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.