ME 413 Systems Dynamics & Control
Chapter 8: Time Domain Analyis of Dynamic Systems Systems
Chapter 8
TimeTime-Domain Analysis of
Dynamic Systems
A. Bazoune
8.1
INTRODUCTION
Poles and Zeros of a Transfer Function
Poles: The poles of a transfer function are those values of s for which the function is undefined
(becomes infinite).
Zeros: The zeros of a transfer function are those values of s for which the function is zero.
Example of the Effect of Pole/Zero Locations
Figure 8-1.
From the development summarized above, the following conclusions can be drawn:
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ME 413 Systems Dynamics & Control
Chapter 8: Time Domain Analyis of Dynamic Systems Systems
A pole of the input function generates the form of the forced response. (i.e., the pole at
the origin generates a step function at the output)
►
A pole of the transfer function generates the form of the natural response (i.e., the pole at
generates
e
−σ t
)
−σ t
►
A pole on the real axis generates an exponential response of the form e , where
pole location on the real axis. Thus, the pole farther the left a pole is on the negative
−σ
is the
1
Pole-Zero Map
1
exp(-t)
exp(-2t)
exp(-3t)
exp(-4t)
0.9
0.8
0.8
0.6
exp(-t)
0.7
exp(-2t)
0.4
Response
0.6
0.2
I m a g A x is
−σ
exp(-3t)
0.5
0
exp(-4t)
0.4
-0.2
0.3
-0.4
0.2
-0.6
0.1
-0.8
0
-1
-4.5
0
1
2
3
4
5
t
-4
-3.5
-3
-2.5
-2
-1.5
-1
-0.5
0
Real Axis
Standard Form of the First Order System Equation
B
1
y +   y = u (t )
τ
τ 
(1)
The transfer function of the previous system is defined by
G(s) =
Y (s)
B /τ
=
U ( s ) s + (1/ τ )
where τ is known as the time constant. It has dimensions of time for all physical systems described
by the first order differential equation above. The above equation can be respresented as
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ME 413 Systems Dynamics & Control
Chapter 8: Time Domain Analyis of Dynamic Systems Systems
G ( s)
U ( s)
Y (s)
Transfer Function
Examples
i)
Spring-Damper System
Equation of motion
x
by + ky = k x
or
y +
1
k
y=  x
b
b
 
k
y
k
b
Comparing the above equation with the standard one, so that
τ = b / k = time constant
with the units of
[τ ] = [b / k ] =
ii)
[ N.s/m ] = s
[]
[ N/m]
RC-Circuit
Equation of motion
RCeo + eo = ei
τ = RC = time constant
R
with the units of
ei i
[ V.s ] × [q ] = s
[τ ] = [ RC ] =
[]
[q] [V]
iii)
eo
C
Liquid Level System
Equation of motion
dh
1
1
+
h= q
dt RC
C
τ = RC = time constant
Q +qi
with the units of
[τ ] = [ RC ] =
iv)
[m]
 m / s 
3
Load valve
×  m 2  = [s ]
H +h
Capacitance C
Q + qο
Resistance R
Thermal System
Equation of motion
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ME 413 Systems Dynamics & Control
Chapter 8: Time Domain Analyis of Dynamic Systems Systems
dθ
1
1
+
θ = qi
dt RC
C
τ = RC = time constant
 o C 
[ Kcal] = s
with the units of [τ ] = [ RC ] =
× o
[]
[ Kcal / s ]  C 
θ = Oven
Temperature
qi
qo =
θ
R
Thus when a first order system is written in the form of
Equation (1) with the coefficient of dy / dt is equal to 1 and the coefficient of the dependant variable
is
(1/ τ ) in which τ
represents the time constant of the system and has always the dimension of
time regardless of the physical system under consideration.
Interest in Analysis of Dynamic Systems
After obtaining a model of a dynamic system, one then needs to apply test signals to see if the system
performs accordingly to certain design specifications put forward by the design of the system. In
general, one requires the system to be
1. stable ( system does not grow out unbounded )
2. with a fast response
3. has a small error as possible (steady sate error).
Other performance criteria may also exist but we will be mainly concerned with the above
mentioned three. Usually the actual input to the dynamic system is unknown in advance; however,
by subjecting the system to standard test signals, we can get an indication of the ability of the system
performance under actual operating conditions. For example, the information we gain by analyzing
the system stability and its speed of response and steady state error due to various types of standard
test signals, will give an indication on the system performance under actual operating condition.
Typical Test Signals
i)
Step Input
Very common input to actual dynamic systems. As it represents a sudden change in the value of the
reference input
u (t )
1
t
ii)
Ramp Input (Constant Velocity)
This represents a situation where the input has a constant rate of increase with time (i.e. constant
velocity)
u (t ) = rt
slope = tan α = r
α
t
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ME 413 Systems Dynamics & Control
iii)
Chapter 8: Time Domain Analyis of Dynamic Systems Systems
Parabolic Input (Constant Acceleration)
Occurs in situation where the input has a constant acceleration.
u (t )
u (t ) = at 2
t
iv)
Impulsive Load Input
A large load over a short duration of time
impulse.
<< 0.1τ
= system time constant can be considered as an
Fˆ δ ( t ) , Fˆ = F ∆t = strength of impulse
u (t )
u (t )
∆t
u (t ) = Fˆ δ t
u (t ) = ∆t F
= impulse
F
t
t
v)
Sinusoidal Load Input
4
 A sin ωt
u (t ) = 
 B cos ωt
A cos(ωt)
B sin(ωt)
3
2
1
0
-1
-2
-3
-4
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
Natural and Forced Response
The solution
y (t )
to the homogeneous differential equation (1) is composed of two parts:
■
Complementary solution:
■
Particular solution :
such that:
y p (t )
yc ( t )
natural response due to initial conditions.
forced response
y ( t ) = y p ( t ) + yc ( t )
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ME 413 Systems Dynamics & Control
Chapter 8: Time Domain Analyis of Dynamic Systems Systems
Transient and Steady State Response
■
Transient response: response from initial state to final state.
Steady sate Response: response as time approaches ∞ .
■
Step Response
0.2
0.18
0.16
0.14
Amplitude
0.12
0.1
0.08
0.06
Steady State
0.04
Transient
0.02
0
0
5
10
15
Time (sec)
Notice that if
lim yc ( t ) = 0 and lim y p ( t ) = a bounded function of time then the system is said
t →∞
t →∞
to be steady state; where the steady state solution is
yss = lim yc ( t ) = lim y p ( t )
t →∞
8.2
t →∞
TRANSIENT RESPONSE ANALYSIS OF FIRST-ORDER
SYSTEM
Rotor mounted in bearings is shown in the figure below. External torque
T (t )
is applied to the
system.
b
ω
T (t )
J
J
ω
bω
T (t )
Apply Newton’s second law for a system in rotation
∑ M = Jθ = J ω
J ω + bω = T ( t ) ⇒ ω + ( b / J ) ω = T ( t ) / J
or
ω +
Define the time constant τ
1
ω = T * (t )
( J / b)
= ( J / b ) , the previous equation can be written in the form
1
ω + ω = T * ( t ) , ω ( 0 ) = ω
τ
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ME 413 Systems Dynamics & Control
Chapter 8: Time Domain Analyis of Dynamic Systems Systems
which represents the equation of motion as well as the mathematical model of the system shown. It
represents a first order system.
Free Response
To find the response
T (t ) = 0
ω ( t ) , take LT of both sides of the previous equation.




 sΩ ( s ) − ω ( 0 )  + 1 Ω ( s )  = 0
 τ  
 L[ω ]
 L[ω ] 


1

 s +  Ω ( s ) = ω
 τ
1

Ω ( s ) = ω  s + 
τ

⇒
Taking inverse LT of the above equation will give the expression of
ω (t )
ω ( t ) = ω e−(b / J ) t = ω e−(t /τ )
It is clear that the angular velocity decreases exponentially as shown in the figure below. Since
− t /τ
lim e ( ) = 0; then for such decaying system, it is convenient to depict the response in terms of a
t →∞
time constant.
A time constant is that value of time that makes the exponent equal to -1. For this system, time
constant τ = J / b . When t = τ , the exponent factor is
e
−( t / τ )
=e
−(τ / τ )
= e −1 = 0.368 = 36.8 %
This means that when time constant = τ , the time response is reduced to
We also have
τ = J / b = time constant
ω (τ ) = 0.37 ω
ω ( 4τ ) = 0.02 ω
ω (t )
ω
0.37 ω
0.02 ω
t
τ
2τ
7/33
3τ
4τ
36.8 %
of its final value.
ME 413 Systems Dynamics & Control
Forced Response
Chapter 8: Time Domain Analyis of Dynamic Systems Systems
T (t ) ≠ 0
Remember
1
ω + ω = T * ( t )
τ
To find the response
ω ( t ) , take LT of both sides of the previous equation for zero initial conditions.
1
s Ω (s ) − ω ( 0) + Ω (s ) = T * (s )
τ
=0
L [ω ]
L [ω ]
1

*
 s + Ω(s) = T (s)
 τ
Ω(s)
1
=
*
1
T ( s) 
s + 
 τ
⇒
The above equation can be written in the more general form as
where
C (s)
is the output or
C (s)
1
=
1
R(s) 
s + 
 τ
the response and R ( s ) is the
input or reference signal. The above
equation can be represented as
R(s)
G (s) =
1
s + (1/τ )
C ( s)
Transfer Function
i)
Impulse Response
In this case, for an unit impulse input of magnitude r ( t ) = Bδ ( t ) ,
R(s) = B
and the above equation can be written in the form
C (s) =
B
1

s + 
 τ
from which
c (t ) = B e
The figure below shows the response
c (t ) = B e
−(1/ τ )t
−(1/ τ )t
( )
. Since we assumed zero I.C’s, the output
( )
must change instantaneously from 0 at time t = 0− to 1 at time t = 0+ .
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ME 413 Systems Dynamics & Control
Chapter 8: Time Domain Analyis of Dynamic Systems Systems
c (t )
B
c (t ) = B e
−(1/ τ )t
t
Figure . Impusle response of a first order system
ii)
Step Response
In this case, for a unit step input of magnitude B ,
R(s) = B s
and the above equation can be written in the form
C (s) =
B
1

s s + 
 τ
a1
a2
+
1
s 
s + 
 τ
=
where
a1 =
Bs
1

s s + 
 τ  s =0
= Bτ
and
1

B s + 
τ
a2 = 
= −B τ
1

s s + 
 τ  s =− 1
τ
Therefore
C (s) =
B
1

s s + 
 τ
=
Bτ
Bτ
−
1
s

s + 
 τ
for which
c (t ) = B τ − B τ e
The response
c (t )
−(1/ τ )t
is represented in the Figure below.
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(
= Bτ 1− e
−(1/ τ )t
)
ME 413 Systems Dynamics & Control
Chapter 8: Time Domain Analyis of Dynamic Systems Systems
c (t )
slope of the tangent at origin =
dc (t )
=B
dt t = 0
Bτ
0.98 B τ
0.9 B τ
Rise Time, Tr : The time for the response
to go from 10% to 90% of its final value.
Tr = 2.2 *τ
0.63 B τ
Settling Time, Ts : The time for the response
to reach, and stay within, ± 2% of its final value.
Ts = 4 *τ
0.1 B τ
τ
3τ
2τ
5τ
4τ
t
Tr
Ts
Figure Step response of a first order system
Time constant: It is the time for
e
−
t
τ
to decay to
e
−
37 %
of its final value, i.e.,
t
= e −1 = 0.37
τ
t =0
Alternatively, the time constant is the time it takes for a step response to rise to
value, i.e.,
−(1/ τ )t
−1
(
)
c ( t ) t =τ = B τ 1 − e
Rise Time: time for the response to go from
t =τ
10 %
= B τ (1 − e
to
90 %
63 %
) = 0.63B τ
of its final value. The rise time
found by solving the expression for the step response for the difference in time
c ( t ) = 0.9
c ( t ) = 0.1 , that is
(
B τ 1− e
−(1/ τ )t
) = 0.9B τ
or
(1 − e ( ) ) = 0.9
− 1/ τ t
e
−(1/ τ )t
= 0.1 ⇒ − ( t / τ ) = ln(0.1) = −2.302 ⇒ t0.9 = 2.302 τ
similarly
(
Bτ 1− e
−(1/ τ )t
) = 0.1B τ
or
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of its final
Tr
is
and
ME 413 Systems Dynamics & Control
Chapter 8: Time Domain Analyis of Dynamic Systems Systems
(1 − e ( ) ) = 0.1
− 1/ τ t
e
−(1/ τ )t
= 0.9 ⇒ − ( t / τ ) = ln(0.9) = −0.105 ⇒ t0.1 = 0.105 τ
Hence , the rise time is
Tr = t0.9 − t0.1 = 2.2 τ
Settling Time: time for the response to reach, and stay within
time
Ts
±2%
of its final value. The settling
c ( t ) = 0.98 . Thus,
is found by solving the expression
(
−(1/ τ )t
−(1/ τ )t
= 0.02 ⇒ − (1/ τ ) t = ln(0.02)
c (t ) = B τ 1 − e
) = 0.98B τ
or
(1 − e ( ) ) = 0.98
− 1/ τ t
⇒ e
or
t = Ts = τ ln(0.02) = 4 τ
Remarks:
1. The smaller the time constant τ , the faster is the response and the furthest is the pole of
B
s ( s + 1/ τ )
C (s) =
2. Steady state error
ess
s − plane
= 0.
lim c ( t ) = lim c p ( t )
due to step input
3. The system is stable, i.e.,
provided that the pole
the complex plane.
iii)
Im
t →∞
s = −1/ τ
t →∞
Re
s = −1/ τ
lies on the left half of
Ramp Response
In this case, for a ramp input of slope B ,
r (t ) = B t
and the expression of
C (s)
⇔
R ( s ) = L  r ( t )  =
B
s2
above can be written in the form
C (s) =
B
1

s2  s + 
 τ
=
a1 a2
b
+ 2+
1
s s

s + 
 τ
where
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ME 413 Systems Dynamics & Control
Chapter 8: Time Domain Analyis of Dynamic Systems Systems






2

d  Bs
−B 

a1 = 
 =
= −B τ 2
2
1

ds  2 
1 
s  s + 
s
+
  τ   s =0   τ  

 s =0




 Bs 2

 B 
a2 = 
 =
 = Bτ
 s 2 s + 1 
 s + 1 
  τ   s =0   τ   s =0
 
1
 B s + τ  
B

b= 
= 2
= Bτ 2
 s  s =−1/ τ
 s2  s + 1  


  τ   s =−1/ τ
Therefore
Bτ 2 Bτ
Bτ 2
C (s) =
=−
+ 2 +
1
1
s
s

2
s s + 
s + 
 τ
 τ
B
Hence
c (t ) = L−1 C ( s )  = −B τ 2 + B τ t + B τ 2e
−(1/ τ ) t
(
= B τ t − τ + τe
c (t )
r (t ) = B t
Bτ
τ
(
c (t ) = B τ t − τ + τ e
−(1/ τ ) t
)
t
Figure
Ramp response of a first order system
12/33
−(1/ τ ) t
)
ME 413 Systems Dynamics & Control
8.3
Chapter 8: Time Domain Analyis of Dynamic Systems Systems
TRANSIENT RESPONSE ANALYSIS OF SECOND-ORDER
SYSTEM
Some Examples of Second Order Systems
Free Vibration without damping
Consider the mass spring system shown in Figure 3-11. The equation of motion can be given
by
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ME 413 Systems Dynamics & Control
Chapter 8: Time Domain Analyis of Dynamic Systems Systems
m x+k x=0
or
x+
k
x = 0 ⇒ x + ωn2 x = 0
m
k
where
k
m
ωn =
m
is the natural frequency of the system and is expressed in rad/s.
Taking LT of both sides of the above equation where
and
x ( 0 ) = x
x ( 0 ) = x
x (t )
Figure 8-11 Mass Spring
System
gives
s 2 X ( s ) − sx ( 0 ) − x ( 0 ) + ωn2 X ( s ) = 0
L [ x]
rearrange to get
X (s) =
sx + x
, ⇒ Remember poles are s = ± jωn
s 2 + ωn2
complex conjugates
X (s) =
and the response
x (t )
ωn
s
+ x 2
2
ωn s + ωn
s + ωn2
x
2
is given by
x (t ) =
x
ωn
sin (ωnt ) + x cos (ωnt )
x ( t ) consists of a sine and cosine terms and depends on the
values of the initial conditions x and x . Periodic motion such that described by the above
It is clear that the response
equation is called simple harmonic motion.
x (t )
slope = x
Im
s − plane
ωn
t
Re
−ωn
Period = T =
Figure 8-12
if
2π
ωn
Free response of a simple harmonic motion and pole location on the s-plane
x ( 0 ) = x = 0,
x ( t ) = x cos (ωnt )
Free Vibration with Viscous damping
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ME 413 Systems Dynamics & Control
Chapter 8: Time Domain Analyis of Dynamic Systems Systems
Damping is always present in actual mechanical systems, although in
some cases it may be negligibly small. Consider the mass spring
damper system shown in the figure. The equation of motion can be
given by
m x + bx + k x = 0
(1)
the characteristic equation of the above equation is
ms 2 + bs + k = 0
k
b
m
(2)
and the two roots of this equation are
s1,2 =
−b ± b 2 − 4 m k
2m
x(t )
(3)
Figure 8-13
We consider three cases:
•
•
•
b2 − 4 m k < 0
b2 − 4 m k = 0
b2 − 4 m k > 0
Roots are complex conjugates (underdamped case)
Roots are real and repeated
(critically damped case)
Roots are real and distinct (overdamped case)
In solving equation (1) for the response
ωn =
s1 = s2
x ( t ) , it is convenient to define
k
= undamped natural frequency, [ rad/s ]
m
and
ζ = dampin gratio =
actual damping value
b
=
critical damping value 2 km
and rewrite equation (2) in the form
s 2 + 2ξωn s + ωn2 = 0
(4)
which is the standard form equation of a second order system.
i)
Underdamped Case 0 < ξ < 1
where
x ( 0 ) = x
and
x ( 0 ) = x , and
X (s) =
Taking LT of both sides of equation (1)
rearrange to get
( s + 2ξωn ) x + x
(5)
s 2 + 2ξωn s + ωn2
knowing that equation (4) can be written as
2
(
s + 2ξωn s + ω = ( s + ξωn ) + ωn 1 − ξ
2
2
n
wich is a complete square equation. The nature of the roots
varying values of damping ratio
ξ
2
s1
)
2
=0
and
s2
of equation (4) with
can be shown in the complex plane as shown in the figure
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ME 413 Systems Dynamics & Control
Chapter 8: Time Domain Analyis of Dynamic Systems Systems
below. The semicircle represents the locus of the roots
in the range
s1
s2
and
for different values of
ξ
0 < ξ <1
ξ =0
ωn
s1
ω
n
ωn
s1 = s 2 = −ωn
ξ =1
−ξω
s2
s1
for ξ > 1
for ξ > 1
1−ξ
2
n
s 1 = −ξω n + ω n 1 − ξ 2
s 2 = −ξω n − ω n 1 − ξ 2
s2
−ω n
ξ =0
Define
ωd = ωn 1 − ζ 2 = damped natural frequency (rad/s)
The relationship between
ζ
and the non-dimensional frequency
(ωd / ωn ) is shown
the figure below.
1
Nondimentional frequency ω /ω
d n
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.2
0.4
0.6
Damping ratio ζ = b/b
0.8
1
cr
Figure
Non-dimensional frequency versus the damping ratio.
Then
X (s) =
( s + ξωn ) x
ξωn x + x
ωd
×
+
2
2
2
2
ωd
( s + ξωn ) + (ωd ) ( s + ξωn ) + (ωd )
from which
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ME 413 Systems Dynamics & Control
Chapter 8: Time Domain Analyis of Dynamic Systems Systems
x ( t ) = L−1  X ( s )  =
ξωn x + x −ξω t
e
sin ωd t + xe −ξω t cos ωd t
ωd
n
n
(6)
or
 ξ

x 
x (t ) = e
x +
 sin ωd t + x cos ωd t 

2
ωd 
 1 − ξ

If the initial velocity x ( 0 ) = 0, the above equation reduces to
−ξωnt
 ξ
x ( t ) = x e −ξωnt 
2

 1 − ξ

 sin ωd t + cos ωd


(7)

t

(8)
or
x ( t ) = C e −ξωnt sin (ωd t + φ )
(9)
where
φ = tan
−1
1− ξ 2
ξ
x
C=
and
1− ξ
x (t )
(10)
2
Im
s − plane
Ce −ξωnt
s1 = −ξ ωn + j ωd
ωn
t
jωd
Re
−ξωn
− jωd
Damped period, Td =
2π
ωd
s2 = −ξ ωn − j ωd
Remarks:
Notice that for this case (undedamped case
0 < ξ < 1)
1. the response is a decaying sinusoid.
2. the frequency of oscillations is
3. For positive damping
(
ωd = ωn 1 − ξ 2
(ξ > 0 ) , the poles
s1
)
.
and
s2
have negative real and lie
entirely on the let half of the complex plane. As a result the transient response decays
with time and the system is said to be stable.
4. The rate at which the transient response decays depends on the coefficient
in
e −ξωnt .
Larger
faster decay of
ξωn
(i.e., smaller
ξωn
of
t
1 / ξωn ) leads to faster transient response (i.e.,
x ( t ) ). The term 1 / ξωn
17/33
is in this case the time constant of the second
ME 413 Systems Dynamics & Control
Chapter 8: Time Domain Analyis of Dynamic Systems Systems
order system. Therefore, the time constant of the second order system can be made
smaller (i.e., its speed faster) by moving the real part
origin of the complex plane.
ii)
Critically damped Case ζ = 1
−ξωn
farther away from the
In this case, the poles the poles
s1 and s2
become
s1 = s2 = −ξωn
and the response
x ( t ) can be obtained from equation (5). Thus
( s + 2ωn ) x + x = ( s + ωn ) x + ωn x + x X (s ) = 2
2
s + 2ωn s + ωn2
( s + ωn )
=
ω x + x x
+ n ( s + ωn ) ( s + ωn ) 2
from which
x ( t ) = x e−ξωnt + (ωn x + x ) te−ξωnt
which is decaying exponentially as shown in the figure below
x (t )
Im
x
s − plane
Re
t
−ξ ωn
iii)
Overdamped Case
both real
ξ >1
In this case, the poles the poles
s1 and s2
s1 = −ξωn + ωn ξ 2 − 1
s2 = −ξωn − ωn ξ 2 − 1
and the response
x ( t ) becomes
 −ξ + ξ 2 − 1
 −(ξωn +ωn
x
x (t ) = 
x −
e
2
2ωn ξ 2 − 1 
 2 ξ − 1
 ξ + ξ 2 − 1
 −(ξωn −ωn
x
+
x +
e
2
2ωn ξ 2 − 1 
 2 ξ − 1
where the response is shown in the figure below
18/33
)
ξ 2 −1 t
)
ξ 2 −1 t
are
ME 413 Systems Dynamics & Control
Chapter 8: Time Domain Analyis of Dynamic Systems Systems
x (t )
Im
Increasing
ξ
x
s − plane
Re
t
− s2
− s1
Remarks:
ξ > 1 ) is similar to that of the first order system
and is the sum of two exponentials. The first has a time constant 1 s1 and the other 1 s2 .
The difference between these two time constants increases as the ξ increases so that the
exponential term corresponding to the smaller one (i.e., 1 s2 ) decays much faster than that
corresponding to 1 s1 . Under such case the second order system may be approximated by a
first order one with time constant equals to 1 s1 . From study of the first order system we
found that the response remains within 2 % of its final value in t > 4 time constants = 4τ
(τ = time constant). For a second order underdamped system τ = 1 ζωn and the time
required for the solution to remain within 2 % of its final value is called the settling time Ts
The response in this case (overdamped case
which from above is given by
Ts = 4τ = ( 4 ζωn )
Free Response of a Second Order System by MATLAB
MATLAB PROGRAM:
>> wn=1;
>> zeta=[0.2 0.5 0.7 1 2 5];
>> for k=1:6
num=[0 0 wn^2];
den=[1 2*zeta(k) wn^2];
sys=tf(num,den)
impulse(sys);
hold on
>> end
19/33
ME 413 Systems Dynamics & Control
Chapter 8: Time Domain Analyis of Dynamic Systems Systems
Free Response of a second order system at different values of the damping ratio ζ
15
ζ = 0.05
(Underdamped )
ζ = 0.5
( Underdamped)
ζ = 0.7
(Underdamped)
10
ζ= 1
(Critically damped)
Free Response x(t)
5
ζ = 2 (Overdamped)
0
ζ= 5
(Overdamped)
-5
-10
-15
0
1
2
3
Time (s)
20/33
4
5
ME 413 Systems Dynamics & Control
Chapter 8: Time Domain Analyis of Dynamic Systems Systems
Experimental Determination of damping ratio
(Logarithmic Decrement)
It is sometimes necessary to determine the damping ratios and damped natural frequencies of
recorders and other instruments. To determine the damping ratio and damped natural
frequency of a system experimentally, a record of decaying or damped oscillations, such as
that shown in the Figure below is needed.
21/33
ME 413 Systems Dynamics & Control
Chapter 8: Time Domain Analyis of Dynamic Systems Systems
x (t )
Td =
x1
2π
ωd
t2 = t1 + Td
x2
x3
xn
t
t1
t2
t3
tn
Td =
2π
ωd
x1 = x1 ( t ) = C e −ξωnt1 cos (ωd t1 − φ )
x2 = x2 ( t ) = C e−ξωnt2 cos (ωd t2 − φ )
The ratio
x2
x1
is equal to
cos (ωd t2 − φ )
x2
ξω t −t
= e n ( 2 1) ×
x1
cos (ωd t1 − φ )
since
t1
and
t2
are selected
Td
seconds apart, one can write
cos (ωd t2 − φ ) = cos (ωd ( t1 + Td ) − φ )
= cos (ωd t1 + ωd Td − φ )
= cos (ωd t1 + 2π − φ )
= cos (ωd t1 − φ )
Hence
x1
= eξωnTd
x2
The Logarithmic Decrement δ is defined as the natural logarithm of the ratio of any
two successive displacement amplitudes , so that by taking the natural logarithm of both
sides of the above equation
 xj 
2π
= ζ ωn Td = ζ ωn
=

2
x 
ωn 1 − ζ
 j +1 
δ = ln 
solving the above equation for
ζ
,
22/33
2πζ
1−ζ
2
(*)
ME 413 Systems Dynamics & Control
Chapter 8: Time Domain Analyis of Dynamic Systems Systems
ζ =
Notice from Eq. (*) that if
ζ << 1
δ
2
( 2π ) + δ 2
(that is, very low damping, which quantitatively means
b << bcr ), 1 − ζ 2 ≈ 1 and thus
δ = 2πζ
(**)
The figure below shows a comparison between Eqs. (*) and (**) versus the damping ratio
ζ
.
12
Logaritmic decrement δ
10
8
2 0.5
δ = 2 π ζ / (1-ζ )
6
δ=2 π ζ
4
2
0
0
0.1
0.2
0.3
0.4
0.5
0.6
Damping ratio ζ
0.7
x1
and
For non-successive amplitudes, say for amplitudes
observe that
0.8
0.9
1
xn+1 , where n
is an integer, we
x1
x x x x
x
= 1 2 3 4 n
xn +1 x2 x3 x4 x5 xn+1
Taking the natural logarithm of both sides of the above equation gives
 x 
x 
 x 
x 
ln  1  = ln  1  + ln  2  + + ln  n 
 x2 
 xn +1 
 x3 
 xn +1 
= δ + δ + + δ = nδ
So
1
n
 x1 

 xn+1 
δ = ln 
Table-1
or
δ=
x 
1
ln  1 
n − 1  xn 
Logarithmic decrement for Various Types of Structures
Types of Structures
Approximate Range of
23/33
ME 413 Systems Dynamics & Control
Chapter 8: Time Domain Analyis of Dynamic Systems Systems
Logarithmic Decrement,
0.02 0.10
0.05 0.15
0.10 0.20
0.1 0 0.30
0.4 0 0.60
Multistory Steel Buildings
Steel Bridges
Multistory Concrete Buildings
Concrete Bridges
Machinery Foundations
δ
Example
Step Response of a Second Order system:
Consider the mechanical system shown in the Figure below. Assume that the system
is at rest for t < 0. At t = 0 , the force u = a • 1(t ) [where a is a constant and 1(t ) is a step force
of magnitude 1 N] is applied to the mass m . The displacement is measured from the
equilibrium position before the input force u is applied. Assume that the system is
underdamped
(ζ
< 1)
x
k
m
b
The equation of motion for the system is
24/33
u
ME 413 Systems Dynamics & Control
Chapter 8: Time Domain Analyis of Dynamic Systems Systems
m x + bx + k x = a i1(t )
The TF for the system is
X (s)
1
=
2
U ( s ) m s + bs + k
Hence
a
X (s)
a
m
=
=
2
b
L 1( t )  m s + bs + k s 2 + s + k
m
m
Define
ωn =
k
= undamped natural frequency, [ rad/s ]
m
and
ζ = dampin gratio =
actual damping value
b
=
critical damping value 2 km
Then
X (s)
L

ωn2
a 
=
2 
2
2 
1( t )  mωn  s + 2ξωn s + ωn 
(8-16)
Hence,

ωn2
a 1
X (s) =


mωn2  s s 2 + 2ζωn s + ωn2 


s + 2ζωn
a 1
=
− 2
2
mωn  s s + 2ζωn s + ωn2

⇓







2
(
= ( s + ζωn ) + ωn 1 − ζ 2
)
2
2
= ( s + ζωn ) + ωd2

ζωn
s + ζωn
a 1


−
−
2
mωn2  s ( s + ζωn )2 + ωd2
( s + ζωn ) + ωd2 

ζωn * ωd
s + ζωn
a 1 1
=
 −
−
2
2
2
mωn  s ωd ( s + ζωn ) + ωd2
s
+
ζω
+ ωd2
(
)
n

=
25/33




ME 413 Systems Dynamics & Control
where
ωd = ωn 1 − ζ 2
x (t ) =
Chapter 8: Time Domain Analyis of Dynamic Systems Systems
. The inverse Laplace transform of the last equation gives

a 
ζ
−ζωnt
−ζωnt
ω
ω
1
−
e
sin
t
−
e
cos
t


d
d

mωn2 
1− ζ 2

a
=
mωn2
=


1 − e −ζωnt 




sin ωd t + cos ωd t  
2

1−ζ

ζ
2

a 
e −ζωnt
−1 1 − ζ

ω
1
sin
tan
−

t
+
2
 d
mωn2 
ζ
1
−
ζ






x ( ∞ ) = a mωn2 . The general shape
The response starts from x(0) = 0 and reaches
of the response curve is shown in the figure below.
Step Response
1.4
1.2
2
a/mω n
Amplitude
0.8
0.6
0.4
Figure 8-11
Step response of a
second order system. The response
curve shown corresponds to the
case where ζ = 0.7 and ωn = 2
0.2
0
0
1
2
3
4
5
6
7
8
Time (sec)
rad/s.
Assume that the system is underdamped
MATLAB PROGRAM:
(ζ
< 1)
>> wn=1;
>> zeta=[0.2 0.5 0.7 1 2 5];
>> for k=1:6
num=[0 0 wn^2];
den=[1 2*zeta(k) wn^2];
sys=tf(num,den)
step(sys);
hold on
x(t )
u (t )
?
1
U
(s)
Input
26/33
t
ωn2
s2 + 2 ζ ωns + ωn2
t
X (s)
Output
ME 413 Systems Dynamics & Control
Chapter 8: Time Domain Analyis of Dynamic Systems Systems
>> end
Step Response
1.6
ζ = 0.2
1.4
0.5
1.2
0.7
1
Amplitude
1
0.8
2
0.6
5
0.4
0.2
0
0
2
4
6
8
10
12
14
16
18
20
Time (sec)
Impulse Response of a Second Order system:
MATLAB PROGRAM:
>> wn=1;
>> zeta=[0.2 0.5 0.7 1 2 5];
>> for k=1:6
num=[0 0 wn^2];
den=[1 2*zeta(k) wn^2];
sys=tf(num,den)
impulse(sys);
hold on
>> end
x(t )
u (t )
?
1
U
( s)
Input
27/33
t
ωn2
s 2 + 2 ζ ωn s + ωn2
t
X (s)
Output
ME 413 Systems Dynamics & Control
Chapter 8: Time Domain Analyis of Dynamic Systems Systems
Impulse Response
1
0.8
ζ = 0.2
0.6
ζ = 0.5
ζ = 0.7
Amplitude
0.4
ζ = 1.0
0.2
ζ = 5.0
ζ = 2.0
0
-0.2
-0.4
0
5
10
15
Time (sec)
28/33
20
25
30
ME 413 Systems Dynamics & Control
Chapter 8: Time Domain Analyis of Dynamic Systems Systems
Table 8.1
y +
Input
1
τ
The Response of the First Order Linear System
y = u ( t ) ,
y ( 0 ) = y
Response
u ( t ) = B t
y (t ) = B τ t − τ + τ e
u ( t ) = B ,
(Step of magnitude
B)
u ( t ) = B * δ ( t )
−( t / τ )
y ( t ) if y ( 0 ) = 0
-
(
y ( t ) = y e
Response
−( t / τ )
y ( t ) = y e
B)
y ( t ) = output and u ( t ) = input
y ( t ) if y ( 0 ) = y
u ( t ) = 0
(Ramp of Slope
where
− t /τ )
(
+ Bτ 1− e
y ( t ) = ( B + y ) e
)+ y e (
−(1/ τ ) t
−(1/ τ )t
)
(
y (t ) = B τ t − τ + τ e
(
y (t ) = B τ 1 − e
−(1/ τ )t
y (t ) = B e
(Impulse of
magnitude B )
29/33
−(1/ τ )t
−(1/ τ )t
−(1/ τ ) t
)
)
ME 413 Systems Dynamics & Control
Chapter 8: Time Domain Analyis of Dynamic Systems Systems
Table 8.2
The Free Response of the Second Order Linear System
mx + bx + kx = 0, x ( 0 ) = x and x ( 0 ) = x
k
rad/s
m
b
b
The damping ratio ζ =
=
bcr 2 mk
The natural frequency
ωn =
The damped frequency
ωd = ωn 1 − ζ 2
Damping
ratio
for
0 ≤ ζ <1
Response
y (t )
  ζ

x 
x ( t ) = e−ζωnt  
x +  sin ωd t + x cos ωd t 
2
ωd 
  1 − ζ

If
0 ≤ ζ <1
x ( 0 ) = x = 0 , this simplifies to
 ζ

sin ωd t + cos ωd t 
x ( t ) = xe−ζωnt 
2
 1 − ζ


1 − ζ 2 
x
=
e−ζωnt sin ωd t + tan −1

ζ 
1− ζ 2

=

ζ 
e−ζωnt cos ωd t − tan −1

1− ζ 2
1 − ζ 2 

x
x ( t ) = x e−ωnt + (ωn x + x ) te−ωnt
ζ =1
If
x ( 0 ) = x = 0 , this simplifies to
x ( t ) = x ( 1 + ωnt ) e−ωnt
 −ζ + ζ 2 − 1
 −(ζωn +ωn
x
x (t ) = 
x −
e
2
2ωn ζ 2 − 1 
 2 ζ − 1
 ξ + ζ 2 − 1
 −(ζωn −ωn
x
+
x +
e
2
2ωn ζ 2 − 1 
 2 ζ − 1
ζ >1
If
)
ζ 2 −1 t
)
ζ 2 −1 t
x ( 0 ) = x = 0 , this simplifies to
−(ζωn +ωn

x (t ) =
−ζ + ζ 2 − 1 e

2 ζ 2 −1 
x
(
)
30/33
)
ζ 2 −1 t
(
2
)
+ ξ + ζ −1 e
(
)
− ζωn −ωn ζ 2 −1 t


ME 413 Systems Dynamics & Control
Chapter 8: Time Domain Analyis of Dynamic Systems Systems
Table 8.3
The Forced Response of the Second Order Linear System
mx + bx + kx = f ( t ) , x ( 0 ) = 0 and x ( 0 ) = 0
k
rad/s
m
b
b
The damping ratio ζ =
=
bcr 2 mk
ωn =
The natural frequency
The damped frequency
ωd = ωn 1 − ζ 2
 ζ
= tan −1 
 1−ζ 2

The phase angle ψ
For overdamped systems

τ 2 = 1/  ζω + ω

n
n




for
for
0 ≤ ζ <1
0 ≤ ζ <1

ζ > 1 , the time constants are τ 1 = 1/  ζω − ω

n
n

ζ 2 −1

and

ζ 2 −1,

ur ( t ) ≡ ramp with slope 1

Inputs us ( t ) ≡ unit step

δ ( t ) ≡ unit impulse
Damping
ratio
0 ≤ ζ <1
Input
y (t )
f ( t ) = ur ( t )
 2ζ 
1  e −ζωnt 
2ζ 2 − 1

y r ( t ) = 2 t +
sin ωd t  −
 2ζ cos ωd t +
2
 ωn 
ωn 
ωn 
1
−
ζ



f ( t ) = us ( t )

1 
e −ζωnt
ys ( t ) = 2 1 −
cos (ωd t − ψ ) 
ωn 
1−ζ 2

f (t ) = δ (t )
f ( t ) = ur ( t )
ζ =1
Response
f ( t ) = us ( t )
yδ ( t ) =
e −ζωnt
ωn 1 − ζ 2
sin ωd t
1 
2
2
t + e −ωnt + te −ωnt − 
2 
ωn  ωn
ωn 
1
ys ( t ) = 2 1 − e −ωnt − ωnte −ωnt 
ω
yr ( t ) =
n
ζ >1
f (t ) = δ (t )
yδ ( t ) = te −ωnt
f ( t ) = ur ( t )
yr ( t ) =
1 
2ζ 
ωn
2 − t τ1
2 −t τ 2
t
+
τ
e
−
τ
e
−


(
)
2
ωn2  2 1 − ζ 2 1
ωn 
31/33
ME 413 Systems Dynamics & Control
f ( t ) = us ( t )
f (t ) = δ (t )
Chapter 8: Time Domain Analyis of Dynamic Systems Systems

1 
ωn
− t τ1
−t τ 2
1
−
e
−
e
τ
τ


(
)
1
2
ωn2  2 ζ 2 − 1

1
yδ ( t ) =
( e − t τ1 − e − t τ 2 )
2
2ωn ζ − 1
ys ( t ) =
32/33
ME 413 Systems Dynamics & Control
Chapter 8: Time Domain Analyis of Dynamic Systems Systems
33/33