so b proportional control action now right
so let me start with a first order system
a proportional controller with the closed
loop transfer function is k t s plus one plus
k ok what is the error if the reference is
a step function right so the steady state
error by ok to evolve this calculations we
had done this before also is one over k plus
one right the first observation is that the
steady state error is non zero ok though i
would ideally want it to be zero but it is
not zero in this case ok now what does this
proportional control do well the first observation
is that the control or the proportional controller
it results in improving the time constant
which means the response is faster on the
flip side there is a steady state error right
now what we could do is now just by observing
this say that well my steady state error can
be made smaller and smaller by ah increasing
the value of the gain k but this may not always
be practically possible or even desirable
because i might amplify unwanted signals and
so on right in some case of cases of higher
order plants which i will i will show you
a example increasing the gain could also have
a tendency to destabilise the system right
for example if i were to look at this root
locus for i just keep on increasing the gain
but i just notice that i am now on the verge
of stability right and that is that is not
desirable ok
so the proportional controller it improves
the response of the system it also helps improve
or lessen the steady state error but only
up to some factor right i just cannot get
it to zero ok now lets see what a pure integral
control action do and this actually go very
well with the with the term side so integral
control action is when i have a k i divide
by s one over s has an integral action right
the the laplace transform of an integral of
a signal just was translated to one over s
of the original signal ok so the output is
ah make something like this so now what is
the error the error is right is something
less happened here right so if i just go through
all these computations the steady state error
becomes zero this we also can see in when
we did the error constant analysis right type
one system if i am tracking a step ah the
steady state error is always zero
so with addition of the integrator as a controller
when the steady state error is zero but the
system order has increased from one to two
and if we keep on dealing with higher and
higher order systems increasing a pole or
adding an additional pole at the origin which
is essentially an integrator might lead the
system to the verge of instability so two
things to observe here right first is proportional
control it improves my transient response
it also helps in reducing the steady state
error but as an eliminate it completely integral
action well i add an integrator to the system
which means i add a pole at the origin it
helps in reducing the steady state error to
zero in this particular example which we are
talking about or in this particular case of
a of of what what is now the type one system
ok higher order systems it may not always
be desirable right that just keep on buying
integrators and dumping into the system it
may not work and we say why it may not work
so if i take this example here well i need
to see well how the system looks like now
right so so total g of s if i call this as
the open loop transfer function g of s here
is k over s square j s plus b ok
now ok lets may ah let me just say that k
equal to one j equal to one b equal to one
and lets see well what is the closed loop
system stable well i can write if you do the
routh analysis but i will do the root locus
right because now we are fairly familiar with
doing that ok so where are the poles where
there is a pole here and a pole here and i
write as equal to zero in the sigma and j
omega axis or the complex plane but another
pole at s equal to minus one now i think we
now remember the rules of the root locus by
heart so this guy will go here and we see
that this guy is actually go this way closed
at a this particular example while you were
doing he root locus parts and you see that
just adding an integrator here doesnt help
me much because the system is now becoming
unstable ok
so we have to be careful in in that way ok
now what is the affect of disturbances right
when when we use these kind of controllers
so again let me let me talk take this kind
of plant here which is a rotational element
with a movement of inertia j and some frictional
coefficient b ok now lets say there is some
disturbance d of s and i will say what is
the affect of this disturbance on the response
of my system and can i get rid of those disturbances
by adjusting the value of k ok so lets say
well i am just interested at the moment in
a step disturbance so i want to see what is
y s over s means the affect of the disturbance
on the output y i can just do this by setting
r equal to zero within this how to derive
this transfer function much much earlier n
the course so so now what happens is well
e s because of d s this becomes the negative
of y so i am just so y goes here there is
no r so e s is simple the negative of y
so i can also see what is the affect of the
disturbance on the error signal i i just stay
forward to see because what we do all when
we do this we just compute this with r of
s equal to zero r of s equal to zero that
is matching here y of s enters here what is
e of s e of s is zero minus the signal what
is coming here so e of s is the negative of
y of s ok ok so what is the steady state error
because of this disturbance well the steady
state error just do all the formulas se the
final value theorem and i can get that this
is minus of t over k now i can take care of
this steady state error i gain by increasing
the value of gain k now what is the affect
of this gain k i am keeping on increasing
gain k
now where does the gain sit in my characteristic
equation the gain is sitting here so k now
has some direct relation to omega n right
s square plus two zeta omega n plus omega
n s right if i just normalise it properly
k so if k increases the system becomes more
and more oscillatory right and or higher order
plants as we saw earlier it can also go to
the verge of instability ok so ext we see
what is the affect of integral action and
we have this integral action addition to the
proportional action and therefore we call
this a proportional plus integral control
action ok so again y s over d s i could be
easily computed to e this way and i see that
you know earlier we had stated that increase
ah or introducing an integral action increases
the order of the system so now i have three
poles ok with r s equal to zero the affect
of of ah the disturbance on the error becomes
something like this ok ok
so what are things which i have in my control
from the control that is t i and i have a
k here ok now i might be tempted directly
to use the steady state error formula from
the final value theorem to see what is a steady
state error to yes stop disturbance ok so
in order to compute what is the affect of
the disturbance on the steady state error
while we are using a proportional plus integral
control we might be tempted to use the final
value theorem directly but only thing which
we need to be careful is that when we use
the final value theorem we must first ensure
that the system is stable which means i need
to choose the k and t i such that the overall
closed loop system is stable or in other words
the routes of the characteristic equation
have negative real parts j and b are given
to me so i am i can just play around with
k and t i
now under this assumption that the system
is stable or that k and t i are chosen such
that this system is stable the steady state
error i just use the formula for the final
value theorem is zero so the observation here
is that proportional plus integral control
action helped us in eliminating the steady
state error now this is the steady state error
due to the disturbance i will just make it
a little more explicit here which means the
affect of disturbance can be nullified in
the system now you may ask a question why
did i not choose only an integral control
action why do i need k plus one over t i s
why not only this guy one over t i s or k
y one over t i s well the answer will be found
in the previous exercise
so just be now just try to write down things
and then well the answer is very obvious there
right so to conclude the what we can say is
that to eliminate the affect of disturbance
we need both the proportional and the integral
terms the proportional term was not enough
as we saw earlier adding an integral term
alone gives us problems which we found in
the previous exercise so the proportional
term here ensures the stability while the
integral terms eliminates the steady state
error and that we could compute ah from here
ok now just look at a derivate control action
so i just have my plant as a rotational element
with a moment of inertia j so i will let me
say i just start with a proportional control
action so why do i even need to use a derivative
action so i just say well the ah put a k here
the closed loop transfer function is k over
j s square plus k if i plot the root locus
it will tell me that the poles are always
on the imaginary axis as the value of the
gain k increases from zero to infinity right
so my response of the system will be purely
oscillatory right and this oscillations wouldnt
die down now i would like these oscillations
to be eliminated right and we see how these
oscillations can be eliminated and when i
say that the oscillations are to be eliminated
it means i must introduce some damping into
the system so this system here does not have
any damping term j s square plus k there is
no term corresponding to the coefficient s
right two zeta omega n s right so the oscillations
can be damped if we incorporate a derivative
term in the controller and incorporation of
a derivative term leads to introducing of
the damping term in the system i will shortly
show you how that looks like ok so can take
a proportional for plus derivative controller
where the controller now looks ah looks like
k one plus k d of s and the output from the
controller y c of s is ah all of these guys
so i have y c is k e of t plus k t d now e
dot of t right we saw this when we were discussing
ah the affect of adding a zero right this
is this is actually like adding a zero to
the system right
so you have the error and the derivative of
the error also so the output now is which
was earlier n the case of the proportional
control only only depending on the error is
now also proportional to the rate of the error
change e dot of t so what does this derivative
term do this derivative term anticipates a
large overshoot with a e dot now how fast
it is moving right e dot of t tells me how
fast my error is moving so it anticipates
the large overshoot based on the rate and
the and when the rate of the change of the
error is too high it takes a corrective action
right so this e dot tells me how fast my error
is growing right and this affect can be noticed
in the improved damping in the system ok so
what is the affect of adding a derivative
or a proportional plus derivative term to
the plant one over j s square which does did
not have any original damping so the output
of the system is now well this k this entire
thing will be numerator now look at the characteristic
equation or the denominator j s square plus
k times t d times s plus k
now this is just a second order system and
we know that for positive values of j k and
t d the system is always stable now what did
the the derivative control do you just added
this extra term here which was absent earlier
right it added a damping term into my characteristic
equation and this damping term is due to the
derivative control action ok now lets combine
all these things we started with proportional
we started with proportional plus integral
control and now we later on saw proportional
plus derivative control well lets combine
all these things together and see what is
the affect of each of these things right ok
so let us look at the affect of p plus d plus
i on a second order unstable plant so let
me say that i am starting with a a plant which
looks like this g s is s square minus b and
for b greater than zero the system is always
unstable because my poles are plus minus square
root of p ok
so the first first thing right so in in an
earlier slide i said proportional control
helps in you know improving the stability
of the system lets verify if its true or not
ok i add a k p here and with this k p i am
just calling this with the with you know ah
subscript p because i am just dealing with
a proportional control action with this proportional
control the closed loop transfer function
is now this one k p s square minus b plus
k p ok now s is some stable well well at this
it could be marginally stable right because
there is no no damping term here right but
at least starting from this unstable system
i could at least get it to a ah marginally
stable configuration by appropriate choice
of the value of k p ok now let us add a derivative
term the derivative term well what we saw
will add some kind of a damping to the system
so the proportional control got an unstable
system to become marginally stable adding
a derivative term will actually make it what
we call also the asymptotically stable system
right
so now the response y s over r s has an additional
damping term in the system and for k p and
k d ah k k d greater than zero and k p greater
than b the system is always stable right and
therefore i can arbitrarily place the closed
loop poles or give an any zeta and omega ah
n i can design the system for any zeta and
omega n this is what i mean by pole placement
and because the poles are decided on the values
of this i can place this where ever i want
right based on appropriate choices of k d
and k p ok so this k p greater than p is also
necessarily necessary here right when this
k p is greater than b then i have marginal
stability ok now what happens to the steady
state error well the steady state error i
just look at the again the formula and i get
that well it is i would like it to be one
because i am tracking this reference signal
well this is not one right therefore there
is a steady state error in the system now
we will see what is the affect of adding an
integral term in one of the earlier slides
we said that adding an integral term eliminates
the steady state error
now lets add a integral term to this guy so
i have a k p the proportional term the derivative
term and now i have an integral term so the
response through this and i compute the steady
state error so limit t tends to infinity this
is one right so what i am so my r of s here
is a unit step but i am not tracking a unit
step here right so my steady state value is
k p k p minus b so i am not computing the
steady state error here but i am just computing
the final value the final value should be
one because i am i am tracking the step signal
it is not one but depending on k p and b so
again appropriate choices of values of k p
we saw earlier also that i can keep on increasing
the k p to be very large value but that may
not help me all the time right ok now adding
an integral controller gives me the output
as one which is what i am tracking right i
am tracking a unit signal ah sorry i am tracking
a unit step right so this adding of an integral
term eliminates my steady state error or it
means that i can actually track a a unit step
right and the so the the conclusion is that
we have perfect tracking of this unit step
moreover we also know that having an integral
control action rejects constant disturbance
input so we individually saw what are the
affects of adding a p term the proportional
controller a derivative term useful in adding
damping to the system or improving the transient
performance and in integral controller which
is useful in improving the steady state performance
in terms of the steady state error
ok so just to summarise what we have learnt
in terms of the proportional derivative and
integral control is when we could start with
an unstable system second order and we could
stabilise it right we could also place the
poles arbitrarily or in a way given any choice
of zeta and omega n or any performance specifications
i could choose my gains k p and k d appropriately
so that i achieve those performance objectives
the system has perfect tracking because of
the integral control and what integral control
also does it it rejects constant disturbance
inputs right so to summarise we define what
are dominant poles and zeros we by just you
know using matlab plots we solve what were
the affect of adding poles and zeros to the
system right that was for second order systems
we also saw the same thing when i add poles
and zeros to an open loop system what is a
consequence on the closed loop system and
we defined three basic control actions proportional
integral and derivative control action so
in the next lecture we will see some ah issues
in implementing p i d controllers
so this stands for proportional integrative
and derivative controllers and we will see
slowly define what are lead and lag compensators
what could be the performance specifications
in the time and frequency domains and the
correlation between these two specifications
if any ok
thank you
