Antenna: three element array, Balanis problem 6.3

Question

I am trying to solve the below Problem (only question (a)), where I am mainly stuck on figuring out how to read this z/y drawing of the magnitude and phases.

The problem has to do with 3 antennas called "three-element array of isotropic sources". We always assume far-field observations.

The problem is from Balanis' Book: Antenna Theory - Analysis and Design, Balanis (Wiley, 2016)

A three-element array of isotropic sources has the phase and magnitude relationships shown. The spacing between the elements is \$ d = λ∕2 \$

Here is how I attempted to solve it.

First of all, it does not strictly define whose "magnitude" it gives me. So I suppose these -1, -j and +1 are the phase and magnitude of the Array factor of each element (antenna). Now, there are two functions to calculate the array factor: \$ AF= \sum_{n=1}^{N} a_{n}e^{j*(n-1)*(k*d*\cos(θ)+β)} \$ and \$ AF_{(n)}=\cos[\dfrac{1}{2}*(k*d*\cos(θ)+β)]\$.

Since the data it gives me include complex numbers, I will use the first function.

Where \$ N = 3 \$ the total number of elements,

\$ a_{n} \$ is the excitation coefficient of each element,

\$ k=\dfrac{2π}{λ} \$ is the wave-number,

\$ d = \dfrac{λ}{2} \$ is already known/given, its the distance between the elements,

and \$ β \$ is the difference in phase excitation between the elements.

The only things are not known, are \$ a_{n} \$ , \$ β \$ and \$ θ \$.

I name \$ ψ = (k*d*\cos(θ)+β) \$ and I do:

\$ AF = a_{1}*e^{j(1-1)*ψ} + a_{2}*e^{j(2-1)*ψ} + a_{3}*e^{j(3-1)*ψ} \$

\$a_{1,2,3} \$ should be \$ 0,-1,1 \$ respectively, according to the drawing.

So it becomes:

\$ AF = 0*e^{j(0)*ψ} -1*e^{j(1)*ψ} + 1*e^{j(2)*ψ} \$

\$ \boxed{AF = -1*e^{j(1)*ψ} + 1*e^{j(2)*ψ} }(1)\$

Note1: Isnt't weird that element \$ \#1 (a_{1}) \$ does not contribute to the AF magnitude?

Now, lets calculate ψ.

\$ ψ = (k*d*\cos(θ)+β) = \dfrac{2π}{λ}*\dfrac{λ}{2}*\cos(θ)+β = π*\cos(θ)+β\$

I am pretty much stuck here. I think θ is where I should put the phases the drawing show me, but I do not know how can I convert them to degrees. The drawing gives me:

\$ 0 - j \$ for the #1 element,

\$ -1 + 0j \$ for the #2 element,

\$ +1 +0j \$ for the #3 element.

If I use a calculator for complex numbers and get their degrees?

Note2: I could also see each complex number as a point in X/Y plane and get its degree based on its slope. For example 0-j is point A(0,-1) , which on X/Y plane is 270°.

I get 90° for #1, 180° for #2 and 0° for #3.

So:

\$ ψ_{1}=π*\cos(90)+β = β \$

\$ ψ_{2}=π*\cos(180)+β =β-π\$

\$ ψ_{3}=π*\cos(0) +β =β+π\$

Now, using \$(1)\$: \$ AF = -1*e^{j(1)*ψ_{2}} + 1*e^{j(2)*ψ_{3}} = -e^{j*(β-π)} + e^{j*2*(β+π)}\$

There is so much uncertainty solving this, like the Note1 and Note2.

Another question: I also think \$ β \$ could be the angles I calculated (90, 180, 0) degrees, since \$ β \$ is called " difference in phase excitation between the elements ". But I used these angles on \$ θ \$ instead.

Answer 1

I think you are probably misinterpreting the drawing and the number of the elements compared to the array factor expression. Here's what I end up with:

$$AF(\theta)=\sum_{n=1}^3 a_n e^{jkz_n\cos\theta}$$

Where \$k=\frac{2\pi}{\lambda}\$ so

$$AF(\theta)=\sum_{n=1}^3 a_n e^{j\frac{2\pi}{\lambda}z_n\cos\theta}$$

From the figure the array element numbering is not in increasing order, but has the first element in the middle so \$z_1=0\$, \$z_2=d\$, and \$z_3=-d\$, with corresponding element excitations \$a_1=-j\$, \$a_2=-1\$, and \$a_3=1\$.

$$AF(\theta)= a_1 e^{j\frac{2\pi}{\lambda}z_1\cos\theta} + a_2 e^{j\frac{2\pi}{\lambda}z_2\cos\theta} + a_3 e^{j\frac{2\pi}{\lambda}z_3\cos\theta}$$

$$AF(\theta)= a_1 e^{j\frac{2\pi}{\lambda}0\cos\theta} + a_2 e^{j\frac{2\pi}{\lambda}d\cos\theta} + a_3 e^{j\frac{2\pi}{\lambda}(-d)\cos\theta}$$

$$AF(\theta)= a_1 + a_2 e^{j\frac{2\pi}{\lambda}d\cos\theta} + a_3 e^{-j\frac{2\pi}{\lambda}d\cos\theta}$$

Using the definition \$d=\frac{\lambda}{2}\$

$$AF(\theta)= a_1 + a_2 e^{j\frac{2\pi}{\lambda}\frac{\lambda}{2}\cos\theta} + a_3 e^{-j\frac{2\pi}{\lambda}\frac{\lambda}{2}\cos\theta}$$

$$AF(\theta)= a_1 + a_2 e^{j\pi\cos\theta} + a_3 e^{-j\pi\cos\theta}$$

Inserting the element excitations

$$AF(\theta)= -j + (-1) e^{j\pi\cos\theta} + (1) e^{-j\pi\cos\theta}$$

$$AF(\theta)= -j - e^{j\pi\cos\theta} + e^{-j\pi\cos\theta}$$

You should be able now to convert this into whatever form you'd want the answer.