Appendix: Spectral response of ARMA(1,1) noise
Let \(x_\alpha\) be the realization of an ARMA(1,1) process, so that
$$\langle x_{\alpha } x_{\beta } \rangle = \left\{ {\begin{array}{*{20}l} {\sigma ^{2} } \hfill & {\alpha = \beta } \hfill \\ {\sigma ^{2} \lambda \rho ^{{|\beta - \alpha |}} } \hfill & {\alpha \ne \beta } \hfill \\ \end{array} } \right.$$
(41)
In the case \(\lambda =1\) we have an AR(1) process.
We define the discrete Fourier transform of \(x_\alpha\) at frequency \(f\) as
$$\begin{aligned} X(f) = N^{-1} \sum _{\alpha =1}^N x_\alpha e^{-i 2\pi f t_\alpha }, \end{aligned}$$
(42)
and we define the periodogram as
$$\begin{aligned} {\mathcal {X}}(f)&= {N \over \sigma ^2} | X(f) |^2\nonumber \\&= {1 \over N \sigma ^2 } \sum _{\alpha =1}^N \sum _{\beta =1}^N x_\alpha x_\beta e^{i 2\pi f (t_\beta - t_\alpha )}. \end{aligned}$$
(43)
When the time series is evenly sampled, \(t_\beta - t_\alpha = (\beta - \alpha ) \tau\) where \(\tau\) is the time spacing. Defining \(\phi = 2\pi f \tau\) we have, we have in this case
$$\begin{aligned} {\mathcal {X}}(f) = {1 \over N \sigma ^2 } \sum _{\alpha =1}^N \sum _{\beta =1}^N x_\alpha x_\beta e^{i (\beta - \alpha ) \phi }. \end{aligned}$$
(44)
When we substitute Eq. (41) into Eq. (44) we get the expected value of the periodogram
$$\begin{aligned} \langle {\mathcal {X}} \rangle&= {1 \over N \sigma ^2} \sum _{\alpha =1}^N \sum _{\beta =1}^N \sigma ^2 \rho _{(\alpha -\beta )} e^{i (\beta -\alpha )\phi }\nonumber \\&= {1 \over N} \sum _{\alpha =1}^N \sum _{\beta =1}^N \rho _{(\alpha -\beta )} e^{i \phi (\beta -\alpha )}. \end{aligned}$$
(45)
We now separate the double sum into three parts: terms for which \(\alpha = \beta\), those with \(\alpha < \beta\), and those with \(\alpha > \beta\). The terms with \(\alpha = \beta\) all have \(\rho _o = 1\), and there are \(N\) such terms, so they contribute
$$\begin{aligned} \left( {1 \over N} \right) N = 1. \end{aligned}$$
(46)
We also note that \(\rho _k = \lambda \rho ^k\), so we are motivated to define the complex number
$$\begin{aligned} z = \rho e^{i \phi } = \rho e^{i 2\pi f \tau }. \end{aligned}$$
(47)
so that the terms with \(\alpha < \beta\) contribute
$$\begin{aligned} S = {\lambda \over N} \sum _{\alpha =1}^{N-1} \sum _{\beta =\alpha +1}^N z^{(\beta -\alpha )} = {\lambda \over N} \sum _{\alpha =1}^{N-1} \sum _{\beta =1}^{N-\alpha } z^\beta . \end{aligned}$$
(48)
Because \(\rho _{-k} = \rho _k\) for all \(k\), the terms with \(\alpha < \beta\) are the complex conjugates of the terms with \(\alpha > \beta\) and their sum is \(\bar{S}\), the complex conjugate of \(S\). Finally we can write the periodogram as
$$\begin{aligned} \langle {\mathcal {X}} \rangle = 1 + S + \bar{S} = 1 + {\lambda \over N} \sum _{\alpha =1}^{N-1} \sum _{\beta =1}^{N-\alpha } z^\beta + {\lambda \over N} \sum _{\alpha =1}^{N-1} \sum _{\beta =1}^{N-\alpha } \bar{z}^\beta , \end{aligned}$$
(49)
where a bar indicates the complex conjugate.
The sums in Eq. (49) are directly calculable, leading first to
$$\begin{aligned} S = {\lambda \over N} \sum _{\alpha =1}^{N-1} {z - z^{N-\alpha +1} \over 1-z} = {\lambda z \over N(1-z)} \sum _{\alpha =1}^{N-1} (1 - z^{N-\alpha }), \end{aligned}$$
(50)
and then directly to
$$\begin{aligned} S&= {\lambda z \over N(1-z)} \sum _{\alpha =1}^{N-1} (1 - z^\alpha ) = {\lambda z \over N(1-z)} \left[ N - 1 - {z - z^N \over 1-z} \right] \nonumber \\&= {\lambda z \over 1-z} \left[ 1 - {1 - z^N \over N(1-z)} \right] . \end{aligned}$$
(51)
We now note that as \(N \rightarrow \infty\), \(S \rightarrow \lambda z/(1-z)\) so the expected value of the periodogram goes to
$$\begin{aligned} \langle {\mathcal {X}} \rangle&= 1 + S + \bar{S} \rightarrow 1 + {\lambda z \over 1-z} + {\lambda \bar{z} \over 1 - \bar{z}}\nonumber \\&= 1 + \lambda \left( {z + \bar{z} - 2 z \bar{z} \over (1 - z) (1 - \bar{z})} \right) . \end{aligned}$$
(52)
Now we can substitute \(z \bar{z} = \rho ^2\) and \(z + \bar{z} = 2 \rho \cos \phi\) to get
$$\begin{aligned} \langle {\mathcal {X}} \rangle&= 1 + 2 \lambda \left( {\rho \cos \phi - \rho ^2 \over 1 + \rho ^2 - 2 \rho \cos \phi } \right) \nonumber \\&= 1 - \lambda + { \lambda (1 - \rho ^2) \over 1 + \rho ^2 - 2 \rho \cos \phi }. \end{aligned}$$
(53)
Finally, when \(\lambda = 1\) we recover the result for an AR(1) process.