Maths: the Laplace Transform of the Dirac's comb?

The Laplace Transform of the Dirac's comb
    +inf             +inf
    Sum d(t - n*T)    is    2*Pi/T Sum d(s/j - 2*Pi/T)     -inf             -inf
I unsuccessfully tried to find the steps between the function and its transform via two paths.
First, since the Dirac's comb is an infinite sequence of Dirac's impulses, some advanced before t=0, and others delayed after t=0, the delay theorem comes to the mind. Moreover, the Laplace Transform of the Dirac's impulse being 1, it could not be simpler!
    ... + exp(-2*s*T) + exp(-2*s*T) + exp(0*s*T) + exp(1*s*T) + exp(2*s*T) + ...
Oops! This series does not converge in the normal sense. It should converge in the sense of distributions, but I do not know how to handle it. :-(
Second, I start from the definition of the Laplace Transform.
    +inf    +inf     Integral Sum d(t - n*T)    exp(-s*t) dt     -inf    -inf
I change the order of the sum and the integral.
    +inf    +inf     Sum     Integral d(t - n*T) exp(-s*t) dt     -inf    -inf
Since the Dirac's impulse differs from 0 only when t=n*T and has an «area» of 1, I can drop the integration.     +inf     Sum exp(-s*t) dt     -inf
Alas! I fall in the same dead-end as I fell above!
Could you help me find the way?
Thank you very much in advance.
--
Jean Castonguay
Électrocommande Pascal
  Click to see the full signature.
Add pictures here
✖
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
Jean Castonguay wrote:

-- snip --

Whenever I get wound up in the infinities of the dirac delta function I start thinking of taking the limit of a pulse of unit area as it's width goes to zero. Usually I use a rectangular pulse.
For the following proof you'll have to accept that the Fourier transform of a repeating waveform is a set of dirac deltas weighted by the Fourier series for the waveform. You can prove this using the similar techniques.
I'm going to prove this for the Fourier transform, and leave it to you to clean it up and extend it to the Laplace domain. I come out different by a factor of 2pi, I suspect because we define the transforms differently.
So say that h(t, a) defines a pulse such that
{ 0 t < -a/2 h(t, a) = { 1/a -a/2 <= t <= a/2 . { 0 t > a/2
Note that h(t, a) has unity area.
Now consider a signal h_r(t, a) which is a repeating train of the pulses defined above:
+inf h_r(t, a) = Sum d(t - n*T) -inf
This has a Fourier series
H_r(n, a) = (T/(2 pi n a))(sin (2 pi n a/T)),
so the Fourier transform of h_r(t, a) is
+inf H_r(w, a) = T / (2 pi a) sum (sin(2 pi n a/T)/n) d(w - 2 pi n/T). n = -inf
_Now_, take the limit of H_r(w, a) as a -> 0, and you'll get the correct expression.
--

Tim Wescott
Wescott Design Services
  Click to see the full signature.
Add pictures here
✖
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
On Sun, 31 Oct 2004 21:08:22 UTC, Tim Wescott
-- snip --

-- snip --
Thank you, Mr. Wescott.
The two words «repeating waveform» woke me up! Of course, the Dirac's comb is a sequence of delta functions which fire (pardon me for the violent word ;-) ) every nT unit of time where T is the period. Being a periodic function, it is legitimate to calculate its Fourier coefficients. Let us assume that our Dirac's comb is made up of unit impulses and let us calculate its Fourier coefficients over the interval [0..T[. In this interval, the delta function fires only once at exactly t = 0.
          T     a[n] = (1/T) Integral[ d(t-n/T) exp(-j*n*2*pi*t/T) dt              0
At t = 0, exp(-j*n*2*pi*t/T evaluates to 1 for all values of n, the delta function integrates to 1, so a[n] = 1/T for all n!
So I am allowed to write: +inf +inf Sum d(t - n*T) = 1/T Sum exp(j*n*2*pi*t/T) -inf -inf
Amazing, is it not? The Dirac's comb is an infinite sum of phasors whose angular speed (in radians) is n*2*pi/T and modulus 1/T.
The plot of the Fourier coefficients versus frequency is repetitive to say the least: a dot of ordinate 1/T at every abscissa n/T. From this plot, we can go intuitively to the Fourier transform. No doubt, this will not satisfy the mathematicians in this group; so let us call them to fill the void. The latter being the «density» of the frequency component in the signal, we can say that there is an infinite density of the frequency component at every frequency n/T. Let us use the delta function to express it.
     +inf     X(f) = 1/T Sum d(f - n/T)      -inf
Most of the time, I prefer to express the spectrum with respect to the natural frequency (2*pi*f - is this the correct word for French word «pulsation»?). Making two substitutions in the above expression does it: replace f by w, and replace T by T/2*pi.
     +inf     X(w) = 2*pi/T Sum d(w - n*2*pi/T)      -inf
P.S.: In French, we use the word «transformation» to describe the Transform mathematical process and the word «transformée» to describe the result. Is there such a distinction in English? The Dirac impulse and the complex exponential are so useful that I am surprised that their «Intellectual Property» is not yet protected by a rain of USPO patents.
--
Jean Castonguay
Électrocommande Pascal
  Click to see the full signature.
Add pictures here
✖
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

Polytechforum.com is a website by engineers for engineers. It is not affiliated with any of manufacturers or vendors discussed here. All logos and trade names are the property of their respective owners.