# Calculating antenna directivity

1. Feb 26, 2009

### JamesGoh

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

Direcitivity is calculating using

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

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

Last edited: Feb 27, 2009
2. Feb 26, 2009

### 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($$\theta$$,$$\phi$$ )$$d\phi$$$$d\theta$$

3. Feb 26, 2009

### JamesGoh

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

4. Feb 27, 2009

### JamesGoh

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

Direcitivity is calculating using

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

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

Last edited: Feb 27, 2009