Difference between revisions of "Rep and Comb functions : Lecture on 26/08/09" - Rhea

Revision as of 09:33, 23 September 2009

Rep Function:

A rep function periodically repeats another function with some specified period T ( basically sampling time). Mathematically a rep operator is a function x(t) convoluted with a summation of shifted deltas:

$ rep_T [x(t)] $ = $ x(t)* P_T (t) $

                 = $ x(t)* \sum_{k=-\infty}^{\infty}\delta(t-kT)  $

                 = $ \sum_{k=-\infty}^{\infty}x(t) * \delta(t-kT)  $

                 = $ \sum_{k=-\infty}^{\infty}x(t-kT)  $

If below is the x(t):

Then the $ rep_T [x(t)] $ looks like:

The resulting function is periodic with period T. To keep x(t) repeating without aliasing, the minimum sampling period should be $ \geqq 2 * T $

Comb Function

A comb function multiplies a function with a periodic train of impulses.

   $  comb_T  ${x(t)} = $  x(t) . \sum_{k=-\infty}^{\infty}\delta(t-kT)  $ = $  x(t) . P_T (t)  $

Using the same x(t) as above as our function to be multiplied with a train of impulses, we get:

Note: The Fourier Transform of a $ rep_T $ x(t) gives the comb function of x(t) but in the frequency domain.

$ \mathfrak{F} (rep_T [x(t)]) = \mathfrak{F} ( x(t) * P_T (t) ) = \mathrm{X}(f).\mathrm{P_T}(f) $

Where

$ \mathrm{P_T}(f) = \mathfrak{F} ( \sum_{k} \delta(t-kT)) $

       $  {{=}}  \mathfrak{F} (\sum_{n=-\infty}^{\infty}  \frac{1}{T} exp \frac{j2nt\pi}{T}  )  $
       
       $  {{=}}  \sum_{n=-\infty}^{\infty}  \frac{1}{T} \delta( f - \frac{n}{T} )  $
       
       $  {{=}}  \frac{1}{T} P_T (f)  $

Therefore:

$ \mathrm{X}(f).\frac{1}{T}P_T (f) {{=}} \frac{1}{T}comb_\frac{1}{T} [ \mathrm{X}(f)] $

To conclude the transform relationship between the comb operator and rep operator surfaces when we take the operator's Continuous Time Fourier Transform :

$ comb_T [x(t)] \iff \frac{1}{T}rep_\frac{1}{T} [ \mathrm{X}(f)] $

$ rep_T [x(t)] \iff \frac{1}{T}comb_\frac{1}{T} [ \mathrm{X}(f)] $