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\section{Exercises}
\label{sec:rn:ex}
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\ifexercises{%
\begin{exercise}{PSD -- I.}
\label{ex:rn:psd1}
Show that the power spectral density of the constant signal $\mathbf{x} : x[n] = \alpha$ (with $\alpha \in \mathbb{R}$) is $P(e^{j\omega}) = \alpha^2\tilde{\delta}(\omega)$.
\end{exercise}
}\fi\ifanswers{%
\begin{solution}{}
The partial DTFT defined in~(\ref{eq:rn:truncDTFT}) is in this case
\[
X_M ( e^{j\omega}) = \sum_{n = -M}^{M} \alpha e^{-j\omega n}
\]
and its squared magnitude is
\begin{align*}
\bigl|X_M(e^{j\omega}) \bigr|^2 &= X^*_M(e^{j\omega}) X_M(e^{j\omega}) \\
&= \sum_{n=-M}^M \alpha e^{j\omega n} \sum_{m=-M}^M \alpha e^{-j\omega m} \\
&= \alpha^2 \sum_{k=-2M}^{2M} c_{2M}(k) e^{-j\omega k}
\end{align*}
where $c_{2M}(k) = 2M + 1 - |k|$ as in~(\ref{eq:rn:ck}). Now, with $w_M(k)$ as in~(\ref{eq:rn:weighFunPSDEq}),
\begin{align*}
P(e^{j\omega}) & = \lim_{M\rightarrow\infty} \frac{1}{2M+1} \bigl| X_M(e^{j\omega}) \bigr|^2 \\
& = \lim_{M\rightarrow\infty} \alpha^2 \sum_{k = -\infty}^{\infty} w_M(k) \, e^{-j\omega k} \\
&= \alpha^2 \, \DTFT{\mathbf{1}} \\
&= \alpha^2\tilde{\delta}(\omega).
\end{align*}
\end{solution}
}\fi
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\ifexercises{%
\begin{exercise}{PSD -- II.}
Show that the power spectral density of the scaled unit step $\alpha\mathbf{u}$ (with $\alpha \in \mathbb{R}$) is $P(e^{j\omega}) = (\alpha^2/2)\tilde{\delta}(\omega)$.
\end{exercise}
}\fi\ifanswers{%
\begin{solution}{}
The partial DTFT in~(\ref{eq:rn:truncDTFT}) is in this case
\[
X_M ( e^{j\omega}) = \sum_{n = 0}^{M} \alpha e^{-j\omega n}
\]
and its squared magnitude is
\begin{align*}
\bigl|X_M(e^{j\omega}) \bigr|^2 &= X^*_M(e^{j\omega}) X_M(e^{j\omega}) \\
&= \alpha^2 \sum_{n=0}^M \sum_{m=0}^M e^{-j\omega (m-n)} \\
&= \sum_{k=-M}^{M} c_M(k) e^{-j\omega k}
\end{align*}
with $c_{M}(k) = M + 1 - |k|$ as for~(\ref{eq:rn:ck}). Now, define
\[
v_M(k) = \begin{cases}
\displaystyle \frac{c_M(k)}{2M + 1} & \ |k| \leq M \\ \\
0 & \ |k| > M
\end{cases}
\]
similarly to~(\ref{eq:rn:weighFunPSDEq}), and note that $\lim_{M\rightarrow\infty} v_M(k) = 1/2$. Then
\begin{align*}
P(e^{j\omega}) & = \lim_{M\rightarrow\infty} \frac{1}{2M+1} \bigl| X_M(e^{j\omega}) \bigr|^2 \\
& = \lim_{M\rightarrow\infty} \alpha^2 \sum_{k = -\infty}^{\infty} v_M(k) \, e^{-j\omega k} \\
&= (\alpha^2/2) \, \DTFT{\mathbf{1}} \\
&= (\alpha^2/2) \tilde{\delta}(\omega).
\end{align*}\end{solution}
}\fi
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\begin{exercise}{AR(1) Processes.}
\label{ex:rn:ar1}
An AR(1) process is defined by the recursion
\[
x[n] = \alpha x[n-1] + w[n]
\]
where $w[n]$ is a white noise process with variance $\sigma^2_w$. Compute the autocorrelation sequence $r_x[k]$. You can derive the expression directly or, equivalently, use~\ref{eq:rn:outputcorr}.
\end{exercise}
\ifanswers{
\begin{solution}{AR(1) Processes.}
From the definition of the AR(1) process
\[
x[n] = \alpha x[n-1] + w[n]
\]
we can multiply both sides of the equation by $x[n-m]$ and take expectations:
\[
\expt{x[n]x[n-m]} = \alpha\expt{x[n-1]x[n-m]} + \expt{w[n]x[n-m]};
\]
Assume $m > 0$; in this case the last term is zero since the past output cannot depend on the current white noise sample. We thus obtain the recurrence relation:
\[
r_x[-m] = \alpha r_x[-m+1].
\]
To start the recurrence we can compute the second moment of the process
\[
\expt{x^2[n]} = \alpha^2 \expt{x^2[n-1]} + \sigma_w^2 + 2\alpha\expt{w[n]x[n-1]};
\]
again, $\expt{w[n]x[n-1]} = 0$ and $\expt{x^2[n]} = \expt{x^2[n-1]} = r_x[0]$ because of stationarity so it is
\[
r_x[0] = \frac{\sigma_w^2}{1 - \alpha^2}.
\]
With this we can say that $r_x[-k] = \alpha^k r_x[0]$ for $k>0$;, because of the symmetry of the autocorrelation, it is finally
\[
r_x[k] = \frac{\alpha^{|k|}}{1-\alpha^2}\sigma_w^2.
\]
\vspace{1em}
Alternatively we can exploit the fact that the AR(1) process is obtained by passing white noise through a filter with transfer function $H(z) = 1/(1 - \alpha z^{-1})$. The impulse response of this filter is $h[n] = \alpha^n u[n]$ and, in order to use~\ref{eq:rn:outputcorr}, we need to compute the sequence $\mathbf{h} \ast \mathcal{R}\{\mathbf{h}\}$. We have, assuming $k>0$,
\begin{align*}
\left(\mathbf{h} \ast \mathcal{R}\{\mathbf{h}\}\right)[k] &= \sum_{n=-\infty}^{\infty} h[n]h[n-k] \\
&= \sum_{n=0}^{\infty} h[n]h[n+k] \\
&= \sum_{n=0}^{\infty} \alpha^n \alpha^{n+k} \\
&= \frac{\alpha^k}{1 - \alpha^2}.
\end{align*}
We can repeat the computation for $k < 0$: we obtain a symmetric sequence which, once convolved with $r_w[n] = \sigma^2_w\delta[n]$ yields again
\[
r_x[k] = \frac{\alpha^{|k|}}{1-\alpha^2}\sigma_w^2.
\]
\end{solution}
\fi
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\begin{exercise}{.}
\label{ex:rn:ar1}
An AR(1) process is defined by the recursion
\[
x[n] = \alpha x[n-1] + w[n]
\]
where $w[n]$ is a white noise process with variance $\sigma^2_w$. Compute the autocorrelation sequence $r_x[k]$. You can derive the expression directly or, equivalently, use~\ref{eq:rn:outputcorr}.
\end{exercise}
\ifanswers{
\begin{solution}{AR(1) Processes.}
From the definition of the AR(1) process
\[
x[n] = \alpha x[n-1] + w[n]
\]
we can multiply both sides of the equation by $x[n-m]$ and take expectations:
\[
\expt{x[n]x[n-m]} = \alpha\expt{x[n-1]x[n-m]} + \expt{w[n]x[n-m]};
\]
Assume $m > 0$; in this case the last term is zero since the past output cannot depend on the current white noise sample. We thus obtain the recurrence relation:
\[
r_x[-m] = \alpha r_x[-m+1].
\]
To start the recurrence we can compute the second moment of the process
\[
\expt{x^2[n]} = \alpha^2 \expt{x^2[n-1]} + \sigma_w^2 + 2\alpha\expt{w[n]x[n-1]};
\]
again, $\expt{w[n]x[n-1]} = 0$ and $\expt{x^2[n]} = \expt{x^2[n-1]} = r_x[0]$ because of stationarity so it is
\[
r_x[0] = \frac{\sigma_w^2}{1 - \alpha^2}.
\]
With this we can say that $r_x[-k] = \alpha^k r_x[0]$ for $k>0$;, because of the symmetry of the autocorrelation, it is finally
\[
r_x[k] = \frac{\alpha^{|k|}}{1-\alpha^2}\sigma_w^2.
\]
\vspace{1em}
Alternatively we can exploit the fact that the AR(1) process is obtained by passing white noise through a filter with transfer function $H(z) = 1/(1 - \alpha z^{-1})$. The impulse response of this filter is $h[n] = \alpha^n u[n]$ and, in order to use~\ref{eq:rn:outputcorr}, we need to compute the sequence $\mathbf{h} \ast \mathcal{R}\{\mathbf{h}\}$. We have, assuming $k>0$,
\begin{align*}
\left(\mathbf{h} \ast \mathcal{R}\{\mathbf{h}\}\right)[k] &= \sum_{n=-\infty}^{\infty} h[n]h[n-k] \\
&= \sum_{n=0}^{\infty} h[n]h[n+k] \\
&= \sum_{n=0}^{\infty} \alpha^n \alpha^{n+k} \\
&= \frac{\alpha^k}{1 - \alpha^2}.
\end{align*}
We can repeat the computation for $k < 0$: we obtain a symmetric sequence which, once convolved with $r_w[n] = \sigma^2_w\delta[n]$ yields again
\[
r_x[k] = \frac{\alpha^{|k|}}{1-\alpha^2}\sigma_w^2.
\]
\end{solution}
\fi
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