I am doing the Coursera course on DSP by Prandoni and Vetterli, and was excited to learn that we can use the tools of vector spaces to analyze discrete signals. At this early point in the course the professors did mention that since the signals live in Hilbert space, we can naturally talk of orthogonality, for one.

I am curious: what other linear algebra operations/concepts do we use with discrete signals, and what is the rationale behind them?


whether or not a signal lives in a Hilbert space depends only on if you say it does. if you say so, then you need to specify an inner product $<x,y>$ (which is pretty easy for finite-length discrete-time signals)

$$ \langle x,y \rangle = \sum_n x[n] y^*[n] $$

and you need to use that inner product definition to define a norm

$$ |x| = \sqrt{\langle x,x \rangle} $$

which should also be the same as the distance measure from $x$ to the $0$ element in that Hilbert space.

and you need to be able to add and subtract elements in the Hilbert space (easy to do for discrete-time signals). why this might be useful would be for defining a signal $x$, an estimation of that signal $\hat{x}$, and an error signal, $e_x$ such that

$$ \hat{x} = x + e_x $$ or $$ e_x = \hat{x} - x $$

we can think of the norm of $e_x$ which is

$$ |e_x| = \sqrt{\langle e_x,e_x \rangle} $$

as a measure of the distance that $\hat{x}$ is from $x$.

if gives you a formal framework for defining error and then maybe help you with algorithms (like a matched filter) for minimizing error, so that a received signal most closely approximates what was transmitted. or so that some other "thing" most closely hits some specified target "thing".

  • $\begingroup$ "whether or not a signal lives in a Hilbert space depends only on if you say it does." Is the Hilbert space a good and natural mathematical abstraction for discrete signals? $\endgroup$
    – Joebevo
    Feb 2 '14 at 3:43
  • $\begingroup$ @Joebevo, i think so. it's better (but more specific) than a Normed-Linear Space (sometimes called a "Banach space) because it has the inner product, which the Banach does not necessarily have. both the Hilbert and Banach spaces have a concept of addition of elements of the space and a "zero vector" or "zero element" that, when added to some other element, does not change the other element. then there are transformations (like the DFT) that sorta have additional meaning in Hilbert space. $\endgroup$ Feb 2 '14 at 14:05

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.