Today, I was trying to deduce by myself the Dolph-Taylor Synthesis. After arriving at the expression:

$$AF(\theta) = P_{N-1} \left( \cos \left( \frac{\psi}{2} \right) \right) = T_{N-1} \left(x_o \cos \left( \frac{\psi}{2} \right) \right) $$

Where \$FA\$ is the Array FActor, \$P_{N-1}\$ is an arbitrary polynomial and \$T_{N-1}\$ is the \$N-1\$th Chebyshev polynomial, I am struck at finding the right value for \$x_o\$

We know that we want a maximum value for the secondary side lobes of SLL. Taking that in linear units, it is equivalent to \$R=10^{-SSL/20}\$. From here on, every single piece of literature I was able to read just magically says that the value we shall give to \$x_o\$ in order to obtain such radiation pattern is:

$$x_0 = \cosh\left( \frac{\cosh^{-1}R}{N-1}\right)$$

What I tried:

We know that \$T_N(\cos \theta) = \cos (n\theta)\$. Hence, for small enought values of x, we can make the change of variables \$\theta = \arccos x\$, so that:

$$T_{N-1} (x) = \cos (n \arccos x)$$

Since we want this to equal $R$:

$$\cos (n \arccos x) = R; \ \ x = \cos\left( \frac{\cos^{-1}R}{N-1}\right)$$

The question: where do the hyperbolic functions arise from?

Thank you in advance!

  • \$\begingroup\$ This is more of a mathematics problem than electrical engineering. \$\endgroup\$
    – Barry
    Jun 2, 2021 at 22:18
  • \$\begingroup\$ Well, it lies in between. I thought about asking it at the Mathematic Stack Exchange forum, but I find it difficult to explain to mathematicians, and I think there are higher chances of finding someone who understands about array synthesis here. \$\endgroup\$ Jun 2, 2021 at 22:30
  • 1
    \$\begingroup\$ The question: where do the hyperbolic functions arise from? — Usually, from having a complex argument to a trigonometric function. \$\endgroup\$
    – Janka
    Jun 2, 2021 at 22:55

1 Answer 1


The Chebyshev polynomials can be expressed as:

\$T_n(u) = cos(m cos^{-1}u) \$ for \$-1 <= u <= 1\$ and

\$T_n(u) = cosh(m cosh^{-1}u)\$ for \$ |u| >= 1\$

In Dolph-Chebyshev synthesis, the region where |u|<1 corresponds to the equi-ripple sidelobes, the main beam occurs for |u|>1. If you want sidelobes of -20dB, these correspond to the Chebyshev peaks of amplitude 1, so you find the value of u that gives the main peak of 10. This requires |u|>1, hence the hyperbolic functions.

  • \$\begingroup\$ Oh, right! Perfect! Nice! Thanks! \$\endgroup\$ Jun 3, 2021 at 0:12

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