This document discusses first and second order systems in the s-domain. It defines a first order system as having the highest power of s in the denominator of the transfer function as 1, while a second order system has the highest power of s as 2. Examples of first order systems include velocity of a car and rotating systems. The transfer function of a first order system is τ/(sτ+1). A second order system transfer function is ωn2/(s2+2ξωns+ωn2). Examples of second order systems include mechanical springs and electrical RLC circuits.

1. 1ST & 2ND Order System In S-Domain
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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
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2. 1ST & 2ND Order System In S-Domain
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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)
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1ST & 2ND Order System In S-Domain
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
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4. 1ST & 2ND Order System In S-Domain
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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)
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