This screencast will discuss the approximation of
higher order transfer functions as FOPDTs or First Order Plus Dead Time transfer functions.
We'll first discuss about the reasons why we want to do this,
discuss one method by which we can do this approximation
and then carry out an example.
So first, the question that might come about is
I have the transfer function on the left which is a sixth order transfer function
why would I want to do this in the first place?
So there are two major reasons to approximate.
The first is mathematical in nature, is the fact the math is easier
So therefore trying to do analysis for the FOPDTs it is much easier to do than trying to deal with a sixth order transfer function.
The second reason is the presence of FOPDTs in tuning.
In a number of situations in order to develop the appropriate tuning parameters
whether it be the controller gain K sub c
the integral time tau sub i
or the derivative time tau sub d
Often times in order to use these tuning parameter methodologies
it requires the development of a FOPDT
and if your process is higher order, this can become problematic.
So there are a number of methods by which higher order transfer functions can be approximated as FOPDTs.
Here we will discuss Skogestad's method. Skogestad is a professor in Norway.
So in order to utilize the Skogestad method, there are a few of things that have to be done.
The first thing that has to be done is the denominator should be in terms of the time constants.
In other words, you have to find the respective time constants you have.
So we want this to become tau one s plus one
times tau two s plus one, etc.
In order to use the Skogestad method,
A couple other things have to be true.
The first is the fact that a dominant time constant must exist.
The dominance of the time constant is the fact that the largest value of tau
must be at least one and a half times greater than the next largest time constant.
How well the Skogestad approximation fits hinges very much on this,
so therefore if the ratio between the largest and the next largest is about one and a half the fit will
work but will not be as good in comparison to a system where the ratio will be five or six or higher.
The third thing to keep into account for the Skogestad method is the fact that the system should be overdamped
In other words this means the fact that there is no oscillations so no imaginary roots
and its a stable system, so in other words there is no positive root in the denominator or all the taus are positive
Why do we have this requirement for an overdamped system?
Well the reason comes from thinking about how a FOPDT process behaves,
So here we do a quick sketch of a graph of how a FOPDT is affected by a step function
So recall the fact that it looks something along these lines
where we define our x axis time and our y axis as our value of interest y.
Because the response for a FOPDT is overdamped it must mean in order for a higher order transfer function
to be approximated by a FOPDT, it should also be overdamped.
Fortunately, from a chemical process standpoint,
this is generally not a problem for us to concern ourselves with so this is generally OK.
Why this function works so well is the fact that as we have a whole bunch of time constants in the denominator
that tends to slow down the process because remember the fact that the denominator indicated dynamics that slow the response down.
So we end up generally as we get to higher order functions
developing at the very beginning something that almost resembles dead time
and then develops some sort of inflection point before reaching the steady state.
So an important point to note about any FOPDT approximation
whether it Skogestad or otherwise
is the fact that the approximation is weakest at short times .
but fortunately, since we are often interested in understanding how the dynamics work from beginning to end
This is generally not too much of a problem, but
if there is a requirement where we must have significance at the very beginning
you would want to make an approximation in the first place.
Now that we have looked at the requirements in order to use the Skogestad method
lets actually describe the Skogestad method in terms of how it is able to approximate higher order transfer functions.
So the way it does it is remember the fact that what we are trying to do is get a FOPDT
So FOPDT is Ke to the negative theta s divided by tau s plus one, where K is the steady state gain
the theta is the dead time, and the tau is the time constant.
So here the Skogestad method needs to find three parameters
K, theta, and tau
We will start with gain. Gain is the simplest because the gain in the Skogestad
is equal to the gain in the higher order
an important point here is making sure the fact that you put the function is in standard form.
And if your denominator has everything in tau s plus tau one s plus one times tau two s plus one times tau three s plus one
it is in standard form
So therefore the gain is the gain.
For the time constant for the FOPDT from the Skogestad
the methodology to find the time constant is you take the largest time constant from your higher order transfer function
and add half of the time constant of the next largest
This is why Skogestad's method is often referred to as Skogestad's half rule.
And the dead time is basically incorporated of everything else
so therefore it represents the other half of the next largest time constant
plus the summation of all other taus
An important point to note here is the fact that any
taus in the denominator would be positive but if by chance if there is a tau in the numerator
and it can be approximated reasonably by a FOPDT that would be a negative.
Finally, we would add in any dead time that is already present in the higher order transfer function.
So the Skogestad is not an exceptionally difficult method to allow us to approximate
a higher order transfer function as a FOPDT.
So what we are now going to do is use these formulas
in order to find a FOPDT for a fifth order transfer function
What we are going to do here is find the FOPDT for the transfer function that we have here.
Well first we have to check the criteria for use of Skogestad's method,
so lets check our criteria.
The first is that the function is overdamped,
That checks out because we have taus that are all positive
Secondly, we want to make sure of the fact that tau of the largest time constant is greater than
or equal to one and a half times the next largest.
In this case our values here are eight over three,
something that is greater than one and a half
so those check, therefore this Skogestad method can be used.
So, we will start with evaluating all variables of interest in a FOPDT
We will start with the gain. The gain is the same gain as our higher order function,
so therefore that is just six.
For the time constant tau that is the largest time constant
plus the half of the next largest time constant from our higher order transfer function
So here that would be eight plus one half times three
or nine and a half.
Finally, theta will represent half of the next largest tau, plus the sum of all other taus, plus any dead time that was present in the initial process.
So here we will take half of three because that is half of the next largest tau.
All of the remaining taus we have, go back to our equation, we have nothing in the numerator
So therefore this will just be two
And we have s plus one squared so have two ones there
two plus on plus one
plus the value of the dead time which is already there, which is two.
When we add these all together we get seven and a half.
So therefore what this means is, is that if our transfer function for that fifth order function
is equal to six e to the negative seven point five s divided by nine point five s plus one
So you may be asking yourself well, OK, you made this approximation, but
how good of an approximation is it?
So to look at this, both transfer functions were put in Simulink to simulate how they would
respond to a step function of magnitude one.
Simulink was not the only program you could use, there are a whole host of other programs,
here we decided to use Simulink.
So when we look at the graph from Simulink what we see is the fact that the
difference between the fifth order transfer function and the FOPDT are relatively small
indicating that the approximation is quite valid.
As noted before, the biggest difference between the two will happen at low times
mainly caused due to the difference in the dead times for the respective processes
recall that the fifth order process had a dead time of two
where as the FOPDT had a dead time of seven and a half
So primarily the reason why you see such a gap at earlier times but note at long times
that the processes both converge to the steady state gain here which is six.
So through this screencast hopefully we are able to understand why approximations of higher order
transfer functions are necessary, and to be able to use a method to make this approximation to develop a FOPDT.