chapter 7. wave motion 7-1 one-dimensional waveicc.skku.ac.kr/~yeonlee/electromagnetics/class...

Wave Motion 7-1 Proprietary of Prof. Lee, Yeon Ho

CHAPTER 7. WAVE MOTION 7-1 ONE-DIMENSIONAL WAVE

A wave → Disturbance of medium It travels with no change in its shape

Longitudinal wave : Displacement of medium Propagation direction Transverse wave : Displacement of medium ⊥ Propagation direction

A wave transports energy in space But medium does not advance → Wave propagates at a great speed.

Fig. 7-1(b) Disturbance varies in time and space

( ),f x tψ = : Wavefunction (7-1) ↑ Wave shape, propagation direction, velocity, and so on.


Hold time or hold position ↑ ↑ Taking a picture ↑ Profile Oscillation in time

At 0t = , ( ) ( ),0x f xψ = → At 1t t= , ( ) ( )1,x t f xψ = ↑ 1x vtΔ = ( )f x translated by 1x vtΔ = → ( )f x

( ) ( )1f x f x vt= − (7-2) A wave traveling in + x -direction ( )f x vtψ = − (7-3) A wave traveling in - x -direction ( )f x vtψ = + (7-4)

Wave motion is explained by differential wave equation

2 2

2 2 2

1x v t∂ ψ ∂ ψ

=∂ ∂

: 1-D (7-5)

↑ Velocity The solution ( )f x vtψ = ± , or ( ) ( )/f vt x g t x vψ = ± = ± (7-6) ↑ Arbitrary


Example 7-1 Sketch wavefunctions, ( )1 f x vtψ = − and ( )2 f vt xψ = − , at time 1t t= for ( )f x given as in Fig. 7-3.

Solution

Both wavefunctions are solutions of the same differential wave equation. If f were a cosine function, which is an even function, we could not distinguish between two waves,

( )1cos x vt− and ( )1cos vt x− , by just looking at their profiles. Example 7-2

Is ψ a traveling wave or not? (a) ( )2ax btψ = − (b) ( )2 2sin ax btψ = − where a and b are constants.

Solution (a) Derivatives of ψ

2

22 2ax∂ ψ

=∂

and 2

22 2bt

∂ ψ=

∂


Inserting them in Eq. (7-5)

2 2

2 22 2

2

/ /

b at x

v

∂ ψ ∂ ψ=

∂ ∂=

Since /v b a= is a constant, ψ is a traveling wave. (b) Derivatives of ψ

( ) ( ) ( )2

22 2 2 22 2 cos 2 sina ax bt ax ax btx∂ ψ

= − − −∂

( ) ( ) ( )2

22 2 2 22 2 cos 4 sinb ax bt bt ax btt

∂ ψ= − − − −

∂

Inserting them in Eq. (7-5)

2 2

2 2/ constantt x

∂ ψ ∂ ψ≠

∂ ∂

Since the ratio, which corresponds to 2v , is not a constant, ψ is not a traveling wave. 7-1.1 Harmonic Wave It has sine or cosine profile ( ) [ ]cos cosA k x vt A kx tψ = − = − ω⎡ ⎤⎣ ⎦ (7-7) ↑ Phase A , amplitude k , propagation constant ω , angular frequency, 2 fω = π

Phase velocity

pv kω

= (7-8)

Harmonic wave is periodic → Spatial period or wavelength, λ Temporal period τ A harmonic wave → Sinusoidal steady-state

Crests represent a phase 2π , not 2 mπ Periodicity requires ( ) ( )cos cosA k x t A kx t± λ − ω = − ω⎡ ⎤⎣ ⎦ (7-9a) ( ) ( )cos cosA kx t A kx t− ω ± τ = − ω⎡ ⎤⎣ ⎦ (7-9b) → 2kλ = π , 2ωτ = π (7-10)

2k π=

λ (7-11a)

2 2 fπω = = π

τ (7-11b)

→ v f= λ (7-12)


Harmonic wave at 1t t= .

The inset → Wave as a function of t ↑ Reversed and delayed

(No leading edge under sinusoidal steady-state condition)

Example 7-3

Given a harmonic wave ( )4cos2 0.2 3x tψ = − π − , find (a) amplitude (b) direction of propagation (c) wavelength (d) temporal period (e) frequency (f) phase velocity

Solution (a) 4 (b) x+ -direction

(c) 2 2 0.2k π= = π ×

λ

5[ ]mλ =


(d) 2 2 3πω = = π ×

τ

[ ]1 sec3

τ =

(e) 2 2 3fω = π = π × [ ]3 1/secf =

(f) [ ]15 /pv f m skω

= = λ =

Example 7-4

A wave generator at 0x = produces a harmonic wave with phase velocity pv . Find the wavefunction in terms of cosine function.

Solution Wavefunction in general form ( )cosA kx tψ = − ω + δ where δ is the initial phase

Consider a reference profile taken at 1t t= ( )1coso A kx tψ = − ω

which has a crest at ( )1 1 1/ px x k t v t= = ω = . ψ in Fig. 8-6 (a) is displaced to the left of oψ by /4λ

( ) ( )[ ]

1 1

1

/4, cos /4

cos /2o x t A k x t

A kx t

ψ = ψ + λ = + λ − ω⎡ ⎤⎣ ⎦= − ω + π

(7-13a)

ψ in Fig. 8-6 (b) is displaced to the right of oψ by /4λ

( ) ( )[ ]

1 1

1

/4, cos /4

cos /2o x t A k x t

A kx t

ψ = ψ − λ = − λ − ω⎡ ⎤⎣ ⎦= − ω − π

(7-13b)

The initial phase corresponds to the initial state of the wave being produced by the generator.


Time variation of ψ at 1x x= .

Let us compare ψ in Fig. (7-7) with the reference wave ( )1coso A kx tψ = − ω , which has a crest at 1 1 1/ / pt t kx x v= = ω = . ψ in Fig. 7-7 (a) is displaced to the right of oψ by /4τ

( ) ( )[ ]

1 1

1

, /4 cos /4

cos /2o x t A kx t

A kx t

ψ = ψ − τ = − ω − τ⎡ ⎤⎣ ⎦= − ω + π

(7-14a)

ψ in Fig. 7-7 (b) is displaced to the left of oψ by /4τ

( ) ( )[ ]

1 1

1

, /4 cos /4

cos /2o x t A kx t

A kx t

ψ = ψ + τ = − ω + τ⎡ ⎤⎣ ⎦= − ω − π

(7-14b)

Since 1x and 1t are arbitrary, and the harmonic wave extends from −∞ to ∞ , they are replaced by x and t . Then, Eq. (7-14) is the same as Eq. (7-13).

7-1.2 Complex Form of Harmonic Wave Complex exponential → Easy to combine harmonic waves Easy to handle phase and impedance

Euler formula cos sinie i± θ = θ ± θ : 1i = − (7-15) Cosine and sine as

1cos Re2

i i ie e eθ θ − θ⎡ ⎤ ⎡ ⎤θ = = +⎣ ⎦ ⎣ ⎦ : real part (7-16a)

1sin Im2

i i ie e ei

θ θ − θ⎡ ⎤ ⎡ ⎤θ = = −⎣ ⎦ ⎣ ⎦ : imaginary part (7-16b)


A harmonic wave in real instantaneous form ( ) ( )cos Re i kx tA kx t Ae −ω +δ⎡ ⎤ψ = − ω + δ = ⎣ ⎦ (7-17)

A harmonic wave in complex form ( ) ( ), i kx tx t Ae −ω +δψ = (7-18) Rewriting Eq. (7-18) ( ) ( ) ˆ, i ikx i t ikx i tx t Ae e e Ae eδ − ω − ω⎡ ⎤ψ = ≡ ⎣ ⎦ (7-19) ↑ Scalar complex amplitude, A , A scalar and a complex number

No change of ω in linear media A harmonic wave in time-independent complex form

ˆ ikxAeψ = (7-20)

The real instantaneous wavefunction is obtained by multiplying ψ with i te − ω and then taking the real part.

Use of i− instead of i → No change in the final real instantaneous wavefunction i− is used more in engineering

To avoid any confusions, we express i− as j− , where 1j = − A harmonic wave in complex form

( ) ( )ˆ, j t kxx t Ae ω −ψ = (7-21)

Ignoring j te ω , a harmonic wave in phasor form ˆ jkxAe −ψ = (7-22)

The real instantaneous wavefunction is obtained by multiplying the phasor with j te ω and taking the real part.

Comparison of two expressions

ˆ ikxAeψ = ˆ jkxAe −ψ =

↑ ↑ Spatial distribution of phase. Time delays of wave Spatial motion of wave. Impedance (Phase increases with increasing x ) (Positive phase means leading in time phase)

An expression should be used consistently in a given problems


Example 7-5 A harmonic wave ikx

oA eψ = travels in free space. After it passes through a lossless medium of 'k nk= in 0 x d≤ ≤ , it is delayed by /4λ , where λ is the wavelength in free space. Determine wavefunction (a) 1ψ in the medium (b) 2ψ after the medium. (c) Find n in terms of d and λ

Solution (a) In the medium

'1

ik x inkxo oA e A eψ = =

(b) 2ψ lags behind ψ by /4λ , or 2ψ is displaced to the left by /4λ ( )/4

2ik x

oA e+λψ =

(c) Across the interface at x d= , the wave should be continuous ( ) ( )1 2x d x dψ = = ψ = or ( )/4ik dinkd

o oA e A e +λ= (7-23) We obtain from Eq. (7-23)

/4nd d= + λ


Example 7-6 A harmonic wave jkx

oA e−ψ = impinges on a material at 3x x= , and undergoes a reflection.

The reflected wave 'ψ propagates in x− -direction. Boundary condition requires that ( ) ( )3 3' / 0.5x xψ ψ = at 3x x= .

(a) Find the phasor of 'ψ (b) Find the total waves at 1x x= and 2x x= (c) What time phase does the total wave at 1x x= lags behind that at 2x x= ?

(a) Incident wave at 3x x= ( ) 3

3jkx

ox A e −ψ = (7-24) Reflected wave in general form ( ) '' jkx j

ox A e + δψ = (7-25) Applying boundary condition at 3x x= ,

( ) ( )3 31'2

x xψ = ψ

or

3 3' 12

jkx j jkxo oA e A e+ δ −= (7-26)

Inserting 3 2.25x = λ and 2 /k = π λ in Eq. (7-26)

' 4.5 4.512

j j jo oA e A eπ+ δ − π=

Ignoring multiples of 2π in the phase

' 12o oA A= and δ = −π (7-27)

Wavefunction of the reflected wave

1'2

jkx joA e

− πψ = (7-28)


(b) Total wave at 1 0.3x x= = λ

( ) ( ) ( )

1 1

1 1 1

0.6 0.4

'1 1 2 2

T

jkx jkx j j jo o o o

x x x

A e A e A e A e− − π − π − π

ψ = ψ + ψ

= + = +

or ( ) 1.68

1 1.43 jT ox A e −ψ = (7-29)

Total wave at 2 1.25x x= = λ

( ) ( ) ( )

2 2

2 2 2

2.5 1.5

'1 1 2 2

T

jkx jkx j j jo o o o

x x x

A e A e A e A e− − π − π π

ψ = ψ + ψ

= + = +

or ( ) 0.5

2 1.5 jT ox A e − πψ = (7-30)

(c) Phase difference between Eqs. (7-29) and (7-30) ( )1 2 1.68 0.5 0.11Δθ = θ − θ = − − − π = − [rad]

The total wave at 1x x= lags behind the total wave at 2x x= by 0.11 radian in time phase. (A negative phase simply means a lag in time phase)

We have used ikxe for a harmonic wave in Example 7-5 and jkxe − in Example 7-6. However, any expression can solve both problems equally well. As may be seen from the examples, the use of ikxe makes it easier to visualize propagation of the wave in space, and the use of jkxe − is

advantageous in specifying time delays of the wave at points in space. 7-2 THREE-DIMENSIONAL PLANE WAVE Three-dimensional differential wave equation

2

22 2

1v t

∂ ψ∇ ψ =

∂ (7-31)

Laplacian operator in Cartesian coordinates

2 2 2

22 2 2x y z

∂ ∂ ∂∇ ≡ + +

∂ ∂ ∂ (7-32)

Solution of the wave equation ( )coso x y zA k x k y k z t⎡ ⎤ψ = + + ± ω + ϕ⎣ ⎦ : ϕ is constant (7-33) In complex form ˆ i i tAe ± ωψ = k ri (7-34) ( )ˆ expoA A j= ϕ , : Scalar complex amplitude x x y y z zk k k= + +k a a a : Wavevector (7-35a)

2 2 2 2x y zk k k π

= + + =λ

k : λ is the wavelength (7-35b)

x y zx y z= + +r a a a :Position vector (7-36)


k ri determines spatial distribution of the phase Points of an equal phase → Phase front or wavefront A wave with plane wavefront → Plane wave.

a=k ri → Phase at r is a (7-37) ↑ Constant o a=k ri → or is one of r ’s satisfying Eq. (7-37) (7-38) Combining Eqs. (7-37) and (7-38) ( ) 0o− =k r ri (7-39) ↑ A plane perpendicular to k and including or

Wavefronts are periodic with a period λ Change position by λ in ka direction

( )ˆ ˆ ˆki i ik iAe Ae e Ae+λ λ= =k r a k r k ri i i (7-40) ↑ 2kλ = π

2k π=

λ : wavenumber (7-41)


A plane wave → Parallel wavefronts moving in ka -direction (Not easy to draw on a paper) A simple cosine ↑ Each point → A plane wavefront Ordinate : magnitude Abscissa : phase k defines the propagation direction and period of the wavefront.


Example 7-7 An incident plane wave ik is partly transmitted across the interface at xy -plane, tk , while the rest is reflected, rk . The boundary condition requires that three waves should have the same spatial period along y -axis, and that the angles of reflection and transmission, rθ and iθ , should be equal. When 2 /i = π λk , determine the wavelength of the transmitted wave

Solution Spatial period of the incident wave along y -axis /sin iλ θ (7- 42) Let tλ be the wavelength of the transmitted wave. Spatial period of the transmitted wave along y -axis /sint tλ θ (7- 43) Equating Eqs. (7- 42) and (7- 43)

sinsin

tt

i

θλ = λ

θ

The wavelength of the transmitted wave is smaller than that of the incident wave, if t iθ < θ .


7-3 ELECTROMAGNETIC PLANE WAVE Maxwell’s equations in free space, 0v = =ρ J

o t∂

∇ × = −μ∂H

E (7-42a)

o t∂

∇ × = ε∂E

H (7-42b)

0∇ =iE (7-42c) 0∇ =iH (7-42d) where E and H , instantaneous electric and magnetic field intensities oε and oμ , permittivity and permeability of free space In Cartesian coordinates ( ) ( ) ( ) ( ), , , ,x x y y z zt t t t= + +r r a r a r aE E E E (7-43a) ( ) ( ) ( ) ( ), , , ,x x y y z zt t t t= + +r r a r a r aH H H H (7-43b) r , position vector xE , a scalar component of E x xaE , a vector component of E Taking the curl of Eq. (7-42a) and inserting Eq. (7-42b)

2

2o o t∂

∇ ×∇ × = −ε μ∂E

E (7-44)

Using a vector identity, 2∇ ×∇ × = ∇∇ − ∇A A Ai , and inserting Eq. (7-42c)

2

22o o t

∂∇ = ε μ

∂E

E : Three-dimensional vector differential wave equation (7-45)

Three vector components of E are mutually exclusive

2

22x

x o o t∂

∇ = ε μ∂E

E (7-46a)

2

22y

y o o t∂

∇ = ε μ∂

EE (7-46b)

2

22z

z o o t∂

∇ = ε μ∂E

E (7-46c)

Solutions are given by plane waves ( ) ( )1, cosx xt E t= − ω + ϕr k riE (7-47a) ( ) ( )2, cosy yt E t= − ω + ϕr k riE → A single solution (7-47b) ( ) ( )3, cosz zt E t= − ω + ϕr k riE ↑ (7-47c) The same k and ω Solution of the three-dimensional vector differential wave equation ( ), Re i i t

ot e − ω⎡ ⎤= ⎣ ⎦k rr E iE (7-48)


oE is the vector complex amplitude ↑ A vector whose scalar components are complex quantities. 1 2 3

yx zii io x y zE e E e E eϕϕ ϕ= + +E a a a (7-49a)

↑ ↑ ↑ x y zϕ = ϕ = ϕ = ϕ , in may cases j

o o EE e ϕ=E a (7-49b) oE , amplitude ϕ , phase angle Ea , unit vector Phase velocity [ ]81 3 10 /p

o o

v m sω= = ≅ ×

ε μk (7-50)

which can be checked by by substituting Eq. (7-48) in Eq. (7-45)

Differential wave equation for magnetic field intensity

2

22o o t

∂∇ = ε μ

∂H

H (7-51)

plane wave solution ( ), Re i i t

ot e − ω⎡ ⎤= ⎣ ⎦k rr H iH (7-52)

E and H are interrelated → Solve either Eq.(7-45) or Eq. (7-51) The other is obtained from Maxwell’s equations.

7-3.1 TRANSVERSE WAVE

E generates H → H generates b ↑ ↑ Perpendicular to E Perpendicular to H E and H are interrelated, symmetric and perpendicular to each other → Propagation direction should be perpendicular to both E and H A plane wave propagating along z -axis. → E has no change along x - and y -axes

( ) ( ), ,0

t tx y

∂ ∂= =

∂ ∂r rE E (7-53)

or ( ) ( ) ( ) ( ), , , ,x x y y z zt z t z t z t= + +r a a aE E E E (7-54)


Inserting Eq. (7-54) in Eq. (7-42c)

0

yx z

z

x y z

z

∂∂ ∂∇ + +

∂ ∂ ∂∂

= =∂

iEEE E

E

Z (7-55)

↑ zE =constant, not a traveling wave Thus, 0z =E (7-56) Then, we have ( ) ( ), ,x x y yz t z t= +a aE E E (7-57) ↑ Perpendicular to the propagation direction of E Inserting Eq. (7-57) in Eq. (7-42a)

y yx x zx y o x o y o zz z t t t

∂ ∂∂ ∂ ∂− + = −μ − μ − μ

∂ ∂ ∂ ∂ ∂a a a a a

E HE H H (7-58)

→ 0z =H (7-59) From Eqs. (7-56) and (7-59) → The electromagnetic wave is a transverse wave. Example 7-8

Find magnetic field intensity of a plane wave that propagates in free space with an electric field intensity ( )coso xE kz t− ωaE Z .

Solution Inserting E in Eq. (7-42a)

( ) ( )cosy o o x x y y z zE kz tz t∂ ∂

− ω = −μ + +∂ ∂

a a a aH H H

It reduces to

( )sin yo okE kz t

t∂

− ω = μ∂

H

Integrating both sides with respect to t

( )cosoy o

o

E kz tε

= − ωμ

H

Thus,

( )cosoo y

o

E kz tε

= − ωμ

aH

H has the same form as E , except for /o oε μ and the direction of the field. Note that E and H are in phase. Also note that E and H are perpendicular to each other, and that ×E H points to the propagation direction of the wave.

chapter 7. wave motion 7-1 one-dimensional waveicc.skku.ac.kr/~yeonlee/electromagnetics/class...

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