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I am solving (d) right now from the below problem, from Balanis Book: Antenna theory.

Problem 6.13. Balanis

I quickly show how to solve the (a), since it will be needed to solve (d).

Directivity (\$D_{o}\$) is: \$ 10*log_{10}(D_{o})=20dB -> D_{o}=100 \$ Now I use the formula for end-fire array to find the number of elements (\$N\$): \$ D_{o}=4*N*\dfrac{d}{λ}=\$ Do=20db, b=λ/4 (given) \$-> N=100 \$

Not, to sub-problem (d). I am looking at the below formula, on the MAXIMA of end-fire arrays. I should find only one Maxima, at 0°.

\$ θ_{0} = cos^{-1}(1-\dfrac{0*λ}{d}) = cos^{-1}(1) = 0° \$

If I try to find any other maxima, it will give me no solution, as expected:

\$ θ_{1}= cos^{-1}(1-\dfrac{1*λ}{d}) = cos^{-1}(1-\dfrac{λ/1}{\dfrac{λ}{4}}) = cos^{-1}(1-\dfrac{4λ}{λ}) = cos^{-1}(1-4) = cos^{-1}(-3)\$ impossible

end-fire array formulas

Now, I will try to find the first minor lobes, using MINOR LOBE MAXIMA formula. On the formula I see the note \$ \dfrac{π*d}{λ}<< 1. \$, (which I think is the reason I am not getting the correct results at the end). my \$ \dfrac{π*d}{λ} \$ is \$ \dfrac{π*\dfrac{λ}{4}}{λ/1}=\dfrac{π*λ*1}{4*λ}=\dfrac{π}{4}=0.78 \$, which is not way less than 1.

I proceed to find some of the angles \$ θ1,θ2,θ3 \$ nevertheless to see what will come up.

\$ θ_{1} = cos^{-1} [ 1-\dfrac{(2*1+1)*λ}{2*100*\dfrac{λ}{4}} ] = cos^{-1}[1-\dfrac{3*λ}{50*λ}] = cos^{-1}(\dfrac{50-3}{50})= 19.94°\$

\$ θ_{2} = cos^{-1} [ 1-\dfrac{(2*2+1)*λ}{2*100*\dfrac{λ}{4}} ] = cos^{-1}[1-\dfrac{5*λ}{50*λ}] = cos^{-1}(\dfrac{10-1}{10})= 25.84°\$

\$ θ_{2} = cos^{-1} [ 1-\dfrac{(2*3+1)*λ}{2*100*\dfrac{λ}{4}} ] = cos^{-1}[1-\dfrac{7*λ}{50*λ}] = cos^{-1}(\dfrac{50-7}{50})= 30.68°\$

Angle results keep increasing, which is logical.

Now, I want to find the amplitude of these angles, I will be put them on the Array factor formula.

The formula for the Array factor is: \$ AF=\dfrac{1}{N}*[\dfrac{sin(\dfrac{N}{2}*ψ)}{sin(\dfrac{1}{2}*ψ)}]|_{N=100} ->=\dfrac{1}{100}*[\dfrac{sin(50*ψ)}{sin(\dfrac{1}{2}*ψ)}] \$

where \$ ψ=k*d*cos(θ)+β \$. Let us ignore \$ β \$ (the difference in phase excitation). k is the wave-number and d=λ/4 (is given) So \$ ψ=\dfrac{2*π}{λ}*\dfrac{λ}{4}=\dfrac{π}{2}*cos(θ) \$.

If I put \$ ψ \$ and \$ AF \$ together, I get:

\$ AF=\dfrac{1}{100}*[\dfrac{sin(50*\dfrac{π}{2}*cos(θ))}{sin(\dfrac{1}{2}*\dfrac{π}{2}*cos(θ))}] = \dfrac{1}{100}*[\dfrac{sin(78.54*cos(θ))}{sin(0.785*cos(θ))}] \$

For \$ θ_{2}=19.94°:\$

\$ AF=\dfrac{1}{100}*[\dfrac{sin(78.54*cos(19.94))}{sin(0.785*cos(19.94))}]=\dfrac{1}{100}*\dfrac{0.960}{0.0128}=0.745\$ amplitude

For \$ θ_{3}=25.84°:\$

\$ AF=\dfrac{1}{100}*[\dfrac{sin(78.54*cos(25.84))}{sin(0.785*cos(25.84))}]=\dfrac{1}{100}*\dfrac{0.943}{0.0123}=0.765\$ amplitude

For \$ θ_{1}=30.68°:\$

\$ AF=\dfrac{1}{100}*[\dfrac{sin(78.54*cos(30.68))}{sin(0.785*cos(30.68))}]=\dfrac{1}{100}*\dfrac{0.924}{0.0117}=0.784\$ amplitude.

It does not make sense to me, the amplitude of the minor lobe maxima keeps increasing instead of decreasing. I must be looking at the wrong formula to find the minor lobe maxima right? The issue with the formula I use must lay to the \$ \dfrac{π*d}{λ}<< 1 \$, which is untrue for my case.

What is the correct formula to use to find the angles of the fire-end array for the minor Lobe maxima?

EDIT: Thanks to @Tony Stewart EE75, I tried another formula, but with no success:

Finding lobes maxima

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  • \$\begingroup\$ p 15 may help ece.mcmaster.ca/faculty/nikolova/antenna_dload/current_lectures/… \$\endgroup\$ Apr 26, 2022 at 20:52
  • \$\begingroup\$ @TonyStewartEE75 Good one I edited the question and tried this formula as well! But the formula output gives me impossible. I get \$ cos^{-1}(X) \$ where X is >1, which makes it impossible. Maybe there are no other lobes? is it possible? \$\endgroup\$ Apr 27, 2022 at 8:30
  • \$\begingroup\$ Nulls & lobe ripple are essential off-centre but attenuate with the number of elements. The current direction must alternate for each neighbouring element for end-fire \$\endgroup\$ Apr 27, 2022 at 13:15

1 Answer 1

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You are on the right track with your array factor:

enter image description here

but I get

\$AF = \frac{sin(\frac{N}{2} kd (1-cos\theta))}{sin(\frac{1}{2} kd (1-cos\theta))}\$ with my \$\theta\$ measured from endfire.

These may be equivalent, but I haven't tried to work that out.
I think your interpretation is wrong. Look at the numerator - it has zeroes whenever

\$\frac{N kd (1-cos\theta )}{2}=n\pi\$

so the first null is at about 16.3 degrees from endfire, next at 23.1 degrees and so on. Between these nulls are the minor lobe maxima (sidelobes), the locations of which are closely approximated by:

\$\frac{N kd (1-cos\theta )}{2}=(n+\frac{1}{2})\pi\$

This is the Dirichlet Kernel ( wikipedia link ) that occurs a lot in discrete fourier theory and array theory.

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  • \$\begingroup\$ Thanks! It looks like this formula gives me the same angles θ as the ones I found before... So the angles I found must be correct... Then I must have calculated the amplitude of those angles wrong... The Array factor is the amplitude on those angles, right? \$\endgroup\$ Apr 27, 2022 at 13:56
  • \$\begingroup\$ I think so (been a while since textbook definitions matter). You normalized your AF to the main peak, whereas mine has a value of N. \$\endgroup\$
    – Tesla23
    Apr 27, 2022 at 22:54

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