Electromagnetics Test 6

For an antenna array:

ψ = βd cos θ + α

For nulls:

\(\frac{{N\psi }}{2} = \pm k\pi ,\;k = 1,2,3 \ldots\)

For broadside array:

\(\left( {\cos {\phi _{null}} - \cos {\phi _{max}}} \right)\frac{{\beta d}}{2} = \pm \frac{{k\pi }}{N}\)

The Beamwidth between first nulls for a broadside long array is given by:

\(\frac{{2\lambda }}{{nd}}\)

For an ordinary end-fire long array, the Beamwidth is given by:

\(\sqrt {\frac{2\lambda }{{nd}}}\)

Concept:

Power Gain of an Antenna is given as:

G = η D 

Where;

η = efficiency and ‘D’ is the Directivity

\(\eta = \frac{{{R_{rad}}}}{{{R_{rad}} + {R_{loss}}}}\)

Rrad = Radiation Resistance and

Rloss = Loss Resistance.

Calculation:

Given:

Rrad = 73 ohm

Rloss = 7 ohm

G = 16

G = η D

\(D = \frac{G}{\eta } = \frac{G}{{{R_{rad}}}}\left( {{R_{rad}} + {R_{loss}}} \right) \)

\(D= \frac{{16 × \left( {73 + 7} \right)}}{{73}}\)

D = 16 × 1.095

D = 17.53

S = Area = Nπr²

When loop radius is halved

\(S' = N\pi {\left( {\frac{r}{2}} \right)^2} = \frac{S}{4}\) 

The loop area becomes one-fourth

\({R_{rad}} = \frac{{320{\pi ^4}{S^2}}}{{{\lambda ^4}}}\) 

Taking log

Log (R_rad) = log(320π⁴) - log(λ⁴) - log(S²)

Differentiating:

\(\frac{{d{R_{rad}}}}{{{R_{rad}}}} = - \frac{1}{{{S^2}}} \times \left( {2s} \right)dS\) 

\(\frac{{d{R_{rad}}}}{{{R_{rad}}}} = - \frac{2}{s}\left( {ds} \right)\) 

\(\frac{{d{R_{rad}}}}{{{R_{rad}}}} = - \frac{2}{s}\left( {\frac{s}{4} - s} \right)\) 

\(= - \frac{2}{s}\left( { - \frac{{3s}}{4}} \right)\) 

\(\frac{{d{R_{rad}}}}{{{R_{rad}}}} = \frac{3}{2} = 1.5\) 

Concept:

\(D = \frac{{4\pi F_{max}^2}}{{\mathop \smallint \nolimits_{\theta = 0}^\pi \mathop \smallint \nolimits_{\phi = 0}^{2\pi } F_\theta ^2\left( {\theta ,\;\phi } \right)\sin \theta d\theta d\phi }}\) 

Calculation:

F_max = A

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

\(D = \frac{{4\pi }}{{\mathop \smallint \nolimits_{\phi = 0}^{2\pi } d\phi \mathop \smallint \nolimits_{\theta = 0}^{\frac{\pi }{2}} {{\sin }^3}d\theta }} = \frac{{4\pi }}{{\left( {2\pi } \right)\left( {\frac{2}{3}} \right)}}\) 

D = 3

Convert to dB

D = 10 log (3) = 4.77 dB

Concept:

Received power in an antenna is given by:

P_r = [P_avg. Flux density].[A_e]

where,

A_e = Effective aperture of the antenna given by

\({{A}_{e}}=\frac{{{\lambda }^{2}}}{4\pi }.G\)

G = gain of the antenna.

λ = operating wavelength.

Calculation:

Given the gain of both antennas are same;

So, \({{A}_{e}}\propto {{\lambda }^{2}}\propto \frac{1}{{{f}^{2}}}\)

Flux illuminated is identical;

i.e. P_r ∝ A_e

so, \({{P}_{r}}\propto \frac{1}{{{f}^{2}}}\)

∴ we can write, ⇒ P_rAf_A² = P_rB. f_B²

\(\Rightarrow \frac{{{P}_{rA}}}{{{P}_{rB}}}=\frac{f_{B}^{2}}{f_{A}^{2}}\)

Putting values,

\(\Rightarrow \frac{{{P}_{rA}}}{{{P}_{rB}}}={{\left( \frac{3\times {{10}^{9}}}{3\times {{10}^{8}}} \right)}^{2}}={{\left( 10 \right)}^{2}}=100\)

\(\frac{{{P}_{A}}}{{{P}_{B}}}=\frac{100}{1}\)