1
$\begingroup$

Let’s say I'm trying to measure a quantity which varies in time as:

$$f(t) = (\text{unknown constant } Q) + (\text{broadband noise})$$

where "broadband noise" has a known power spectrum with a finite bandwidth.

I want to measure the constant $Q$, so I repeatedly take measurements $f(0), f(1), f(2), \ldots$ to formulate a guess: $$Q_N = (f(0) + f(1) + \cdots + f(N-1))/N$$ Obviously, if I take enough measurements, I'll get the right answer, i.e. $\lim_{N\rightarrow \infty} Q_N = Q$.

My question is: As $N \rightarrow \infty$, does $\langle |Q_N - Q|\rangle \propto N^{-1/2}$ or $\propto N^{-1}$? Paradoxically, I have seemingly-convincing arguments for both.

Case for $N^{-1}$: If you break down the vibration into sinusoids, each gives an error contribution that looks like something proportional to $\frac1N(\cos(\phi) + \cos(\phi + ω) + \cos(\phi + 2ω) + \cdots)$. Although the term in parentheses looks kinda like a random walk, it's actually bounded, so the error goes as 1/N. (Unless ω is 2π times an integer, which is vanishingly unlikely; anyway you can assume the noise is at ω < 0.1.) Well, if that's true for a sinusoid, it should be true for broadband noise which is just a sum of sinusoids. So we conclude that the error goes as $N^{-1}$.

Case for $N^{-1/2}$: When you approximate $Q$ with $f(t)$, there is obviously an error in this approximation. The error depends on the instantaneous noise environment. Beyond the correlation time of the noise, the errors are all uncorrelated with each other. When you add up uncorrelated errors, it's a random walk, so the error goes as $N^{-1/2}$.

At least one of these two arguments must be wrong! Thanks in advance!!

$\endgroup$

2 Answers 2

1
$\begingroup$

If I understand your question correctly, notice that while the vibration has bounded error, it still fluctuates, something like $\cos(N\omega)$, in a similar fashion to a random walk. In particular, $\cos(\phi+\omega n)$ will tend to stick near $\pm 1$, due to the weird properties of $\cos(U)$ where $U$ is uniformly random and hence be similar to a standard random walk. These fluctuations will cause the increase of error from $1/N$ to $1/\sqrt{N}$.

Let $X_n=\cos(\phi+\omega n)$. Then you're looking at $\frac{1}{N}\sum_{n=1}^N{X_n}$. Now I'm just computing in an adhoc way the variance which for each term would be $E[X_n^2]-E[X_n]^2=E[X_n^2]=\cos^2(\phi+n\omega)\approx 1$. You can make these arguments rigorous by treating $X_i$ as a random variable $X_i=\cos(U_i)$ where $U_i$ is uniform random on $[0,\pi]$.

Thus the average variance of the error contribution would be something like:

$\frac{1}{N^2}\sum_{k=1}^N \mbox{Var}(X_k)=\frac{1}{N},$

which will look like an error of $N^{-1/2}$.

$\endgroup$
5
  • $\begingroup$ Where did the $k^2$ in the denominator come from? $\endgroup$ May 18, 2016 at 21:47
  • $\begingroup$ @SteveB: I'm just squaring it as if calculating variance. $\endgroup$
    – Alex R.
    May 18, 2016 at 22:01
  • $\begingroup$ I still don't understand. The sum you wrote is $\cos^2(ω) + \cos^2(2ω)/4 + \cos^2(3ω)/9+\cdots$. Thanks to the $k^2$ denominator, each term has a lower and lower weight. But in $Q_N$ as I defined it, I'm just averaging everything with equal weight. Right? Thanks in advance. $\endgroup$ May 18, 2016 at 23:30
  • $\begingroup$ @SteveB: I've reworded the explanation without the $k$s. Does it make sense now? $\endgroup$
    – Alex R.
    May 19, 2016 at 0:00
  • $\begingroup$ I think you missed the key surprise that motivated my question. It's not that $\cos(Nω)$ is bounded (which is obvious). The surprise is that if φ is random but ω is fixed, then $E[(\sum_{n=1}^N \cos(φ+nω))^2]$ does not grow with N, but has a finite bound independent of N. (Except for special values of ω.) Ref. The sum looks like a random walk but it really isn't. Evidently the correlations between $X_i$ and $X_j$ -- which you neglect -- may be important. $\endgroup$ May 19, 2016 at 12:46
0
$\begingroup$

I figured it out eventually.

If you look at the PSD of the noise, you get $N^{-1/2}$ to the extent that the PSD is nonzero at $\omega,2\omega,\ldots$. You get $N^{-1}$ from the noise power at every other frequency away from the integer multiples of $\omega$.

If you assume (as I did in the question) that the noise bandwidth is strictly limited to frequencies away from integer multiples of $\omega$, then that's a very restrictive assumption about the time-domain profile. It mandates relationships between points that are arbitrarily far apart in time. So that undermines the "Case for $N^{-1/2}$" that I wrote down. The answer in this case is $N^{-1}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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