How to identify impulse response of noisy system?
If a linear time-invariant system has (deterministic) input $x(t)$ and output $y(t)$, then the cross-correlation function $R_{y,x}$ of the output and input has value $h \star R_{x,x}$ where $R_{x,x}$ is the autocorrelation function of the input signal and $h$ is the impulse response of the linear system.  For completeness, the autocorrelation function $R_{y,y}$ of the output is given by
$$R_{y,y} = h\star \tilde{h}\star R_{x,x} = (h\star\tilde{h})\star R_{x,x} = R_{h,h}\star R_{x,x}$$
where $\tilde{h}(t) = h(-t)$ is the impulse response $h$ with time reversed, and $R_{h.h}$
is the autocorrelation function of the impulse response signal $h$.  In the frequency domain, this relation is the familiar power spectral density relationship
$$S_y(f) = |H(f)|^2 S_x(f).$$

Coming back to the business at hand, suppose that $x$ is a periodic signal with
a very long period, much longer than the measurable duration of the impulse
response $h$. Then, the output signal $y$ also has the same period, and essentially looks like periodic repetitions of $h\star \hat{x}$ where $\hat{x}$ denotes one period
of $x$..  The input autocorrelation function
$R_{x,x}$ is also a periodic signal and thus $R_{y,x}$ looks like periodic 
repetitions of $h\star \hat{R}_{x,x}$ where $\hat{R}_{x,x}$ is the periodic
autocorrelation function of $\hat{x}$.

Finally, suppose that $\hat{R}_{x,x}$ is an impulse and $R_{x,x}$ is
thus a periodic impulse train with very long period.  Then,  $R_{y,x}$
is just periodic repetitions of $h$, the impulse response that we seek. 
Now, this is all fine and dandy but everyone knows that impulses
don't exist in real life and money can't buy happiness. But as
the vulgar rich know, and all others suspect, money
can buy the most remarkable substitutes for happiness. Similarly, there exist
signals whose periodic autocorrelation functions approximate impulses.
If $\hat{R}_{x,x}(0)$ is very large while $\hat{R}_{x,x}(\tau)$
is very small, possibly even $0$, for all other $\tau$,
then $\hat{R}_{x,x}$ is essentially impulse-like. Even better,
such signals can even be chosen to be two-valued. They are, of course,
the binary $m$-sequences, also known as maximal-length linear feedback
shift register sequences. A $m$-bit linear feedback shift register
whose feedback connections are specified by a primitive binary
polynomial of degree $m$ produces a sequence of period $2^m-1$.
If the clock cycle duration is $T$ seconds, the output waveform
from the shift register has period $(2^m-1)T$ seconds. Regarded
as a two-level waveform with levels $\pm 1$, the periodic autocorrelation
function has a peak value of $(2^m-1)T$ which is very large
compared to the off-peak value of $-T$, and indeed the difference
can be made very large even with moderate values of $m$.



So, how does all this work? The input is a periodic two-level $m$-sequence
signal whose periodic autocorrelation is the proverbial inverted
thumbtack function. We compute the cross-correlation of the output
and input signals, effectively getting the impulse response of the 
system as the result. True, the result is actually the response to
periodic thumbtacks but the discrepancy can be made small. Note also
that we get the impulse response without having an impulse
applied to the system. This is important: most linear systems
are actually nonlinear systems operated in a regime where the
signals are small in amplitude, and the excursions from the
nominal operating point are small enough that the tangent to
the (nonlinear) operating curve works well enough as a substitute
for the exact response (small-signal analysis). So, the input
is low-level but by a miracle of modern mathematics, we are
able to compute the response of the linearized version
of the nonlinear system to an impulse without using a large
signal that will blow fuses and drive the system into saturation,
overload, etc.