Dirac Delta: Finite Height in Fourier Analysis

[frenzal_dude]

Hi,
if the definition of a dirac delta (impulse) function is just a sinc function with an infinite height and 0 width, why is it that they are shown and used in Fourier analysis as having a finite height?

for example g(t) = cos(2*PI*f0*t) has two impulses of height = 1/2 at f=+/-f0

 

[yungman]

I don't know that Dirac Delta function has anything to do with sync function.

The definition of one dimention ( say x ) Dirac Delta function [itex]\delta(x-a)[/itex] is equal zero everywhere except equal to infinity at x=a and

[tex] \int_{-\infty}^{\infty}\delta(x-a)dx=\int_{a^-}^{a^+}\delta(x-a)dx=1[/tex]

Sync function is just [itex]sync(x)=\frac{\sin x}{x}[/itex] and equal to 1 when x=0.

 

[frenzal_dude]

ok maybe the sinc was a bad example,
but still, if it is equal to infinity at x=a, why do they represent the impulse as having a finite height of 1/2 in Fourier analysis?

 

[rbj]

i would suggest going to the Wikipedia site for the Dirac delta function.

for a strict mathematician, the definition is not simple. in fact, mathematicians would say that the Dirac delta is not even a function. they like to call it a "distribution" or a "generalized function". i am not all that sure of what they mean, but one thing i remember from my college daze was, in the EE class [itex] \delta(t) [/itex] is a function that is zero everywhere except t=0 and its integral is 1. but in a Real Analysis class (where you learn the difference between Riemann and Lebesgue integration) any function that is zero "almost everywhere" has an integral that is zero. so there's some kinda disconnect there.

for practical purposes, treat the dirac impulse as one of those "nascent" delta functions (the spikes with unit area that get thinner and thinner), but be careful when you find yourself talking about it with your math prof.

 

[MisterX]

ok maybe the sinc was a bad example,
but still, if it is equal to infinity at x=a, why do they represent the impulse as having a finite height of 1/2 in Fourier analysis?

You may have seen something like an arrow. This arrow notation would not mean the impulse has a finite height equal to 1/2. The arrow may have a specific height to indicate the magnitude of the coefficient of the [delayed] Dirac delta it represents.

 

[yungman]

ok maybe the sinc was a bad example,
but still, if it is equal to infinity at x=a, why do they represent the impulse as having a finite height of 1/2 in Fourier analysis?

Have not touch Fourier Transform for a long time. Far as I remember a dirac delta function is represented by and infinite amount of frequency components after Fourier Transform which means it is flat response on the FT graph with no distinct frequency peak stand out. You have to add all the frequencies together so the final amplitude is high.

I think you'll have better luck going to either Classical Physics or go to ODE/PDE sub-forum here where people there don't worry about producing tangible result and spend all the effort talking about definition and theory!

 

[frenzal_dude]

how can I move this thread to that forum?

 

[yungman]

how can I move this thread to that forum?

Re-post and bag for forgiveness when you get a warning for posting the same post in two different area

OR

Change the wording and pray the moderator do not recognize it!

BTW, do you follow about the broad band frequency components in FT so even if individual frequency component is only 1/2 height, but if summing infinite frequencies together, it is possible to get an infinite amplitude at one point.