Fundamentals of Automatic Control
Instructors: Dr. Helmy El-Zoghby
helmy_028123288@yahoo.com
2024/3/21 1
Lecture 9
Time response of control systems
(2-Second order systems)
2
2) Time Response of Second Order Control Systems
• We have already discussed the affect of location of poles and zeros on the
transient response of 1st order systems.
• Compared to the simplicity of a first-order system, a second-order system exhibits
a wide range of responses that must be analyzed and described.
• Varying a first-order system's parameter (T, K) simply changes the speed and
offset of the response
• Where as, changes in the parameters of a second-order system can change the
form of the response.
• 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.
3
2) Time Response of Second Order Control Systems
A general second-order system is characterized by the following
transfer function.
C(s) n2
2
R(s) s 2 n s n2
n un-damped natural frequency of the second order system,
which is the frequency of oscillation of the system without
damping.
damping ratio of the second order system, which is a
measure of the degree of resistance to change in the
system output.
2) Time Response of Second Order Control Systems
• For example : Determine the un-damped natural frequency and
damping ratio of the following second order system.
C( s ) 4
2
R( s ) s 2 s 4
• Compare the numerator and denominator of the given transfer
function with the general 2nd order transfer function.
C( s ) n2
2
R( s ) s 2 n s n2
2
n 4 n 2 rad / sec
2 n s 2s
2 2 2
n 1
s 2 n s n s 2s 4
0.5
2) Time Response of Second Order Control Systems
Examples of second order systems R L
1-RLC circuit: A second-order system example.
u C y
Y (s ) 1 1/ LC
2
U (s ) LCs RCs 1 s (R / L)s 1/ LC
2
2-Mass-Spring-Damper System: A second-order system
example.
y (t )
k
u (t )
Y (s ) 1 1 k /m f m
2 2
U (s ) ms fs k k s (f / m )s k / m
6
2) Time Response of Second Order Control Systems
- 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):
7
2) Time Response of Second Order Control Systems
- 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!
8
2) Time Response of Second Order Control Systems
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 (natural)- when the system has two imaginary poles (ζ = 0).
4. Critically damped - when the system has two real but equal poles (ζ = 1).
9
2) Time Response of Second Order Control Systems
Increasing ξ
Effect of different damping ratio, ξ
10
2) Time-Domain Specification of second order system
Under damping case and unit step input
Given that the closed loop TF
C (s) n 2
T (s) 2
s 2 n s n
2
R (s)
- Given a step input R(s) =1/s, then the system output
(or step response) is;
or
B cos 1 ( )
11
2) Time-Domain Specification of second order system
Under damping case and unit step input
1 s 2 n
C( s )
s s n 2 n2 1 2
• Above equation can be written as
1 s 2 n
C( s )
s s n 2 d2
• Where d n 1 2 , is the frequency of transient oscillations and is called
damped natural frequency.
• The inverse Laplace transform of above equation can be obtained easily if C(s) is
written in the following form:
1 s n n
C (s)
s s n 2
2
d s n
2
2
d
12
2) Time-Domain Specification of second order system
Under damping case and unit step input
1 s n n
C(s)
s s n d s n 2 d2
2 2
2
n 1
2
1 s n 1
C (s)
s s n 2 2
d s n 2 2
d
1 s n d
C( s )
s s n d
2 2
1 2 s 2 2
n d
n t
c(t ) 1 e cos d t e nt sin d t
1 2
13
2) Time-Domain Specification of second order system
Under damping case and unit step input
n t
c(t ) 1 e cos d t e nt sin d t
1 2
c(t ) 1 e nt cos d t sin d t
1 2
• For undamping case 0
2
d n 1
n
c(t ) 1 cos n t
14
2) Time-Domain Specification of second order system
Under damping case and unit step input
1.4
n t
c( t ) 1 e cos d t sin d t 1.2
1 2
1
0.8
1.8
0.6
1.6
0.4
1.4
0.2
1.2
0
0 2 4 6 8 10
1
0.8
if 0 . 5 and n 3
1.4
0.6
1.2
0.4 1
0.2 0.8
0 0.6
0 2 4 6 8 10
if 0.1 and n 3
0.4
0.2
0
0 2 4 6 8 10
if 0 . 9 and n 3
15
2) Time-Domain Specification of second order system
Under damping case and unit step input
n t cos d t
c(t ) 1 e sin d t
1 2
16
2) Time-Domain Specification of second order system
Under damping case and unit step input
For 0< <1 and ωn > 0, the 2nd order system’s response due to a unit step input is
as follows.
Important timing characteristics: delay time, rise time, peak time, maximum
overshoot, and settling time.
17
2) Time-Domain Specification of second order system
Under damping case and unit step input
• The delay (td) time is the time required for the response to reach half the final
value the very first time.
• The rise time is the time required for the response to rise from 10% to 90%, 5% to
95%, or 0% to 100% of its final value. For underdamped second order systems,
the 0% to 100% rise time is normally used. For overdamped systems, the 10% to
90% rise time is commonly used.
• The peak time is the time required for the response to reach the first peak of the
overshoot.
• The maximum overshoot is the maximum peak value of the response curve
measured from unity. If the final steady-state value of the response differs from
unity, then it is common to use the maximum percent overshoot. It is defined The
amount of the maximum (percent) overshoot directly indicates the relative
stability of the system.
• The settling time is the time required for the response curve to reach and stay
within a range about the final value of size specified by absolute percentage of
the final value (usually 2% or 5%). 18
2) Time-Domain Specification of second order system
Under damping case and unit step input
(a) Tr – rise time = d
n 1 2
(b) Tp – peak time = d
n 1 2
(c) %MP=OS% – percentage maximum overshoot = 2
1
e
3
(d) Ts – settling time (5% error) = n
Ts – settling time (2% error) = 4
n
19
2) Time-Domain Specification of second order system
Effect of poles location in transient response
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2) Time Response of Second Order Control Systems
Example.1
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2) Time Response of Second Order Control Systems
Example.1
22
2) Time Response of Second Order Control Systems
Example.2
23
2) Time Response of Second Order Control Systems
Example.2
24
2) Time Response of Second Order Control Systems
Example.2
25
2) Time Response of Second Order Control Systems
Example.3
• For the system shown in Figure, determine the values of gain K and
velocity-feedback constant Kh so that the maximum overshoot in the
unit-step response is 0.2 and the peak time is 1 sec. With these values of
K and Kh, obtain the rise time and settling time. Assume that J=1 kg-m2
and B=1 N-m/rad/sec.
26
2) Time Response of Second Order Control Systems
Example.3
27
2) Time Response of Second Order Control Systems
Example.3
Since J 1 kgm 2 and B 1 Nm/rad/sec
C( s ) K
2
R( s ) s (1 KK h )s K
• Comparing above T.F with general 2 nd order T.F
C( s ) n2
2
R( s ) s 2 n s n2
(1 KK h )
n K
2 K
28
2) Time Response of Second Order Control Systems
Example.3
(1 KK h )
n K 2 K
• Maximum overshoot is 0.2. • The peak time is 1 sec
tp
d
3.141
1
n 1 2
) ln0.2
1 2
ln( e
3.141
n
1 0.456 2
n 3.53
29
2) Time Response of Second Order Control Systems
Example.3
n 3.53
(1 KK h )
n K
2 K
3.53 K
0.456 2 12.5 (1 12.5 K h )
3.532 K K h 0.178
K 12.5
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2) Time Response of Second Order Control Systems
Example.3
tr 4
n 1 2
ts
n
t r 0.65s t s 2.48s
3
ts
n
t s 1.86s
31
2) Time Response of Second Order Control Systems
Step response by matlab
Unit step response of the transfer function system
25
Consider the system: T s
s 2 4 s 25
Unit Step Response of H(s)
1.4
%*****Numerator & Denominator of T(s)
>>num = [0 0 25];den = [1 4 25]; 1.2
%*****Specify the computing time
1
>>t=0:0.1:7;
>>step(num,den,t) 0.8
Amplitude
%*****Add grid & title of plot
0.6
>>grid
>>title(‘Unit Step Response of T(s)’) 0.4
0.2
0
0 1 2 3 4 5 6 7
Time (sec)
32
2) Time Response of Second Order Control Systems
Step response by matlab
Alternative way to generate Unit step response of the transfer function, T(s)
%*****Numerator & Denominator of T(s)
>>num = [0 0 25];den = [1 4 25];
%*****Create Model
>>T=tf(num,den);
>>step(T)
>>step(10*T)
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2) Time Response of Second Order Control Systems
Impulse response of the transfer function system
25
Consider the system: T s
s 2
4 s 25
Impulse Response of H(s)
3
%*****Numerator & Denominator of T(s)
>>num = [0 0 25];den = [1 4 25];
2.5
%*****Specify the computing time 2
>>t=0:0.1:7; 1.5
>>impulse(num,den,t)
A m p litu d e
1
%*****Add grid & title of plot 0.5
>>grid
0
>>title(‘Impulse Response of T(s)’)
-0.5
-1
0 1 2 3 4 5 6 7
Time (sec)
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THANKS
2024/3/21 35
