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Prof. Eleftheriades desribed a Chebychev antenna array design method that looks different than the one of the text [1].

Portions of that procedure are like that of the text. For example, if a side lobe level of $20 \log_{10} R$ is desired, a scaling factor

$$\begin{equation}\label{eqn:chebychevSecondMethod:20} x_0 = \cosh\lr{ \inv{m} \cosh^{-1} R }, \end{equation}$$

is used. Given $N$ elements in the array, a Chebychev polynomial of degree $m = N – 1$ is used. That is

$$\begin{equation}\label{eqn:chebychevSecondMethod:40} T_m(x) = \cos\lr{ m \cos^{-1} x }. \end{equation}$$

Observe that the roots $x_n’$ of this polynomial lie where

$$\begin{equation}\label{eqn:chebychevSecondMethod:60} m \cos^{-1} x_n’ = \frac{\pi}{2} \pm \pi n, \end{equation}$$

or

$$\begin{equation}\label{eqn:chebychevSecondMethod:80} x_n’ = \cos\lr{ \frac{\pi}{2 m} \lr{ 2 n \pm 1 } }, \end{equation}$$

The class notes use the negative sign, and number $n = 1,2, \cdots, m$. It is noted that the roots are symmetric with $x_1′ = – x_m’$, which can be seen by direct expansion

$$\begin{equation}\label{eqn:chebychevSecondMethod:100} \begin{aligned} x_{m-r}’ &= \cos\lr{ \frac{\pi}{2 m} \lr{ 2 (m – r) – 1 } } \\ &= \cos\lr{ \pi – \frac{\pi}{2 m} \lr{ 2 r + 1 } } \\ &= -\cos\lr{ \frac{\pi}{2 m} \lr{ 2 r + 1 } } \\ &= -\cos\lr{ \frac{\pi}{2 m} \lr{ 2 ( r + 1 ) – 1 } } \\ &= -x_{r+1}’. \end{aligned} \end{equation}$$

The next step in the procedure is the identification

$$\begin{equation}\label{eqn:chebychevSecondMethod:120} \begin{aligned} u_n’ &= 2 \cos^{-1} \lr{ \frac{x_n’}{x_0} } \\ z_n &= e^{j u_n’}. \end{aligned} \end{equation}$$

This has a factor of two that does not appear in the Balanis design method. It seems plausible that this factor of two was introduced so that the roots of the array factor $z_n$ are conjugate pairs. Since $\cos^{-1} (-z) = \pi – \cos^{-1} z$, this choice leads to such conjugate pairs

$$\begin{equation}\label{eqn:chebychevSecondMethod:140} \begin{aligned} \exp\lr{j u_{m-r}’} &= \exp\lr{j 2 \cos^{-1} \lr{ \frac{x_{m-r}’}{x_0} } } \\ &= \exp\lr{j 2 \cos^{-1} \lr{ -\frac{x_{r+1}’}{x_0} } } \\ &= \exp\lr{j 2 \lr{ \pi – \cos^{-1} \lr{ \frac{x_{r+1}’}{x_0} } } } \\ &= \exp\lr{-j u_{r+1}}. \end{aligned} \end{equation}$$

Because of this, the array factor can be written

$$\begin{equation}\label{eqn:chebychevSecondMethod:180} \begin{aligned} \textrm{AF} &= ( z – z_1 )( z – z_2 ) \cdots ( z – z_{m-1} ) ( z – z_m ) \\ &= ( z – z_1 )( z – z_1^\conj ) ( z – z_2 )( z – z_2^\conj ) \cdots \\ &= \lr{ z^2 – z ( z_1 + z_1^\conj ) + 1 } \lr{ z^2 – z ( z_2 + z_2^\conj ) + 1 } \cdots \\ &= \lr{ z^2 – 2 z \cos\lr{ 2 \cos^{-1} \lr{ \frac{x_1′}{x_0} } } + 1 } \lr{ z^2 – 2 z \cos\lr{ 2 \cos^{-1} \lr{ \frac{x_2′}{x_0} } } + 1 } \cdots \\ &= \lr{ z^2 – 2 z \lr{ 2 \lr{ \frac{x_1′}{x_0} }^2 – 1 } + 1 } \lr{ z^2 – 2 z \lr{ 2 \lr{ \frac{x_2′}{x_0} }^2 – 1 } + 1 } \cdots \end{aligned} \end{equation}$$

When $m$ is even, there will only be such conjugate pairs of roots. When $m$ is odd, the remainding factor will be

$$\begin{equation}\label{eqn:chebychevSecondMethod:160} \begin{aligned} z – e^{2 j \cos^{-1} \lr{ 0/x_0 } } &= z – e^{2 j \pi/2} \\ &= z – e^{j \pi} \\ &= z + 1. \end{aligned} \end{equation}$$

However, with this factor of two included, the connection between the final array factor polynomial $\ref{eqn:chebychevSecondMethod:180}$, and the Chebychev polynomial $T_m$ is not clear to me. How does this scaling impact the roots?

Example: Expand $\textrm{AF}$ for $N = 4$.

The roots of $T_3(x)$ are

$$\begin{equation}\label{eqn:chebychevSecondMethod:200} x_n’ \in \setlr{0, \pm \frac{\sqrt{3}}{2} }, \end{equation}$$

so the array factor is

$$\begin{equation}\label{eqn:chebychevSecondMethod:220} \begin{aligned} \textrm{AF} &= \lr{ z^2 + z \lr{ 2 – \frac{3}{x_0^2} } + 1 }\lr{ z + 1 } \\ &= z^3 + 3 z^2 \lr{ 1 – \frac{1}{x_0^2} } + 3 z \lr{ 1 – \frac{1}{x_0^2} } + 1. \end{aligned} \end{equation}$$

With $20 \log_{10} R = 30 \textrm{dB}$, $x_0 = 2.1$, so this is

$$\begin{equation}\label{eqn:chebychevSecondMethod:240} \textrm{AF} = z^3 + 2.33089 z^2 + 2.33089 z + 1. \end{equation}$$

With

$$\begin{equation}\label{eqn:chebychevSecondMethod:260} z = e^{j (u + u_0) } = e^{j k d \cos\theta + j k u_0 }, \end{equation}$$

the array factor takes the form

$$\begin{equation}\label{eqn:chebychevSecondMethod:280} \textrm{AF} = e^{j 3 k d \cos\theta + j 3 k u_0 } + 2.33089 e^{j 2 k d \cos\theta + j 2 k u_0 } + 2.33089 e^{j k d \cos\theta + j k u_0 } + 1. \end{equation}$$

This array function is highly phase dependent, plotted for $u_0 = 0$ in fig. 1, and fig. 2.

fig 1. Plot with u_0 = 0, d = lambda/4

fig 2. Spherical plot with u_0 = 0, d = lambda/4

This can be directed along a single direction (z-axis) with higher phase choices as illustrated in fig. 3, and fig. 4.

fig 3. Plot with u_0 = 3.5, d = 0.4 lambda

fig 4. Spherical plot with u_0 = 3.5, d = 0.4 lambda

These can be explored interactively in this Mathematica Manipulate panel.

References

[1] Constantine A Balanis. Antenna theory: analysis and design. John Wiley \& Sons, 3rd edition, 2005.