[PD] unit impulse without [dirac~]

Mathieu Bouchard matju at artengine.ca
Sun Nov 18 02:15:48 CET 2007

On Sat, 17 Nov 2007, Andy Farnell wrote:

> Okay, one for the mathematicians, as Ugur mentioned the proper name this 
> is Kronecker Delta.

a Dirac impulse and Kronecker Delta are not the same thing. Kronecker 
Delta is a function of two variables, which is the indicatrix of equality: 
for example, [==] is the Kronecker Delta, as it gives 1 when two inputs 
are equal, and 0 when they aren't. (See below for the link between Dirac 
and Kronecker)

a Dirac impulse is defined as something whose ordinary integral is 1, but 
which is zero everywhere except at point zero. This makes it impossible as 
an ordinary real function, so over a real domain it has to be defined as a 
special operator that pretends to be a function. Over a discrete domain 
you will approximate it using a spike, but the spike will not be of height 
1 unless you consider 1 sample to be the unit of the time axis. If you 
count 1 second = 44100 samples as being the unit, then you have to make a 
spike of 44100 high, so that when multiplied with 1/44100 duration it 
gives an integral of 1.

> Is it correct to say that the Dirac impulse preserves energy, as it 
> tends towards zero time length the amplitude goes to infinity and if we 
> squashed its amplitude to zero it would be infinitely long?

I don't understand what this means. A signal does not "preserve" anything. 
An operator may preserve a property, and a two-input operator may preserve 
a property if it has a particular input on the left or right side. But 
what is it that you mean?

> In which case Dirac impulses are theoretical and not practical digital 
> signals?

It's possible that you have a continuous ("real") version of the Dirac 
impulse, which is only theoretical, but if you make an extended definition 
that is both compatible with the original continuous definition and with 
discrete domains, then the discrete version will be an ordinary function 
(a true digital signal) whereas the continuous version cannot be an 
ordinary function.

You could think of the Dirac impulse as being the identity element of the 
convolution operator, or as being the limit of impulses of integral 1 
where the width is reduced as close as possible to 0. With continuous 
functions there is no "closest possible" so you get a freaky result, but 
for continuous functions you get something which is a particular special 
case of Kronecker Delta (with $2 = the time of the impulse) times a 
compensation constant.

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