Question

The radiation intensity of a given antenna is U = 2(sin θ sin ϕ) in the range 0 ≤ θ ≤ π and 0 ≤ ϕ ≤ π and 0 elsewhere. The directivity is:

1. 3 dB
2. 6 dB
3. 8 dB
4. 9 dB

Answer 1

Correct Answer - Option 2 : 6 dB

Concept:

The directivity of an antenna is defined as:

\(D = \frac{{{U_{max}}}}{{{U_{avg}}}}\)

Where U_avg is evaluated as:

\({U_{avg}} = \frac{1}{{4\pi }}\mathop \smallint \limits_U\left( {\theta ,\phi } \right) \cdot d{\rm{\Omega }}\)

dΩ is the solid angle.

Calculation:

For the given U(θ, ϕ), the average radiation intensity will be:

\({U_{avg}} = \frac{1}{{4\pi }}\mathop \smallint \limits_{0 = 0}^\pi \mathop \smallint \limits_{\phi = 0}^\pi \left( {2\left( {\sin \theta \sin \phi } \right)} \right)\sin \theta d\theta d\phi \)

\( = \frac{2}{{4\pi }}\mathop \smallint \limits_{0 = 0}^\pi {\sin ^2}\theta d\theta \mathop \smallint \limits_{\phi = 0}^\pi \sin \phi d\phi \)

\( = \frac{1}{{2\pi }}\mathop \smallint \limits_{0 = 0}^\pi \left( {\frac{{1 - \cos 2\theta }}{2}} \right)d\theta \left. {\left( { - \cos \theta } \right)} \right|_0^\pi \)

\( = \frac{1}{{4\pi }}\left. {\left[ {0 - \frac{{\sin 2\theta }}{2}} \right]} \right|_0^\pi \left[ { - \left( {\cos \pi - \left( 0 \right)0} \right)} \right]\)

\( = \frac{1}{{4\pi }}\left[ {\pi - \left( {\left( { - 1 - 1} \right)} \right)} \right] = \frac{1}{{2\pi }} = 0.5\)

Also, the maximum radiation intensity for the given antenna occurs at \(\theta = \frac{\pi }{2}\) and \(\phi = \frac{\pi }{2}\), i.e.

\({U_{max}} = 2\sin \frac{\pi }{2} \cdot \sin \frac{\pi }{2}\)

U_max = 2

∴ The directivity of the antenna will be:

\(D = \frac{{{U_{max}}}}{{{U_{avg}}}} = \frac{2}{{0.5}} = 4\)

In dB, this can be expressed as:

D(dB) = 10 log₁₀ (4)

= 6 dB