My lecture slides mention that two-level ("2-tasoiset" in Finnish) signals are good in identification with linear systems. I am trying to understand this in noisy system. I cannot find any method really to identify the impulse response, perhaps good pre-processing and some other checks?

This is a sub-problem 2F here for the course Mat-2.4129 in Aalto University.

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    $\begingroup$ Translation of the actual question of the problem would be helpful. $\endgroup$
    – Naresh
    Dec 18, 2012 at 6:00
  • $\begingroup$ Also, based on the notation in the problem, I don't think $Y(s)$ would be the system's impulse response. It is most likely the Laplace transform of the impulse response; if you let $s=j\omega$, then the resulting $H(\omega)$ is the system's frequency response. $\endgroup$
    – Jason R
    Dec 18, 2012 at 14:13

1 Answer 1


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.

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    $\begingroup$ Very nice answer. $\endgroup$
    – Jason R
    Jan 9, 2013 at 14:19
  • $\begingroup$ My teacher Jouni stressed that you can get the impulse response by Laplance-inverse conversion. I can understand this $y(t)=g(t)u(t)+v(t)$ so $Y(s)=G(s)U(s)+V(s)$ where $y,Y$ for output, $g,G$ for control and $v,V$ for noise. You could also take inverse Fourier transform. Is this detail in your answer? What is "proverbial inverted thumbtack function" -- moved here. $\endgroup$
    – hhh
    Jan 9, 2013 at 18:23
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    $\begingroup$ Your opening remark about two-level signals being useful in system identification is what led to my answer. If two-level signals are not important, and inversion of LaPlace transforms must be used, then several other methods can be used. It is clear from your recent edit (after I had posted my answer) of your own answer that you are looking for something quite different from the method described in my answer. I suggest you wait for someone else to answer your question. Meanwhile, I will consider whether I should delete my own answer as being non-responsive to the question asked. $\endgroup$ Jan 9, 2013 at 18:47
  • $\begingroup$ @DilipSarwate do not delete, it is probably great. I am just dumb to undertand it, taking a while to undertand your meaning. You have some good points such as the end. "Recent" edit? $\endgroup$
    – hhh
    Jan 9, 2013 at 19:00

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