Engineering Electromagnetics
Engineering Electromagnetics
9th Edition
ISBN: 9781260029963
Author: Hayt
Publisher: MCG
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Chapter 12, Problem 12.9P
To determine

(a)

The percentage of the energy incident on the boundary is reflected.

Expert Solution
Check Mark

Answer to Problem 12.9P

The percentage of the energy incident on the boundary reflected is =6.25×102.

Explanation of Solution

Given:

Two regions with region 1 : z<0 and region 2 : z>0 are perfect dielectrics.

Radian frequency of uniform plane wave =3×1010rad/s.

Wavelengths in two regions

   λ1=5cm,λ2=3cm

Concept Used:

Use below formula to calculate the value of reflection coefficient which gives the value of the percentage of the energy incident on the boundary is reflected.

   Γ=η2η1η2+η1=η01/ε r2'η01/ε r1'η01/ε r2'+η01/ε r1'=ε r1'/ε r2'1ε r1'/ε r2'+1

Calculation:

We first use below formula:

   εr1'=( 2πc λ 1 ω)2 and εr2'=( 2πc λ 2 ω)2

So, we have εr1'/εr2'=(λ2/λ1)2

And with the μ=μ0 in both the regions, we will calculate:

   Γ=η2η1η2+η1=η0 1/ ε r2 ' η0 1/ ε r1 ' η0 1/ ε r2 ' +η0 1/ ε r1 ' = ε r1 ' / ε r2 ' 1 ε r1 ' / ε r2 ' +1=( λ 2 / λ 1 )1( λ 2 / λ 1 )+1=λ2λ1λ2 1=353+5=28=14

The fraction of the incident energy that is reflected is then |Γ|2=1/16=6.25×102.

To determine

(b)

The percentage of the energy incident on the boundary is transmitted.

Expert Solution
Check Mark

Answer to Problem 12.9P

The percentage of the energy incident on the boundary transmitted is =0.938.

Explanation of Solution

Given:

Two regions with region 1 : z<0 and region 2 : z>0 are perfect dielectrics.

Radian frequency of uniform plane wave =3×1010rad/s.

Wavelengths in two regions

   λ1=5cm,λ2=3cm

Concept Used:

Use below formula to calculate the value of reflection coefficient which gives the value of the percentage of the energy incident on the boundary is reflected.

   Γ=η2η1η2+η1=η01/ε r2'η01/ε r1'η01/ε r2'+η01/ε r1'=ε r1'/ε r2'1ε r1'/ε r2'+1

Then calculatethe percentage of the energy incident on the boundary transmitted by using formula 1|Γ|2.

Calculation:

We first use below formula.

   εr1'=( 2πc λ 1 ω)2 and εr2'=( 2πc λ 2 ω)2

So, we have εr1'/εr2'=(λ2/λ1)2

And with the μ=μ0 in both the regions, we will calculate:

   Γ=η2η1η2+η1=η0 1/ ε r2 ' η0 1/ ε r1 ' η0 1/ ε r2 ' +η0 1/ ε r1 ' = ε r1 ' / ε r2 ' 1 ε r1 ' / ε r2 ' +1=( λ 2 / λ 1 )1( λ 2 / λ 1 )+1=λ2λ1λ2 1=353+5=14

The fraction of the transmitted energy that is reflected is then 1|Γ|2=16.25×102=0.938.

To determine

(c)

The standing wave ratio in region 1.

Expert Solution
Check Mark

Answer to Problem 12.9P

The standing wave ratio in region 1 is s=1.67.

Explanation of Solution

Given:

Two regions with region 1 : z<0 and region 2 : z>0 are perfect dielectrics.

Radian frequency of uniform plane wave =3×1010rad/s.

Wavelengths in two regions

   λ1=5cm,λ2=3cm

Concept Used:

Using the below formula to calculate the value of reflection coefficient which gives the value of the percentage of the energy incident on the boundary is reflected.

   Γ=η2η1η2+η1=η01/ε r2'η01/ε r1'η01/ε r2'+η01/ε r1'=ε r1'/ε r2'1ε r1'/ε r2'+1

Use s=1+|Γ|1|Γ| to calculate the standing wave ratio.

Calculation:

We first use below formula.

   εr1'=( 2πc λ 1 ω)2 and εr2'=( 2πc λ 2 ω)2

So, we have εr1'/εr2'=(λ2/λ1)2

And with the μ=μ0 in both the regions, we will calculate

   Γ=η2η1η2+η1=η0 1/ ε r2 ' η0 1/ ε r1 ' η0 1/ ε r2 ' +η0 1/ ε r1 ' = ε r1 ' / ε r2 ' 1 ε r1 ' / ε r2 ' +1=( λ 2 / λ 1 )1( λ 2 / λ 1 )+1=λ2λ1λ2 1=353+5=14

The standing wave ratio is:

   s=1+|Γ|1|Γ|=1+1/411/4=53=1.67

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