Program No. 2

Object-: Write a program to find out the unit step response for the given function-: S(s2+9s+19)/s3+7s2+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 a common analysis tool used to determine certain metrics about a system. Typically, when a new system is designed, the step response of the system is the first characteristic of the system to be analyzed.

Convolution

However, the impulse response cannot be used to find the system output from

the system input in the same manner as the transfer function. If we have the

system input and the impulse response of the system, we can calculate the

system output using the convolution operationas such:

y(t) = h(t) * x(t)Remember: an asterisk meansconvolution, not multiplication!

Where " * " (asterisk) denotes the convolution operation. Convolution is a

complicated combination of multiplication, integration and time-shifting. We can

define 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 operation

can be difficult to perform. Therefore, many people prefer to use the Laplace

Transform (or another transform) to convert the convolution operation into a

multiplication operation, through the Convolution Theorem.

Time-Invariant System Response

If the system in question is time-invariant, then the general description of the

system can be replaced by a convolution integral of the system's impulse

response 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 it

can 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, called

the Convolution Theorem, that makes the operation of convolution easier:

Convolution Theorem

Convolution in the time domain becomes multiplication in the complex Laplace

domain. Multiplication in the time domain becomes convolution in the complex

Laplace 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 and

Frequency response can all be taken from this specific function. Nyquist and

Bode plots can be drawn from the open loop Transfer Function. These plots

show 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 system

can be derived.For all these reasons and more, the Transfer function is an

important 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's

input, represented in the complex Laplace domain.If the complex Laplace

variable is s, then we generally denote the transfer function of a system as

either G(s) or H(s). If the system input isX(s), and the system output is Y(s), then

the transfer function can be defined as such:

If we know the input to a given system, and we have the transfer function of the

system, we can solve for the system output by multiplying:

Y(s) = H(s)X(s)

Solution of the question:-

C(s)=R(s).G(s)

C(s)1/s.[(s2+9s+19)/s3+7s2+14s+8]

= s2+9s+19/s3+7s2+14s+8

Break the equations by partial fraction

C(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 :-

num=[1 7 8]den=[1 14 12 4]g=tf(num,den)grid onsubplot(2,1,1)step(g)t=0:0.01:6c=11/2*exp(-t)-5/2*exp(-2*t)-1/6*exp(-4*t)grid onsubplot(2,1,2)plot(t,c)grid on

graph:-

0 5 10 150

1

2

3Step Response

Time (sec)

Ampli

tude

0 1 2 3 4 5 60

1

2

3

Solution of the problem:

The given transfer Function : S2+9s+9/s3 +7s2+14s+8

Taking first the denominator:s3 +7s2+14s+8s2(s+1)+6s(s+1)+8(s+1)(s+1)( s2+6s+8)(S+1)(s2+4s+2s+8)(s+1)(s(s+4)+2(s+4)(s+1)(s+4)(s+2)

S2+9s+9/s3 +7s2+14s+8=[a/s+1]+[b/s+4]+[c/s+2] S2+9s+9=A(s+4)(s+2)+b(s+1)(s+2)+c(s+1)(s+4)

Put s=-1We get ,A =11/3Putting S=-4We get,B=-5/2Putting S=-2We get,B=-1/6

Putting these values and taking the laplace transform of that we get, 11/3e-t -5/2e-2t-1/6e-4t

Result:- we have successfully found out the unit step respose of the given

transfer function .

Questions:-

1)Area under a unit impulse function ifAns) unity

2) Laplace transform of doublet is Ans) s

3) If the poles of a control system lie on the imaginary axis in s-plane the system will be :

Ans)marginally stable.

4) Roots of the charcterstics equations are nothing but: Ans) poles of a ststem.

5) In Mechanical system,the spring force is proportional to:Ans) Displacement