Let's assume that we have a dynamical system

$$G(s) = \frac{Y(s)}{U(s)}$$

And we have measure the input $u(t)$

enter image description here

And we have measure the output $y(t)$.

enter image description here

Here we can see that if we increase the frequency at the input of the system, the output is going to decrease.

The question is simple: How can I find the amplitudes over time of by just knowing $y(t)$ and $u(t)$. Nothing more.

My goal is to estimate a transfer function from arbitrary frequency input and output and I going to use least square to curve fit on a transfer function.

Can I just drag a line of all peaks/tops of the sine waves?

  • $\begingroup$ Maybe it is just me but: can you please clarify what you are working with? The notation makes me think that it is $Y(s)=\mathcal L\{y(t)\}$ and $U(s)=\mathcal L\{u(t)\}$ but you have fft in the tags $\endgroup$ – Holo Jan 3 '18 at 4:37
  • $\begingroup$ I did a simple System ID in Python suggestions-for-fitting-noisy-exponentials-with-scipy-curve-fit which massages the data to be smoothly periodic, and takes ratio of ffts of U and Y - for your frequency sweep I would time reverse, invert and concatenate the time series before the fft, and fit the ratio to a Laplace domain transfer function for a shelving low pass filter $\endgroup$ – f5r5e5d Jan 3 '18 at 5:25
  • $\begingroup$ @Holo I doing system identification of a transfer function into frequency plane. $\endgroup$ – Daniel Mårtensson Jan 3 '18 at 6:58
  • $\begingroup$ @f5r5e5d Did it work well ? $\endgroup$ – Daniel Mårtensson Jan 3 '18 at 6:58
  • $\begingroup$ @DanielMårtensson both transform are doing this... $\endgroup$ – Holo Jan 3 '18 at 7:07

Since you have $u(t)$ and $y(t)$, you can compute the Laplace transform of each and get, by your first equation,

$$ G(s) = \frac{L(y(t))}{L(u(t))}. $$

If your measurements are continuous time and perfect then this is exact. If you have noisy measurements, then this is approximate. If you have discrete measurements you need to look at Hankel matrices and realization theory.


Your Answer

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

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