3
$\begingroup$

I just started on DSP and I have a question that I would like to ask.

I have a zero-mean uncorrelated wide sense stationary discrete-time random process {$x[n]$, $n$ is a set of integers}. $x[n]$ is input to a causal linear time-invariant system below, denoted by its transfer function $H(z)$.

$$x[n]{\longrightarrow}\boxed{H(z)}{\longrightarrow }y[n]$$

Since the output is a random process, suppose that its autocorrelation function, $R_y[k]$, is measured without approximation error.

From the information given, is there a method to identify the system transfer function, $H(z)$ or, equivalently, its impulse response, $h[n]$?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ i wanted to fix up some notation (first started with the answer which was necessary because something was falsely stated). we leave the "$k"$ in $R_x[k]$ because it's the lag, not the time index $n$. $$ R_x[k] \triangleq \lim_{N \to \infty} \frac{1}{2N+1} \sum\limits_{n=-N}^{+N} x[n] \cdot x[n+k] $$ $\endgroup$
    – robert bristow-johnson
    Nov 22, 2016 at 17:37
  • 1
    $\begingroup$ and if $x[n]$ is ergodic, then $$ R_x[k] = \mathrm{E} \{ x[n] \cdot x[n+k] \} $$ where $\mathrm{E} \{ \cdot \}$ is the probabilistic Expectation value or Expected value and is derived from the probability density functions for the random process $x[n]$. if it's "ergodic" then all of the time averages equal probabilistic averages (or are taken to be equal). $\endgroup$
    – robert bristow-johnson
    Nov 22, 2016 at 17:43

1 Answer 1

Reset to default
3
$\begingroup$

From only the process statistics, you cannot derive the exact impulse response/transfer function. However, by the Wiener-Khinchin-Theorem you know, that the power spectrum density (PSD) of an input signal to an LTI system is transformed with the magnitude-square of the transfer function:

Let $S_x(\omega)$ be the PSD of the input and $S_y(\omega)$ be the PSD at the output of the system. Note that

$$\begin{align} S_x(\omega) &= \mathrm{DTFT} \{ R_x[n] \} \\ &= |X(e^{j\omega})|^2 \\ \end{align}$$

where $X(e^{j\omega})= \mathrm{DTFT} \{x[n]\}$ and similarly for $S_y(\omega)$.

Then, you have

$$ S_y(\omega)=|H(e^{j\omega})|^2 S_x(\omega) $$

and you can at least calculate the magnitude of $H(e^{j\omega})$ by inverting the equation.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.