- #1

phinfinity

- 4

- 6

- Homework Statement
- A double line of N equally spaced oscillating dipoles is situated as shown in figure (below). All dipoles in row A are driven in the same phase, and all those in row B lag 90° in phase behind those of row A. Sketch the radiation pattern I(θ) in the equatorial plane at a great distance from the array

- Relevant Equations
- Based on Interference chapter in feynman lectures: https://www.feynmanlectures.caltech.edu/I_29.html

Although problem says sketch, what I really want is the formulae by θ for the radiation. After trying this for a while I cheated and looked at the solution. I still can't figure out the steps on how to get to the solution. the answer is:

I(\theta) = 2I_0 [ 1 + cos(\frac{\pi}{2}(1-cos\theta)) ] \frac{sin^2(\frac{N\pi}{2}sin\theta)}{sin^2(\frac{\pi}{2}sin\theta)}

$$

What I tried:

I'll assume that the intensity from a single antenna at a great distance is ##I_0## and I need to find what is the combined intensity after accounting for interference. To start with let's find out the phase shift for each antenna with respect to the top-right most antenna in row B as 0 phase shift. Ignoring the phase shift from distance, Every antenna in row A is ##\pi/2## ahead of row B.

Let us number the antenna in both rows starting with the right end as k=0 , k = 1,... k=N-1.

In row B, after doing the geometry the phase-shift per antenna in terms of distance is : ##k \frac{\lambda}{2} \sin\theta##. (Separated from top right antenna by a horizontal distance of ##k\frac{\lambda}{2}## & and target point is at an angle θ). So the total phase shift in row A is ##k\pi sin \theta##.

In row A, we need to do a slightly more complicated geometry, where the antenna is ##\frac{\lambda}{4}## vertically below & ##k\frac{\lambda}{2}## horizontally to the left. The distance between the two antenna is ##\frac{\lambda}{4}\sqrt{4k^2+1}##. The angle to the target point with respect to the line between the two antenna is ##tan^{-1}(\frac{1}{2k}) + \theta - \pi/2##. So the phase-shift in terms of distance is : ##\frac{\lambda}{4}\sqrt{4k^2+1} cos(tan^{-1}(\frac{1}{2k}) + \theta - \pi/2)## . Simplifying & in terms of phase shift in radians & adding the phase shift difference it is driven at, total phase shift: ##\frac{\pi}{2}(\sqrt{4k^2+1} sin(tan^{-1}(\frac{1}{2k}) + \theta)-1)##

So using complex numbers the combined Electric field would be represented by:

$$

\hat{E} = \sum_{k=0}^{N-1} e^{i k\pi sin \theta} + \sum_{k=0}^{N-1} e^{i \frac{\pi}{2}(\sqrt{4k^2+1} sin(tan^{-1}(\frac{1}{2k}) + \theta)-1)}

$$

The intensity should be the magnitude of ##\hat{E}## squared, which we can get by multiplying it by it's complex conjugate. But at this point the math become so complicated, that I couldn't figure out how to take it further.

Can someone please help me with the next steps on how to get to the answer from here? I feel like there might be some simple trick or such that I have overlooked in solving this. Thanks!