Assume that we have our discrete transfer function $G(e^{j\omega_k})$ where, $k = \frac{\pi k}{M} , k = 0, \dots , M$.

If we want to find Markov-parameters $g_k$ from the impulse response

$$g_k = \left\{\begin{matrix} D, & k = 0\\ CA^{k-1}B & k > 0 \end{matrix}\right.$$

one way to do it is to use the Inverse Discrete Fourier Transform (IDFT) according to

$$g_i = \frac{1}{2M} \sum_{k=0}^{2M-1}G(e^{\frac{j2\pi k}{2M}})\frac{j2\pi ik}{2M} , i = 0, \dots , M$$

But how would be in real life if I don't know my transfer function. The only thing I know are input and output data.

Let's say that we have a beam which one side is welded into a wall. Then I take a hammer and jack the other side of the beam. The beam is swaying. I measure the impulse response. After I have measure my impulse response. I take my hammer again and jack the beam even harder and measure the swaying impulse response again. Repeat with harder impulse.


Is this the right method to create markov-parameters from impulse-frequency responses?

I know that the Inverse Discrete Fourier Transform transform frequency data to time domain data. But I don't know how to create the frequency data from a impulse response. That's why I'm asking you how to do that.

If you still don't understand what I mean. Please look at heading "3 The algorithm" at page 4 here.


It's a document how to do system identification of impulse responses.

  • 1
    $\begingroup$ Your question is unclear. If you know $x ,y = x \ast g$ then $Y(k) = X(k) G(k)$ so you can deduce $G$ and $g$ from $X,Y$. If $x,y$ are noisy (ie. $y \approx x \ast g$ which is the case in real life) then you'll set $G(k)=\frac{Y(k)}{X(k)+\epsilon e^{i \text{arg}(X(k))} }$ instead of $G(k)=\frac{Y(k)}{X(k)}$. $\endgroup$ – reuns Dec 17 '17 at 2:30
  • $\begingroup$ @reuns I don't understand what x and y are. Let's say that I can measure my impulse response. How can I then create the markov parameters from frequency domain ? $\endgroup$ – Daniel Mårtensson Dec 17 '17 at 12:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.