Chapter 5
Transient Response
When trying to determine the response of a system to a specied input, the response can be broken
up into two main components:
transient
which denes the behavior of the response from intial to
nal state and steady-state is the system output as t goes to ∞. The goal is to determine if there is a
way system parameters to visualize graphically how the system responds. Based on the application,
there could be specic requirements which limit the deviation of the transient and steady-state.
5.1
First Order Systems
Step Input
C(s)
R(s)
=
C(s) =
1
T s+1
(assume R(s) is a unit step)
1 1
s T s+1
h
c(t) = L
−1
1
s
=
a1
s
+
a2
where a1 =1 and a2 =-T
iT s+1
t
T
− T s+1 = 1 − e− T
...where T is a time constant that dictates the response speed (smaller T means faster response)
1
Figure 5.1: Step input response of rst order system
Ramp Input
1
C(s)= s12 T s+1
=
−T
s
1
s2
+
+
T2
T s+1
t
→ c(t) = t − T + T e− T
Figure 5.2: Ramp input response of rst order system
Impulse Input
Impulse: input over a short time (e.g. hitting a gong)
C(s) =
1
T s+1
→ c(t) =
1
T
−t
eT
2
Figure 5.3: Impulse input response of rst order system
5.2
Second Order Systems
For a mass-spring-damper the dierential equation can be written as:
mẍ + bẋ + kx = 0 with initial conditions x(0) = x0 , ẋ(0) = ẋ0
The solution is linear and in the form of: x(t) = c1 eλ1 t + c2 eλ2 t + · · · + cn eλn t
where λ =
λ depends on the parameters of the system
−b ±
√
b2 − 4mk
2m
m , b , k
and can contain real (ai ) or imaginary (di )
parts. It can take on three dierent cases which aect the shape of the system response:
Case 1) b2 − 4mk < 0 −→ 2 complex roots
Taking one complex root to intuitively understand the response shape: λ1 = −a1 + d1 i where a1 is
a variable substitute for the real portion and d1 is the imaginary portion.
Therefore x(t) = c1 e−a1 +d1 i + · · · = c1 e−a1 t ed1 it + · · · When plotting x(t), the e−a1 t decays or amplies the amplitude of an oscillating sine wave determined by ed1 it . The oscillating sine wave can
be equated by using Euler's formula to convert an imaginary exponential to a cosine/sine expression.
2) b2 − 4mk = 0 −→ 2 repeated real roots
One of the repeated poles: λ1 = −a2 containing only a real component which leads to a solution
x(t) = c1 e−a2 t which takes on the form of a decaying exponential curve.
3
3) b2 − 4mk > 0 −→ 2 distinct real roots
One of the distinct poles: λ1 = −a3 containing only a real component which leads to a solution
x(t) = c1 e−a3 t which takes on the form of a decaying exponential curve.
Plotting each of the three cases, each can be labeled as such: the underdamped response oscillates
(due to having an imaginary component) before settling to a value while the
critically damped
and overdamped approach an asymptote at the desired value (Fig. 5.4).
Figure 5.4: All three cases and their general response shape
To understand why these responses are categorized as such, observations can be drawn from each
response by breaking it down into two parts: a) oscillation occuring at a specic frequency b)
exponential decay of the amplitude or value towards a desired value. These two characteristics
can be dened in parallel by two variables: a) ωn (natural frequency) b) ζ (damping ratio). The
above example for the three cases of types of poles that can result from specied system parameters
is written in terms of
, however for systems that are not mass-spring-dampers, a more
m, b, k
generalized 2nd order system can be dened using the two parameters ωn and ζ above:
ẍ + 2ζωn ẋ + ωn2 x = 0
Transforming the original dierential equation for the mass-spring-damper to the generalized 2nd
order form yields equations to nd natural frequency and damping ratio in terms of m,
b
k
ẋ + x = 0
m
m
r
k
k
2
→ ωn =
ωn =
m
m
ẍ +
4
b, k
.
2ζωn =
b
b
→ζ= √
m
2 mk
The three cases described above can be dened separately by observing the damping ratio ζ : in the
underdamped case ζ <1, the critically damped system has ζ =1, and the overdamped system has
ζ >1. Finding the poles of the 2nd order system in terms of ωn and ζ :
p
p
−2ζωn ± 4ζ 2 ωn2 − 4ωn2
=
...the term inside the square root simplies to: ωn ζ 2 − 1 = ωd j
2
Dening a new variable ωd as damped natural frequency as the eect the damping ratio has on
the natural frequency: ωd = ωn 1 − ζ 2
p
To verify this equation two cases of ζ can be observed:
...if ζ = 0 then ωd = ωn there is no damping, purely a sine wave with the same natural frequency.
...if ζ = 1 then ωd = 0 there are no oscillations there is no frequency.
Finding the transfer function for the 2nd order system to eventually derive the equation for the cart
displacement x(t):
ẍ + 2ζωn ẋ + ωn2 x = 0 x(0) = x0 , ẋ(0) = ẋ0
2
s X(s) − sx(0) − ẋ(0) + 2ζωn [sX(s) − x(0)] + ωn2 X(s) = 0
X(s) =
(s + 2ζωn )x(0) + ẋ(0)
s2 + 2ζωn s + ωn2
To nd the inverse Laplace transform of the above expression, it needs to be broken up into the
summation of sine and cosine functions with unknown coecients c1 , c2 :
X(s) = c1
k
s−a
+ c2
→ x(t) = c1 eat sin kt + c2 eat cos kt
(s − a)2 + k 2
(s − a)2 + k 2
"
x(t) = e
−ζωn t
!
#
1
p
x(0) +
ẋ(0) sin ωd t + x(0) cos ωd t
ωd
1 − ζ2
ζ
The following example substitute values for m,b,k and solves for the displacement of the cart when
the damper coecient b is altered:
5
Example 5.1: Eect of Damper Coecient on Response Characteristics
mẍ + bẋ + kx = 0 x(0) = 2, ẋ(0) = 0, m = 1, k = 2
···
When altering parameters of a system which in term aect the transient response, there is information that can be extracted from the curve itself that may need to meet specic design requirements
based on the application. Taking an underdamped system as an example:
Figure 5.5: Transient response characteristics for an underdamped system
Mp (maximum percent overshoot): maximum peak value measured from the reference
=
√−ζπ
x(tp ) − x(∞)
= e 1−ζ2
x(∞)
tp (peak time): time required to reach rst peak of overshoot
=
π
p
ωn 1 − ζ 2
ts (settling time): time for response to reach and stay within ±2% of nal value
=
4
ζωn
tr (rise time): time required for response to rise from 10% → 90% (overdamped) or 0% → 100%
(underdamped) of nal value
6
5.3
Pole-Zero Plotting
By identifying the poles and zeros of a transfer function and graphing their location, it can intuitively
show the response behavior. A way to do this is to produce a pole-zero plot which graphically
represent the imaginary and real parts of the pole or zero on the complex plane.
Let's say there is a transfer function:
H(s) =
(s2
s−2
+ 1)(s + 3)
zeros : 2, poles : ±j, −3
Mark poles with (x) and zeros with (o) in the complex plane:
Figure 5.6: Pole-zero plot of H(s)
The location of poles along the real axis can also signicantly change the response and how it can
be characterized:
a)H1 (s) =
1
→ h1 (t) = e2t
s−2
c)H3 (s) =
b)H2 (s) =
1
→ h2 (t) = e−2t
s+2
1
1
→ h3 (t) = sin 2t
s2 + 4
2
7
Figure 5.7: Pole-zero plot and responses depending on location of the pole
A positive pole leads to a positive coecient in the exponent which is commonly described as an
unstable
system since as t goes to ∞, the response also approaches innity. Contrarily, a negative
pole leads to a negative coecient in the exponent which decays the value to zero and seen as a
stable
system. For a pole that has only imaginary components, this is called marginally stable since
the response oscillates but doesn't either settle to a certain value or approach innity.
While the above goes through three simple examples, in 2nd order systems which may have both
imaginary and real components, relating the transient response specications to pole location is key
to visualizing the response itself. The poles of a 2nd order system are written as:
= −ζωn ± ωd j
Plotting these in a pole-zero plot:
8
Figure 5.8: Pole-zero plot of poles that range from purely real to purely imaginary
β is the angle between the real axis and the location of the pole and can be related to the damping
ratio:
ζωn
=ζ
ωn
cos β =
β = cos−1 (ζ)
The three locations listed on Fig. 5.8 can be associated with dierent damping ratio values which
aect the location of the pole:
1) ζ = 0 (no damping) : −ζω
n ± ωn
q
1 − ζ
2 j = ±ωn j
(imaginary only)
2) ζ < 1 (underdamped) : −ζωn ± ωn 1 − ζ 2 j (real + imaginary)
p
1 − ζ 2 j = −ωn (repeated real)
3) ζ = 1 (critically damped) : −ζωn ± ωn
p
The eect of strategically moving the poles in the complex plane is shown in Fig. 5.9, where
certain design requirements may need to be maintained while reducing or amplifying another. The
three examples shown below observe keeping the same amplitude descent, frequency, and overshoot
parameters.
9
Figure 5.9: Moving the poles while maintaining a certain design requirement
5.4
Transient Response of Higher Order Systems
In the previous section, responses for rst and second order systems were reviewed. However, the
increased complexity of a system can create higher order transfer functions in the form of:
n
C(s) =
a0 X ai
+
s
s + pi
i=1
Finding a way to see if breaking up the high order transfer function into a series of smaller order
transfer functions, understanding the eect certain poles have on the transient response may be
small enough to ignore and reduce the order of the system. To illustrate this, the following example
10
goes through the process and comparison of this eect:
125
1
5
25
H(s) =
= H1 (s) + H2 (s) + H3 (s) =
(s + 1)(s + 5)(s + 25)
s + 1 s + 5 s + 25
Poles: − 1, −5, −25
Drawing the pole-zero plot and the transient response for each subsystem in addition to the total
system h(t) (Fig. 5.10):
Figure 5.10: High order system transient response compared to smaller subsystems
h1 (t) has a transient response most similar to h(t), therefore H(s) can be simplied into a lower
order transfer function:
1
s+1
. A rule commonly used when trying to reduce high order transfer
functions to lower order ones can be the following:
If the ratio of real components ≥ 5 or the real component of the next closest pole to the imaginary
axis is ≥ 5,the pole closest to the imaginary axis will dominate and the higher order system can be
assumed to behave similarly to a lower order system containing only this pole(s).
The reason for this assumption is that the higher the negative real component of a pole, the faster
it decays, therefore its eect on the transient is minimized.
11
Example 5.2: High Order Transient Response
For each of the transfer functions below, nd which pole(s) dominate the transient response
and plot to compare.
G1 (s) =
G2 (s) =
50
(s + 10)(s2 + 2s + 5)
(s +
1)(s2
···
12
29
+ 10s + 29)