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In this video, we will find the step response of a control system so we have a control system
with its transfer function we apply a step input and we have to find the total response
of the system the total response consists of two parts the forced response and the natural
response and we have to find both of them then we will also see how we can obtain the
form of a control system's response just by looking at the transfer function so lets get
started so let us say that we have a system denoted by G(s) and we apply an input which
is denoted by R(s) and the output of the system is denoted by C(s) and let us say that this
input is a step input so which is represented by 1/s so this is our step input so lets start
with the transfer function so the system has a transfer function equal to s+2 divided by
s+5, so this is a simple first order system which has a pole at -5 and a zero at -2 so
if you plot this on sigma jw 1,2 and a zero here this is -5 so there is a pole here so
now when we apply this step input so as you know the output is the product of the input
and the transfer function so output of the system C(s) so this is called the output also
called the response of the system or output response of the system so when we say response
so it means what is C(s) or the output of the system so we can also find this in t-domain
which we will do next but we start from the s-domain because the transfer function is
always in s-domain so this is the transfer function so when we apply the step input the
output will be s+2, s+5 multiplied by 1/s the step input so this is the output of the
system now here this is the a polynomial fraction and in order to go from s-domain to t-domain
what we do we use partial fractions and we decompose this into two smaller fractions
so lets sa A divided by s plus B divided by s+5 so this is what we have to find these
constants in order to as you will see that at the end of this video, we you dont have
to find these constants, without finding the constants you can write down the form of the
output response and this is what we will do just by inspection of this transfer function
we will be able to write the response of the system both in s-domain or in t-domain.
so this is what we are going to learn here but we will start from the mathematics just
to let you know that how we find these constants and there is a video about Heaviside method
which we using here to find these constants which is also called the cover up method so
there is video you can find in the list of the videos so there will be a video available
so what we do in Heaviside method is if we have to find A, what is the denominator of
A?
s, so we put s=0, we find that value we put that value into this we ignore this term which
is the denominator of A. similarly in order to find B we put s+5=0 which will give us
s=-5 we put s=-5 into this whole and we ignore the denominator of B s+5, so we put a finger
here this is also called a cover up method you just cover this with your finger so this
is what we are going to do so A= s+2, s+5 at s=0 and
we are ignoring this which will give us if I put s=0 in this these two, we get A=2/5
similarly for B we write down s+2 ignore this thing so divided by s at s equals -5, what
we get here equals to -5+2 is -3 divided by -5 it will make it 3 divided by 5. this is
how we get A and B. Now these are the two constants so now here I want to show you something
that if you look at the evaluation of these constants you will see that zero and the pole
both are involved in finding these constants so now I will go forward and I will come back
to this point now the output response of the system equals to 2 divided by 5 plus 3 divided
by 5 and this is in the s-domain now if you use the Laplace transform table from going
from s to t-domain you will find that this term 2 plus 5 divided by s is basically in
time-domain now we are going from s-domain to time domain so the output in time domain
will be equal to this is simply two divided by five and plus this is simply 3 divided
by 5 e raise to the power minus 5t so this is the Inverse laplace transform of this s-domain
output of the system so now we get this so this is the total response of the system and
this has two components the first which is 2 divided by 5 if you look at s-domain this
is s this 1/s was out input so the first part of this total response is due to the input
function the step function and the second part is due to the transfer function or the
poles of the system so this is pole of the system so the part of the response which is
due to the input function is called the forced response because we are trying to get a response
we are trying to input something and the part of the response that is dependent on that
input is basically a forced because we are forcing it so this is called the forced response
and this is called the natural response because pole is a characteristic of the system so
the response due to the pole or poles of the system if there are more than one poles there
will be more terms here so that is called the natural response and the response which
is due to the input is called the forced response lets see in the beginning we have this and
we want to find the response of this transfer function so as you can see that this zero
only effects the constant because when we were finding the constants we were putting
this in the numerator so the zero of the system only contributes to the constant only the
poles contribute to the response of the system the output of the system as here so we can
we can say just by looking at this that there will be some constant lets say we say k1 lets
say we say in the s-domain so there will be some constant k1 divided by if the input is
step s plus some constant k2 k2 and divided by s+5 so this is how we can by looking at
the transfer function we can write down this equation and from this equation we can also
write down in time-domain because only this will be in time-domain so C in the time domain
would be k1+k2 e raise to the power minus whatever is the pole which is 5t so this is
how we can write the response straight away in s-domain or in t-domain in both ways we
can easily find so in s we have to write the input function , in time-domain we have to
only write the constant so let us do one or two exercises so you can quickly see that
how can we quickly find the response of the system just by looking at the transfer function
of the system so you should be able to write down this equation by looking at the transfer
function of the system let us do some examples so this is an example so we have G(s) and
again R(s) = step input and the C(s) we have to find out and the transfer function of the
system is equal to lets say s+3,4 and 5 so first let us write down in s-domain so we
will say the output of the system in s-domain would be equal to some constant k1 multiplied
by the input so k1multiplied by 1 divided by s so it is simply k1 divided by s + second
constant and the first pole k2 divided by this one plus k3 divided by s+4 plus k4 divided
by s+5 so this is the response of the system in s-domain but normally when we say the output
response of the system we are talking about in time-domain not in s-domain because in
time-domain it is easier for us to visualize so let us write down this in t-domain so c(t)
would be only k1 here plus k2 e raise to the power -3t plus k3 e raise to the power -4t
and k4 e raise to the power -5t. so this is the response of the system now if you have
to find these constants then you can use the heaviside method which I described earlier
and you can find these constants watch that Heaviside method video and you will be able
to find the constants if you have to find the constants otherwise this is just called
the form of the response so this is the total response only this part is due to the input
so this is the forced response and all these terms they are the natural response of the
system because these terms are due to the characteristics of the system due to the poles
of the system so therefore this is natural and this is forced.
so very simple so now what if there are more than one zeroes.
the input is step but the transfer function has more than one zeroes s+2 s+3 s+4 so even
in that situation because the input is step so you will get one s here I have not written
the output equation here because C(s) would be equal to 1 divided by s into the transfer
function so if for example there are more zeroes still
you have only step input is coming here so you will only get one constant and all the
effect of other zeroes will be in calculations of these constants so that will be the effect
of those zeroes otherwise the form of the output this is called the form of the output
will be just like this one constant due to zeroes and all other constants due to poles
so the form of the system response would still be the same even if there are more zeroes
so I hope that after this you should be able to write down the form of the response of
the system and both components tghe forced and the natural just by looking at the transfer
function of the system so I hope you find this video useful if so please subscribe to
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on control systems or on Latex and other subjects which I teach thanks for watching bye for
now.