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Welcome to the lecture titled model-free controller design. In this lecture, we will discuss about
the tuning of PID controllers, without obtaining a transfer function model.
So, the parameters of the controller will be explicitly expressed in terms of the limit
cycle output parameters. Ziegler and Nichols in 1942, suggested tuning method based on
the concepts of the critical point on the process Nyquist curve, and since then many
more methods have been proposed and suggested to design controllers without finding the
transfer function models, one such method will be discussed in this lecture. And the
advantages of the method, we are going to discuss in this lecture estimates the critical
gain and critical frequency more accurately than the existing conventional relay auto
tuning methods, because the relay will not be an ideal relay alone, it will be connected
in series with some integral controller thus giving us certain advantages.
Again, the controller will remain in operation throughout the operation of the closed loop
control system. Auto tuning method gives a symmetrical and smooth limit cycle output
in the presence of static load disturbance and measurement noise, because of the integral
controller with along with the relay or the modified relay, because of the presence of
the modified relay, what we will have? We will have clean not only clean limit cycle
output, but also the limit cycle output not disturbed by any static load disturbance.
Now also the method does not require prior information about the process so, we need
not find the transfer function model rather the parameters of the controllers will be
functions of the transfer function model parameters or indirectly speaking directly using the
limit cycle parameters we shall find explicit expressions for the parameters of a PID controller.
So, no need of finding the transfer function model and no more process information is required
for tuning the parameters of a controller. Needs to design only two controller parameters
by this method we are going to discuss, we need to design only two controller parameters;
namely a proportional gain and a derivative time constant from the modified relay experiment.
So when a controller of the form G c (s) is equal to K p 1 plus 1 upon T i s times q plus
T d s is considered a series form of PID controller is considered, what I mean by the two controller
two controller parameters, we are going to find explicit expressions for K p and T d
of course, T i will be also found ultimately, but we need to design only two controller
parameters such as K p and T d and we need not go for T i, because T i will be set during
the relay experiment. Let us see in detail what type of auto tunings scheme we have;
so initially what is done? A modified relay is employed to induce limit cycle output.
So, the switch is connected to this point and limit cycle oscillation or relay test
is carried out; now, the modified relay is made up of an ideal relay so, the modified
relay is made up of an ideal relay in series with in series with an integral controller integral controller. So, what is
the form of the integral controller G C N (s) will be up the form 1 plus 1 upon T i
s so when this integral controller is connected in series with an ideal relay we do get the
modified relay and when the modified relay is connected with the process in closed loop
feedback then limit cycle output is obtained so, limit cycle output is obtained.
So, after obtaining the limit symmetrical limit cycle output for symmetrical relay we
will make measurements of the peak amplitude A p and the period T u which will give you
the angular frequency omega equal to 2 pi upon T u. So, the frequency and peak amplitude
of the limit cycle output will be measured then based on the information of frequency
and peak amplitude we shall set the parameters of the controller and then the switch will
be moved to this sub point for normal operation of the closed loop system so that is what
we are going to do now, we have to find explicit expressions for parameters of not only the
i controller, but also the remaining P D controller which will be added to this giving us a controller
PID controller of the form G C (s) is equal to K p 1 plus 1 upon T i s times 1 plus T
d s. So, initially what will be done when you are
starting a process or commissioning a process then you have no information about the process
so, during that time to initiate the relay test what will be done the T i the integral
time constant can be set to certain values like choose from 10 to 20 when of course,
the process G (s) is stable so, when G (s) is a stable process then at that time please
choose or set the value for T i from 10 to 20 choose any value from 10 to 20 and then
you conduct the relay test then after conducting the relay test you will get the limit cycle
output measure the frequency of the limit cycle output and then you reset update the
value of the integral parameter and the how it will be updated that we shall discuss after
sometime. Now, then that updated value of T i will be
put in the integral controller and again a relay test will be conducted to finally, find
the peak amplitude and frequency of limit cycle output and that peak amplitude and frequency
will be employed to find the parameters of the PID controller or basically the to find
the parameter K p and T D and T i has already been found thus we will be able to get a PID
controller and then we can resume the normal operation of a process that is how the model
free tuning is done model free online tuning or model free design of controller is done.
Now, I will go to the mathematics of the way we can derive explicit expressions for parameters
of the PID controller. Now, when the modified relay is in the loop when the modified relay
in the loop let the dynamics of dynamics of process is described by
described by a transfer function G (s) and that of the controller is of the controller
is G C N (s) and of the relay is N so we shall use the describing function for the relay.
So, I am using the N for that I am using the N for that then under limit cycle condition
or to obtain a limit cycle condition the loop gain has to be minus 1 then G (s) G C N (s)
N will be equal to minus 1, why that is so? So if you draw the nyquist diagram; so, this
is your nyquist diagram nyquist diagram where we have the real and imaginary then the limit
cycle condition occurs this is the operating point operating point when the relay test
is conducted so the net gain will be minus 1 plus j 0 so the net gain loop gain will
be minus 1 so this point corresponds to minus 1 plus j 0. Now, therefore, the loop gain
will be equal to G (s) times G C N (s) times N is equal to minus 1.
So, we know the form of G C N (s) what is G C N (s) it is given as one plus 1 upon T
i s then G C N in frequency domain, now G C N (j omega) will be equal to 1 plus 1 upon
j omega T i which can ultimately be expressed in magnitude and phase angle form as 1 plus
j omega T i upon j omega T i giving us 1 plus omega T i square root upon omega T i with
phase angle tan inverse of omega T i minus pi by 2 so this is how I get the frequency
domain representation for G C N (j omega) or the integral controller. Now let me use
some function phi let phi is equal to tan inverse omega T i also that implies tan phi
is equal to omega T i then now that will enable us to get expression for G C N (j omega) as
1 plus tan square phi root upon tan phi with angle tan inverse omega T i is of course,
phi So with angle phi minus pi by 2 so further simplification gives in the numerator sec
theta divided by tan theta with angle phi minus pi by 2 which can be simplified as 1
upon sin theta with angle phi minus pi by 2.
So, what we have got, the G C N G C N (j omega) gives us 1 upon sin phi with angle phi minus
pi by 2 where phi equal to tan inverse omega T i so please keep in mind the in frequency
domain G C N can be expressed by this form. Now, we know that we know that during the
relay experiment the loop gain is minus 1 so, I have already written therefore, the
loop gain G (s) G C N (s) times N is equal to minus 1 implies, in frequency domain now
G (j omega) G C N (j omega) times N is equal to minus 1 implies G (j omega) the dynamics
of the process can be expressed in the form of minus 1 upon N into 1 upon G C N (j omega)
then what you get substitute G C N (j omega) over here, that will give us now minus sin
phi upon N with angle of course pi by 2 minus phi, which again can be expressed as sin phi
divided by N with angle due to this minus 1 minus pi will appear. So I will I can write
this as minus pi plus pi by 2 minus phi and upon simplification you get minus pi by 2
minus phi. So, what we have got G (j omega) is equal to sin phi divided by N with angle
minus pi by 2 minus phi. Now, this is what you get during the relay
experiment, now when the PID controller is injected or put in the loop when the PID controller
comes into picture during normal operation of the system, that time what happens we will
get a new Nyquist diagram.
So, let the controller controlling the process be G (s) G C (s) is equal to K p times 1 plus
T d s times 1 plus 1 upon T i s, now this dynamics this P D controller will push the
operating point from minus 1 plus j 0.2 another operating point which will be the operating
point during normal operation of the closed loop system. So, the Nyquist diagram for the
new operating point will be shown now, so, let this be the nyquist diagram having real
and imaginary parts and let the plot be given like this so we are going from the minus 1
plus j 0.2 some normal operating point of the closed loop system.
So, let that operating point shifted operating point be this one, because you are operating
the system with a controller now or and the job of the controller is to get a new operating
point such that, the phase and gain margins of a systems are improved that is the objective
of posing the operating point from minus 1 plus j 0 to some new point where you will
get improved phase and gain margins. So, let this be the new operating point at which you
will have a an unit circle of the form it is not to the scale this the unit circle suppose
then I will have a phase margin of phi for this one and corresponding gain margin gives us the magnitude 1 upon
psi over here. So, this is the new operating point operating point due to the controller
G C (s). So, what you have got the loop gain here in this case is, how much now, it is
nothing but simply your G C (s) G (s) So, the loop gain at this is G C (s) G (s) and
the phase margin is having phi and the gain margin is having psi giving us the magnitude
1 upon psi minus 1 upon psi at this point corresponding to this. So, little bit of analysis
will give you the magnitude for this vector and the phase angle for this vector.
So, how this point can be represented, I can represent this point by this operating point
is giving us a magnitude of suppose G C (s) G (s) will have in a magnitude with phase
angle of course, how much this is minus pi plus phi, how to find M how to find M now,
so, M can be found if you take the cosine of phi so cosine of phi will give you minus
1 upon psi cosine of phi times M; this is the M. So, divided by M or M can be obtained
as minus 1 upon psi cost phi so, this is how the magnitude and phase of that point is found.
Now, this magnitude as far as magnitude is concerned I will have 1 upon psi cost phi
so the new operating point will give us G C G( j omega) can be written as 1 upon xi
cost phi this is the magnitude of the vector you have got and of course, with phase angle
minus pi plus phi. So, once you have correctly tracked the new operating point with the phase
angle of minus pi plus phi and the magnitude of 1 upon xi cos phi then further analysis
can be carried out easily why how you have got this operating point I do believe, when
you have got earlier the operating point here, at that time you had a relay in the loop and
you were on the verge of instability or I mean to say the system oscillates or the output
of the system becomes oscillatory when the when the phase is minus 180 degree and at
that time you have got oscillation in the system.
So, this point has been pushed to some new operating point new operating point with the
help of a controller so, when the controller dynamics is added you go to this new operating
point and the new operating point results in results in the phase and gain margin so,
you can easily find the phase and gain margins associated with these with the help of what
the vector this vector with the help of this vector. So, as I have said if the if the gain
margin is psi then 1 upon psi will be the span over here and if the phase margin is
phi then you have got the angle minus pi plus phi over here so this vector the new operating
point is denoted by a magnitude of 1 upon xi cos phi with angle minus pi plus phi this
is very important to get correct expression for the new operating point.
So, after describing all these things let me proceed with the analysis now, the analysis
will be very simple now, because you know the magnitude and phase angle of the new operating
point or when the controller is in the loop. Now how much will be G (j omega) from, here
G (j omega) will be equal to 1 upon xi cos phi with angle minus pi plus phi into 1 upon
G C (j omega) and what is G C (j omega)? G C (j omega) the controller dynamics in frequency
domain is now K p times 1 plus j omega T d times 1 plus 1 upon j omega T i this is what
you have got so, G (j omega) is this much.
Now, I will go to little bit of analysis of this system like G C G (j omega) is equal
to 1 upon xi cos phi times sorry with angle minus pi plus phi, but G (j omega) is equal
to sin phi upon N with angle minus pi by 2 plus phi how I have got this one during the
limit cycle we had put the loop gain to minus 1 and that gave us the expression for G (j
omega) as G (j omega) is equal to sin phi upon N with angle minus pi by 2 plus phi let
me show you we have already derived that so G j omega G j omega is equal to sin phi divided
by N with angle minus pi by 2 plus phi minus pi so so G j omega is equal to minus pi by
2 minus phi that implies G C G (j omega) from here using that expression G C G (j omega)
will be equal to 1 upon xi cos phi with angle minus pi plus phi into 1 upon G (j omega).
So substitute the value for G (j omega) here, which is nothing but sin phi divided by N
so N will appear here with angle of course, it will go to the numerator, so pi by 2 plus
phi. So, thus giving us G C (j omega) is equal to N divided by xi times cos phi times sin
phi with net angle minus pi by 2 plus 2 phi so G C (j omega) is found to be this much.
So, let me rewrite once more G C (j omega) G C j omega is equal to N by xi cos phi sin
phi N by xi cos phi sin phi with angle 2 phi minus pi by 2 or the angle is minus pi by
2 plus 2 phi, further G C (j omega) is equal to K p times 1 plus j omega T d times 1 plus
1 by j omega T i, but 1 plus 1 upon j omega T i is how much that already we have found
let me show you that, how much you have found so that is equal to 1 upon sin phi with angle
phi minus pi by 2 so 1 upon sin phi upon with angle phi minus pi by 2 1 upon sin phi with
angle phi minus pi by 2 so G C (j omega) is also equal to 1 plus j omega T d times 1 upon
sin phi with angle phi minus pi by 2. So, equate the expression for G C (j omega) because
you have got two expressions for G C (j omega).
So, when you equate the two what you get equating the two will result in equating G C (j omega)
as G C (j omega) and writing the both sides, now will give us N upon xi time cos phi sin
phi with angle 2 phi minus pi by 2 this is equal to K p times 1 plus j omega T d with
1 by sin phi of course, with angle phi minus pi by 2. So, what I have done I am equating
this with this so, equating with the right half of this so equating with the right half
of this so that will give you now further simplification giving us K p times 1 plus
j omega T d is equal to N divided by xi cost phi xi cost phi with angle of course, 2 phi
minus pi by 2 and then minus phi plus pi by 2 so this is equal to N divided by xi cost
phi with net angle of phi. So, finally, what we have obtained for the P D controller; the
P D controller has to have this much in frequency domain.
So, K p times 1 plus j omega T d has to be equal to N divided by xi cos phi with angle
phi so, K p times 1 plus j omega T d is equal to N divided by xi cost phi with angle phi.
So, please allow me again to find the phase angle of the left hand side in the form of
let phi is equal to tan inverse omega T d now when phi is equal to tan inverse omega
T d implies tan phi is equal to omega T d. Then this expression in magnitude and phase
angle form because the right half is expressed in the magnitude and h form will give us K
p times root of 1 plus omega square T d square with angle tan inverse omega T d is equal
to N by xi cos phi with angle phi. Now, substitution of omega T d over here will
give you K p root of 1 plus tan square phi is equal to sorry with angle with angle phi
is equal to N divided by xi cos phi with angle phi so, that implies K p into 1 by cos phi
so root of 1 plus tan square phi will be sec phi so giving us 1 upon cos phi with angle
phi is equal to N by xi cos phi with angle phi.
So, this will cancel out this will cancel out angle will cancel out angle will cancel
out and giving us K p is equal to N by xi so K p is equal to finally, we have got K
p equal to N by ziti, what is N? N is the describing function describing function for
the ideal relay ideal relay what is that? So, N is equal to 4 h by pi A p where, h is
the relay setting and A p is the peak amplitude of the output signal, which output signal
limit cycle output signal. So, you get the final expression for K p one of the important
parameter of the PID controller as K p is equal to 4 h upon pi A p times ziti.
So, this is how you have got expression for the explicit expression for the parameter
of the PID controller. So, k p is equal to 4 h upon pi A p xi this the final expression
for the parameter of the PID control whatever do the remaining parameters T d T d is equal
to if you look at, what is T d by definition tan phi is equal to omega T d. So, T d is
equal to tan phi by omega so, T d equal to tan phi by omega this the explicit expression
for the second parameter of the PID controller again what is the remaining controller parameter
T i; so, the T i has got also the expression tan phi upon omega because we know that tan
phi is equal to omega T i. So, these are the three explicit expressions
we have for the parameters of the PID controller and what are the unknowns we have in the right
half of the all these three explicit parameters of the PID controller we have got A p and
we have got omega so, A p and omega are the peak amplitude and frequency of frequency of limit cycle
output. So, conduct the relay test obtain the symmetrical
output measure this peak amplitude A p and measure the period T u which will give you
omega equal to 2 pi by T u this is how you obtain the peak amplitude and frequency from
the limit cycle output substitute over here, substitute in the one, two and three and estimate
the parameters of the PID controller this is how tuning of PID controller, model free
tuning of PID controller is done.
So, what is phi here and xi here let me explain so phi and xi or user or operator defined
so the planned operator or system operator he decides, what will be this phi and ziti.
So, phi is basically the a value with is greater than the phase margin of a closed loop system
so analysis will show that phi is greater than phase margin of a closed loop system
and similarly xi is greater than the gain margin gain margin of a closed loop control
system. So, phi and xi is user defined so, these are also known thus we obtain the three
parameters of the PID controller using formula the given over here, which uses the user defined
parameters psi and xi and which also uses the parameters obtain from the limit cycle
output such as peak amplitude and frequency.
So, we will go to simulation study now and before going to the simulation studies let
me once more explain the steps for the automatic tuning of the PID controller without finding
transfer function models for the dynamics of a system or process or plant.
So, auto tuning test starts with the initial choice of T i which is between 10 to 20 for
stable processes for unstable processes you have check, what suitable values you have
to choose for that when you have no information available for the process dynamics, but once
the process is in operation this parameter can be easily obtained, because you can use
the default value. Now T i is updated for a user defined phase angle of phi greater
than equal to 30 before beginning the second stage of relay test using the updating formula
of course, T i is equal to tan of 30 degree or greater than equal to 30 degree upon omega.
So, omega is the frequency you obtain from the fastest substitute over here then find
T i substitute T i while conducting the second stage of relay test amplitude A p and frequency
omega of the limit cycle output are output are measured in the second stage of the auto
tuning test. So, those values will be used in the formula to find the PID parameters
and the parameters of the PID controller are then obtained from set of the formula for
a chosen value of psi and phi find tuning of the controller if necessary can be done
with different user defined phase angles phi.
We will go to the simulation study, now consider a second order plus dead time process dynamics
given by G (s) is equal to e to the power minus s upon s plus 1 square with the initial
choice of T i is 20 seconds and as the relay setting of h equal to 0.5 relay test is conducted,
how this is the simulation diagram is given. So, the relay setting is plus minus 0.5 then
a p i a not p i, i controller as 1 plus 1 by 20 s which gives you 20 plus 1 divided
by 20 is employed for the process and relay test is conducted then the critical frequency
found or the frequency of oscillation is found to be 1.2925 radiant per second and choosing
a phase of 30 degree T i is now updated to 0.4467, how do you get this T i as I have
said T i is equal to T i is equal to tan 30 degree divided by omega.
So, 1.2925 this will give you T i as 0.4467 and then another relay test is conducted so
another relay test is conducted further where the i controller is now 0.4467 is plus 1 divided
by 0.4467. Then the test yields a critical frequency
of omega is equal to 0.675 radiant per second so this is the final value we will be using
in the formula for PID parameters amplitude of the peak amplitude of the output is of
the magnitude 1.602. So, omega is 0.6575 and A p is equal to 1.602 then the T i is calculated
using the formula T i is equal to tan 30 degree tan 30 degree divided by omega so, 0.6575
giving us T i as T i is equal to 0.8781 thus the T i one parameter of the PID controller
is finally estimated then the remaining parameters are calculated h, K c is equal to 4 h divided
by pi A p ziti, where xi is the user defined value we have chosen a gain margin of xi is
equal to 2.
So, K c is equal to 4 h divided by pi A p xi is equal to 4 into h is 0 divided by pi
into A p is 1.602 and xi is 2 giving us K p sorry not K c K p as K p is equal to 0.1987
and the other parameters T i which is equal to T d as the expression for both are same
so T i is equal to T d is equal to tan phi divided by omega is equal to tan 30 degree
divided by 0.6567 is equal to 0.8781. So, T i is equal to T d is equal to 0.8781 thus
all the parameters of the PID controller are estimated.
Let us see what sought of result we get from the PID controller, how the PID controller
performs? So, a simulation diagram is employed to see the performance of the PID controller
this is the process your G (s). So, the process dynamics is now G (s) is equal to e to the
power minus s divided by s plus 1 square this is what we have considered please keep in
mind we had using a process G (s) is equal to e to the power minus s divided by s plus
1 square. So, the process is there this is the process.
Now, the controllers are present here, now we have got the controller 1 plus 1 divided
by 0.8781 which is given in this form is presented and gain up K p is found to be 0.1987 and
the P D controller is realized here, if you carefully see observe then what you have got
from here it is nothing but one plus T d s this is realized using the derivate block
with a gain and you have of course 1 plus T d s realized by this.
So, the series form of PID controller is employed now for the system G (s) given by G (s) is
equal to e to the power minus s upon s plus 1 square. So, the PID controller is now G
C (s) is equal to 0.1987 times 1 plus 1 divided by 0.8781 times 1 plus 0.8781 so this PID
controller is employed now. So, whole PID controller is from here to here please keep
in mind this is the PID controller G C (s) can be realized in this form then you need
a applied at time t equal to zero unit step is the reference input and a step load disturbance
of magnitude 0.5 occurs at time t equal to 40 second.
So, the system is subjected to a static load disturbance of magnitude t equal to 40 seconds
then what sort of performance we get this is the response we are getting so the closed
loop response to step input and load disturbance is shown over here so we have got quite satisfactory
time response for the closed loop system of course, with the PID controller using the
model free controller design technique, why I am telling we have got quite successful
or quite useful response time response, because the rise time rise time and settling time
the two important parameters associated with the time response after process are found
to be satisfactory. So, rise time and settling time are less and
the disturbance response is also satisfactory because the disturbance static load disturbance
magnitude l is equal to 0.5 so we have got satisfactory reference input as well as load
disturbance rejection from the closed loop system.
Now, to test the robustness of the method it is assumed that there are plus 10 percent
variations in all parameters of the process; that means, the process was G (s) was e to
the power minus s upon s square plus 2 plus 1, I have varied the different parameters
the steady state gain was 1, I have made it to 1.1 the delay is made to 1.1 seconds then
the time constants due to the variation in the time constant I have got 1.21 is square
plus 2.2 s plus 1 so this is the perturbed plant.
So, assuming plus 10 percent variations in all parameters of the process, now we will
see the response we get, we get a performance given by the dashed line. So, this is the
response we get from the perturbed plant so the two responses are not different not significantly
different from each other giving us or informing us that we do get robust responses provided
by the PID controller.
So, let me summarize the lecture now we have discussed a method for tuning of PID controllers
without using parametric model of a process dynamics directly we have found formulated
some tuning rules based on the limit cycle parameter. So, K p, T i and T d are now functions
of K p, T i, T d all the three parameters of the PID controller are functions of peak
amplitude and frequency of the limit cycle output and user defined xi and phi the phase
and gain of a system gain margin of a system. The model free control technique is also found
to be simple and robust it does give robust performances because the design is based on
phase and gain margins.
Let us come to the point to ponder any guideline we have for choosing the phase and the gain
values, generally it has been found from analysis that the gain margin of the loop gain margin
is found to be greater than the user defined xi and the phase margin phase margin is similarly
found to be greater than phi. Therefore, one can choose xi to be greater than equal to
2 and phi to be greater than equal to 30 degree that will give not only robust performances
rather satisfactory time domain specific performances also for the closed loop system. Thanks.