Calculating antenna directivity

[JamesGoh]

Homework Statement

An antenna has a normalised E-filed pattern, En where $$\theta$$ = vertical angle as measured from z-axis and $$\phi$$ = azimuth angle measured from x-axis.
Calculate the exact directivity

En has a non-zero value whenever $$0 <= \theta <= \pi$$ and $$0 <= \phi <= \pi$$. Elsewhere, En is zero


The correct answer is 6 , but I cannot get this number

Homework Equations

$$En = sin(\theta)sin(\phi)$$

Direcitivity is calculating using

$$D = 4\pi/ ( \int\intPn(\theta,\phi )d\phid\theta )$$

The Attempt at a Solution

Ok, first of all $$Pn = En^{2}$$

Therefore $$Pn = sin(\theta)^{2}sin(\phi)^{2}$$

Now $$sin^{2}(\theta) = 0.5(1 - cos(2\theta))$$ with respect to $$\theta$$

If we perform the integration of the $$sin^{2}(\theta)$$ terms we should get

$$\theta/2 - sin(2\theta)/4$$

Applying the $$0 <= \theta <= \pi$$ limits, I got $$\pi/2$$ for the first integral. If we integrate the $$sin(\phi)$$ term, we should get $$\pi/2$$ as well, making $$\pi^{2}/4$$ the actual answer for

$$\int\intPn(\theta,\phi )d\phid\theta )$$

Substituting $$\pi^{2}/4$$ into the denomiator part of D, I get $$16/\pi$$ which obviously is not right. I am not sure where I could be going wrong

 

[JamesGoh]

If we perform the integration of the $$sin^{2}(\theta)$$ terms we should get

$$\theta/2 - sin(2\theta)/4$$

Applying the 0 <= $$\theta$$ <= $$\pi$$ limits, I got $$\pi$$/2 for the first integral. If we integrate the $$sin(\phi)$$ term, we should get $$\pi$$/2 as well, making $$\pi^{2}/4$$ the actual answer for

$$\int$$$$\int$$Pn($$\theta$$,$$\phi$$ )$$d\phi$$$$d\theta$$

 

[JamesGoh]

Guys, please refer to my 2nd most outlining the integration process

cheers and thanks in advance

 

[JamesGoh]

Homework Statement

An antenna has a normalised E-filed pattern, En where $$\theta$$ = vertical angle as measured from z-axis and $$\phi$$ = azimuth angle measured from x-axis.
Calculate the exact directivity

En has a non-zero value whenever $$0 <= \theta <= \pi$$ and $$0 <= \phi <= \pi$$. Elsewhere, En is zero


The correct answer is 6 , but I cannot get this number

Homework Equations

$$En = sin(\theta)sin(\phi)$$

Direcitivity is calculating using

$$D = 4\pi/$$ $$\int$$$$\int$$$$Pn$$($$\theta$$,$$\phi$$ )$$d\phi$$$$d\theta$$

The Attempt at a Solution

Ok, first of all $$Pn = En^{2}$$

Therefore $$Pn = sin(\theta)^{2}sin(\phi)^{2}$$

Now $$sin^{2}(\theta) = 0.5(1 - cos(2\theta))$$ with respect to $$\theta$$

If we perform the integration of the $$sin^{2}(\theta)$$ terms we should get

$$\theta/2 - sin(2\theta)/4$$

Applying the $$0 <= \theta <= \pi$$ limits, I got $$\pi/2$$ for the first integral. If we integrate the $$sin(\phi)$$ term, we should get $$\pi/2$$ as well, making $$\pi^{2}/4$$ the actual answer for

$$\int$$$$\int$$$$Pn$$($$\theta$$,$$\phi$$ )$$d\phi$$$$d\theta$$

Substituting $$\pi^{2}/4$$ into the denomiator part of D, I get $$16/\pi$$ which obviously is not right. I am not sure where I could be going wrong