hello when learning control systems the most important step is to find out the
transfer function of any system because once you know the transfer function you
can apply a lot of principles and a lot of methods to understand the system but
to find out the transfer function there are essentially only four critical steps
so in this video we will learn how to find the transfer function of a simple
mechanical system but it doesn't mean that it is only limited to a simple
systems you can apply this principle to any system how complex it might be so
let's get started and see how to find the toss of function so this is our
system in this system we has a mass we have a mass M which is attached to a
spring in the damper we apply a force F in this mass and the mass moves a
distance X so this is the situation know one of whenever you want to find out a
transfer function the first step the first step is to decide what ratio you
want what is your output and what is your
input in between these two in between what two values you want to find out the
transfer function because the function is essentially a ratio of the output of
a system to the input but what what parameter is our prime interest for us
so in this situation as you can see that our input is the force applied to the
system and our output is the displacement of the mass so therefore we
say that we want a ratio of the displacement to the applied force as our
transfer function but this ratio have to have we have to have find this ratio in
s domain so a transfer function of this system which we are interested in which
is represented by G of s equals the ratio of the displacement as output to
the force as a input so the ratio of the displacement in s domain to force in s
domain is our transfer function so once you have decided these two values the
next step is to apply the Newton's laws and to find
the equation of motion so in this situation we are saying that we are
applying the force F T so all the rest of the forces opposing the motion so
they impede the motion of the force so there are three forces here one is the
spring force one is the damper force and the next force is the M into a the mass
time acceleration so mass into acceleration so all these three forces
are opposing this force so we invite us a simple Freebody diagram here so let us
this is F so the force on the right is force T means in time domain so the
opposing force the force due to spring is K into X X of T the force due to dead
bodies always damping coefficient multiplied by velocity so the damping
coefficient is FB and the velocity can be written as DX by DT because the
velocity is the first derivative of displacement the third force is due to
the mass of the system which is M into a and F is M and E is the derivative of
the velocity and the second derivative of displacement so and we can write
acceleration as d square X by DT square so these three forces are acting on the
left hand side and the applied force is all right so the first step was to
decide our put and it put this is your own choice
the second important step is to write the equation of motion so let's write
down these equations so we say that all these forces impede the movement of
applied force so therefore we can write on the left hand side we have K of X T
the spring force plus FV DX by DT the damper force plus second derivative d
square X by DT times the mass equals that lag force F T so this is our
equation of motion so this is the second critical step two
right the equation the third step is to go from time domain to estimate because
our transfer function is always at a showing s domain so this is your third
critical step so step number three is to move from time domain to estimate so in
order to move from time domain to estimate
there are three quantities in time domain the displacement the velocity and
the acceleration and we have to find out what are the Laplace transforms of X of
T DX by DT and what is the Laplace transform of d square X by DT square so
let me write down here then Laplace transform of any function of time equals
the function in de as domain so for example if this is force so you can
write the Laplace transform of f TS f of s and if it's XT displacement so Laplace
transform of XT will be simply X s so X s is the Laplace transform of this and K
is the constant now what is the Laplace transform of the first derivative so
Laplace transform of derivative his F of s minus because it's a first derivative
so there will be one initial condition some initial condition but for linear
system we know all in with all initial conditions all zero conditions are
ignored so we will not consider this initial condition so this is the Laplace
transform of first derivative the Laplace transform of second derivative
of any function equal to s this coil as the F of s and minus now there will be
two initial conditions plus something else but again we ignore these initial
conditions and we only take this value similarly this value and this value now
we can put these values here so we can have plus FB DX by DT so this is
X instead of F so we will simply write as X of s plus s square X of s
multiplied by M and equal to f of s so this is our equation in s domain so now
we have moved this was our third step to move from time domain to s domain so the
first was to decide output input second was to write the equation of motion
third is move from time to estimate and the fourth in the last step is to take
this ratio so first we have to simplify a little bit if you can see here that we
can take XS s comma so if we take X s as common so we get K plus FB into s plus M
into X square equals F of s so now the last step is to take the ratio so we
have to take the ratio of X s / f s so we have to bring this F s here and this
thing on the other hand side so we write x s divided by FS equal to 1 divided by
all this and normally it's a good practice to write the highest S term
first because this is the order of the characteristic equation so we write em s
square plus FB s plus K so this is okay this this step you can stop here you can
say this is the transfer function of the system but it's more often a common
practice to keep this the coefficient of the highest term has 1 so we can divide
numerator denominator with em and that will be our
final step so X of s divided by F of s equal to 1 divided by M and then we have
s square plus FP m times s plus T divided by M so both of these equations
are correct so this is the transfer function of this
system and as you might expect that all the critical parameters are in this the
K spring constant the damping coefficient and the mass all of these
values are here so this is our transfer function for this simple mechanical
system no we can do this we can apply this principle to find out the transfer
function of more complex systems and this we will learn in our next videos