Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN: 9780190698614
Author: Sadiku, Matthew N. O.
Publisher: Oxford University Press
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Chapter 13, Problem 5P

(a)

To determine

Prove that the radiation resistance is Rrad=20π2[lλ]2.

(a)

Expert Solution
Check Mark

Explanation of Solution

Calculation:

Using the mentioned equation given in the textbook,

Azs=ejβr4πrl2l2Io(12|z|l)ejβzcosθdz=ejβr4πrIo[l2l2(12|z|l)cos(βzcosθ)dz+jl2l2(12|z|l)sin(βzcosθ)dz]

Consider the real part of Azs. Therefore,

Azs=ejβr4πrIo[l2l2(12|z|l)cos(βzcosθ)dz]=ejβr4πr2Io[0l2(12zl)cos(βzcosθ)dz]=Ioejβr2πr(2l){1cos[β(l2)cosθ]β2cos2θ}

Azs=Ioejβrπrlβ2cos2θ[1cos(βl2cosθ)]        (1)

The general expression to calculate the electric field intensity in relationship with the magnetic vector potential.

Es=jωμAs

From the above expression,

Eθs=jωμsinθAzs=jβηsinθAzs        (2)

Substitute equation (1) in equation (2)

Eθs=jβηsinθ{Ioejβrπrlβ2cos2θ[1cos(βl2cosθ)]}

Eθs=(jηIoejβrπrl){sinθ[1cos(βl2cosθ)]βcos2θ}        (3)

In equation (3), if βl21 then cos(βl2cosθ)=1(βl2cosθ)22!. Therefore, equation (3) becomes,

Eθs=(jηIoejβrπrl){sinθ[1(1(βl2cosθ)22!)]βcos2θ}=(jηIoejβrπrl){sinθ[(βl2cosθ)22]βcos2θ}=(jηIoejβrπrl)(βl28sinθ)

Eθs=jηIo8πrβlejβrsinθ        (4)

Write the general expression for the average radiated power.

Pave=|Eθs|22η        (5)

Here,

η is the intrinsic impedance, and

Eθs is the electric field intensity.

Substitute equation (4) in (5).

Pave=|jηIo8πrβlejβrsinθ|22η=η2Io264π2r2β2l2sin2θ2η=ηIo2128π2r2β2l2sin2θ

Write the general expression to calculate the radiated power.

Prad=SPavedS        (6)

Here,

dS is the differential surface which is equal to r2sinθdθdϕ.

Substitute ηIo2128π2r2β2l2sin2θ for Pave and r2sinθdθdϕ for dS in equation (6) with integration limits.

Prad=SηIo2128π2r2β2l2sin2θr2sinθdθdϕ=02π0πηIo2128π2r2β2l2sin2θr2sinθdθdϕ=(ϕ)02π0πηIo2128π2β2l2sin3θdθ=2π0πηIo2128π2β2l2sin3θdθ

Simplify the equation as follows,

Prad=2π0π(120π)Io2128π2(2πλ)2l2sin3θdθ{β=2πλ,η=120π}=120Io264(2πλ)2l20πsin3θdθ

Prad=480π2Io264(lλ)20πsin3θdθ        (7)

On solving the integration, 0πsin3θdθ=43. Therefore, equation (7) becomes,

Prad=480π2Io264(lλ)2(43)

Prad=10π2Io2(lλ)2        (8)

Consider the expression to calculate the radiated power.

Prad=12Io2Rrad        (9)

Here,

Io is the peak value of input current, and

Rrad is the radiation resistance.

Equating equation (8) and (9).

12Io2Rrad=10π2Io2(lλ)212Rrad=10π2(lλ)2Rrad=20π2(lλ)2

Conclusion:

Thus, the radiation resistance Rrad=20π2[lλ]2 is proved.

(b)

To determine

Find the length (l) of the dipole.

(b)

Expert Solution
Check Mark

Answer to Problem 5P

The length (l) of the dipole is l=0.05λ.

Explanation of Solution

Calculation:

Refer to part (a),

The radiation resistance for a short dipole antenna is,

Rrad=20π2[lλ]2

Substitute 0.5 for Rrad in above equation to find the length of the dipole l.

0.5=20π2[lλ]2[lλ]2=0.520π2lλ=0.520π2l=0.05λ

Conclusion:

Thus, the length (l) of the dipole is l=0.05λ.

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