Introduction to Transfer Function

hello everyone and welcome back to the

next lecture of control systems

in this presentation we are going to

discuss the transfer function

so let's get started the transfer

function is an important parameter of an

lti system

like impulse response we use the

transfer function to define the lti

system

we will first see the definition of

transfer function the transfer

function is the ratio of laplace

transform of

output to the laplace transform of input

when all the initial conditions are

assumed to be 0.

this point all the initial conditions

are assumed to be 0

is a very very important point we cannot

define the transfer function

without assuming the initial conditions

to be zero

and it is because we define the transfer

function for lti system

and in case of lti systems all the

initial conditions

are zero we have discussed the reason in

the previous lecture

if the initial conditions are not zero

then the system will not be a linear

system

and hence the system will not be lti

that's why

to define the transfer function all the

initial conditions

are assumed to be 0. now suppose we are

having a system

which is of course an lti system and its

impulse response

is ht we are giving an input xt to the

system and its response

is y t now we all know that y t

is equal to the convolution of these two

functions

so we can write y t is equal to x t

convolution with h t and from the

convolution property

we know that the convolution in time

domain

is the multiplication in frequency

domain so by convolution property

we can write y of s is equal to x of s

multiplied with h of s where y s is the

laplace transform of y t

x s is the laplace transform of x t and

h s is the laplace transform of h t

now from this equation if i take the

ratio of y of s

to x of s then we will have h of s is

equal

to y of s over x of s and this is the

transfer function

of this system now here we have defined

the transfer function

for this system and we know this system

is an lti system

so by default in this case the initial

conditions

are equal to zero but if we define the

transfer function for an arbitrary

system

then in that case the initial conditions

must be equal to zero

and to understand this in a better

manner we will take one example

we can also notice that hs is the

laplace transform of h t

so we can say that the transfer function

for any ldi system

is the laplace transform of the impulse

response of that lti system

suppose we are given the transfer

function of an lti system

and we are asked to calculate the

impulse response then we can calculate

it easily

by taking the inverse laplace transform

of hs we are going to solve many such

questions in the upcoming lectures

so now we are done with the introduction

of transfer function

let's take one example to understand it

in a better manner

and the example is given as find the

transfer function

of the system given by d square y t over

dt square

plus 3 multiplied d y t over dt

plus 2 y t is equal to x t where

x t is the input and yt is the output

so we are given a system which is

defined by this differential equation

where xt is the input to the system and

yt

is the output and we need to calculate

the transfer function

and we know the transfer function is the

laplace transform of

output to the laplace transform of input

now moving on to the solution

if we take the laplace transform on both

the sides then the laplace transform of

d

square y t over dt square by time

differentiation property

is given as s square y s minus y

0 minus minus y dash 0 minus

similarly the laplace transform of d y t

over dt

will be s5s minus y of

0 minus and it is multiplied with 3

so by homogeneity principle 3 will be

multiplied in the laplace transform also

similarly the laplace transform of 2 y t

is 2 y s and now the laplace transform

of

x t is equal to xs and in this way we

have converted this differential

equation

to its laplace domain and now we need to

define the transfer function

but remember to define the transfer

function the initial conditions of the

system

must be equal to zero so the initial

condition terms

in this equation must be equal to zero

so if we put

these three terms equal to zero we will

have

s square multiplied by s plus 3

multiplied

s multiplied by s plus 2 multiplied by s

is equal to xs now we have y s

common on the left hand side so taking y

of s

is common we will get y of s multiplied

s square plus 3 s plus 2 is equal to x s

now if we take the ratio y s to x s we

will get

y of s over x of s is equal to 1 over s

square plus three s

plus two also if we factorize this we

will get

s plus one multiplied s plus two so the

transfer function of this system is

given as

h of s is equal to one over s plus one

multiplied s plus 2 and in this way we

have calculated the transfer function

for this system

and now we are done with this lecture we

will take some more problems on transfer

function

in the upcoming lectures thank you for

watching this lecture

i'll end this lecture here see you in

the next one