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Week 10

Intrinsic Impedance

Plane Wave Reflection

Reflection, Transmission, Refraction

Polarization

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Any medium through whichplane waves propagate have aproperty called intrinsicimpedance or electromagneticimpedance, denoted by .

The intrinsic impedance isanalogous to the characteristicimpedance of a cable intransmission line theory.

For a given frequency, theintrinsic impedance of ahomogeneous material isgiven by the ratio of the E fieldphasor to the H field phasor.

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Consider a plane wave with fields

E= Eie jte jkzaxB= Bie jte jkzay

Recall that these fields are related by

jk Eiejte jkzay= j Bie jte jkzay

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Since H = B / it follows that

jk Eiejte jkzay= j Hie jte jkzay

kEi= Hi

= Ei/Hi= /k

= (/)1/2

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For a region with slight electrical conductivity( > 0, e.g. seawater), the equation becomes

This means that the E and H fields will be outof phase.

Exercise:What is the intrinsic impedance of aperfect conductor?

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When a plane wave passesfrom one dielectric medium (A)to another medium (B) withdifferent intrinsic impedance, afraction of its energy will bereflected and the remainderwill be transmitted through theinterface.

In this case, a reflectioncoefficient is used to describe

the amplitude of the reflectedpart of the wave relative to theamplitude of the incident wave.

The arrows indicate the

direction of propagation

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In particular, the reflection coefficient is thecomplex ratio of the electric field phasor ofthe reflected wave (E) to that of the incidentwave (E+).

This is typically represented with a (capitalgamma) and can be written as:

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Let the intrinsic impedances of the first andsecond media be 1and 2, respectively.

At normal incidence the reflection coefficient of

medium 2 relative to 1 is given by

Exercise:A plane wave hits a mirror at normalincidence. Determine the reflection coefficient.

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Incident wave

Ei= Eie jte jkzaxHi= Hie jte jkzayand 1= Ei/ Hi

Reflected wave

Er= Ere jte jkzaxHr= Hre jte jkzayand 1= - Er/Hr

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At the interface, the components of the electric field andmagnetic field parallel to the surface must be continuous.

Consequently the following boundary condition must besatisfied:

Ei+ Er= Et Ei+ Ei= Et

Hi+ Hr= Ht Ei/1Ei/1= Et/2

This yields

(1 )/1= (1 + )/ 2

= (21) / (2+ 1)

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When the electric field Eiis incident upon a perfectly conducting platelocated a z = 0, the reflected field isEr= Eie +jte +jkzx.

The total field in region 1

E1= Ei Er=Eie +jt(e jkz+ e+jkz) x Since = 1, the expression simplifies to

E1= Ei Er=Eie +jt(e jkz e +jkz) xE1= 2jEie +jtsin (kz) x

Taking of the real part of E1yieldsE1=2Eisin (t) sin (kz) x

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Exercise: Derive an expression for the total

electric field if the reflection coefficient is =1.

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Standing Wave in a Microwave Oven

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Consider the stationary electric field in amicrowave oven:

E= Eoe jtsin (kz) ax V/ m The oven operates at a typical frequency of 2.45

GHz. The width of the oven W is 2.5.

The electric field is zero on the side walls of theoven (i.e. at z = 0 and z = W).

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What is the width of the oven? Is this a traveling wave or a standing wave?

What are the boundary conditions in theoven?

How many hot spots are there?

Where is |E| the strongest? In other words,where do you put the turkey?

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The transmission coefficient is used when a planewave encounters a change in the intrinsic impedance.

The transmission coefficient describes the amplitudeof the transmitted wave relative to the incident wave.

The transmission coefficient is a measure of howmuch of an electromagnetic wave passes through asurface.

The transmission coefficient is defined as the ratioof the transmitted electric field to the ratio of theincident electric field.

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When the electric field Eiis incident upon a surface located a z =0, the reflected field isEt= Eoe +jte jkzx.

At the interface, the component of the electric field parallel to

the surface must be continuous.

Consequently the following boundary condition must besatisfied:

Ei(0) Er(0)= Et(0)Ei+ Ei= Ei

or 1 + =

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When an electromagnetic field movesfrom a medium of a given refractiveindex n1into a second medium withrefractive index n2, both reflection andrefraction of the plane wave mayoccur.

An incident plane wave POstrikes atpoint Othe interface between twomedia of refractive indexes n1and n2.

Part of the wave is reflected as waveOQand part refracted as ray OS. Theangles that the incident, reflected andrefracted waves make to the normal ofthe interface are given as i, rand t,respectively.

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The law of reflection states that thedirection of the incident plane waveand the direction of the reflectedplane wave make the same angle withrespect to the surface normal.

That is, the angle of incidence equalsthe angle of reflection:i= r

Exercise: An electromagnetic planewave is reflected by a perfectconductor. The angle of incidence is45o. What percentage of the wavewill be reflected at 30o?

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Snell's law (also known asDescartes' law or the law ofrefraction) describes therelationship between the

angles of incidence andrefraction, when referringan electromagnetic wave,passing through aboundary between twodifferent isotropic media,such as water and glass.

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The law says that the ratio of the sines ofthe angles of incidence and of refractionis a constant that depends on the media.

The refraction of an electromagneticwave occurs at the interface between two

media of different refractive indices,with n2> n1.

Since the velocity is lower in the secondmedium (v2< v1), the angle of refraction2is less than the angle of incidence 1.

That is, the ray in the higher-indexmedium is closer to the normal.

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When light moves from a dense toa less dense medium, such asfrom water to air, Snell's lawcannot be used to calculate therefracted angle when the resolvedsine value is higher than 1.

At this point, light is reflected inthe incident medium, known asinternal reflection.

Before the ray totally internallyreflects, the light refracts at the

critical angle; it travels directlyalong the surface between thetwo refractive media, without achange in phases like in otherforms of optical phenomena.

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In order to calculate the critical angle, let 2=90oand solve for crit:

When 1> crit, no refracted ray appears, andthe incident ray undergoes total internal

reflection from the interface medium.

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The index of refraction of water at opticalwavelengths is approximately 1.33.

Suppose that a swimmer is underwaterlooking up at the water-air interface.

Determine the minimum angle at which she

starts seeing the bottom of the pool reflectedon the interface.

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Polarization is a property ofwaves that describes theorientation of their oscillations.

A polarizing filter, such as apair of polarizing sunglasses,

can be used to observe thiseffect by rotating the filterwhile looking through it at thereflection off of a distanthorizontal surface.

At certain rotation angles, thereflected light will be reducedor eliminated.

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For a simple harmonic wave, where the amplitude of the electricvector varies in a sinusoidal manner in time, the two componentshave exactly the same frequency:

Ex= Exexp [jx] exp [ j(tkzz) ]axEy= Eyexp [jy] exp [ j(tkzz) ]ay

The two components may not have the same amplitude:

Ex Ey

The two components may not have the same phase, that is theymay not reach their maxima and minima at the same time:

x y

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In the figure below, the Efieldis propagating in the + zdirection.

The field has two componentsExand Ey. The components are both in

phase and orthogonal(perpendicular) to each other.

http://en.wikipedia.org/wiki/Image:Linear_polarization_schematic.png

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In the case of circularpolarization, the twoorthogonal components Exand Ey have exactly the sameamplitude and are exactlyninety degrees out of phase.

In this case one component is

zero when the othercomponent is at maximum orminimum amplitude.

http://en.wikipedia.org/wiki/Image:Circular_polarization_schematic.png

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The direction the field rotates in,depends on which of the twophase relationships exists.

Depending on which way the

electric vector rotates these casesare called

right-hand circular polarizationand left-hand circular polarization.

In electrical engineering, therotation is defined as seen fromthe source, such as from atransmitting antenna.

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More generally, the twocomponents are not in phaseand do not have the sameamplitude.

In this case the polarization iscalled elliptical polarizationbecause the electric vector

traces out an ellipse in theplane (the polarizationellipse).

http://en.wikipedia.org/wiki/Image:Elliptical_polarization_schematic.png

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Read book sections 7-1, 7-2, 7-3, 8-1, 8-2 Solve end-of-chapter problems 7.1, 7.5,

7.10, 8.1, 8.17