In this screencast we'll take a look at how the selection of your controller type in the
PID family can influence the final steady state value of your process due to a set point
change. So we've provided a block diagram for a generic feedback control system. The
first-order process being controlled by a particular PID family controller. The dynamics
of the process are slow enough so that the dynamics of the sensor/transmitter and the
valve can assumed to be negligible. It can be assumed that there are no disturbances present
in the process, hence why theres no disturbance value seen in our block diagram. So what we're
going to look at here is the offset that occurs if the set point of the controlled variable
is instantaneously increased by a magnitude M for, we'll look at a P only controller and
a PI controller. What we'll first do is evaluate what our blocks are, so here we're told that
the process is a first-order transfer function, so that we'll write it here as K(p) over tau
s plus one, and we're told that the valve and the sensor/transmitter are much faster than,
the dynamics are much faster than the process, which is another way of saying that the time
dynamics are negligible, so therefore we'll just have a gain for the valve and a gain
for the transmitter. For the controller we're given that we have a P only controller and
a PI controller, so for the P only controller we'll be looking at K(c), and for a PI controller
we'll be looking at K(c) times one plus one over tau(i) s, and these are the transfer
functions for a P only and a PI controller respectively. So the problem is asking us
to find the offset. What the offset represents is is the value of a set point of the variable
y minus the value of y as time goes to infinity when all the time dynamics have been settled
out. The goal here is to have an offset that is equal to zero, because in this case here
if I change my set point by three degrees Celsius I am going to want my output variable
changed by that exact same amount, because that is the value I want. If I have a set
point change of magnitude M that would again say, for example, if I am increasing the temperature
from eighty to eighty five degrees Celsius the change that I am seeing here would be
five degrees, or in this case the magnitude of the step change M. The important to note
here that we are doing this analysis based in the deviation variables for simplicity.
So in the Laplace domain our step input can be represented as M over s. So what we now
need to do is find out what y prime would be, and then once we can subtract these off
we can determine what our offset is. So what we now need to do is we need to find y as
a function of s in the Laplace domain because that can then lead us to finding y at time
infinity. So what this requires is we need to do some block diagram algebra to evaluate
y over y(sp). So if we carry out this block diagram algebra what we can see is that we can
have y as we go through the loop here we'll see that y multiplies by G(m), and goes through
it'll be G(c), G(v), G(p), and there will be a minus sign because of the minus that
we see here, then we also have plus y(sp) multiplied by K(m), G(c), G(v), G(p), and
that when we come to here will equal y. So we can rearrange this a bit. We can isolate
the y, and then finally just rearrange a few things to get y over y(sp) equals K(m), G(c),
G(v), G(p), divided by one plus G(m), G(c), G(v), G(p). So what we'll now do is we'll
substitute in the values which we know. So we'll substitute in K(m), the controller will
leave for now, G(v) will be K(v), G(p) will be K(p) over tau s plus one, and we'll have
something similar in the denominator. We can further isolate y by moving y(sp) over and
remembering that y(sp) is just the step change M over s. Now that we have our value of the
transfer function for the output variable what we can now do is begin to analyze for
the different control schemes. So first we'll start with a P only control scheme, and there
are a number of ways in which we can determine the steady state value of this, the method
that we'll use here is we'll take advantage of the final value theorem. So the final value
theorem states that the limit of a function in the time domain as it goes to infinity
is equal to the limit of that function in the s domain as s approaches zero multiplied
by s. So here, as for a P only controller, what that means is that G(c) will equal K(c),
that means that our function will be the limit as s approaches zero of s multiplied by our
function. What we'll notice here is that the s's cancel, and that both tau s plus ones
will go to one. So therefore this leaves us with the final value here as MK(m)K(c)K(v)K(p)
divided by one plus K(m)K(c)K(v)K(p). As you can see this value clearly does not equal
just M, so that means that we have an offset, so the offset here will be equal to M, what
we want the value to be, minus this value that we have here, which we can factor out
and have one minus K(m)K(c)K(v)K(p) divided by one plus K(m)K(c)K(v)K(p). We can do a
little bit of simplification here, and that leaves us with our final answer, which can
be written as M over one plus K(m)K(c)K(v)K(p), which is often represented as M over one plus
K sub OL, where K sub OL is the open loop gain because it represents the gain of the
parameters in the open loop. So here what we're showing is that P only control, although
it helps with the speed of a process, response, and is a little less susceptible to stability
issues than other control schemes, the major issue with P only controls is that you will
always have an offset in terms of where your process will operate, so it will not operate
at your new set point M, it will be off by M over one plus the open loops gain, and as
you can see here by looking inside this is as K(c), the controller gain goes up, the
offset goes down. There is also a trade off with that because as K(c) goes up stability
problems increase, so the chance that we'll have an issue with stability also increases,
so there is a give and take there. For part (b) we'll do the exact same analysis except
for a PI controller, and for a PI controller our transfer function here is going to be
K(c) times one plus one over tau(i) s. If we put that into our transfer function analysis
that we just did using the same approach as what we did for the P only we'll have that
the limit as t approaches infinity of our function will equal the limit as s approaches
zero of s times our function, M over s, multiplied by K(m), the controller transfer function,
we'll multiply this by K(v) and by our process function, and then our denominator will be
one plus the exact same values we had in the numerator. So before we do any analysis what
we're going to do is simplify our numerator and denominator by finding a least common
denominator. So to do that we're going to first rewrite whats in the parenthesis for
the i control as tau(i) s plus one over tau(i) s. Now we'll find the least common denominator
for the denominator, and now our denominator will have tau(i) s times tau s plus one, plus
K(m), K(c), K(v), K(p), times tau(i) s plus one divided by the tau(i) s times tau s plus
one. So what we'll now do is begin some cancellations, what we'll notice here is that our s's cancel
out in the front, the denominators of both the numerator and denominator cancel out.
What we also see is that the first term of the denominator goes to zero because of the
fact that tau(i) s will equal zero as s goes to zero, and similarly the tau(i) s terms
will go away for the same reason. So what this leaves us with after some algebra is
that M times K(m), K(c), K(v), K(p) over K(m), K(c), K(v), K(p), which in fact explicitly
cancels, so therefore our limit here is M. Recall that this is the value that we would
want to have if we have a change in our set point of magnitude M, we would want our output
variable to change by that exact same value, so therefore the PI control does what we would
want it to do as it gives us zero offset. So in this screencast we have shown how P
and PI control respond differently to a process when there is a change in the set point.