Gain Calculation for a Lossless End-Fire Antenna Array

Answer:

To solve the problem, we need to calculate the gain ( G ) of the end-fire array of isotropic point sources, and then compare it with the gain calculated from an approximate equation.

Given Data:

Number of sources ( n = 10 )
Spacing ( d = \frac{2}{4} = 0.5 ) (assuming this is in wavelengths)
The normalized field pattern is given as: [ E(\phi) = \pi \sin\left(\frac{n}{2}\right) \frac{\sin\left(\frac{kd}{2}\right)}{k/2} ] where ( k = \frac{2\pi}{\lambda} ).

(a) Calculate the Gain ( G )

For an end-fire array, the gain ( G ) can be calculated using the formula: [ G = n \cdot \text{Directivity} ] The directivity ( D ) of an end-fire array can be approximated as: [ D \approx \frac{n^2}{2} ] Thus, the gain ( G ) can be calculated as: [ G = n \cdot D = n \cdot \frac{n^2}{2} = \frac{n^3}{2} ] Substituting ( n = 10 ): [ G = \frac{10^3}{2} = \frac{1000}{2} = 500 ]

(b) Calculate the Gain from the Approximate Equation (9)

The approximate equation for the gain of an end-fire array can also be expressed as: [ G \approx n^2 \cdot \frac{1}{\sin(\frac{\pi d}{\lambda})} ] For ( d = 0.5 \lambda ): [ G \approx n^2 \cdot \frac{1}{\sin\left(\frac{\pi \cdot 0.5}{1}\right)} = n^2 \cdot \frac{1}{\sin\left(\frac{\pi}{2}\right)} = n^2 ] Substituting ( n = 10 ): [ G \approx 10^2 = 100 ]

(c) What is the Difference?

Now we can compare the two calculated gains:

Gain from the directivity calculation: ( G = 500 )
Gain from the approximate equation: ( G \approx 100 )

The difference between the two gains is: [ \text{Difference} = 500 - 100 = 400 ]

Summary of Results:

(a) Gain ( G = 500 )
(b) Gain from approximate equation ( G \approx 100 )
(c) Difference ( = 400 )