>how to get at u(k)? Can I run time backwards in some way so the H is
>stable in reverse time?
Flipping an impulse response in time will not change its absolute sum,
therefore stability is independent of the "direction" of your time.
Emre
Reply by glen herrmannsfeldt●September 8, 20092009-09-08
HardySpicer <gyansorova@gmail.com> wrote:
< If I have a random signal u(k) and a (known) transfer function H then
< y(k)=Hu(k) and if I know H and H is minimum phase then I ca neasily
< find u(k).
When the subject is Deconvolution, I always recommend Jansson's
"Deconvolution of Images and Spectra."
I had the first edition from the library some years ago. When
I wanted to buy my own, I couldn't find one. It turned out that
the second edition was about to come out, so I waited and bought that.
You might also find one in a nearby engineering library.
-- glen
Reply by Vladimir Vassilevsky●September 7, 20092009-09-07
HardySpicer wrote:
> If I have a random signal u(k) and a (known) transfer function H then
>
> y(k)=Hu(k) and if I know H and H is minimum phase then I ca neasily
> find u(k).
>
> Suppose H is nonminimum phase eg
>
>
> y(k)=u(k)-2u(k-1)
>
> how to get at u(k)? Can I run time backwards in some way so the H is
> stable in reverse time?
What is the goal of this exercise?
Vladimir Vassilevsky
DSP and Mixed Signal Design Consultant
http://www.abvolt.com
Reply by illywhacker●September 7, 20092009-09-07
On 7 Sep, 01:56, HardySpicer <gyansor...@gmail.com> wrote:
> If I have a random signal u(k) and a (known) transfer function H then
>
> y(k)=Hu(k) and if I know H and H is minimum phase then I ca neasily
> find u(k).
>
> Suppose H is nonminimum phase eg
>
> y(k)=u(k)-2u(k-1)
>
> how to get at u(k)? Can I run time backwards in some way so the H is
> stable in reverse time?
In general you need to assign a prior probability to u. If the
operator H is invertible and the noise is small, this will probably
not make too much difference, but for large SNR and non-invertible
operators it is essential.
illywhacker;
Reply by Rune Allnor●September 7, 20092009-09-07
On 7 Sep, 01:56, HardySpicer <gyansor...@gmail.com> wrote:
> If I have a random signal u(k) and a (known) transfer function H then
>
> y(k)=Hu(k) and if I know H and H is minimum phase then I ca neasily
> find u(k).
>
> Suppose H is nonminimum phase eg
>
> y(k)=u(k)-2u(k-1)
>
> how to get at u(k)? Can I run time backwards in some way so the H is
> stable in reverse time?
Deconvolution is among the hardest quantitative problems
in DSP. A minimum phase transfer function H(z) can easily
be inverted in frequency domain since 1/H(z) will be causal
and stable. A non-minimum phase H(z) is in the worst case
impossible to invert: Zeros on the unit circle can not be
'undone'. As for mixed-phase filters, it's anybody's guess
how to first of all - in the case of MA models estimated
from autocovariance sequences - identify what roots are
inside and outside the unit circle, and then merge the causal
and anticausal signals to form a useful deconvolved signal.
And of course, most of the above are problems one encounter
even when the system function H(z) is known. In the case of
unknown H(z) you also get all kinds of model mismatch problems,
and so on and so forth.
Rune
Reply by Tim Wescott●September 6, 20092009-09-06
On Sun, 06 Sep 2009 16:56:29 -0700, HardySpicer wrote:
> If I have a random signal u(k) and a (known) transfer function H then
>
> y(k)=Hu(k) and if I know H and H is minimum phase then I ca neasily find
> u(k).
>
> Suppose H is nonminimum phase eg
>
>
> y(k)=u(k)-2u(k-1)
>
> how to get at u(k)? Can I run time backwards in some way so the H is
> stable in reverse time?
>
> hardy
Yes, but.
For the case you present the transfer function will be stable in 'reverse
time', and in general a transfer function with zeros that are strictly
inside and outside of the stability regions can be separated into parts
that are stable in forward and reverse time, respectively.
But consider the case of a transfer function with a zero or more right on
the stability boundary, i.e.
y(k) = u(k) + u(k-1),
or y(k) = u(k) - 2 b u(k-1) + u(k-2), with |b| <= 1,
or y(k) = u(k) + u(k - n), with n > 0.
In all of these cases you will have to assume a starting state for u(k),
and if you add any noise at all then the inverse transfer function's
response to the noise will be infinite.
Even if your forward transfer function just has zeros that are relatively
_close_ to the stability boundary then you're in the same pickle, as your
inverse transfer function may _theoretically_ have a finite noise
response, but in practice it may well be too big to be worthwhile.
--
www.wescottdesign.com
Reply by HardySpicer●September 6, 20092009-09-06
If I have a random signal u(k) and a (known) transfer function H then
y(k)=Hu(k) and if I know H and H is minimum phase then I ca neasily
find u(k).
Suppose H is nonminimum phase eg
y(k)=u(k)-2u(k-1)
how to get at u(k)? Can I run time backwards in some way so the H is
stable in reverse time?
hardy