Colorado School of Mines CHEN403
Underdamped Systems
2nd Order Systems
Output modeled with a 2nd order ODE:
d2 y
dy
a2 2 + a1
+ a0 y = b ⋅ f ( t )
dt
dt
If a0 ≠ 0 , then:
2
a2 d 2 y a1 dy
b
dy
2 d y
+
+
y
=
f
t
⇒
τ
+ 2ζτ + y = K p f ( t )
(
)
2
2
a0 dt
a0 dt
a0
dt
dt
where: τ is the natural period of oscillation.
ζ is the damping factor.
K p is the steady state gain.
For deviation variables, where y ( 0) = f ( 0 ) = 0 , the Laplace transform will be:
(τ s
2 2
)
+ 2ζτs + 1 y ( s ) = K p f ( s ) ⇒ G ( s ) =
y(s)
f (s)
=
Kp
τ s + 2ζτs + 1
2 2
Dynamic Response of Underdamped 2nd Order System
If ζ < 1 , then the poles in the transfer function are complex conjugates. Let’s look at
response to a unit step change. f ( t ) = α ⋅ H ( t ) ⇒ f ( s ) = α / s . So, for a 2nd order system:
y(s) =
α
τ s + 2ζτs + 1 s
Kp
2 2
⋅
1
τ2 s + 2ζτ 
y ( s ) = αK p  − 2 2

 s τ s + 2ζτs + 1 
2ζ


s
+
1

τ
y ( s ) = αK p  −
2ζ
1
 s s2 + s + 2 
τ
τ 


ζ ζ 

s+ +



1
τ τ 
∴ y ( s ) = αK p  − 
2
s 
ζ  1 − ζ2 

 s + τ  + τ2 




Inverting:
Underdamped Systems
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December 21, 2008
Colorado School of Mines CHEN403

 t 1 − ζ2
y ( t ) = αK p 1 − e −t ζ / τ cos 

τ



 t 1 − ζ2
ζ
−
e −t ζ / τ sin 


τ
1 − ζ2





 
The sine & cosine terms can be combined into a single sine term with a phase angle offset.
Remember:
p cos θ + q sin θ = r sin ( θ + φ )
r = p2 + q2 and tan φ =
where:
So:
r = 1+
ζ2
1
=
2
1−ζ
1 − ζ2
tan φ =
1 − ζ2
ζ
p
q
Defining:
ω≡
then:
1 − ζ2
τ


e −t ζ / τ
y ( t ) = αK p 1 −
sin ( ωt + φ ) 


1 − ζ2
Underdamped Systems
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December 21, 2008
Colorado School of Mines CHEN403
A
C
y/Kp
Response Time
Rise Time
t
Figure 1. Response Curve for Underdamped System
There are several terms defined to describe the characteristics of an underdamped system’s
response. These are also shown in the Figure 1:
•
Ultimate Value. This is the value that the response settles down to at very large
times. This can easily be determined as:


 
e −t ζ / τ
lim
y∞ = t lim
y
t
=
α
K
1
−
sin
ω
t
+
φ

(
)   = αK p
→ ∞ { ( )} t → ∞  p
2

 
−
ζ
1


•
Period of oscillation. The time between crossing of the ultimate value. This will
also be the time between the peaks and valleys. Since the frequency of oscillation
is:
ω=
1 − ζ2
τ
then the period of oscillation is:
T=
1 2π
2πτ
=
=
f ω
1 − ζ2
Underdamped Systems
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December 21, 2008
Colorado School of Mines CHEN403
•
Rise time. The time it takes the response to first get to the ultimate value. This
can be easily determined as:


e −triseζ / τ
y ( t rise ) = αK p 1 −
sin ( ωt rise + φ )  = αK p


1 − ζ2
1−
e −t riseζ / τ
1 − ζ2
e −t riseζ / τ
1 − ζ2
sin ( ωt rise + φ ) = 1
sin ( ωt rise + φ ) = 0
sin ( ωt rise + φ ) = 0
ωt rise + φ = π
π−φ
τ
∴ t rise =
=
ω
1 − ζ2
•

 1 − ζ2
 π − tan −1 
 ζ





 
Overshoot. A measure of how far the response exceeds the ultimate value. On the
figure, this is A / α . The formula given for this in the text is:
 −πζ 

Overshoot = exp 
 1 − ζ2 


This is calculated by finding the time at which the first maximum occurs and then
finding the corresponding value in the response curve. The 1st derivative of the
response curve is:
αK p d − t ζ / τ
dy
e
=−
sin ( ωt + φ ) 
dt
1 − ζ 2 dt
αK p
dy
=−
dt
1 − ζ2
 ζ −t ζ / τ

sin ( ωt + φ ) + ωe −t ζ / τ cos ( ωt + φ ) 
− τ e


αK pe − t ζ / τ  ζ
dy

=−
− sin ( ωt + φ ) + ω cos ( ωt + φ ) 

2
dt

1−ζ  τ
The sine & cosine terms can be combined into a single sine term to give:
αK p e − t ζ / τ

 ωτ  
ζ2
dy
=−
ω2 + 2 sin  ωt + φ + tan −1  −  
τ
dt
 ζ 
1 − ζ2

Underdamped Systems
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December 21, 2008
Colorado School of Mines CHEN403

 1 − ζ2
αK p e − t ζ / τ 1 − ζ 2 ζ 2
dy
−1


sin
tan
=−
+
ω
+
t
 ζ
τ2
τ2
dt

1 − ζ2


 τ 1 − ζ2
 − tan −1 

ζ
τ





 
α K pe − t ζ / τ
dy
∴
=−
sin ( ωt )
dt
τ 1 − ζ2
So, the time t * at which the 1st maximum occurs is at:
*
αK p e − t ζ / τ
dy
=−
sin ωt * = 0 ⇒ sin ωt * = 0
2
dt t =t *
τ 1−ζ
(
∴ t* =
)
(
)
π
ω
Then:


e −πζ / ωτ
ymax = y t * = αK p 1 −
sin ( π + φ ) 


1 − ζ2
( )

1
 πζ  
sin ( φ ) exp  −  
ymax = αK p 1 +
 ωτ  

1 − ζ2


1 − ζ2
1
sin  tan −1
ymax = αK p 1 +

ζ

1 − ζ2


1
ymax = αK p 1 +

1 − ζ2
(

 πζ
τ
 exp  −
2



 τ 1−ζ

πζ
1 − ζ 2 exp  −
 1 − ζ2

)


πζ
∴ ymax = αK p 1 + exp  −
 1 − ζ2





 



 



 
Finally,


πζ
αK p 1 + exp  −
 1 − ζ2

ymax − y∞

Overshoot =
=
y∞
αK p
•

  − αK p

 −πζ 
 

= exp 
 1 − ζ2 


Decay Ratio. The ratio of the overshoot of two successive peaks, C / A . The text
gives this expression as:
 −2πζ
2
Decay Ratio = ( Overshoot ) = exp 
 1 − ζ2

Underdamped Systems
-5-




December 21, 2008
Colorado School of Mines CHEN403
We know from the derivation for the overshoot that the peaks and valleys will
occur when:
(
)
sin ωt * = 0 ⇒ ωt * = 0, π , 2π , … , nπ , …
The peaks will occur at the odd multiples:
ωt * = π , 3π , … , ( 2n − 1) π , …
So, the n-th peak will have a response value of:
 e −(2n−1)πζ / ωτ

ypeak ,n = y t n* = αK p 1 −
sin ( ( 2n − 1) π + φ ) 


1 − ζ2
( )

 ( 2n − 1) πζ  
1
ypeak ,n = αK p 1 −
sin ( π + φ ) exp  −

ωτ

1 − ζ2

 

 ( 2n − 1 ) πζ  
1

sin ( φ ) exp  −
ypeak ,n = αK p 1 +
2



1 − ζ2
1
−
ζ

 


1 − ζ2
1
sin  tan −1
ypeak ,n = αK p 1 +

ζ

1 − ζ2


 ( 2n − 1 ) πζ  
 exp  −

2



1
−
ζ


 

 ( 2n − 1) πζ  

∴ ypeak ,n = αK p 1 + exp  −
2



1
−
ζ

 
Note that the valleys will occur at the even multiples:
ωt * = 2π, 4π , … , 2nπ , …
so the n-th valley will have a response value of:
 e −2nπζ / ωτ

yvalley ,n = αK p 1 −
sin ( 2nπ + φ ) 


1 − ζ2

1
 2nπζ  
yvalley ,n = αK p 1 −
sin ( φ ) exp  −

 ωτ  

1 − ζ2

 2nπζ
∴ yvalley ,n = αK p 1 − exp  −
 1 − ζ2





 
Now, the decay ratio, R , will be:
Underdamped Systems
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December 21, 2008
Colorado School of Mines CHEN403
 ( 2n + 1) πζ 
αK p exp  −


ypeak ,n+1 − y∞
1 − ζ 2 

=
R=
ypeak ,n − y∞
 ( 2n − 1 ) πζ 
αK p exp  −

2


−
ζ
1


 ( 2n + 1) πζ 

exp  −
2


−
ζ
1


R=
 ( 2n − 1) πζ 
exp  −

2


−
ζ
1


 ( 2n + 1) πζ ( 2n − 1 ) πζ 

R = exp  −
+
2
2


1
1
−
ζ
−
ζ



2πζ
∴ R = exp  −
 1 − ζ2





y/AKp
Response Time
t
Figure 2. Occurrence of Response Time in an Underdamped System
•
Response time. The time it takes the response to stay within ±5% of the ultimate
value. Figure 2 shows that this is a much more complicated value to figure out,
Underdamped Systems
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December 21, 2008
Colorado School of Mines CHEN403
since the response curve may cross over the threshold values many times before it
final settles down within the range. But we can bracket the value by finding the
1st peak and the 1st valley that remains within the tolerance. If a peak is the 1st
extremum to stay within the tolerance, then the response time will be between
this 1st peak within the tolerance and the last valley out of the tolerance.
Similarly, if a valley is the 1st extremum to stay within the tolerance, then the
response time will be between this 1st valley within the tolerance and the last
peak out of the tolerance.
Let us denote the tolerance as ε , where by tradition ε = 0.05 . Then the response
time will be before the lowest value of n which satisfies:
ypeak ,n − y∞ < εy∞
ypeak ,n < (1 + ε ) y∞

 ( 2n − 1) πζ  
  < ( 1 + ε ) αK p
αK p 1 + exp  −
2



−
ζ
1


 ( 2n − 1 ) πζ 
<1+ε
1 + exp  −
2


−
ζ
1


 ( 2n − 1 ) πζ 
<ε
exp  −
2


1
−
ζ


−
( 2n − 1) πζ < ln
1 − ζ2
(ε)
 1 ln ( ε ) 1 − ζ 2 

∴ np = min n > −
2
2πζ


The response time will also be before the lowest value of n which satisfies:
y∞ − yvalley ,n < εy∞
yvalley ,n > (1 − ε ) y∞

 2nπζ
αK p 1 − exp  −
 1 − ζ2


 2nπζ
exp  −
 1 − ζ2

Underdamped Systems

  > ( 1 − ε ) αK p

 

<ε


-8-
December 21, 2008
Colorado School of Mines CHEN403
−
2nπζ
1 − ζ2
< ln ( ε )

ln ( ε ) 1 − ζ2 

∴ nv = min n > −
2πζ


So, if np ≤ nv , then the response time will be between the peaks represented by np
and nv − 1 , or times (2np − 1)π / ω and 2(nv − 1)π / ω . However, if nv < np , then the
response time will be between the peaks represented by nv and np − 1 , or times
2nv π / ω and (2np − 3)π / ω .
For example, Figure 1 was generated using ζ = 0.15. So:
−
ln ( ε ) 1 − ζ2
2πζ
= 3.1 ⇒ nv = 4
2
1 ln ( ε ) 1 − ζ
−
= 3.6 ⇒ np = 4
2
2πζ
So, the response time is found from where the response curve crosses the lower
limit between the 3rd valley and the 4th peak.
One further thing to note is that the response can be before the rise time if the 1st
peak does not exceed the tolerance for the response time. This will happen if ζ
stays larger then a threshold value of:
*2
 π 2 
1 ln ( ε ) 1 − ζ


*
−
= 1 ⇒ ζ = 
 + 1
*
2
2πζ
  ln ( ε ) 

−1/2
For the traditional tolerance, this threshold value is ζ * = 0.690 .
Underdamped Systems
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December 21, 2008