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JamesGoh
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Homework Statement
An antenna has a normalised E-filed pattern, En where [tex]\theta[/tex] = vertical angle as measured from z-axis and [tex]\phi[/tex] = azimuth angle measured from x-axis.
Calculate the exact directivity
En has a non-zero value whenever [tex]0 <= \theta <= \pi[/tex] and [tex]0 <= \phi <= \pi[/tex]. Elsewhere, En is zero
The correct answer is 6 , but I cannot get this number
Homework Equations
[tex]En = sin(\theta)sin(\phi)[/tex]
Direcitivity is calculating using
[tex]D = 4\pi/ ( \int\intPn(\theta,\phi )d\phid\theta )[/tex]
The Attempt at a Solution
Ok, first of all [tex]Pn = En^{2}[/tex]
Therefore [tex]Pn = sin(\theta)^{2}sin(\phi)^{2}[/tex]
Now [tex]sin^{2}(\theta) = 0.5(1 - cos(2\theta))[/tex] with respect to [tex]\theta[/tex]
If we perform the integration of the [tex]sin^{2}(\theta)[/tex] terms we should get
[tex]\theta/2 - sin(2\theta)/4[/tex]
Applying the [tex]0 <= \theta <= \pi[/tex] limits, I got [tex]\pi/2[/tex] for the first integral. If we integrate the [tex]sin(\phi)[/tex] term, we should get [tex]\pi/2[/tex] as well, making [tex]\pi^{2}/4[/tex] the actual answer for
[tex]\int\intPn(\theta,\phi )d\phid\theta )[/tex]
Substituting [tex]\pi^{2}/4[/tex] into the denomiator part of D, I get [tex]16/\pi[/tex] which obviously is not right. I am not sure where I could be going wrong
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