wave packets+solution of em wave eq

8/2/2019 Wave Packets+Solution of EM Wave Eq

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Superposition of Waves & Wave packets


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y(t) = Sin t 0 < t < 200

Superposition of waves and wave packet formation


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y(t) = [Sin t + Sin (1.08 t)]/2 0 < t < 200


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y(t) = [Sin t + Sin(1.04 t) + Sin (1.08 t)]/3

0 < t < 400


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y(t) = [Sin t + Sin(1.02 t) + Sin (1.04 t)

+ Sin(1.06 t) + Sin (1.08 t)]/5

0 < t < 400


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y(t) = [Sin t + Sin(1.01 t) + Sin (1.02 t)

+ Sin(1.03 t) + Sin (1.04 t) + Sin (1.05 t)

+ Sin (1.06 t) + Sin (1.07 t)+ Sin (1.08 t)]/9

0 < t < 800


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y(t) = [Sin t + Sin(1.01 t) + Sin (1.02 t)

+ Sin(1.03 t) + Sin (1.04 t) + Sin (1.05 t)

+ Sin (1.06 t) + Sin (1.07 t)+ Sin (1.08 t)]/9

0 < t < 400


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Wave Equations: EM Waves


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Electromagnetic waves

for E field

for B field


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In general,electromagnetic waves

2

2

2

2 1tc

Where represents E or Bor their components


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#A plane wavesatisfies waveequation in Cartesian coordinates

#Aspherical wavesatisfies wave

equation inspherical polarcoordinates

#

A cylindrical wavesatisfies waveequation in cylindrical coordinates


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Solution of 3D wave equation

In Cartesian coordinates

2

2

22

2

2

2

2

22

1 tczyx

Separation of variables

)()()()(),,,( tTzZyYxXtzyx


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Substituting forwe obtain

2

2

22

2

2

2

2

211111

t

T

Tcz

Z

Zy

Y

Yx

X

XVariables are separated out

Each variable-term independentAnd must be a constant


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So we may write

2

2

22

2

2

2

2

22

2

2

1;1

;1;1

t

T

Tk

z

Z

Z

ky

Y

Yk

x

X

X

z

yx

where we use222222

kkkkc zyx


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Solutions are then

tizik

yikxik

etTezZ

eyYexX

z

yx

)(;)(

;)(;)(

Total Solution is

)()()()(),,,( tTzZyYxXtzyx

)]([ zkykxkti zyxAe

].[ rktiAe

plane wave


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Traveling 3D plane wave


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spherical waves


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spherical waves

2

2

222

2

222

2

2 1sin

cossin12

rrrrrr


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rr

rrrrr

2

22

22 12

2

2

2

2

2

11

tcrr

rr

Alternatively

The wave equation becomes


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2

1

r

u

r

u

rr

Put r

rur

)()(

Then ur

urr

r

2

ru

rur

ru

ur

urrr

rr

2

2

2 Hence


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2

22

2

11

r

u

rrr

rr

Therefore

Wave equation transforms to

2

2

22

2 111tu

rcru

r

2

2

22

2 1tu

cru


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)()(),( tTrRtru

22

2

22

2

111 ktT

TcrR

R

kcetTerR tiikr

)(;)(

Which follows that

Separation of variables

)()(

krtieru

Solutions are

Total solution is


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)(1)(krti

er

r

)()( 11)(krtikrti

e

r

e

r

r

outgoingwaves

incomingwaves

Final form of solution

General solution

spherical wave


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Cylindrical waves

2

2

2

2

22

2

2 11

zrrrr


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with angular and azimuthal symmetry, theLaplacian simplifies and the wave equation


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The solutions are Bessel functions.For large r, they are approximated as


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A plane wave satisfies one-dimensionalwave equation in Cartesian coordinates

The position vector must remainperpendicular to the given plane


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The wave then satisfies the generalizationof the one-dimension wave equation


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Plane EM waves in vacuum


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Wave vector k is perpendicular to E

Wave vectorkis perpendicular toB


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B is perpendicular to E


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B, kandE make a right handed

Cartesian co-ordinate system


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Plane EM waves in vacuum

wave packets+solution of em wave eq

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