Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN: 9780190698614
Author: Sadiku, Matthew N. O.
Publisher: Oxford University Press
Question
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Chapter 13, Problem 9P

(a)

To determine

Prove that the generated far-field expressions for a thin dipole of length l carrying sinusoidal current Iocosβz are Hϕs=jIoeβr2πrcos(βl2cosθ)cosβl2sinθ, and Eθs=ηHϕs.

(a)

Expert Solution
Check Mark

Explanation of Solution

Calculation:

Consider the mentioned equation given in the textbook,

dAzs=μIocosβzdz4πr'ejβr'        (1)

Refer to the mentioned Figure given in the textbook, the integration limits must be ±l2.

Azs=μIo4πrl2l2ejβ(rzcosθ)cosβzdz{r'=rzcosθ}

Azs=μIo4πrejβrl2l2ejβzcosθcosβzdz        (2)

Refer to the integral tables of Appendix A.8,

eazcosbzdz=eaz(acosbz+bsinbz)a2+b2+c        (3)

Apply equation (3) in (2),

Azs=μIoejβrejβzcosθ4πr(jβcosθcosβz+βsinβz)β2cos2θ+β2|l2l2=μIoejβrejβzcosθ4πr(jcosθcosz+sinβz)β(1cos2θ)|l2l2=μIoejβrejβzcosθ4πr(jcosθcosz+sinβz)β(sin2θ)|l2l2{1cos2θ=sin2θ}=μIoejβr[cos(βzcosθ)jsin(βzcosθ)]4πr(jcosθcosz+sinβz)β(sin2θ)|l2l2

Applying the limits to the above equation as follows,

Azs=μIoejβr4πr1βsin2θ[sinβl2cos(βl2cosθ)cosθcosβl2sin(βl2cosθ)]

Consider the general relationship between magnetic flux density and magnetic field intensity.

Bs=μHs=×As

Rearrange the equation as follows,

Hϕs=1μ(×As)

Hϕs=1μr[r(rAθ)Arθ]        (4)

Where, Aθ=Azsinθ and Ar=Azcosθ.

Therefore, equation (4) can be rewritten as follows,

Hϕs={Io2πrejβrβ(jβsinθ)[sinβl2cos(βl2cosθ)cosθcosβl2sin(βl2cosθ)]+Io2πr2ejβr(sinβl2cos(βl2cosθ)cosθcosβl2sin(βl2cosθ))}

Consider for a far-field expression, only 1r terms remains. Therefore, above expression becomes,

Hϕs=jIoejβr2πr[sinβl2cos(βl2cosθ)cosθcosβl2sin(βl2cosθ)sinθ]

and the relationship between electric field intensity and magnetic field intensity is,

Eθs=ηHϕs

Conclusion:

Thus, the generated far-field expressions for a thin dipole of length l carrying sinusoidal current Iocosβz are Hϕs=jIoeβr2πrcos(βl2cosθ)cosβl2sinθ, and Eθs=ηHϕs is proved.

(b)

To determine

Sketch the group pattern f(θ) in Part (a) for l=λ,3λ2,and2λ.

(b)

Expert Solution
Check Mark

Answer to Problem 9P

The group pattern f(θ) in Part (a) for l=λ,3λ2,and2λ is plotted.

Explanation of Solution

Calculation:

Refer to Part (a),

Hϕs=jIoejβr2πr[sinβl2cos(βl2cosθ)cosθcosβl2sin(βl2cosθ)sinθ]

From the above expression, the group pattern is,

f(θ)=cos(βl2cosθ)cosβl2sinθ        (5)

Consider the expression for the phase constant.

β=2πλ

Here,

λ is the wavelength.

For l=λ:

Substitute 2πλ for β and λ for l in equation (5).

f(θ)=cos((2πλ)λ2cosθ)cos(2πλ)λ2sinθ=cos(πcosθ)cosπsinθ=cos(πcosθ)+1sinθ

MATLAB code to plot the group pattern f(θ) for l=λ.

theta=-360:pi./180:360;

c_n1=(cos(pi.*cos(theta)))+1;

s_n1=sin(theta);

f1=abs(c_n1./s_n1);

polar(theta,f1)

MATLAB output:

Figure 1 shows the polar plot for the group pattern f(θ) for l=λ.

Elements Of Electromagnetics, Chapter 13, Problem 9P , additional homework tip  1

For l=3λ2:

Substitute 2πλ for β and 3λ2 for l in equation (5).

f(θ)=cos((2πλ)(3λ2)2cosθ)cos(2πλ)(3λ2)2sinθ=cos(3π2cosθ)cos3π2sinθ=cos(3π2cosθ)sinθ

MATLAB code to plot the group pattern f(θ) for l=3λ2.

theta=-360:pi./180:360;

c_n2=cos((3.*pi./2).*cos(theta));

s_n2=sin(theta);

f2=abs(c_n2./s_n2);

polar(theta,f2)

MATLAB output:

Figure 2 shows the polar plot for the group pattern f(θ) for l=3λ2.

Elements Of Electromagnetics, Chapter 13, Problem 9P , additional homework tip  2

For l=2λ:

Substitute 2πλ for β and 2λ for l in equation (5).

f(θ)=cos((2πλ)(2λ)2cosθ)cos(2πλ)(2λ)2sinθ=cos(2πcosθ)cos2πsinθ=cos(2πcosθ)1sinθ

MATLAB code to plot the group pattern f(θ) for l=2λ.

theta=-360:pi./180:360;

c_n3=cos(2.*pi.*cos(theta))-1;

s_n3=sin(theta);

f3=abs(c_n3./s_n3);

polar(theta,f3)

MATLAB output:

Figure 3 shows the polar plot for the group pattern f(θ) for l=3λ2.

Elements Of Electromagnetics, Chapter 13, Problem 9P , additional homework tip  3

Conclusion:

Thus, the group pattern f(θ) in Part (a) for l=λ,3λ2,and2λ is plotted.

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