3
$\begingroup$

Whenever we need to find component of one signal along another signal we can do so by the help of dot product. I want to as if this method is valid for all kind of signals. Are there any restrictions upon either of the signals?

$\endgroup$
4
  • 1
    $\begingroup$ >"Whenever we need to find component of one signal along another signal we can do so by the help of convolution." If you would explain just a little bit what you mean by "component of one signal along another signal", and how you believe convolution is used to find the answer, it would help people in formulating an answer. Also, what kinds of restrictions are you concerned that your signals might not meet? Physically generated? finite duration? finite bandwidth? $\endgroup$ Jan 15 '12 at 20:41
  • 3
    $\begingroup$ I think bubble might have convolution confused with correlation, which can be interpreted as an inner-product operator on two equal-length vectors. The inner product can be used to calculate the projection of one vector onto another, which might be what is being asked about. $\endgroup$
    – Jason R
    Jan 15 '12 at 21:00
  • $\begingroup$ Indeed, it looks like the OP asks about correlation since the above definition matches somewhat definition of an inner product. $\endgroup$
    – Phonon
    Jan 15 '12 at 22:45
  • $\begingroup$ @Jason thanks!! Corrected the question. Please once again consider this question. Sorry for the grand error. $\endgroup$
    – bubble
    Jan 16 '12 at 15:28
5
$\begingroup$

Generally speaking, about the only restriction that the signals need satisfy is that they have finite energy or power. Since you mention vectors by which is presumably meant $N$-tuples of real or complex numbers and dot products, the restriction is the trivial one that each component of the $N$-tuple has finite value.

Practically, there are several issues that you need to think about depending on what you are trying to do. Obviously $$\langle \mathbf x, \mathbf y \rangle = \sum_{n=1}^N x[n](y[n])^*$$ is the dot product but how accurate a measure it is of the inner product $$\langle x(t), y(t) \rangle = \int_{-\infty}^{\infty} x(t)(y(t))^* \mathrm dt$$ of the continuous-time signals $x(t)$ and $y(t)$ from which these samples came depends on whether the signals were filtered before sampling, the sampling rate, whether the samples span the entire time interval where the signals are nonzero, whether windowing was applied to the samples, etc. You might be interested in the cross-correlation function $R_{x,y}(\cdot)$ instead of just the inner product (which is $R_{x,y}(0)$) in which case FFTs might be used and you have to make sure that the signals are properly zero-padded so that you get the right answers. Finally, we began with the "trivial" restriction that each component is finite which always holds in practice, but if the components are large in magnitude, you need to worry about the possibility of overflow in the inner product computation.

So, as you can see, in theory, theory and practice are the same, but in practice, they are different.

$\endgroup$
2
  • $\begingroup$ Thanks for your time Dilip. So it seems that inner product will give me the desired information as long as the signals are finite. I could not understand what you meant by cross-correlation. Can you elaborate further ? Or point to some link ? $\endgroup$
    – bubble
    Jan 16 '12 at 17:26
  • $\begingroup$ $$R_{x,y}(\tau)=\int_{-\infty}^{\infty} x(t)(y(t-\tau))*\mathrm dt$$ measures how well $x(t)$ is aligned with a delayed version of $y(t)$. It might be that for any number of reasons, $x(t)$ and $y(t)$ are misaligned in time by $\tau$, and so $x(t)$ has a larger projection on $y(t-\tau)$ than on $y(t)$. If you find $R_{x,y}(\tau)$ for all $\tau$, you can determine the best fit. It is easier to calculate all of $R_{x,y}$ instead of one point at a time if you use FFT; same reason that entire convolution $x*y$ is easier than to do it for one sample of $x*y$ at a time: $O(N\log N)$ vs $O(N^2)$. $\endgroup$ Jan 16 '12 at 18:15

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.