Chebychev polynomial

Antenna array design with Chebychev polynomials

March 23, 2015 ece1229 No comments , , , ,

[Click here for a PDF of this post with nicer formatting]

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.

ChebychevSecondMethodPolarFig3pn

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

ChebychevSecondMethodSPolarFig4pn

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.

 

ChebychevSecondMethodPolarFig1pn

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

ChebychevSecondMethodSPolarFig2pn

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

March 22, 2015 ece1229 No comments , , , ,

[Click here for a PDF of this post with nicer formatting]

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

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

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

\begin{equation}\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}
\end{equation}

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

\begin{equation}\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}
\end{equation}

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

\begin{equation}\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,
\end{equation}

or

\begin{equation}\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}
\end{equation}

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

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

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

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

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

\begin{equation}\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} }.
\end{equation}

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.

ChebyChevThreeScaledFig2pn

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

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

or

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

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

\begin{equation}\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 }.
\end{equation}

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

ChebychevT3FittingFig3pn

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

ChebychevT3FittingDbFig4pn

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

March 13, 2015 ece1229 No comments , , ,

[Click here for a PDF of this post with nicer formatting]

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

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

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

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

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

or

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

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

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

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

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 \( m \).

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

To examine the general case

\begin{equation}\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}
\end{equation}

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

\begin{equation}\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.
}
\end{equation}

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.

Plots

ChebychevFig1pn

Properties

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

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

Recurrance relation.

Prove \ref{eqn:tschebyscheff:100}.

Answer.

To show this, let

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

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

Recall the cosine addition formulas

\begin{equation}\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}
\end{equation}

Applying this gives

\begin{equation}\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}
\end{equation}

First order LDE relation.

Prove \ref{eqn:tschebyscheff:420}.

Answer.

To show this, again, let

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

Observe that

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

so

\begin{equation}\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}
\end{equation}

Plugging this in gives

\begin{equation}\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}
\end{equation}

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

\begin{equation}\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.
\end{equation}

} % answer

Second order LDE relation.

Prove \ref{eqn:tschebyscheff:400}.

Answer.

This follows the same way. The first derivative was

\begin{equation}\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}
\end{equation}

so the second derivative is

\begin{equation}\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}
\end{equation}

Putting all the pieces together gives

\begin{equation}\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}
\end{equation}

Orthogonality relation

Prove \ref{eqn:tschebyscheff:440}.

Answer.

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

\begin{equation}\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}
\end{equation}

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

\begin{equation}\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}
\end{equation}

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

\begin{equation}\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}
\end{equation}

References

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