short time Fourier transform

Hey, I have a question about localization a signal in time domain by windowing.

would anyone explain to me these phrases:

1) The basis functions sin(ωt) and cos(ωt) are not localized in time!(support region in frequency = 0)

2) The δ(t) is not localized in frequency! (support region in time = 0)

cheers.

Re: short time Fourier transform

1) The basis functions are nonzero for arbitrarily large times. For example, $\displaystyle \sin \left( {(4n+1)\pi\over 2}\right) = 1 $ for arbitrarily large $\displaystyle n\in\mathbb N$.

2) The delta distribution is nonzero for artitrarily large frequencies. Note that $\displaystyle \delta(t) = \int d\omega e^{i\omega t}$, and hence in the freqency domain, $\displaystyle \delta(t)$ is the constant function 1 which does not have compact support (take nonzero values over an infinite range).

Re: short time Fourier transform

Quote:

Originally Posted by

**vincisonfire** 1) The basis functions are nonzero for arbitrarily large times. For example, $\displaystyle \sin \left( {(4n+1)\pi\over 2}\right) = 1 $ for arbitrarily large $\displaystyle n\in\mathbb N$.

2) The delta distribution is nonzero for artitrarily large frequencies. Note that $\displaystyle \delta(t) = \int d\omega e^{i\omega t}$, and hence in the freqency domain, $\displaystyle \delta(t)$ is the constant function 1 which does not have compact support (take nonzero values over an infinite range).

Thanks for reply, but I don't get the point of view :(

what it means: support region in frequency = 0 for sin & cos?

Re: short time Fourier transform

The support is the set of times $\displaystyle t$ such that the function takes nonzero values $\displaystyle \{t|f(t)\neq 0\}$. For a $\displaystyle \delta$-function, the support is $\displaystyle \{0\}$.

So what the statement above is saying is: note that even though a $\displaystyle \delta$-function's support is a single number, its Fourier transform is completely unbounded.