# signals that are band-limited are not time-limited

Discussion in 'Homework Help' started by screen1988, May 20, 2013.

1. ### screen1988 Thread Starter Member

Mar 7, 2013
310
3
I have read "signals that are band-limited are not time-limited". Is there a way to prove this?

2. ### Tesla23 Active Member

May 10, 2009
323
67
Yes.

It is quite a deep result and I don't know how much complex analysis you have done, but the shortest proof I know starts with a bandlimited signal f(t). It is straightforward to show from the fourier transform that being bandlimited means that f(t) is analytic over the entire complex plane, hence if it vanishes on any interval (e.g. f(t) = 0 for t>T), then it vanishes everywhere. Hence a bandlimited signal cannot be time-limited. The converse is identical. You will find this stuff in more rigorous texts (e.g. Papoulis).

A more handwaving argument would be that if it is bandlimited then you don't change it by multiplying the spectrum by a rectangular window, hence you don't change it in the time domain by convolving with a sinc function - this tends to spread it outside any time window you assume it is contained in.

screen1988 likes this.

Mar 7, 2013
310
3
4. ### Tesla23 Active Member

May 10, 2009
323
67
I don't know what you mean, your original question was can it be proved and the answer is precisely yes. If you want a statement of what is proved, then a bandlimited function is one where:
F(ω) = 0 for |ω| > B for some B

and a time-limited function is one where:
f(t) = 0 for |t| > T for some T

A precise statement of the theorem is that a signal cannot be both time-limited and bandlimited. e.g. see here: http://dsp-book.narod.ru/HFTSP/8579ch38.pdf

screen1988 likes this.
5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Intuition & precision don't always go hand in hand. It might be worth asking what the Op understands by the terms time limited & bandwidth limited. Starting with some common ground might be helpful.

6. ### screen1988 Thread Starter Member

Mar 7, 2013
310
3
Thanks, that is what I want to prove.
To be true, I am not good enough to track your answer. Now I will ask you in more detail about it.
Then in the time domain we have f(t)*sinc. I know that sinc function is non-timelimited function but I can't see why f(t) is non-timelimited function.
Could you explain it more detail?
I didn't know about analytic function and consulted it here:http://en.wikipedia.org/wiki/Analytic_function
But after know this seems not straightforward for me to show that? Can you give some more detail?

7. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Presumably, if the windowing range in the frequency domain is less than the actual bandlimited signal frequency domain range then one would obtain different results in the time domain with respect to the original signal and its aforementioned frequency domain windowed version.

For instance - suppose I take as my original time limited signal an ideal impulse. This will map onto the frequency domain as a constant value over an infinite bandwidth. If I now truncate that original bandwidth by applying a unit window about the origin, I end up with a rectangular frequency distribution. When the inverse Fourier transform is applied to this bandwidth limited version of the original impulse signal spectrum the result is a sinc function - which (as already pointed out by Tesla23) is by definition not a time limited function. If I now apply a second more tightly constrained unit rectangular window to the already truncated spectrum, I will obtain a corresponding inverse Fourier transform also having the form of a sinc function but different to the first sinc function obtained with the less constraining spectrum window.

Last edited: May 23, 2013