# inverse laplace transform

Started by December 21, 2005
```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. I guess I'm trying to
follow the impulse invariance IIR filter design method.

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)

and the corresponding Z transform for h(t) as
H(z) = T*z*e^(-aT) / (z - e^(-aT))^2

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? I haven't done z
or laplace transforms since school...so I'd appreciate some help with this
math.

Cheers

```
```Bhaskar Thiagarajan wrote:
> I'm working on trying to model a non-linear system (described by a second
> order differential eqn) into a discrete IIR filter. I guess I'm trying to
> follow the impulse invariance IIR filter design method.
>
> 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)
>
> and the corresponding Z transform for h(t) as
> H(z) = T*z*e^(-aT) / (z - e^(-aT))^2
>
> 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 Laplace and Z transform are linear.
```
```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.

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

> 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?

Unfortunately this is only exact for things that are sitting inside the
sampled-time domain -- for things outside the sampled time domain its
just an approximation.  If you're trying to make a discrete-time thingie
replicate the behavior of a continuous-time thingie then you're probably
better off using something like the Tustin approximation and tweaking
things after the fact to get the response right.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
```
```"Tim Wescott" <tim@seemywebsite.com> wrote in message
news:wYGdnbPA6obFKjTe4p2dnA@web-ster.com...
> > 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

```
```"Bhaskar Thiagarajan" <bhaskart@deja.com> wrote in message
news:43a9f98a\$0\$15781\$14726298@news.sunsite.dk...
> "Tim Wescott" <tim@seemywebsite.com> wrote in message
> news:wYGdnbPA6obFKjTe4p2dnA@web-ster.com...

> 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.

Never mind this Tim...just needed to work through the algebra (twice).
friend who could do that and I wasn't sure if I should admire him or hate
him :-)

Cheers

```
```Bhaskar Thiagarajan wrote:

> "Tim Wescott" <tim@seemywebsite.com> wrote in message
> news:wYGdnbPA6obFKjTe4p2dnA@web-ster.com...
>
>>
>>>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
>>
>>
>>>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

Yes, those were typos and you're right.
>
> 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.
>
Good -- 'cause I took MathCad's word for it.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
```
```Bhaskar Thiagarajan wrote:

> news:43a9f98a\$0\$15781\$14726298@news.sunsite.dk...
>
>>"Tim Wescott" <tim@seemywebsite.com> wrote in message
>>news:wYGdnbPA6obFKjTe4p2dnA@web-ster.com...
>>
>
>
>
>>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.
>
>
> Never mind this Tim...just needed to work through the algebra (twice).
> friend who could do that and I wasn't sure if I should admire him or hate
> him :-)
>
I long ago stopped doing this kind of stuff in my head or anywhere else
related to my brain.  I use MathCad, because I have to try the algebra
_more_ than twice to be sure.

In this case I got the H_s(z) in a form that I could work with, adjusted
my math until MathCan indicated it was equal to the mess it came up
with, and called it good.

and it's not as user friendly.  MatLab will do it for \$3K, plus another
\$3K for the control systems toolbox, and also not symbolically.  For
some reason I use MathCad and SciLab...

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
```
```Bhaskar Thiagarajan wrote:

> "Tim Wescott" <tim@seemywebsite.com> wrote in message
> news:wYGdnbPA6obFKjTe4p2dnA@web-ster.com...
>
>>
>>>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.
>
You can have an any-order (up to infinite) differential equation that is
linear.  The order of the diff. eq and it's linearity are orthogonal
properties.

You may want to do a bit of digging and remind yourself of the
properties of linearity and time-invariance (they are two different
properties).  The Laplace transform only works on systems that are
linear and time invariant.  The z transform only works on systems that
are linear and who's time variance is cyclical at the reference rate.

If you're clever and determined you can use Fourier analysis on some
systems that are time varying, like systems with sampling and

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
```
```"Tim Wescott" <tim@seemywebsite.com> wrote in message
news:wYGdnbPA6obFKjTe4p2dnA@web-ster.com...
> > 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.
>

You can get special dispensation from the Pope.

HH

```