Re: inverse laplace transform

"Tim Wescott" <tim@xxxxxxxxxxxxxxxx> wrote in message
news:wYGdnbPA6obFKjTe4p2dnA@xxxxxxxxxxxxxxx
> Bhaskar Thiagarajan wrote:
> > Hi all
> >
> > I'm working on trying to model a non-linear system (described by a
second
> > order differential eqn) into a discrete IIR filter.
>
> Whoa! Stop right there!
>
> The Laplace transform (and the z transform) only work with linear
> systems. You simply cannot do a Laplace transform of a nonlinear
> system: it doesn't work. You can dink with Volterra series and all that
> fun stuff, but then you're not really doing a Laplace any more.

Hmm...I think I mis-spoke. My instinct tells me that a second order
differential equation is non-linear. However, I just looked up the Rick
Lyons' book and I see that he calls it a linear system. I'm just going to
take his word for it for now and revisit it later.

> Now, you _can_ get past the Linearity Nazi by claiming that you're
> _approximating_ your nonlinear system with a linear one, and taking the
> Laplace transform of that. I'll allow you through, but only if you tell
> me you've checked to make sure that your approximation is accurate
> enough for your expected inputs.
>
> > I guess I'm trying to
> > follow the impulse invariance IIR filter design method.
>
> (sniping about nonexistance of impulse response of nonlinear systems
> ommited (sorta))
> >
> > In the laplace domain, the transfer function looks like this
> > H(s) = a^2/(s+a)^2
> >
> > I tried looking up laplace transform tables to find the time domain form
for
> > this and the closest pair I could find was
> > H(s) = 1/(s+a)^2 -> h(t) = t*e^(-at)
>
> That's correct.
> >
> > and the corresponding Z transform for h(t) as
> > H(z) = T*z*e^(-aT) / (z - e^(-aT))^2
>
> Looks good from here.
> >
> > My question is, can I assume my extra term a^2 in my numerator will
transfer
> > to the z domain as a constant multiplier in the numerator?
>
> Yes. Both the z transform and the Laplace transform are linear
> operations, so superposition holds.
>
> > I haven't done z
> > or laplace transforms since school...so I'd appreciate some help with
this
> > math.
> >
> The impulse-invariance method is not usually a good way to do this,
> because you have this funny problem with scaling with the sampling rate.
> Impulse-invariance makes sense in the high-falutin' world of
> mathematics, but it only translates if you treat the discrete-time stuff
> as a true chain of impulses that get run through a reconstruction
> filter, etc. The first reconstruction filter that your real-world
> signal passes through is a DAC, which acts like a zero-order hold, which
> (if you treat your DAC inputs as impulses) has a DC gain that varies
> with sampling rate.
>
> For control systems design there is an exact transformation from the
> point of view of the software. It goes from the input to the DAC to the
> output of the ADC, and it recognizes that the "impulse response" of the
> DAC is a unit pulse that is as wide as the sampling rate. To calculate
> it you find the step response of your system:
>
> a^2 1 1 a
> H_s(s) = ---------- = - - ----- - --------- ,
> s(s + a)^2 a s + a (s + a)^2
>
> convert it to the time domain
>
> h_s(t) = 1 - e^-at - a t e^-at ,
>
> find the z transform of that at a sample interval of T_s:
>
> z (1 - d - d * a*T_s)z - (1 - d - a*T_s)d
> H_s(z) = ----- ---------------------------------------
> z - 1 (z - d)^2
>
> where d = e^-aT_s.
>
> Finally, multiply H_s(z) with the inverse of the z transform of the step
> function to get the z domain transfer function:
>
> (1 - d - d * a*T_s)z - (1 - d - a*T_s)d
> H_s(z) = --------------------------------------- .
> (z - d)^2
>
> Now that was simple and direct, wasn't it?


Yes - very nice. Thanks for helping out Tim. That last equation should've
been H(z) and not H_s(z) and I *think* the first equation should've had the
first term on the right as 1/s and not 1/a

I applied this technique to model my system and so far it mostly makes sense
(whereas using the earlier scheme, my step response of the filter was
growing exponentially).
However, I can't seem to match 'a' (in my case this is the inverse of the
critically damped time constant of my system) to it's definition, which is,
the step response should reach 0.35 the max value in this duration. After
creating my filter, examining the step response shows this time constant to
be larger than what I used to get the filter.

I couldn't find anything in the z transform tables I have to match what you
did to get from h_s(t) to H_s(Z)...but I'll take your word for it.


> Tim Wescott
> Wescott Design Services
> http://www.wescottdesign.com


.

Relevant Pages

  • Re: Why exp(-st) in the Laplace Transform?
    ... If you take the Laplace transform, ... equivalent to multiplying it by a real number ). ... distinguishes the Laplace Transform from the Fourier Transform. ...
    (sci.physics.research)
  • Re: Why exp(-st) in the Laplace Transform?
    ... If you take the Laplace transform, ... equivalent to multiplying it by a real number ). ... distinguishes the Laplace Transform from the Fourier Transform. ...
    (sci.math.research)
  • Re: minimum phase system - poles
    ... and impulse response is the convolution kernel. ... Is it the Fourier transform of the ... I did not see the laplace transform in math major courses except ...
    (comp.dsp)
  • Re: Why exp(-st) in the Laplace Transform?
    ... The existence of both the Laplace and Fourier transforms ... come from the symmetry of the real line under translations (t -> t + a, ... transform kernels, exp, for w any real number. ... If you take the Laplace transform, ...
    (sci.physics.research)
  • Re: Why exp(-st) in the Laplace Transform?
    ... The existence of both the Laplace and Fourier transforms ... come from the symmetry of the real line under translations (t -> t + a, ... transform kernels, exp, for w any real number. ... If you take the Laplace transform, ...
    (sci.math.research)