Second – Order System

 Second-order systems exhibit a wide range of

responses which must be analyzed and described.
• Whereas for a first-order system, varying a
single parameter changes the speed of response,
changes in the parameters of a second order
system can change the form of the response.

 For example: a second-order system can display

characteristics much like a first-order system or,
depending on component values, display damped
or pure oscillations for its transient response.
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 Second – Order System

- A general second-order system is characterized by

the following transfer function:

- We can re-write the above transfer function in the

following form (closed loop transfer function):

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 Second – Order System

- referred to as the un-damped natural

frequency of the second order system, which
is the frequency of oscillation of the system
without damping.

- referred to as the damping ratio of the

second order system, which is a measure of
the degree of resistance to change in the
system output.

Poles;
Poles are complex if ζ< 1!

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 Second – Order System

- According the value of ζ, a second-order system

can be set into one of the four categories:
1. Overdamped - when the system has two real
distinct poles (ζ >1).
2. Underdamped - when the system has two
complex conjugate poles (0 <ζ <1)
3. Undamped - when the system has two
imaginary poles (ζ = 0).
4. Critically damped - when the system has two
real but equal poles (ζ = 1).

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 Time-Domain Specification
Given that the closed loop TF

C ( s) n 2
T ( s)   2
R( s) s  2 n s  n 2
The system (2nd order system) is parameterized by ς and ωn

For 0< ς <1 and ωn > 0, we like to investigate its response

due to a unit step input

Two types of responses that

are of interest:
(A)Transient response
(B)Steady state response

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Transient Steady State
(A) For transient response, we
have 4 specifications:
 
(a) Tr – rise time =
n 1   2

(b) Tp – peak time =
n 1   2

(c) %MP – percentage maximum overshoot = 
1 2
e x100%
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(d) Ts – settling time (2% error) =
 n

(B) Steady State Response

(a) Steady State error

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 Question : How are the performance
related to ς and ωn ?
- Given a step input, i.e., R(s) =1/s, then the system output
(or step response) is;

- Taking inverse Laplace transform, we have the step

response;

Where; or   cos 1 ( )
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Second – Order System
C ( s) n 2
T ( s)   2
R( s) s  2 n s  n 2

Mapping the poles into s-plane

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Lets re-write the equation for c(t):

Let:   1  2
and

d  n 1   2 Damped natural frequency

n   d

Thus:
1
c(t )  1  e  n t
sin d t   

where   cos 1 ( )
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 Transient Response Analysis

1) Rise time, Tr. Time the response takes to rise from

0 to 100%
1
c(t ) t T r  1  e  n t
sin d t     1

0 0

sin(d Tr   )  0  
Tr 
d Tr    sin (0)  
1
n 1   2
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 Transient Response Analysis
2) Peak time, Tp - The peak time is the time required for
the response to reach the first peak, which is given by;

c(t ) 0
t T p


c(t )
t T p

1


(n )e nt sin(d t   )  1 e nt cos(d t   ) n 1   2  0 
n  T  1 2  nT p
sin(d Tp   )   
cos(d Tp   )
n
e n p
 e

1 
1 2
tan  
tan(d Tp   )   

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We know that tan( )  tan(   )
So, tan(d Tp   )  tan(   )

From this expression:

d T p      
d T p  

 
Tp  
d n 1   2
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 Transient Response Analysis
3) Percent overshoot, %OS - The percent overshoot is
defined as the amount that the waveform at the peak time
overshoots the steady-state value, which is expressed as a
percentage of the steady-state value.

C (Tp )  C ()
%MP  x100%
C ( )

OR

C max  Cfinal
%OS  x 100
Cfinal
C (Tp )  1 1
x100%   e  nt sin d t   x100%
1    
 n  
1   1 2      
 e  n 
sin  d      x100%

   d  


1
 e 1 2
sin    x100%

 
sin( ) 
1 2

1 2
 e x100%  e x100%


From slide 24 sin   1   2

  1  2

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 

Therefore, 1 2
%MP  e x100%

- For given %OS, the damping ratio can

be solved from the above equation;

 ln %MP / 100

  ln %MP / 100
2 2

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 Transient Response Analysis
4) Setting time, Ts - The settling time is the time
required for the amplitude of the sinusoid to decay
to 2% of the steady-state value.
To find Ts, we must find the time for which c(t) reaches & stays
within +2% of the steady state value, cfinal. The settling time is
the time it takes for the amplitude of the decaying sinusoid in c(t)
to reach 0.02, or

 nTs 1
e  0.02
1  2

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Thus, Ts 
 n
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 UNDERDAMPED

Example 2: Find the natural frequency and damping

ratio for the system with transfer function
36
G( s)  2
Solution: s  4.2s  36
Compare with general TF

•ωn= 6

•ξ =0.35

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 UNDERDAMPED
Example 3:
3: Given the transfer function

find Ts , %OS, T p
Solution:

n  10   0.75

Ts  0.533s, %OS  2.838%, T p 0.475s

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UNDERDAMPED

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Overdamped Response

a=9

 1
9 9
C ( s)  
s( s  9s  9) s( s  7.854)(s  1.146)
2

s= 0; s = -7.854; s = -1.146 ( two real poles)

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 7.854t 1.146t
c(t )  K1  K 2 e  K 3e

OVERDAMPED RESPONSE !!!

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 Underdamped Response

a=3

0   1

c(t )  K1  e 1.5t ( K 2 cos 2.598t  K 3 sin 2.598t )

s = 0; s = -1.5 ± j2.598 ( two complex poles)
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UNDERDAMPED RESPONSE !!!
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Undamped Response

a=0

 0

c(t )  K1  K 2 cos 3t
s = 0; s = ± j3 ( two imaginary poles)
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UNDAMPED RESPONSE !!!
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 Critically Damped System

a=6

 1
3t 3t
c(t )  K1  K 2 e  K 3te
S = 0; s = -3,
3,--3 ( two real and equal poles)
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CRITICALLY DAMPED RESPONSE !!!
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Second – Order System

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Effect of different damping ratio, ξ

Increasing ξ

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 Second – Order System
Example 4: Describe the nature of the second-order
system response via the value of the damping ratio for
the systems with transfer function

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1. G ( s)  2
s  8s  12

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2. G( s)  2 Do them as your
s  8s  16 own revision

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3. G ( s)  2
s  8s  20

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