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Re: test this in matlab for my please

From: Doug Stewart
Subject: Re: test this in matlab for my please
Date: Fri, 15 Jul 2016 06:50:21 -0400

On Thu, Jul 14, 2016 at 2:56 PM, Ozzy Lash <address@hidden> wrote:

Now Matlabe gives

ans =
  0.1 z
  -----          This is wrong!!!
  z - 1

which has the zero at the origin but has a gain of 1/10th of what it should!
Matlab the does a correction factor of 1/T when they plot it so that it comes out correct,
but this does not help anyone who just wants the formula not the plot.

I think Matlab uses this convention for the impulse response because they are using the dirac delta function in the continous domain and the kronecker delta in the discrete time domain.  The dirac delta has area 1 under the curve, while the kronecker delta has an area of 1/fs under the curve (assuming a zero order hold, and defining the kronecker delta to have value 1 at index 0). 

I think I have seen books teach this either the way Matlab is doing it, or by assuming a fs scale factor on the delta function that they use.  I think in either case, once you get past the early chapters, you just forget about any scale factor.


Thanks for the clarification.
I can live with the 1/Fs factor.
But I have one question.

when i do   [ 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0...] in octave, did
i make a kronecker delta or a dirac delta ?

The bigger problem is the fact that the numerator and denominator are wrong.

For the last example I did all the math by hand on paper, and found that Matlab 
agreed with my work and not with impinvar().


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