Antenna: three element array, Balanis problem 6.3

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.

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.

• I see. So, the exercise should say "phase and magnitude of element excitations (or excitation coefficients) are shown" instead of "phase and magnitude relationships are shown". Using the word "relationship" made me think it would be angle differences between the elements. That is why I tried to find the angle of each element and then use it on θ. Apr 24, 2022 at 11:57
• Also, the formula for $AF$ includes $β$ on the exponent. This would make element #1 to be $-j*e^{β}$ instead of $-j$. Why is $β$ not included in your calculations? Apr 24, 2022 at 12:06
• exp(j*beta) can be combined with the excitation amplitude making it complex. Its a matter of preference if you want a complex excitation coefficient or a real one and an additional phase term. Apr 24, 2022 at 14:46