In this screencast we will take advantage of the use of transfer functions in order
to determine a good start for tuning parameters for a process. So in this screencast we will
look at a system with given transfer functions for the valve, sensor/transmitter, and process,
and what we want to do here is develop tuning parameters for a PID controller. So provided
on the right hand side is a table of the Ziegler-Nichols tuning parameters. So for a PID controller
we would want to take advantage of this row. So just a few important things to remember
about Ziegler Nichols, one is that the two variables of interest here are K(cu), which
is the ultimate gain, and P sub u, which is the ultimate period. Experimentally the ultimate
gain and ultimate period can be found using the continuous cycling method, and in the
continuous cycling method you are running your controller as only a P only controller
and your goal is to get a response that is undamped. However, if transfer functions are
known, if we can model the process, the valve, and the sensor/transmitter with good accuracy,
if the transfer functions are known we can find K(cu) and P(u), the ultimate gain, ultimate
period, via the method of direct substitution. So that is what we're going to take advantage
of in this situation. So recall that direct substitution involves an analysis of the characteristic
equation, which is one plus the open loop gain which equals zero. So with the characteristic
equation we'll have one plus G(v)G(m)G(p) and the controller. So we'll substitute those
transfer functions into our characteristic equation. The important point here for the controller
is that even though we're looking for a PID controller, we have to assume that when we're
doing this analysis that G(c) equals K(c). The reason why is because the continuous cycling
method, which is used for the ultimate gain and the ultimate period, runs on only a P
only controller, so therefore in order to complete that analysis we have to assume that
our controller is as such. So from here we can do a little bit of math. So the next step
in direct substitution is we let s equal i omega, and then we substitute that in for
all our values here for s in the equation. We then take advantage of the fact that the
imaginary i squared equals negative one. What we then do here is we break this into the
real and imaginary part for we have some terms which have i in it and we have some terms
that do not. The important point here is that since the right hand side equals zero the
real part of the equation and the imaginary part of the equation must equal zero. Our
real term has two variables in it, omega and K(c), but our imaginary term only has one
variable, omega, so will solve for that. If we move one of the terms over we'll end up
with ten i omega cubed equals eight i omega. One of the omegas drops out, its zero, its
a negligible answer for this, having a frequency of zero doesn't really help anything. The
i's will cancel so this will leave us with omega squared equals eight tenths, or that
omega is the square root of eight over ten, which is approximately point eight nine four.
So that gives us our omega value, this here represents the ultimate frequency, so we can
then find the ultimate gain using the second equation. So using the real equation we'll
have negative seventeen times eight over ten, which is omega squared plus one plus four
K(c) equals zero. So after some math we see that K(c), in this case this would be the
ultimate gain, equals three point one five. So we have found the ultimate gain, now in
order to find the ultimate period we take advantage of the relationship between the
ultimate gain and the ultimate frequency, which is two pi over omega, so this would
be two pi over about point eight nine four, which will be seven point zero three, and
this will have units of minutes because of the fact that our time constants had units
of minutes. So with our values of the ultimate gain and the ultimate period found we can
now use the table we showed at the begining order to find our parameters, so therefore the K(c) value
and how you would begin to control the system from the P standpoint represents the ultimate
gain divided by one point seven, so this will be three point one five divided by one point
seven, and what that division is done we have one point eight five. The units are generally
percent controller output over percent transmitter output, although that can depend based on
some of the units you had for your transfer functions. Tau(i) is the ultimate period divided by two,
so this will be seven point zero three divided by two, and that will be three point five
one five, or three point five two minutes, and the derivative time will be the ultimate
period divided by eight, or seven point zero three divided by eight which is point eight
seven nine minutes. So an important point to note here is that these parameters are
not the be all and end and perfect parameters for this process. These tuning
parameters serve as a starting point for our analysis in terms of where we want to potentially
being the tuning scheme. Another important here issue is that this is tuning parameters for
a Zielger-Nichols, which assumes that the desired response is a quarter decay, so therefore
if quarter decay is not exactly what you want you could potentially use a whole host of
other tuning parameters or correlations, taking advantage of various landmarks about some
combination of your actual process, your sensor/transmitter, your valve, and potentially your controller.
So in this screencast we showed how we can use transfer functions to us the Ziegler-Nichols
method in order to determine a good start for tuning parameters to control your process.