Why use cross-spectral density to calculate frequency response?

Question

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 $$G=\frac{S_{yu}}{S_{uu}}$$ where $S_{uu}$ is the Fourier transform of the autocorrelation of the input, and $S_{yu}$ is the Fourier transform of the cross-correlation of the output and the input (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!

Answer 1

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.

Answer 2

A section from this answer is helpful to your question and it complements the answer provided by robert bristow-johnson:

• In Frequency domain, we have: $$Y = G\cdot U$$ where U is the input frequency spectrum, so you'd be tempted to simply re-arrange it to get the frequency response: $$G = \dfrac{Y}{U}$$ This is problematic if there's noise $N_0$ at the output: the resulting estimate is: $$\tilde{G} = \frac{Y+N_0}{U} = G + \frac{N_0}{U} = > G+\frac{1}{\texttt{SNR}}$$ So, the lower the $\texttt{SNR}$ (equivalent to the stronger the noise at the output), the higher the estimator's bias.

• Here comes the frequency averaging estimator: Let's look at the definition of the Cross Power Spectral Density: \begin{align} S_{yu} &= \mathbb{E}\left[ \overline{U}\cdot(Y+N_0) \right]\\ &= \mathbb{E}\left[ \overline{U}\cdot Y + \overline{U}\cdot N_0\right]\\ &= \mathbb{E}\left[ \overline{U}\cdot GU + \overline{U}\cdot N_0\right]\\ &= G\cdot\mathbb{E}\left[\overline{U}U\right] + \mathbb{E}\left[\overline{U}\cdot N_0\right]\\ &= GS_{uu} + \mathbb{E}\left[\overline{U}\cdot N_0\right] \tag{1} \end{align}

• Assuming the noise in the output, $N_0$, is un-correlated with the input, we have $$\mathbb{E}\left[\overline{U}\cdot N_0\right] = 0$$ so (1) becomes: $$S_{yu} = GS_{uu}$$ and we get $$G = \frac{S_{yu}}{S_{uu}} = \tilde{G} $$ i.e., the estimator is perfect.

Of course this is just theory, since the Expected Value Operator $\mathbb{E}$ is a statistical operator based in infinite time. In a practical measurement scenario (finite time, noisy), the cross term $\overline{U}\cdot N_0$ is never actually $0$, but by using frequency averaging (Welch's method) to compute $S_{yu}$, the cross term $\overline{U}\cdot N_0$ does go towards 0 since $U$ and $N_0$ are un-correlated, resulting in a better estimate of $G(\omega)$ than simply dividing the output spectrum by the input spectrum.

Answer 3

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."