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Subject: Re: Relationship between a functions derivitives and BW?
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Date: Tue, 06 Nov 2012 14:21:36 -0600
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>>>>Quick question for you guys...
>>>>
>>>>Is there any relationship between constraining a functions derivitives
>>>and
>>>>constraining its bandwith?
>>>>
>>>>For example, if we were to consider the output of a lowpass random
>>>>process... call it x_lp[n]. One could argue that that sequence must be
>>>>'smooth' by some account. (how smooth has to do with the lowpass
>>bandwith
>>>>of the process)
>>>>
>>>>
>>>>x_lp[n] 'smoothness' should also be able to be described by the
>>functions
>>>>derivitives either existing or also being continuous.  Is there a
>>>straight
>>>>forward relationship?
>>>>
>>>>thanks in advance
>>>>
>>>
>>>Hi,
>>>
>>>There is a fundamental inequality, called Bernstein's inequality, that
>>>seems to be what you are looking for. If a signal s(t) has two-sided
>>>bandwidth B and is bounded on the time domain by a constant A>0, then
it
>>>holds that
>>>
>>>  |s'(t)|<= pi * A B/2.
>>>
>>>The signal attaining the bound is A*sin(pi*B*t+phi), where phi is any
>>>phase. For the n-th order derivative, it is
>>>
>>>  |s^(n)(t)|<= (pi*B/2)^(n-1) A.
>>>
>>>I hope this helps.
>>>
>>>Jesus. 
>>>
>>>
>>
>>Ups, the last inequality is
>>
>>|s^(n)(t)|<= (pi*B/2)^n A.
>>
>>Jesus.
>>
>>
>
>Sorry, another correction. The inequalities are
>
>|s'(t)|<= pi * A B/2
>
>and
>
>|s^(n)(t)|<= (pi*B)^n A.
>
>Jesus.
>
>


Thanks!!
I will explore this.

There was a little talk as to whether this result only holds for the
continuous case or not. Does this result hold for the discrete case also
where i can conclude x[n+1] - x[n] is bounded by (pi * A B/2)??

If so, this is a pretty neat result.