RS ENE 428 Microwave Engineering Lecture 2 Uniform plane waves.

RS

ENE 428Microwave

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 = 3108 m/s (speed of light)

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

1

pu

p

r

cu

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Propagation in lossless-charge free media

• intrinsic impedance

• wavelength

2

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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, finda) phase constant

b) wavelength in the polyethelene

= 295 rad/m

= 2.13 cm

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c) propagation velocity

d) Intrinsic impedance

e) Amplitude of the magnetic field intensity

v = 2x108 m/s

= 250.77

H = 1.99 A/m

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Propagation in dielectrics• Cause

– finite conductivity– polarization loss ( = ’-j” )

• Assume homogeneous and isotropic medium

' "( ) 777777777777777777777777777777777777777777H E j j E

" '[( ) ] 7777777777777777777777777777H j E

" effDefine

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Propagation in dielectrics

From2 ( ) j j

and2 2( ) j

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Propagation in dielectrics

We can derive2

( 1 1)2

2

( 1 1)2

and 1

.1 ( )

j

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Loss tangent

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

"

' 'tan

eff

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Low loss material or a good dielectric (tan « 1)

• If or < 0.1 , consider the material

‘low loss’ , then

1

2

(1 ).2

jand

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Low loss material or a good dielectric (tan « 1)

• propagation velocity

• wavelength

1

pu

2 1

f

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

• In this case or > 10, we can

approximate

1

2 f

45 .

jje

therefore

2

1 1)

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-1 or 0.368 times of

the initial field

• propagation velocity

• wavelength

1 1 1m

f

pu

22

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Ex2 Given a nonmagnetic material having r = 3.2 and = 1.510-4 S/m,

at f = 30 MHz, find a) loss tangent

b) attenuation constant

tan = 0.03

= 0.016 Np/m

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c) phase constant

d) intrinsic impedance

= 1.12 rad/m

= 210.74(1+j0.015)

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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

= 117.21 rad/m

= 5.36 cm

v = 3.22 m/s

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c) compare these answers with the same wave propagating in a free space

= 1.26x10-6 rad/m

= 5000 km

v = 3x108 m/s

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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/ =

E77777777777777 1

0E e

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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

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Currents in conductor

• To understand a concept of sheet resistance

1L LR

A wt

1 LR

t w Rsheet () Lw

1sheetR

t sheet resistance

from

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

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0

0

zx x

zx x

E E e

J E e

Therefore the current that flows through the slab at t is

;xI J dS ds dydz

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;xI J dS ds dydz

00 0

wz

xz y

I E e dydz

0

0

zxw E e

0 .xI w E A

From

Jx or current density decreases as the slab gets thicker

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0xV E L

0

0

1xskin

x

E LV L LR R

I w E w w

For distance L in x-direction

For finite thickness,

R is called skin resistanceRskin is called skin-effect resistance

0 00 0

(1 )t w

z tx x

z y

I E e dydz w E e

/

1

(1 )skin tRe

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Current is confined within a skin depth of the coaxial cable

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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

b) The skin resistance

= 7.66x10-4 m

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c) The dc resistance

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The Poynting theorem and power transmission

2 21 1( )

2 2E H d S J E dV E dV H dV

t t

7777777777777777777777777777777777777777777777777777777777777777777777

Poynting theorem

Total power leavingthe surface

Joule’s lawfor instantaneouspower dissipated per volume (dissi-pated by heat)

Rate of change of energy storedIn the fields

2W/mS E H 777777777777777777777777777777777777777777

Instantaneous poynting vector

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Example of Poynting theorem in DC case

2 21 1( )

2 2E H d S J E dV E dV H dV

t t

7777777777777777777777777777777777777777777777777777777777777777777777

Rate of change of energy storedIn the fields = 0

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2 z

IJ a

a

77777777777777

By using Ohm’s law,

From

2 z

J IE a

a

7777777777777777777777777777

2 2

2 20 0 0( )

a LId d dz

a

2 22

1 LI I R

a

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E H d S777777777777777777777777777777777777777777

From Ampère’s circuital law,

Verify with

H dl I7777777777777777777777777777

2 aH I 77777777777777

2

IH a

a

77777777777777

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2

2 32

IS d S a d dz

a

7777777777777777777777777777

2

2 2 32 2z

I I IS E H a a a

aa a

777777777777777777777777777777777777777777

2 222

2 3 20 02

LI a I Ld dz I R

a a

Total power

W

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Uniform plane wave (UPW) power transmission• Time-averaged power density

1Re( )

2avgP E H

777777777777777777777777777777777777777777

amount of power avgP P d S7777777777777777777777777777

for lossless case, 00

12

77777777777777j z j zx

avg x yx

EP E e a e a

201

2x

avg zE

P a 77777777777777

W/m2

W/m2

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Uniform plane wave (UPW) power transmission

0

z j z jxxE E e e e a

77777777777777

intrinsic impedance for lossy medium nje

0

1 1 z j z jz xxH a E a E e e e a

7777777777777777777777777777

0 njz j z jxy

Ee e e e a

for lossy medium, we can write

RS ENE 428 Microwave Engineering Lecture 2 Uniform plane waves.

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