## Antenna array design with Chebychev polynomials

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

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

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

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

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

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

or

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

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

\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}

The next step in the procedure is the identification

\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}

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

\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}

Because of this, the array factor can be written

\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}

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

\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}

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

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

so the array factor is

\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}

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

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

With

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

the array factor takes the form

\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.

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.

## Chebychev antenna array design

In our text [1] is a design procedure that applies Chebychev polynomials to the selection of current magnitudes for an evenly spaced array of identical antennas placed along the z-axis.

For an even number $$2 M$$ of identical antennas placed at positions $$\Br_m = (d/2) \lr{2 m -1} \Be_3$$, the array factor is

\label{eqn:chebychevDesign:20}
\textrm{AF}
=
\sum_{m=-N}^N I_m e^{-j k \rcap \cdot \Br_m }.

Assuming the currents are symmetric $$I_{-m} = I_m$$, with $$\rcap = (\sin\theta \cos\phi, \sin\theta \sin\phi, \cos\theta )$$, and $$u = \frac{\pi d}{\lambda} \cos\theta$$, this is

\label{eqn:chebychevDesign:40}
\begin{aligned}
\textrm{AF}
&=
\sum_{m=-N}^N I_m e^{-j k (d/2) ( 2 m -1 )\cos\theta } \\
&=
2 \sum_{m=1}^N I_m \cos\lr{ k (d/2) ( 2 m -1)\cos\theta } \\
&=
2 \sum_{m=1}^N I_m \cos\lr{ (2 m -1) u }.
\end{aligned}

This is a sum of only odd cosines, and can be expanded as a sum that includes all the odd powers of $$\cos u$$. Suppose for example that this is a four element array with $$N = 2$$. In this case the array factor has the form

\label{eqn:chebychevDesign:60}
\begin{aligned}
\textrm{AF}
&=
2 \lr{ I_1 \cos u + I_2 \lr{ 4 \cos^3 u – 3 \cos u } } \\
&=
2 \lr{ \lr{ I_1 – 3 I_2 } \cos u + 4 I_2 \cos^3 u }.
\end{aligned}

The design procedure in the text sets $$\cos u = z/z_0$$, and then equates this to $$T_3(z) = 4 z^3 – 3 z$$ to determine the current amplitudes $$I_m$$. That is

\label{eqn:chebychevDesign:80}
\frac{ 2 I_1 – 6 I_2 }{z_0} z + \frac{8 I_2}{z_0^3} z^3 = -3 z + 4 z^3,

or

\label{eqn:chebychevDesign:100}
\begin{aligned}
\begin{bmatrix}
I_1 \\
I_2
\end{bmatrix}
&=
{\begin{bmatrix}
2/z_0 & -6/z_0 \\
0 & 8/z_0^3
\end{bmatrix}}^{-1}
\begin{bmatrix}
-3 \\
4
\end{bmatrix} \\
&=
\frac{z_0}{2}
\begin{bmatrix}
3 (z_0^2 -1) \\
z_0^2
\end{bmatrix}.
\end{aligned}

The currents in the array factor are fully determined up to a scale factor, reducing the array factor to

\label{eqn:chebychevDesign:140}
\textrm{AF} = 4 z_0^3 \cos^3 u – 3 z_0 \cos u.

The zeros of this array factor are located at the zeros of

\label{eqn:chebychevDesign:120}
T_3( z_0 \cos u ) = \cos( 3 \cos^{-1} \lr{ z_0 \cos u } ),

which are at $$3 \cos^{-1} \lr{ z_0 \cos u } = \pi/2 + m \pi = \pi \lr{ m + \inv{2} }$$

\label{eqn:chebychevDesign:160}
\cos u = \inv{z_0} \cos\lr{ \frac{\pi}{3} \lr{ m + \inv{2} } } = \setlr{ 0, \pm \frac{\sqrt{3}}{2 z_0} }.

showing that the scaling factor $$z_0$$ effects the locations of the zeros. It also allows the values at the extremes $$\cos u = \pm 1$$, to increase past the $$\pm 1$$ non-scaled limit values. These effects can be explored in this Mathematica notebook, but can also be seen in fig. 1.

fig 1. T_3( z_0 x) for a few different scale factors z_0.

The scale factor can be fixed for a desired maximum power gain. For $$R \textrm{dB}$$, that will be when

\label{eqn:chebychevDesign:180}
20 \log_{10} \cosh( 3 \cosh^{-1} z_0 ) = R \textrm{dB},

or

\label{eqn:chebychevDesign:200}
z_0 = \cosh \lr{ \inv{3} \cosh^{-1} \lr{ 10^{\frac{R}{20}} } }.

For $$R = 30$$ dB (say), we have $$z_0 = 2.1$$, and

\label{eqn:chebychevDesign:220}
\textrm{AF}
= 40 \cos^3 \lr{ \frac{\pi d}{\lambda} \cos\theta } – 6.4 \cos \lr{ \frac{\pi d}{\lambda} \cos\theta }.

These are plotted in fig. 2 (linear scale), and fig. 3 (dB scale) for a couple values of $$d/\lambda$$.

fig 2. T_3 fitting of 4 element array (linear scale).

fig 3. T_3 fitting of 4 element array (dB scale).

To explore the $$d/\lambda$$ dependence try this Mathematica notebook.

# References

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

## Tschebyscheff polynomials

In ancient times (i.e. 2nd year undergrad) I recall being very impressed with Tschebyscheff polynomials for designing lowpass filters. I’d used Tschebyscheff filters for the hardware we used for a speech recognition system our group built in the design lab. One of the benefits of these polynomials is that the oscillation in the $$\Abs{x} < 1$$ interval is strictly bounded. This same property, as well as the unbounded nature outside of the $$[-1,1]$$ interval turns out to have applications to antenna array design.

The Tschebyscheff polynomials are defined by

\label{eqn:tschebyscheff:40}
T_m(x) = \cos\lr{ m \cos^{-1} x }, \quad \Abs{x} < 1 \label{eqn:tschebyscheff:60} T_m(x) = \cosh\lr{ m \cosh^{-1} x }, \quad \Abs{x} > 1.

### Range restrictions and hyperbolic form.

Prof. Eleftheriades’s notes made a point to point out the definition in the $$\Abs{x} > 1$$ interval, but that can also be viewed as a consequence instead of a definition if the range restriction is removed. For example, suppose $$x = 7$$, and let

\label{eqn:tschebyscheff:160}
\cos^{-1} 7 = \theta,

so
\label{eqn:tschebyscheff:180}
\begin{aligned}
7
&= \cos\theta \\
&= \frac{e^{i\theta} + e^{-i\theta}}{2} \\
&= \cosh(i\theta),
\end{aligned}

or

\label{eqn:tschebyscheff:200}
-i \cosh^{-1} 7 = \theta.

\label{eqn:tschebyscheff:220}
\begin{aligned}
T_m(7)
&= \cos( -m i \cosh^{-1} 7 ) \\
&= \cosh( m \cosh^{-1} 7 ).
\end{aligned}

The same argument clearly applies to any other value outside of the $$\Abs{x} < 1$$ range, so without any restrictions, these polynomials can be defined as just

\label{eqn:tschebyscheff:260}
\boxed{
T_m(x) = \cos\lr{ m \cos^{-1} x }.
}

### Polynomial nature.

Eq. \ref{eqn:tschebyscheff:260} does not obviously look like a polynomial. Let’s proceed to verify the polynomial nature for the first couple values of $$m$$.

• $$m = 0$$.\label{eqn:tschebyscheff:280}
\begin{aligned}
T_0(x)
&= \cos( 0 \cos^{-1} x ) \\
&= \cos( 0 ) \\
&= 1.
\end{aligned}
• $$m = 1$$.\label{eqn:tschebyscheff:300}
\begin{aligned}
T_1(x)
&= \cos( 1 \cos^{-1} x ) \\
&= x.
\end{aligned}
• $$m = 2$$.\label{eqn:tschebyscheff:320}
\begin{aligned}
T_2(x)
&= \cos( 2 \cos^{-1} x ) \\
&= 2 \cos^2 \cos^{-1}(x) – 1 \\
&= 2 x^2 – 1.
\end{aligned}

To examine the general case

\label{eqn:tschebyscheff:340}
\begin{aligned}
T_m(x)
&= \cos( m \cos^{-1} x ) \\
&= \textrm{Re} e^{ j m \cos^{-1} x } \\
&= \textrm{Re} \lr{ e^{ j\cos^{-1} x } }^m \\
&= \textrm{Re} \lr{ \cos\cos^{-1} x + j \sin\cos^{-1} x }^m \\
&= \textrm{Re} \lr{ x + j \sqrt{1 – x^2} }^m \\
&=
\textrm{Re} \lr{
x^m
+ \binom{ m}{1} j x^{m-1} \lr{1 – x^2}^{1/2}
– \binom{ m}{2} x^{m-2} \lr{1 – x^2}^{2/2}
– \binom{ m}{3} j x^{m-3} \lr{1 – x^2}^{3/2}
+ \binom{ m}{4} x^{m-4} \lr{1 – x^2}^{4/2}
+ \cdots
} \\
&=
x^m
– \binom{ m}{2} x^{m-2} \lr{1 – x^2}
+ \binom{ m}{4} x^{m-4} \lr{1 – x^2}^2
– \cdots
\end{aligned}

This expansion was a bit cavaliar with the signs of the $$\sin\cos^{-1} x = \sqrt{1 – x^2}$$ terms, since the negative sign should be picked for the root when $$x \in [-1,0]$$. However, that doesn’t matter in the end since the real part operation selects only powers of two of this root.

The final result of the expansion above can be written

\label{eqn:tschebyscheff:360}
\boxed{
T_m(x) = \sum_{k = 0}^{\lfloor m/2 \rfloor} \binom{m}{2 k} (-1)^k x^{m – 2 k} \lr{1 – x^2}^k.
}

This clearly shows the polynomial nature of these functions, and is also perfectly well defined for any value of $$x$$. The even and odd alternation with $$m$$ is also clear in this explicit expansion.

### Properties

In [1] a few properties can be found for these polynomials

\label{eqn:tschebyscheff:100}
T_m(x) = 2 x T_{m-1} – T_{m-2}

\label{eqn:tschebyscheff:420}
0 = \lr{ 1 – x^2 } \frac{d T_m(x)}{dx} + m x T_m(x) – m T_{m-1}(x)

\label{eqn:tschebyscheff:400}
0 = \lr{ 1 – x^2 } \frac{d^2 T_m(x)}{dx^2} – x \frac{dT_m(x)}{dx} + m^2 T_{m}(x)

\label{eqn:tschebyscheff:440}
\int_{-1}^1 \inv{ \sqrt{1 – x^2} } T_m(x) T_n(x) dx =
\left\{
\begin{array}{l l}
0 & \quad \mbox{if $$m \ne n$$ } \\
\pi & \quad \mbox{if $$m = n = 0$$ } \\
\pi/2 & \quad \mbox{if $$m = n, m \ne 0$$ }
\end{array}
\right.

### Recurrance relation.

Prove \ref{eqn:tschebyscheff:100}.

To show this, let

\label{eqn:tschebyscheff:460}
x = \cos\theta.

\label{eqn:tschebyscheff:580}
2 x T_{m-1} – T_{m-2}
=
2 \cos\theta \cos((m-1) \theta) – \cos((m-2)\theta).

\label{eqn:tschebyscheff:540}
\begin{aligned}
\cos( a + b )
&=
\textrm{Re} e^{j(a + b)} \\
&=
\textrm{Re} e^{ja} e^{jb} \\
&=
\textrm{Re}
\lr{ \cos a + j \sin a }
\lr{ \cos b + j \sin b } \\
&=
\cos a \cos b – \sin a \sin b.
\end{aligned}

Applying this gives

\label{eqn:tschebyscheff:600}
\begin{aligned}
2 x T_{m-1} – T_{m-2}
&=
2 \cos\theta \Biglr{ \cos(m\theta)\cos\theta +\sin(m\theta) \sin\theta }
– \Biglr{
\cos(m\theta)\cos(2\theta) + \sin(m\theta) \sin(2\theta)
} \\
&=
2 \cos\theta \Biglr{ \cos(m\theta)\cos\theta +\sin(m\theta)\sin\theta) }
– \Biglr{
\cos(m\theta)(\cos^2 \theta – \sin^2 \theta) + 2 \sin(m\theta) \sin\theta \cos\theta
} \\
&=
\cos(m\theta) \lr{ \cos^2\theta + \sin^2\theta } \\
&= T_m(x).
\end{aligned}

### First order LDE relation.

Prove \ref{eqn:tschebyscheff:420}.

To show this, again, let

\label{eqn:tschebyscheff:470}
x = \cos\theta.

Observe that

\label{eqn:tschebyscheff:480}
1 = -\sin\theta \frac{d\theta}{dx},

so

\label{eqn:tschebyscheff:500}
\begin{aligned}
\frac{d}{dx}
&= \frac{d\theta}{dx} \frac{d}{d\theta} \\
&= -\frac{1}{\sin\theta} \frac{d}{d\theta}.
\end{aligned}

Plugging this in gives

\label{eqn:tschebyscheff:520}
\begin{aligned}
\lr{ 1 – x^2} &\frac{d}{dx} T_m(x) + m x T_m(x) – m T_{m-1}(x) \\
&=
\sin^2\theta
\lr{
-\frac{1}{\sin\theta} \frac{d}{d\theta}}
\cos( m \theta ) + m \cos\theta \cos( m \theta ) – m \cos( (m-1)\theta ) \\
&=
-\sin\theta (-m \sin(m \theta)) + m \cos\theta \cos( m \theta ) – m \cos( (m-1)\theta ).
\end{aligned}

Applying the cosine addition formula \ref{eqn:tschebyscheff:540} gives

\label{eqn:tschebyscheff:560}
m \lr{ \sin\theta \sin(m \theta) + \cos\theta \cos( m \theta ) } – m
\lr{
\cos( m \theta) \cos\theta + \sin( m \theta ) \sin\theta
}
=0.

### Second order LDE relation.

Prove \ref{eqn:tschebyscheff:400}.

This follows the same way. The first derivative was

\label{eqn:tschebyscheff:640}
\begin{aligned}
\frac{d T_m(x)}{dx}
&=
-\inv{\sin\theta}
\frac{d}{d\theta} \cos(m\theta) \\
&=
-\inv{\sin\theta} (-m) \sin(m\theta) \\
&=
m \inv{\sin\theta} \sin(m\theta),
\end{aligned}

so the second derivative is

\label{eqn:tschebyscheff:620}
\begin{aligned}
\frac{d^2 T_m(x)}{dx^2}
&=
-m \inv{\sin\theta} \frac{d}{d\theta} \inv{\sin\theta} \sin(m\theta) \\
&=
-m \inv{\sin\theta}
\lr{
-\frac{\cos\theta}{\sin^2\theta} \sin(m\theta) + \inv{\sin\theta} m \cos(m\theta)
}.
\end{aligned}

Putting all the pieces together gives

\label{eqn:tschebyscheff:660}
\begin{aligned}
\lr{ 1 – x^2 } &\frac{d^2 T_m(x)}{dx^2} – x \frac{dT_m(x)}{dx} + m^2 T_{m}(x) \\
&=
m
\lr{
\frac{\cos\theta}{\sin\theta} \sin(m\theta) – m \cos(m\theta)
}
– \cos\theta m \inv{\sin\theta} \sin(m\theta)
+ m^2 \cos(m \theta) \\
&=
0.
\end{aligned}

### Orthogonality relation

Prove \ref{eqn:tschebyscheff:440}.

First consider the 0,0 inner product, making an $$x = \cos\theta$$, so that $$dx = -\sin\theta d\theta$$

\label{eqn:tschebyscheff:680}
\begin{aligned}
\innerprod{T_0}{T_0}
&=
\int_{-1}^1 \inv{\lr{1-x^2}^{1/2}} dx \\
&=
\int_{-\pi}^0 \lr{-\inv{\sin\theta}} -\sin\theta d\theta \\
&=
0 – (-\pi) \\
&= \pi.
\end{aligned}

Note that since the $$[-\pi, 0]$$ interval was chosen, the negative root of $$\sin^2\theta = 1 – x^2$$ was chosen, since $$\sin\theta$$ is negative in that interval.

The m,m inner product with $$m \ne 0$$ is

\label{eqn:tschebyscheff:700}
\begin{aligned}
\innerprod{T_m}{T_m}
&=
\int_{-1}^1 \inv{\lr{1-x^2}^{1/2}} \lr{ T_m(x)}^2 dx \\
&=
\int_{-\pi}^0 \lr{-\inv{\sin\theta}} \cos^2(m\theta) -\sin\theta d\theta \\
&=
\int_{-\pi}^0 \cos^2(m\theta) d\theta \\
&=
\inv{2} \int_{-\pi}^0 \lr{ \cos(2 m\theta) + 1 } d\theta \\
&= \frac{\pi}{2}.
\end{aligned}

So far so good. For $$m \ne n$$ the inner product is

\label{eqn:tschebyscheff:720}
\begin{aligned}
\innerprod{T_m}{T_m}
&=
\int_{-\pi}^0 \cos(m\theta) \cos(n\theta) d\theta \\
&=
\inv{4} \int_{-\pi}^0
\lr{
e^{j m \theta}
+ e^{-j m \theta}
}
\lr{
e^{j n \theta}
+ e^{-j n \theta}
}
d\theta \\
&=
\inv{4} \int_{-\pi}^0
\lr{
e^{j (m + n) \theta}
+e^{-j (m + n) \theta}
+e^{j (m – n) \theta}
+e^{j (-m + n) \theta}
}
d\theta \\
&=
\inv{2} \int_{-\pi}^0
\lr{
\cos( (m + n)\theta )
+\cos( (m – n)\theta )
}
d\theta \\
&=
\inv{2}
\evalrange{
\lr{
\frac{\sin( (m + n)\theta )}{ m + n }
+\frac{\sin( (m – n)\theta )}{ m – n}
}
}{-\pi}{0} \\
&= 0.
\end{aligned}

# References

[1] M. Abramowitz and I.A. Stegun. Handbook of mathematical functions with formulas, graphs, and mathematical tables, volume 55. Dover publications, 1964.