control systems nptel lecture 30

1. Control Systems
Prof. C. S. Shankar Ram
Indian Institute of Technology, Madras
Department of Engineering Design
Lecture – 30
Performance Specifications
Part 2
So, that is one concept which I wanted to introduce, yes.
Student: [Inaudible]
(Refer Slide Time: 00:34)
All right, so, see what I mean is that see let us say you have a tenth order system for example,
right. So, you need to design a closed loop control system which is order of 10, but still I can
always say look I will specify performance in this manner; see the reason is a following,
right. So, we have explicit expressions for specific performance specification for systems of
order 1 and 2 and by and large the dominant dynamics is going to be of that order, all right.
So, consequently, I can essentially create regions like this and if I am able to place all my
poles in this region, I am done ok, we will double check ok. Obviously, I would place the
dominant pair as close to the imaginary axis as possible right is it not, then only they will
remain dominant see think about it. We will do all this design process when we learn about
root locus and do the design examples; see when I am using maximum peak overshoot and

2. settling time for an under damped second order system, I am implicitly assuming that my
dominant poles are going to be a pair of a complex conjugate poles, right, which should be
the closest to the imaginary axis right. So, when I design I should ensure that the dominant
poles basically are complex conjugate and they are the closest to the imaginary axis that is
something which I should ensure, right. So, by and large it works that way. You can say look I
do not want dominant complex conjugate poles, I will have a dominant poles on the negative
real axis fine you are still satisfying the performance requirements.
But what will happen is that the system response will become sluggish that is what will
happen because like your rise time value will go up as we discussed earlier right, yeah. So,
this is just what to say a I would say an engineering solution right to visualize a domain in the
s plane where you can place the closed loop poles; that is the application right. Of course, if
someone says look design a closed loop control system whose time constant is let us say less
than 4 seconds what will we do? The moment I say time constant I am telling you that the
dominant dynamics will that will be that of a first order system and if time constant is 4
seconds what should happen what was the pole of a first order system: -1/T. right for a stable
first order system. So, if I want time constant of at most four seconds,
1
T
=
−1
4
. So, I want
all my poles to be to the left of -0.25 line ok. So, that is how we should visualize, is it clear.
So, that is an alternative example, right.
So, if I want my performance specification I am saying that look I want a time constant of let
us say less than 4 seconds; then automatically we look at the step response of a stable first
order system whose pole is at -1/T. So, time constant of at most four means my poles should
be to the left of minus 0.25 line; that is it, right. So, that will ensure that I satisfy that criteria
ok. So, that is the way we construct and then like we look at perform specification ok.

3. (Refer Slide Time: 04:13)
So, one more thing which we are going to do today at least start today is analysis of steady
state errors ok. So, before we go we put everything together and then like start doing
controller design. So, this is something which we are going to discuss today ok. So, how do
you analyze steady state errors because by and large you know like we design controllers to
do some let us say regulation or tracking right; follow a desired behavior. See for example, I
would design the air conditioner in this room to meet a set temperature right. So, I will come
and set the temperature as 25 degree Celsius.
I want my controller to maintain the temperature of air near to 25 degree Celsius right;
obviously, I would be happy if as time progresses I stay near 25 degree Celsius, yes, right is it
not, that is when I will tell that my system or controller is performing well, right. So, steady
state errors become very important, right; so, and they become very critical, right.
So, say for example, I am developing an autonomous vehicle, right, I wanted to essentially
follow a particular trajectory I wanted to be an autonomous car which has to travel within a
lane, alright. So, I do not want the car to oscillate in the lane and also like deviate from the
path, right; that is going to have serious implications right. So, as time progresses what is
going to happen to that vehicle right. So, that is something which we should figure out. So,
let us let us look at steady state errors. So, how do we analyze steady state errors? Let us
construct the standard closed loop negative feedback system that we have been looking at.

4. So, this is the feedback path transfer function and this is the feedback diagram right. So,
please note what was G(s) called that is what is called as a forward path transfer function
right H(s) is the feedback path transfer function, right. So, recall that the closed loop transfer
function is essentially equal to
Y (s)
R(s)
=
G(s)
1+G (s)H (s)
So, that was the expression for the closed loop transfer function ok. So, we are going to
essentially make an important assumption ok, let us assume that the closed loop system is
stable ok. So, of course, then only we can talk about steady state errors right think about it
right, let us say if I do not stabilize the closed loop system there is no point in talking about
steady state performance right.
So, let us assume that the closed loop system is stable ok. So, that first requirement of
stability the critical requirement of stability has been met right that is an assumption we will
make before we proceed right. So, immediately we note that
E(s)
R(s)
=
1
1+G(s)H (s)
So, what is the name that we associate to the function G(s)H(s) ; open loop transfer
function, right. So, G(s) is what is called as a forward path transfer function right H(s) is what
is called as a feedback path transfer function right. So, G(s)H (s) is what is called as the
open loop transfer function.
Please remember these terms. see of course, in your quizzes and exams I will not give an nice
block diagram and then like give you expressions right I would say design a negative
feedback system and if I say unity; that means, H(s) = 1 right, I may say the plant transfer
function is something controller transfer function is something. So, then you should be able to
figure out G(s) and so on right correct.
So, please remember these terms, right. So, this implies that the steady state error ok, let me
call it as ess=lim
t→∞
e(t )=lim
s→0
sE(s)=lim
s→0
sR(s)
1+G(s) H (s)
So, that is what we have for the steady state error .

5. (Refer Slide Time: 10:29)
Now let G(s)H (s)=
b0 s
m
+b1 s
m−1
+…+bm−1 s+bm
a0 s
n
+a1 s
n−1
+…+an−1 s+an
So, let us say G(s)H (s) comes out to be in this generic form. Please note that I have
already have closed loop stability ok, I am reiterating that point again and again yeah. So,
now, let us say we can rewrite this as
a
¿
(¿0¿s
p
+ ^a1 s
p−1
+…+ ^ap−1 s+1)
s
n ^¿
G(s)H (s)=
b0 s
m
+b1 s
m−1
+…+bm−1 s+bm
a0 s
n
+a1 s
n−1
+…+an−1 s+an
=
K (^b0 s
m
+ ^b1 s
m−1
+…+^bm−1 s+1)
¿
So, I what I
am doing is that I am doing two things, first thing is and I am seeing whether there are some
open loop poles at the origin.
See for example, what are open loop poles? There the poles are the open loop transfer
function right. So, if I look at the denominator, if I have poles at the origin then an is 0;
obviously, right. So, I can pull some s term out, common. So, I can have 1 pole at the origin 2
poles at the origin 3 poles of the origin or none ok.

6. So, the value of capital N can be from 0, 1, 2, 3 and so on right. So, I am sure all of us agree
correct, once I pull it out what I do is that remaining factors also what I am doing is that I am
making the constant term as 1, I do some algebraic manipulation and then introduce a factor
K such that the constant term becomes 1 we will see why right shortly.
So, this is the general structure of the open loop transfer function.
When, N=0 → Type 0 System
N=1 → Type 1 System
N=2 → Type 1 System
So, when a type 0 system does not have any open loop poles at the origin ok, yes.
Student: [Inaudible]
Yes, they are different coefficients right that is why I am putting a hat over it, right.
So, I am just doing some algebraic operations I am just writing in a general form; so, that you
will see why I am doing it because I want the constants to be 0.
But because in the two polynomials when I substitute s = 0 because that is where I want how
I would use my final value theorem right I want to end up with just 1 and 1. So, consequently
I am putting whatever that ratio which comes out as K right I would say ok. So, depending on
the value of capital N, we have type 0, type 1, type 2 type 3 and so on and see for example, if
you use an integral controller we automatically get a type 1 system.
I am sure all of us agree right see what is the transfer function of an integral controller?
K i
s
If you use
Ki
s
in our open loop transfer function, I automatically have a s in the
open loop transfer function denominator right. So, I automatically get a type 1 system, I am
just giving an example ok. So, please remember this ok. So, that is the terminology which we
use.
Student: (Refer Time: 17:46).

7. see denominator of course, I am assuming that there are no pole 0 cancellation to the origin,
right. So, I am assuming that there is no 0 at the origin ok.
Student: (Refer Time: 17:58).
Yeah yes.
Student: (Refer Time: 18:03).
So, for type 1 and higher my original an should have been 0, as you can see from here ok, I
am only writing a pth order polynomial here, right, the only thing is an I am making the
constant term of the pth order polynomial as 1.
Student: (Refer Time: 18:33).
We will see.
Student: (Refer Time: 18:43).
Yeah. So, I pre-emptively told you why am I taking out let me once again do as do another
example then we will I think, we will stop here and then we will continue right let me do an
example right. So, let us say, I have
G(s)H (s)=
s+4
s
3
+3s
2
+5 s
So, what I do is that I take 4 out in the numerator and then I take 5s common in the
denominator. Therefore,
G(s)H (s)=
s+4
s
3
+3s
2
+5 s
=
4(0.25s+1)
5s(0.2s
2
+0.6s+1)
=
0.8(0.25 s+1)
s(0.2 s
2
+0.6 s+1)
So, here N is 1, p is 2 , m is 1 and K happens to be 0.8; that is what is called as the open loop
gain like we will come back shortly next class.