Normalized correlation is used to measure how similar are two signals. I can understand similarity visually but I haven't seen a mathematical definition for the similarity of 2 signals anywhere. So I have 2 big questions that are bothering me:

  1. Is there a mathematical definition for how similar two signals are?(similar in same fashion of visually similar)

  2. With that definition, is there a proof that shows correlation is maximum between similar signals and not any other signal?


Okay so I thought about 1 and I have a mathematical definition for similar signals. Basically s1 is similar to s2 if they are proportional, that is s2 = a.s1. So we can focus on question 2 now :). A proof that says out of all functions out there, the correlation of s1 and s2 is maximum only if s1=k.s2

  • 2
    $\begingroup$ en.wikipedia.org/wiki/Cross-correlation $\endgroup$ – Andy aka Aug 5 '17 at 17:40
  • $\begingroup$ I couldn't comment due to me low reputation, but you may find this interesting. It is called wavelet-based semblance analysis - link.It is a signal processing technique used to compare to time series signals and has a range of visual outputs. $\endgroup$ – tomdertech Aug 5 '17 at 19:03
  • $\begingroup$ I've read the wikipedia article. It doesn't define similarity or provide a proof why the correlation is max between 2 similar signals. $\endgroup$ – doubleE Aug 5 '17 at 19:21

As you correctly noted, similarity is not a rigorously defined mathematical term. However, "distance" can be defined mathematically. Quoting Wikipedia:

In statistics and related fields, a similarity measure or similarity function is a real-valued function that quantifies the similarity between two objects. Although no single definition of a similarity measure exists, usually such measures are in some sense the inverse of distance metrics...

Here's a hand-wavy explanation, the math isn't totally precise, but hopefully it's enough to get some intuition:

Let's say we have real valued signals $x(t)$ and $y(t)$ defined on the unit interval $t \in [0,1]$. We will say that these signals are similar if the distance $||x-y||$ is small. Note that there are various choices of how to measure distance. A commonly used distance measure is the squared difference a.k.a. L2 distance defined as $$ ||x-y||^2_{L^2} \triangleq \int_0^1 (x(t)-y(t))^2 dt \ . $$

Let's say we have three signals $x$, $y$ and $z$. Normalize all signal energies to 1 so that $\int_0^1 (x(t))^2 dt = 1,$ etc. Suppose $x$ is more similar to $y$ than $z$. This means

$$||x-y||_{L^2} < ||x-z||_{L^2}$$


$$\int_0^1 (x(t)-y(t))^2 dt < \int_0^1 (x(t)-z(t))^2 dt$$

or, rearranging terms,

$$\int_0^1 x(t) y(t) dt > \int_0^1 x(t) z(t) dt \ .$$

In other words, the correlation between $x$ and $y$ is higher than that between $x$ and $z$. This shows that in order to find a "more similar" signal, maximizing the correlation is indeed the right thing to do (assuming L2 distance).

  • $\begingroup$ That was a really insightful derivation...I tried to formulate the problem in a more precise way. Please see my edit to question. $\endgroup$ – doubleE Aug 5 '17 at 20:08
  • 1
    $\begingroup$ Note that because I normalized all signal energies to 1, the constant k in your edited question is 1. In other words - the signal that is most similar to a given signal is the given signal itself. The maximum value of the correlation is then equal to the signal energy i.e. 1. $\endgroup$ – Atul Ingle Aug 5 '17 at 20:14
  • $\begingroup$ as far as i can tell, the math is fine. there's nothing hand-wavy or imprecise about it. $\endgroup$ – robert bristow-johnson Aug 5 '17 at 22:35
  • $\begingroup$ actually the L2 norm or "distance" is the square root of the expression above. but that fact does not change any of the reasoning, which is correct as is. $\endgroup$ – robert bristow-johnson Aug 5 '17 at 23:28
  • $\begingroup$ @robertbristow-johnson thanks for the edits. I think it's a bit hand-wavy because I didn't precisely define these functions as being in the space $L^2[0,1]$. The choice of the interval $[0,1]$ is also arbitrary. I also assumed unit norm signals without much justification. $\endgroup$ – Atul Ingle Aug 5 '17 at 23:48

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.