(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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

2. Relevant equations

[tex]En = sin(\theta)sin(\phi)[/tex]

Direcitivity is calculating using

[tex]D = 4\pi/ ( \int\intPn(\theta,\phi )d\phid\theta )[/tex]

3. 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. Im not sure where I could be going wrong

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# Calculating antenna directivity

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