0
$\begingroup$

I am taking a class about system identification and currently learning about cross-spectral density. My textbook says that the frequency response, G, of a system can be determined as $$G=\frac{S_{uy}}{S_{uu}}$$ where $S_{uu}$ is the Fourier transform of the autocorrelation of the input, and $S_{uy}$ is the Fourier transform of the cross-correlation of the input and the output (the cross-spectral density).

Now, in basic control systems classes you lean that $G=\frac{Y}{U}$.

The former method, using the cross-spectral densities, seems to be related to this more basic calculation, but I'm unsure why we bother using the cross-spectral densities at all. Is it simply to help eliminate noise from the signals or is there some other reason for it?

Thanks!

$\endgroup$
1
$\begingroup$

in the audio biz, we call this the "Two-channel FFT". the cool thing about it is that you can measure the magnitude response of a room or something using music (that is decently broadbanded) as the test signal (and divide the output spectrum by the input spectrum). you don't need to pollute peoples' ears with an impulse train or a maximum-length-sequence or pink noise or with a swept-frequency chirp.

$\endgroup$
  • $\begingroup$ Neat! Hadn't though about that! $\endgroup$ – SirCumference Jun 27 '17 at 1:25
0
$\begingroup$

Perhaps not so much about noise reduction but more about posing system ID as a hypothesis in a probabilistic context, with confidence regions about the systems parameters.

From Wikipedia,

https://en.wikipedia.org/wiki/System_identification

"The field of system identification uses statistical methods to build mathematical models of dynamical systems from measured data. System identification also includes the optimal design of experiments for efficiently generating informative data for fitting such models as well as model reduction."

$\endgroup$

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.