CONTROL ENGNEERING 1ST & 2ND Order System In S-Domain

WAQAR AHMED BETL/H/F10/0111

1st & 2nd Order System in S-Domain

Introduction:

The order of a system is defined as being the highest power of derivative in the differential equation, or being the highest power of s in the denominator of the transfer function. A first-order system only has s to the power one in the denominator, while a second-order system has the highest power of s in the denominator being two.

Response Analysis of First Order System Many systems are approximately first-order. The important feature is that the

storage of mass, momentum and energy can be captured by one parameter. Examples of first-order systems are velocity of a car on the road, control of the velocity of a rotating system, electric systems where energy storage is essentially in one capacitor or one inductor

If the dynamic relation between the reference input r(t) and output c(t) of a system is the form of

By taking a0 as a common, it can be written as

---------------- (i)

where r = 1/ 0 = time constant and K = 0/ 0 = gain of the system. Taking the Laplace transform of Eq. (i) and assuming all initial conditions = 0 we have

(Sτ + 1)C(s) = KR(s) This yields the transfer function as

G(s) =

τ

The order of the transfer function is 1. So, such systems are called first order systems. The block diagram and the signal flow graph of a first order system are given in Figure

CONTROL ENGNEERING 1ST & 2ND Order System In S-Domain

WAQAR AHMED BETL/H/F10/0111

Example 1: Mechanical system

m is the mass, u(t) is the external force, y(t) is the velocity and b is the friction coefficient. By Newton’s law, we have the following differential equation:

By taking Laplace, assuming initial conditions to zero, mSY(s)+bY(s) = U(S) G(s) = Y(s)/U(s) = 1/mS+b Example 2: Electrical system

R is the resistance, C is the capacitance, u(t) is the input voltage and Y(t) is the output voltage. By Kirchhoff’s law:

Thus

By taking laplace, assuming initial conditions to zero ,we have

RC SY(s) + Y(s)= U(s)

G(s)=Y(s)/U(s)= 1/(1+RCS)

Response Analysis of Second Order System

Second order system is given by,

By taking a0 as a common,

----------(ii)

CONTROL ENGNEERING 1ST & 2ND Order System In S-Domain

WAQAR AHMED BETL/H/F10/0111

Where

ωn , ξ, and k are called natural frequency of oscillation , damping ratio and static sensitivity

or (gain ) respectively . Now by taking Laplace Transform of equation (ii), assuming initial

conditions to zero, we get

By rearranging terms , we get the transfer function for the second order system as,

OR

This is a second order system , because power of s in the denominator is two.

Example 1: Mechanical system For the mechanical system shown in the figure, m is the mass, k is the spring

constant, b is the friction coefficient, u(t) is the external force and y(t) is the displacement. From Newton’s second law Σforce = ma.

By taking Laplace transform, assuming initial

conditions to zero, we get,

G(s)= Y(s)/U(s) = 1/mS2+bS+k

CONTROL ENGNEERING 1ST & 2ND Order System In S-Domain

WAQAR AHMED BETL/H/F10/0111

Example 2: Electrical system: RLC circuit

Using Kirchhoff’s law:

--------------(iii)

Where i(t)= C

Hence equation (iii) will be

By taking Laplace Transform , set initial conditions to zero,

we have transfer function G(s),

G(s) = Y(s)/U(s)= 1/( LCS2 + RCS + 1)