theorem in DC case free space

Verify with

From Ampère’s circuital law,

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Example of Poynting theorem in DC case free space

Total power

W

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• Time-averaged power density

W/m2

amount of power

for lossless case,

W

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for lossy medium, we can write

intrinsic impedance for lossy medium

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