Lecture 3

EE-344 Wave Propagation and Antennas

Dr. Muhammad Anis Chaudhary

Review of Lecture 2

• A medium is characterized by its constitutive parameters σ , ε, µ

– Lossy Dielectrics
* σ ̸= 0, ε = εr ε0 , µ = µr µ0
– Lossless Dielectrics
* σ ≪ ωε or σ → 0, ε = εr ε0 , µ = µr µ0
– Free Space
* σ = 0, ε = ε0 , µ = µ0
– Good Conductors
* σ ≫ ωε or σ → ∞, ε = ε0 , µ = µr µ0

• Lossy Dielectrics, is the general case

• Solve Maxwell’s equations for this general case

• other cases can be described in terms of solutions for Lossy Dielectric

Review of Lecture 2
Summary

Vector Wave Equations

¯ 2 Ēs − γ 2 Ēs = 0 , ∇
• ∇ ¯ 2 H̄s − γ 2 H̄s = 0

• where
p
– γ = α + jβ = jω µ(σ + jωε)=propagation constant of the medium
s q 
µε  σ 2
– α =ω 2 1 + ωε − 1 = attenuation constant of the medium (Np/m)
s q 
µε  σ 2
– β =ω 2 1 + ωε + 1 = phase constant (rad/m)

EE-344 Wave Propagation and Antennas, Dr. M Anis Ch, Lecture 3 Page 1 of 17
Review of Lecture 2
Summary Continued
• if Ēs has only x-component i.e. Ēs = Exs (z)âx
 2 
• the solution of the scalar wave equation, ∂∂z2 − γ 2 Exs (z) = 0 is

– Ē(z,t) = E0 e−αz cos(ωt − β z)aˆx

E0 −αz
– H̄(z,t) = |η| e cos(ωt − β z − θη )ây
q √
jω µ µ/ε
– where η = σ + jωε = |η| e jθη ,|η| = h i1 and tan (2θη ) = σ
ωε
σ 2 4
1+( ωε )

• As the wave propagates along +z direction, it decreases in amplitude by a factor e−αz

Review of Lecture 2
Summary Continued
|J¯cs | σ Im[εc ] ε”
• Loss Tangent of the medium = tanθ = |J¯ds |
= ωε = Re[εc ] = ε′

• where εc is called complex permittivity of the medium

jσ
• εc = ε ′ − jε” = ε − ω

• ε ′ = ε , ε” = σ
ω

Plane Waves in Good Conductors

Wave Propagation in Good Conductors

Plane Waves in Good Conductors

• Constitutive Parameters for Good Conductor are

σ
– σ ≫ ωε or ωε ≫ 1, ε = ε0 , µ = µ0 µr

• Using the results for Lossy Dielectrics

s q  s q 
σ 2
µε   µε  σ 2 σ
– α =ω 2 1 + ωε − 1 = ω 2 ωε ∵ ωε ≫1
q
µε  σ
 q
µσ
q
ω µσ √
– or α = ω 2 ωε =ω 2ω = 2 = π f µσ
s
√
q  q
µε  σ 2
– β =ω 2 1 + ωε + 1 = ω µσ
2 = π f µσ
q
ω 2ω 2π
– u= β = µσ ,λ = β

EE-344 Wave Propagation and Antennas, Dr. M Anis Ch, Lecture 3 Page 2 of 17
Plane Waves in Good Conductors

• Intrinsic Impedance can be given as

√
jθ µ/ε σ
– η = |η| e where|η| = h
η
i 1 and tan (2θη ) = ωε
2 4
1+( ωε
σ
)
√ √ q q q
µ/ε µ/ε
– As ωε σ
≫ 1, |η| = h i1 = √ σ = µ
ε
ωε
σ = ωµ
σ
2
( ωε
σ
) 4 ωε

σ σ
– Also as ωε → ∞ ⇒ tan((2θη ) = ωε →∞
– or 2θη → 90o ⇒ θη = 45o
q π
– so η = |η| e = ωσµ e j 4
jθ η

• Ē(z,t) = E0 e−αz cos(ωt − β z)aˆx

• H̄(z,t) = √Eω0 µ e−αz cos(ωt − β z − 45o )ây

σ

– thus Ē leads H̄ by 45 degrees.

Plane Waves in Good Conductors

Skin Depth or Penetration Depth

• the distance δ after which the EM wave amplitude is reduced by a factor e−1 = 0.3679 ≈
0.37 (37% of the original value)

– i.e. E0 e−αz |z=δ = E0 e−αδ = E0 e−1

1
– or αδ = 1 ⇒ δ = α

EE-344 Wave Propagation and Antennas, Dr. M Anis Ch, Lecture 3 Page 3 of 17
Plane Waves in Good Conductors

• For a good conductor η can be given as

q π
q   q h i q
– η = |η| e jθη = ωσµ e j 4 = ωσµ cos π4 + jsin π4 = ωσµ √12 + j √12 = ω2σµ [1 + j]
q
– but δ = α1 = ω µσ 2
⇒ ω µσ 2
= δ 2 ⇒ ω µ = σ2δ 2
q
– η = 2σ22 δ 2 [1 + j] = 1+σδ
j

z
• Ē(z,t) = E0 e− δ cos(ωt − δz )aˆx ∵ α = β = 1
δ
z
• H̄(z,t) = q E0
2
e− δ cos(ωt − δz − 45o )ây
σ 2δ 2

Plane Waves in Good Conductors

Plane Waves in Good Conductors

Skin Depth
• is the depth to which an EM wave can penetrate the given medium
• the fields and associated currents are confined to a very thin layer (skin) of the conductor.
hence the name skin depth.

Practical Importance of Skin Depth

EE-344 Wave Propagation and Antennas, Dr. M Anis Ch, Lecture 3 Page 4 of 17
 • At very high frequencies a very thin layer of conductor carries most of the current

• so if we coat the conductor with silver (the best available conductor) then all the current
will flow through the silver coating.

• the electrical performance would be the same as if the entire conductor is made up of
solid silver.

Plane Waves in Good Conductors

Practical Importance of Skin Depth; continued

• Silver oxidizes very easily, so Gold is often preferred as it is inert

– e.g. attenuation in Waveguides is minimized by silver plating

• FM and TV antennas

– are hollow tubular conductors rather than solid cylinders

• EM Shielding

– Sensitive microwave components can be shielded by conductive casings whose thick-

ness need to be a few skin depths only

Plane Waves in Good Conductors

DC and AC Resistance

• DC Resistance is
ℓ
– Rdc = σS
– which is the direct consequence of ohm’s law (J¯ = σ Ē)

• Skin Resistance (Surface Resistance)

q
– Rs = σ1δ = πσf µ
– is the resistance of a unit width and unit length of the conductor

• AC Resistance

– for a given width w and length ℓ,

– Rac = wℓ Rs = ℓ
σδw

EE-344 Wave Propagation and Antennas, Dr. M Anis Ch, Lecture 3 Page 5 of 17
Plane Waves in Good Conductors
DC and AC Resistance; Continued
• for a cylindrical conductor having radius a and δ ≪ a

• w = 2πa

Plane Waves in Good Conductors

Example 10.4
• A uniform plane wave propagating in a medium has

Ē = 2e−αz sin 108t − β z ây V /m

If the medium is characterized by εr = 1, µr = 20, and σ = 3S/m, find α, β and H̄.

Power and the Poynting Vector

Power and the Poynting Vector

Power and the Poynting Vector

• In a wireless communication system, EM fields are used to transport information over

long distances

• So energy must be associated with EM waves

• The transport of energy is possible in any type of medium even in the absence of medium
(vacuum)

• To derive an expression of the rate of such energy transportation, let’s start with Maxwell’s
Curl Equations

EE-344 Wave Propagation and Antennas, Dr. M Anis Ch, Lecture 3 Page 6 of 17
 ¯ × Ē = −µ ∂ H̄
– ∇ ∂t
¯ × H̄ = σ Ē + ε ∂ Ē
– ∇ ∂t

• Dotting both sides of 1st equation with H̄ and 2nd Eq with Ē , we have
¯ × Ē = −µ H̄. ∂ H̄


– H̄. ∇ ∂t
¯ × H̄ = σ Ē.Ē + ε Ē. ∂ Ē


– Ē. ∇ ∂t
¯ × Ē − Ē. ∇ ¯ × H̄ = −µ H̄. ∂ H̄ − σ Ē.Ē − ε Ē. ∂ Ē



– or H̄. ∇ ∂t ∂t

Power and the Poynting Vector

¯ × H̄ = −µ H̄. ∂ H̄ − σ Ē.Ē − ε Ē. ∂ Ē

¯ × Ē − Ē. ∇



• or H̄. ∇ ∂t ∂t

• Using the vector identity

¯ Ā × B̄ = B̄. ∇ ¯ × B̄
¯ × Ā − Ā. ∇




– ∇.

• Above equation can be re-written as

¯ (Ē × H̄) = −µ H̄. ∂ H̄ − σ Ē.Ē − ε Ē. ∂ Ē
– ∇. ∂t ∂t

• As ∂ n
∂t u = nun−1 ∂∂tu
2
– ⇒ ∂ 2
∂t E = 2E ∂∂tE or 12 ∂t
∂ ∂
Ē.Ē = Ē. ∂t Ē ∵ Ē.Ē = |Ē| = E 2

• Substitute this in above relations

¯ (Ē × H̄) = −ε 1 ∂ [Ē.Ē] − µ 1 ∂ [H̄.H̄] − σ Ē.Ē
– ∇. 2 ∂t 2 ∂t
2
• As Ē.Ē = |Ē| = E 2 and H̄.H̄ = H 2
¯ (Ē × H̄) = −ε 1 ∂ E 2 − µ 1 ∂ H 2 − σ E 2
– ∇. 2 ∂t 2 ∂t
¯ (Ē × H̄) = − ∂
 ε 2 µ 2 2
– ∇. ∂t 2 E + 2 H − σ E

Power and the Poynting Vector

¯ (Ē × H̄) = − ∂ 2 + µ H2 − σ E2
ε 
• ∇. ∂t 2 E 2

• Taking Volume integral of both sides

´ ´  ε 2 µ 2 ´
– v ∇.¯ (Ē × H̄) dv = − ∂ E + H dv − 2
∂t v 2 2 v σ E dv
¸ ´
• Applying the Divergence Theorem i.e. s Ā.d S̄ = v ∇. ¯ Ādv
¸ ∂
´  ε 2 µ 2 ´ 2
– s (Ē × H̄) .d S̄ = − ∂t v 2 E + 2 H dv − v σ E dv

Poynting Theorem

EE-344 Wave Propagation and Antennas, Dr. M Anis Ch, Lecture 3 Page 7 of 17
 • Poynting Theorem is the mathematical expression of the law of conservation of energy as
applied to electromagnetic fields
¯ (Ē × H̄) = − ∂ ε E 2 + µ H 2 − σ E 2 Differential Form
 
– ∇. ∂t 2 2
¸ ´  ε 2 µ 2 ´
– s (Ē × H̄) .d S̄ = − ∂t v 2 E + 2 H dv − v σ E 2 dv Integral Form
∂

• Total Power Leaving the volume = rate of decrease in energy stored in E and H fields -
Ohmic power loss

Power and the Poynting Vector

Poynting Vector
• the quantity Ē × H̄ is called the poynting vector P̄
• is the instantaneous power density vector associated with the EM field at any given point
• is directed along the direction of wave propagation âk , and is thus normal to both Ē and
H̄ for a plane wave

• Now for a lossy dielectric, Ē can be given as

– Ē(z,t) = E0 e−αz cos (ωt − β z) âx
E0 −αz
– H̄(z,t) = |η| e cos (ωt − β z − θη ) ây
âx ây âz
– P̄(z,t) = Ē(z,t) × H̄(z,t) = Ex 0 0 = âz Ex Hy
0 Hy 0
E0 −αz
– P̄(z,t) = âz E0 e−αz cos (ωt − β z) |η| e cos (ωt − β z − θη )
E02 −2αz
– P̄(z,t) = |η| e cos (ωt − β z) cos (ωt − β z − θη ) âz

Power and the Poynting Vector

• Time average Poynting vector (W /m2 ) can be given as

´T
– P̄ave (z) = T1 0 P̄(z,t)dt
– P̄ave (z) = 21 Re (Ēs × H̄s∗ )
• Total time average power crossing a given surface S is
´
– Pave = s P̄ave .d S̄

• P̄(x, y, z,t) is Poynting Vector in Watts/square meter and is time varying

• P̄ave (x, y, z) is the time average of the Poynting Vector in Watts/square meter
• Pave is a total time-average power through a surface in watts and is a scalar

EE-344 Wave Propagation and Antennas, Dr. M Anis Ch, Lecture 3 Page 8 of 17
Power and the Poynting Vector
Example 10.8

• In a nonmagnetic medium

Ē = 4 sin 2π × 107t − 0.8x âz V /m

Find

• (a.) εr , η

• (b.) The time-average power carried by the wave.

Reflection of a Plane Wave at Normal Incidence

Reflection of a Plane Wave at Normal Incidence

Reflection of a Plane Wave at Normal Incidence

• When a plane wave from one medium strikes another medium, it is partly reflected and
partly transmitted

• the part of the incident wave that is reflected or transmitted depends on the constitutive
parameters (σ , ε, µ) of the two media

Normal Incidence

• direction of wave propagation is normal to the interface between the two media

• Assume the plane wave propagating along +z-direction in medium 1 (σ1 , ε1 , µ1 ) is inci-
dent normally on the boundary z=0 between medium 1 and medium 2 (σ2 , ε2 , µ2 )

– medium 1 (z<0) and medium 2 (z>0). interface is at z=0

– subscript
* i → incident wave , r → reflected wave , t → transmitted wave

Reflection of a Plane Wave at Normal Incidence

Incident Wave

• wave (Ēi , H̄i ) is travelling along +âz in medium 1 (assuming time factor e jωt ), if wave is
polarized in +x direction
Ei0 −γ1 z
– Ēis (z) = Ei0 e−γ1 z âx and H̄is (z) = Hi0 e−γ1 z ây = η1 e ây

EE-344 Wave Propagation and Antennas, Dr. M Anis Ch, Lecture 3 Page 9 of 17
Reflection of a Plane Wave at Normal Incidence
Reflected Wave

• Reflected wave (Ēr , H̄r ) is travelling along −âz in medium 1 ,assuming wave is polarized
in +x direction

– Ērs (z) = Er0 e+γ1 z âx

– H̄rs (z) = Hr0 e+γ1 z (−ây ) = − Eηr01 e+γ1 z ây

Transmitted Wave

• Transmitted wave (Ēt , H̄t ) is travelling along +âz in medium 2 ,assuming wave is polar-
ized in +x direction

– Ēts (z) = Et0 e−γ2 z âx

Et0 −γ2 z
– H̄ts (z) = Ht0 e−γ2 z ây = η2 e ây

EE-344 Wave Propagation and Antennas, Dr. M Anis Ch, Lecture 3 Page 10 of 17
Reflection of a Plane Wave at Normal Incidence

• So we see in medium 1

– total field in medium1 = incident field + reflected field

– Ē1 = Ēi + Ēr and H̄1 = H̄i + H̄r

• in medium 2

– total field in medium 2 = transmitted field

– Ē2 = Ēt and H̄2 = H̄t

Reflection of a Plane Wave at Normal Incidence

• At the interface z=0, the boundary conditions require the tangential components of Ē and
H̄ must be continuous

• As the waves are transverse Ē and H̄ are entirely tangential to the interface

at z=0

• Ē1tan = Ē2tan

• Ēi (z = 0) + Ēr (z = 0) = Ēt (z = 0)

• Ēi (0) + Ēr (0) = Ēt (0)

• Ēi0 âx + Ēr0 âx = Ēt0 âx ⇒ Ēi0 + Ēr0 = Ēt0

EE-344 Wave Propagation and Antennas, Dr. M Anis Ch, Lecture 3 Page 11 of 17
at z=0

• H̄1tan = H̄2tan

• H̄i (0) + H̄r (0) = H̄t (0)

Ēi0 Ēr0 Ēt0 Ēi0
• η1 ây − η1 ây = η2 ây ⇒ η1 − Ēηr01 = Ēt0
η2

Reflection of a Plane Wave at Normal Incidence

Ēi0
• Ēi0 + Ēr0 = Ēt0 and η1 − Ēηr01 = Ēt0
η2

• eliminating Ēt0 , implies

Ēi0 Ēi0 +Ēr0
– η1 − Ēηr01 = η2
– η2 Ēi0 − η2 Ēr0 = η1 Ēi0 + η1 Ēr0
– η2 Ēi0 − η1 Ēi0 = η2 Ēr0 + η1 Ēr0
η2 −η1
– η2 +η1 Ēi0 = Ēr0 = ΓĒi0 , Γ is called the Reflection Coefficient

• eliminating Ēr0 , implies

– Ēi0
η1 − Ēt0η−1Ēi0 = Ēt0
η2
– η2 Ēi0 − η2 Ēt0 + η2 Ēi0 = η1 Ēt0
– 2η2 Ēi0 = (η2 + η1 )Ēt0
2η2
– Ēt0 = (η2 +η1 ) Ēi0 = τ Ēi0 , τ is called the Transmission Coefficient

Reflection of a Plane Wave at Normal Incidence

Reflection Coefficient and Transmission Coefficient
η2 −η1
• Reflection Coefficient =Γ = η2 +η1

2η2
• Transmission Coefficient =τ = (η2 +η1 )

• other properties are

– 1 + Γ = τ i.e. 1 + ηη22 −η1

+η1 =
η2 +η1 +η2 −η1
η2 +η1 = 2η2
η2 +η1
– 0 ≤ |Γ| ≤ 1
– both reflection and transmission coefficients are dimensionless and may be complex

EE-344 Wave Propagation and Antennas, Dr. M Anis Ch, Lecture 3 Page 12 of 17
Reflection of a Plane Wave at Normal Incidence
Case A
• Medium 1: A perfect dielectric (σ1 = 0, ε1 , µ1 ) ⇒ α1 = 0, γ1 = jβ1
q
• Medium 2: A perfect conductor (σ2 = ∞, ε2 , µ2 ) ⇒ η2 = ωσµ ∠450 = 0

η2 −η1 0−η1 2η2

• Γ= η2 +η1 = 0+η1 = −1 and τ = (η2 +η1 ) =0

• So the wave is totally reflected which is expected for a perfect conductor as the fields
must vanish in the perfect conductor

• the totally reflected wave combines with the incident wave to form a standing wave

• Standing Wave

– it stands, doesnot travel

– it consists of two travelling waves Ēi and Ēr of equal amplitude but opposite direc-
tion

Reflection of a Plane Wave at Normal Incidence

Case A; continued
• so Γ = −1 and τ = 0

• ⇒ Ē2 = Ēt = 0

• Ē1s = Ēis + Ērs = (Ei0 e−γ1 z + Er0 eγ1 z )âx

• Ē1s = Ei0 (e− jβ1 z − e jβ1 z )âx ∵ Γ = −1 ⇒ Er0 = −Ei0 and γ1 = jβ1

• Ē1s = −2 jEi0 sin (β1 z) âx

• Ē1 = Re Ē1s e jωt = Re −2 jEi0 sin (β1 z) âx e jωt

 
• or Ē1 = Re 2Ei0 sin (β1 z) âx e jωt e− jπ/2 ∵ − j = e− jπ/2

• Ē1 = 2Ei0 sin(β1 z)cos(ωt − π2 )âx = 2Ei0 sin(β1 z)sin(ωt)âx

Reflection of a Plane Wave at Normal Incidence

Case A; continued
• Similarly, H̄1s = H̄is + H̄rs = ( Eηi01 e−γ1 z − Eηr01 eγ1 z )ây

• using Er0 = −Ei0 and γ1 = jβ1

 
• H̄1s = Eηi01 e− jβ1 z + e jβ1 z ây = Ei0
η1 2cos(β1 z)ây

Ei0
• H̄1 = Re H̄1s e jωt =


η1 2cos(β1 z)cos(ωt)ây

EE-344 Wave Propagation and Antennas, Dr. M Anis Ch, Lecture 3 Page 13 of 17
Reflection of a Plane Wave at Normal Incidence
Case A; continued

Reflection of a Plane Wave at Normal Incidence

Standing and Travelling Waves

Reflection of a Plane Wave at Normal Incidence

Case B

• Media 1 and 2 are lossless i.e

q
µ1
– Medium 1: (σ1 = 0, ε1 , µ1 ) ⇒ α1 = 0, γ1 = jβ1 , η1 = ε1
q
µ2
– Medium 2: (σ2 = 0, ε2 , µ2 ) ⇒ α2 = 0, γ2 = jβ2 , η2 = ε2

• η1 and η2 are real

• Γ and τ are real

• we will consider 2 cases

– Case B.1: η2 > η1 , Γ > 0

– Case B.2: η2 < η1 , Γ < 0

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Reflection of a Plane Wave at Normal Incidence
Case B.1

• η2 > η1 , Γ > 0

– here standing wave plus travelling wave in medium 1

– travelling wave in medium 2
– here |Eis | ̸= |Ers |

• Ē1s = Ēis + Ērs = (Ei0 e−γ1 z + Er0 eγ1 z )âx

• Ē1s = Ei0 (e− jβ1 z + Γe jβ1 z )âx ∵ γ1 = jβ1 , Er0 = ΓEi0

• using Euler’s Identity

• Ē1s = Ei0 (cosβ1 z − jsinβ1 z + Γcosβ1 z + jΓsinβ1 z)âx

• Ē1s = Ei0 [(Γ + 1)cosβ1 z + j(Γ − 1)sinβ1 z] âx

p
• |Ē1s | = Ei0 (Γ + 1)2 cos2 β1 z + (Γ − 1)2 sin2 β1 z
p
• |Ē1s | = Ei0 (Γ2 + 1 + 2Γ)cos2 β1 z + (Γ2 + 1 − 2Γ)sin2 β1 z
p p
• |Ē1s | = Ei0 Γ2 + 1 + 2Γ(cos2 β1 z − sin2 β1 z) = Ei0 Γ2 + 1 + 2Γcos (2β1 z)

Reflection of a Plane Wave at Normal Incidence

Case B.1; continued
p
• |Ē1s | = Ei0 Γ2 + 1 + 2Γcos (2β1 z)

• |Ē1s | is maximum when cos (2β1 z) = 1, 2β1 zmax = 2nπ, n = 0, 1, 2, 3, ....

√
– |Ē1s |max = Ei0 Γ2 + 1 + 2Γ = Ei0 (1 + Γ)

• thus 2β1 zmax = 2nπ, n = 0, 1, 2, 3, .... or −2β1 zmax = 2nπ, n = 0, 1, 2, 3, ....

• Similarly

• |Ē1s | is minimum when cos (2β1 z) = −1 and 2β1 zmin = π + 2nπ, n = 0, 1, 2, 3, .... or
−2β1 zmin = π + 2nπ, n = 0, 1, 2, 3, ....
√
– |Ē1s |max = Ei0 Γ2 + 1 − 2Γ = Ei0 (1 − Γ)

EE-344 Wave Propagation and Antennas, Dr. M Anis Ch, Lecture 3 Page 15 of 17
Reflection of a Plane Wave at Normal Incidence
Case B.2
• η2 < η1 , Γ < 0 ⇒ Γ = − |Γ|
p
– substitute Γ = − |Γ| into |Ē1s | = Ei0 Γ2 + 1 + 2Γcos (2β1 z)
q
– |Ē1s | = Ei0 |Γ|2 + 1 − 2 |Γ| cos (2β1 z)
– thus |Ē1s | is maximum when cos (2β1 z) = −1
– and|Ē1s | is minimum when cos (2β1 z) = 1

Reflection of a Plane Wave at Normal Incidence

Animations
Travelling Wave
Standing Wave
Combination of the two

Reflection of a Plane Wave at Normal Incidence

Notes
• In medium 1, the total field is comprised of a travelling part as well as the standing wave
– as the incident and reflected waves have amplitudes that are not equal in magnitude
• |H1 | minimum occurs whenever there is |E1 |maximum and vice versa
• the transmitted wave is a purely tranvelling wave

EE-344 Wave Propagation and Antennas, Dr. M Anis Ch, Lecture 3 Page 16 of 17
Reflection of a Plane Wave at Normal Incidence
Standing Wave Ratio

• is defined as the ratio of |E1 |max to |E1 |min or (|H1 |max to |H1 |min )
|E1 |max |H1 |max 1+|Γ|
– i.e. s = |E1 |min = |H1 |min = 1−|Γ|
s−1
– rearranging⇒ |Γ| = s+1

• As |Γ| ≤ 1 ⇒ 1 ≤ s ≤ ∞

• the standing wave ratio is dimensionless

• in decibels (dB)

– s dB = 20log10 s

Reflection of a Plane Wave at Normal Incidence

Example 10.9

• In free space (z ≤ 0), a plane wave with

H̄i = 10 cos 108t − β z âx mA/m

is incident normally on a lossless medium (ε = 2ε0 , µ = 8µ0 ) in region z ≥ 0. Determine

the reflected wave H̄r , Ēr and the transmitted wave H̄t , Ēt .

Reflection of a Plane Wave at Normal Incidence

Example 10.10

• Given a uniform plane wave in air as

Ēi = 40 cos (ωt − β z) âx + 30 sin (ωt − β z) ây V /m

• (a) Find H̄i

• (b) If the wave encounters a perfectly conducting plate normal to the z-axis at z = 0, find
the reflected wave Ēr and H̄r

• Figures in this lecture are from

– Sadiku MN. Elements of electromagnetics

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