inverse laplace transform

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inverse laplace transform

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Bhaskar Thiagarajan
Guest

Posted: Thu Dec 22, 2005 12:40 am Post subject: inverse laplace transform

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) = te^(-at) and the corresponding Z transform for h(t) as H(z) = Tz*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

Bhaskar Thiagarajan
Guest

Posted: Thu Dec 22, 2005 1:17 am Post subject: Re: inverse laplace transform

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

Quote:

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.

Quote:

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.

Quote:

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

Andreas Schwarz
Guest

Posted: Thu Dec 22, 2005 1:17 am Post subject: Re: inverse laplace transform

Bhaskar Thiagarajan wrote:

Quote:

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.

Tim Wescott
Guest

Posted: Thu Dec 22, 2005 1:17 am Post subject: Re: inverse laplace transform

Bhaskar Thiagarajan wrote:

Quote:

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
enough for your expected inputs.

Quote:

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))

Quote:

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.

Quote:

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

Looks good from here.

Quote:

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.

Quote:

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

Bhaskar Thiagarajan
Guest

Posted: Thu Dec 22, 2005 7:26 am Post subject: Re: inverse laplace transform

"Bhaskar Thiagarajan" <bhaskart@deja.com> wrote in message
news:43a9f98a$0$15781$14726298@news.sunsite.dk...

Quote:

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

Quote:

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).
Hopefully you didn't jump from one line to the other in your head...I had a
friend who could do that and I wasn't sure if I should admire him or hate
him :-)

Cheers
Bhaskar

Tim Wescott
Guest

Posted: Thu Dec 22, 2005 7:31 am Post subject: Re: inverse laplace transform

Bhaskar Thiagarajan wrote:

Quote:

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

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

Yes, those were typos and you're right.

Quote:

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

Tim Wescott
Guest

Posted: Thu Dec 22, 2005 7:36 am Post subject: Re: inverse laplace transform

Bhaskar Thiagarajan wrote:

Quote:

"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...

Bhaskar Thiagarajan wrote:

thread details snipped

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).
Hopefully you didn't jump from one line to the other in your head...I had a
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.

SciLab (www.scilab.org) will also do this for free, but only numerically
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

Tim Wescott
Guest

Posted: Thu Dec 22, 2005 7:42 am Post subject: Re: inverse laplace transform

Bhaskar Thiagarajan wrote:

Quote:

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
superheterodyne radio receivers.

--

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

Horatio Hornblower
Guest

Posted: Thu Dec 22, 2005 9:16 am Post subject: Re: inverse laplace transform

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

Quote:

You can get special dispensation from the Pope.

HH

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