Re: Inverse of Non Minimum Phase System

On Thu, 14 Jun 2007 09:15:02 +0000, mobi wrote:
(top posting fixed)
On Jun 13, 1:48 am, Tim Wescott <t...@xxxxxxxxxxxxxxxx> wrote:
gokul_s1 wrote:
Suppose one takes a simple non minimum phase system with the z transform

H(z) = 1 - 2*z. This system can obviously not be inverted as the pole will
lie outside the unit circle (inversion will lead to unstable impulse
response).

However if one tries to invert this system using Y(w) = 1/H(w) and then
take the IFFT I do end up with a time domain impulse response y(t) (y(t) =
IFFT(Y(w))) which is bounded.

I am wondering where is the discrepancy? Would appreciate any insight into
this problem

Do you mean H(z) = 1 - 2/z? H(z) = 1 - 2*z describes a non-causal
system, but one with a stable zero.

I'll assume you mean the former, and that 1/H(z) = z / (z - 2).

The Fourier transform* is agnostic to causality, and a pole at |z| > 1
is only unstable if it is causal -- if you take the poles that lie
outside the unit circle as running backward in time then they are
perfectly stable. The only poles that really give the Fourier transform
trouble are the ones that lie right _on_ the stability boundary, and
even their effects can be side stepped.

So when you find the inverse Fourier transform of your "unstable" system
the math takes it as a stable, non-causal system, i.e. h(k) =
u(-k+1)2^(-k), where u(k) = 0 for all k < 0 and 1 for all k >= 0.

* Note that this is also the case with the continuous-time Fourier
transform -- causality means nothing, but every pole with a real part is
stable.

Can any one clarify a confusion i have,
In this particular example, if I use a pole at 0.5 (inside the unit
circle), shouldnt i get the desired result? I mean I will get a valid
inverse! This inverse is stable also.

To invert the system it must have a transfer function with the following
properties:

* a numerator with the same order as the denominator
* the numerator roots (i.e. the transfer function zeros) must be stable

Then you can invert the system. In your case if you have some stable
pole and a stable _zero_ (say at z = 0.5) then you could invert the
system.

--
Tim Wescott
Control systems and communications consulting
http://www.wescottdesign.com

Need to learn how to apply control theory in your embedded system?
"Applied Control Theory for Embedded Systems" by Tim Wescott
Elsevier/Newnes, http://www.wescottdesign.com/actfes/actfes.html
.

Relevant Pages

  • Re: Inverse of Non Minimum Phase System
    ... take the IFFT I do end up with a time domain impulse response y= ... The Fourier transform* is agnostic to causality, and a pole at |z|> 1 is only unstable if it is causal -- if you take the poles that lie outside the unit circle as running backward in time then they are perfectly stable. ... The only poles that really give the Fourier transform trouble are the ones that lie right _on_ the stability boundary, and even their effects can be side stepped. ... Note that this is also the case with the continuous-time Fourier transform -- causality means nothing, but every pole with a real part is stable. ...
    (comp.dsp)
  • Fourier transform of a signal having a pole on the unit circle
    ... How can we claculate the fourier transform of a signal which has a pole ... exist if there exists a pole on the unit circle, ...
    (comp.dsp)
  • Re: Inverse Fourier Transform by Residue Theorem and Contour Integration
    ... inverse Fourier transform of the function ... My current strategy is to integrate about a closed contour around the pole ... This would then equal the integral along the real axis between -R ... and so need to close in the lower half plane in order to make ...
    (sci.math)
  • Re: Inverse of Non Minimum Phase System
    ... if I use a pole at 0.5 (inside the unit ... lie outside the unit circle (inversion will lead to unstable impulse ... The only poles that really give the Fourier transform ... transform -- causality means nothing, but every pole with a real part is ...
    (comp.dsp)
  • Re: Fourier transform
    ... its Fourier coefficients do not lie in l^p for any p < 2. ... such that the Fourier transform f^ does ... "Understanding Godel isn't about following his formal proof. ...
    (sci.math)