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Unit Step function in matlab

# Unit Step function in matlab

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to find a unitstep of a given transfer function with matlab programming
to find a unitstep of a given transfer function with matlab programming

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06/18/2013

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Program No. 2Object-:

Write a program to find out the unit step response for the givenfunction-:S(s
2
+9s+19)/s
3
+7s
2
+14s+8
Apparatus-:
Matlab software 7.0.4
Theory-:Step Response

Similarly to the impulse response, the
step response
of a system is the output of the system when a unit step function is used as the input. The step response is acommon analysis tool used to determine certain metrics about a system.Typically, when a new system is designed, the step response of the system is thefirst characteristic of the system to be analyzed.
Convolution
However, the impulse response cannot be used to find the system output fromthe system input in the same manner as the transfer function. If we have thesystem input and the impulse response of the system, we can calculate thesystem output using the
convolution operation
as such:
(
) =
h
(
) *
x
(
)Remember: an asterisk means
convolution
, not
multiplication
!Where " * " (asterisk) denotes the convolution operation. Convolution is acomplicated combination of multiplication, integration and time-shifting. We candefine the convolution between two functions,
a(t)
and
b(t)
as the following:(The variable τ (greek tau) is a dummy variable for integration). This operationcan be difficult to perform. Therefore, many people prefer to use the LaplaceTransform (or another transform) to convert the convolution operation into amultiplication operation, through the
Convolution Theorem
.
Time-Invariant System Response

If the system in question is time-invariant, then the general description of thesystem can be replaced by a convolution integral of the system's impulseresponse and the system input. We can call this the
convolution description
of a system, and define it below:
Convolution Theorem
This method of solving for the output of a system is quite tedious, and in fact itcan waste a large amount of time if you want to solve a system for a variety of input signals. Luckily, the Laplace transform has a special property, calledthe
Convolution Theorem
, that makes the operation of convolution easier:
Convolution Theorem
Convolution in the time domain becomes multiplication in the complex Laplacedomain. Multiplication in the time domain becomes convolution in the complexLaplace domain.The Convolution Theorem can be expressed using the following equations: This also serves as a good example of the property of
Duality
.
Using the Transfer Function
The Transfer Function fully decribes a control system. The Order, Type andFrequency response can all be taken from this specific function. Nyquist andBode plots can be drawn from the open loop Transfer Function. These plotsshow the stability of the system when the loop is closed. Using the denominator of the transfer function, called the characteristic equation the roots of the systemcan be derived.For all these reasons and more, the Transfer function is animportant aspect of classical control systems. Let's start out with the definition:
Transfer Function
The Transfer function of a system is the relationship of the system's output to it'sinput, represented in the complex Laplace domain.If the complex Laplace

variable is
s
, then we generally denote the transfer function of a system aseither
G(s)
or
H(s)
. If the system input is
X(s)
, and the system output is
Y(s)
, thenthe transfer function can be defined as such:If we know the input to a given system, and we have the transfer function of thesystem, we can solve for the system output by multiplying:
(
s
) =
(
s
)
X
(
s
)
Solution of the question:-
C(s)=R(s).G(s)C(s)1/s.[(s
+9s+19)/s
3
+7s
+14s+8] = s
+9s+19/s
3
+7s
+14s+8 Break the equations by partial fractionC(s)=11/3(s+1)-5/2(s+2)-1/6(s+4)Taking laplace inverse of each term,we get

11/3e
-t
-5/2e
-2t
-1/6e
-4t
Coding :-
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