I am working on the "System identification : theory for the user" by Lennart Ljung (freely available here) and it is one of these books which contains exercises but no answers... My exercise is the 2G2 p47.

The context is as follows :

s is a discrete scalar real quasi-stationary signal, a quasi-stationary signal being a stochastic signals that verifies : $$ \exists C>0\quad\text{such that :} $$ $$\forall t\in \mathbb{N}\quad\vert E[s(t)]\vert \leq C \tag{1}\label{eq1}$$ $$\forall t,r\in \mathbb{N}\quad\vert E[s(t)s(r)]\vert \leq C \tag{2}\label{eq2}$$ $$\forall\tau\in\mathbb{Z}\quad R_s(\tau) = \lim_{N\to\infty}\frac{1}{N}\sum_{t=1}^{N}E[s(t)s(t-\tau)]\quad\text{exists}\tag{3}\label{eq3}$$

We assume the spectrum exists. In this context it is defined as :

$$\Phi_s(\omega) = \sum_{\tau=-\infty}^{+\infty}R_s(\tau)e^{-i\omega\tau}\tag{4}\label{eq4}$$

My question bears on question (ii) of the following exercise : $$ \begin{split} &\textbf{(i) : }\quad \forall\omega\in\mathbb{R}\quad\Phi_s(\omega) \quad\text{is real }\\ &\textbf{(ii) : }\quad \forall\omega\in\mathbb{R}\quad\Phi_s(\omega)\geq 0\\ &\textbf{(iii) : }\quad \forall\omega\in\mathbb{R}\quad\Phi_s(\omega)=\Phi_s(-\omega)\\ \end{split} $$

Quickly, the answer for (i) and (iii) :

Remark how in \ref{eq3} the sum starts at $t=1$ so, by convention we can set $s(t)=0$ for $t\leq 0$. Then we may note that : $$\forall N\geq 1\quad \sum_{t=1}^{N}s(t)s(t+\tau)=\sum_{k=\tau+1}^{N+\tau}s(k)s(k-\tau)$$ then via the definition \ref{eq3}, it comes easy that :

$$\forall\tau\in\mathbb{Z}\quad R_s(\tau)=R_s(-\tau)\tag{5}\label{eq5}$$

from that, by using this symmetry in the sum of \ref{eq4}, we get :

$$\Phi_s(\omega) =R_s(0) +2\sum_{\tau=1}^{+\infty}R_s(\tau)cos(\omega\tau)\tag{6}\label{eq6}$$

which instantly solves (i) and (iii).

Now, how to solve (ii)? Here is what I tried :

I thought that probably $\Phi_s$ is continuous. Then if we can show that, except if $s=0$ $\Phi_s$ does not cancel then $\Phi_s$ is of constant sign. Just one evaluation (maybe $\Phi_s(0)\leq 0$?) and then we're good but I don't know how to proceed.

Another idea I had is that if I have $$\sum_{\tau=-\infty}^{+\infty}\vert R_s(\tau)\vert < R_s(0)\tag{7}\label{eq7}$$ then I can conclude. But I cannot even show that the sum in \ref{eq7} exists. All I have is that I'm pretty sure that the relation \ref{eq3} defines a scalar product and therefore, by Cauchy Schwarz, we should get that :

$$\forall \tau\in\mathbb{Z}\quad R_s(\tau)\leq R_s(0)$$

Does anybody know how to solve this? I'm sure I missed something stupid.

NOTE : I already asked this question in math stackexchange where it had very little success but given the question it felt legit to ask it here. I hope it's ok.


1 Answer 1


Note that since the power spectrum is the DTFT of the auto-correlation function, we have


Since $R_s[0]$ equals the power of the the process $s(t)$, it follows that

$$R_s[0]=\frac{1}{2\pi}\int_{-\pi}^{\pi}\Phi_s(\omega)d\omega\ge 0\tag{2}$$

We can define a new process $\tilde{s}(t)$ which is a bandpass filtered version of $s(t)$ with no frequency components outside the interval $[\omega_1,\omega_2]$. The power of $\tilde{s}(t)$ is

$$R_{\tilde{s}}[0]=\frac{1}{\pi}\int_{\omega_1}^{\omega_2}\Phi_s(\omega)d\omega\ge 0\tag{3}$$

where I've used the fact that $\Phi_s(\omega)=\Phi_s(-\omega)$.

Eq. $(3)$ must be true for any interval $[\omega_1,\omega_2]$, which is only possible if $\Phi_s(\omega)\ge 0$ holds.

  • $\begingroup$ I might be missing something but isn't the inverse DTFT supposed to be $$R_s[n] = \frac{1}{2\pi}\int_{2\pi}\Phi_s(\omega)e^{jn\omega}d\omega$$? (note the integrale limits) $\endgroup$
    – NokiYola
    Apr 12, 2023 at 11:32
  • 1
    $\begingroup$ @NokiYola: You're right, I was thinking about continuous-time. Fixed. $\endgroup$
    – Matt L.
    Apr 12, 2023 at 11:33
  • $\begingroup$ I think you actually answered the question. I'm not sure that's what the author had in mind but unless I missed something it works. Thank you ! $\endgroup$
    – NokiYola
    Apr 12, 2023 at 12:57
  • $\begingroup$ I was just wondering, is it ok to say that for any given arbitrary $2\pi$-periodic there is a unique associated quasi-stationary signal? $\endgroup$
    – NokiYola
    Apr 12, 2023 at 13:17
  • 1
    $\begingroup$ @NokiYola: I would say that for any non-negative $2\pi$-periodic function one can find a discrete-time process with a power spectrum that equals the given function, but I'm pretty sure that such a process is not unique. $\endgroup$
    – Matt L.
    Apr 12, 2023 at 15:42

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