Chapter 3

Time-Domain Analysis of Control

Systems

3.1 Introduction
This chapter refers to the time-domain analysis of linear time-invariant con-
trol systems. The problem of time-domain analysis may be briefly stated as
follows: given the system (i.e., given a specific description of the system) and
its input, determine the time-domain behavior of the output of the system.

In the analysis problem, we will use selected input signals to test the response
of control systems. This response will be characterized by a selected set of
response measures. The basic motivation for system analysis is that one can
predict (theoretically) the system’s behavior.

3.2 System Time Responses

The manner in which a dynamic system responds to an input, expressed as
a function of time, is called the time response. It is possible to compute the
time response of a system if the following is known:

• the nature of the input, expressed as a function of time

• the mathematical model of the system.

The time response of any control system has two components:

(a) Transient response: Is that particular part of the response of the system
which tends to zero as time increases. It is a function only of the system
dynamics, and is independent of the input signal.

(b) Steady-state response: Is that particular part of the response of the sys-
tem that remains after the transient component has reached zero. It is a
function of both the system dynamics and the input signal.

49
50 CHAPTER 3. TIME-DOMAIN ANALYSIS OF CONTROL SYSTEMS

The total response of the system is always the sum of the transient and steady-
state components. In the design problem, specifications are usually given in
terms of the transient and steady-state performance, and controllers are de-
signed so that the specifications are all met by the design system.

For control system design, we need a basis of comparison of performance of

different designs. One way of setting up this basis is to specify particular test
signals and compare the response of different designs to these signals.

3.3 Common test Signals

All the following signals are defined for t ≥ 0.

3.3.1 The unit impulse function

The unit impulse function is based on a rectangular function δ (t) such that

1
δ (t) = , 0 ≤ t ≤ , >0


As  → 0, δ (t) approaches the unit impulse δ(t). The major properties of this
function are
Z ∞
1. δ(t)dt = 1
0
Z ∞
2. g(t)δ(t − a)dt = g(a)
0

3. L[δ(t)] = 1

It is very useful for modeling shock inputs.

3.3.2 The unit step function

The unit step function is defined as:

1
r(t) = 1, t≥0 =⇒ L[r(t)] =
s

and is useful for modeling sudden disturbances.

3.3.3 The unit ramp function

The unit ramp function is defined as:

1
r(t) = t, t≥0 =⇒ L[r(t)] =
s2

It is useful for modeling gradually changing inputs.

3.4. RESPONSE OF FIRST ORDER SYSTEMS 51

3.3.4 The sinusoidal functions

The sinusoidal functions
s
cos ωt = Re ejωt =⇒ L[cos ωt] =
s2 + ω 2
ω
sin ωt = Im ejωt =⇒ L[sin ωt] = 2
s + ω2
are very important in frequancy response techniques.

3.4 Response of first order systems

3.4.1 Standard form
Consider a first-order differential equation
dy(t)
a + by(t) = cr(t)
dt
Take Laplace transoform with zero initial conditions
asY (s) + bY (s) = c R(s)
(as + b)Y (s) = c R(s)
The first order transfer function is
Y (s) c
G(s) = =
R(s) as + b
To obtain the standard form, divide by b
c/b
G(s) =
1 + (a/b)s
which is written as
K
G(s) =
1 + τs

3.4.2 Step response

The response of the control system to the unit-step input is called the unit-step
response. When r(t) is a unit step
 
K 1 1
Y (s) = =K −
s(τ s + 1) s s + 1/τ

=⇒ y(t) = K(1 − e−t/τ ) = K − Ke−t/τ (3.1)

The first term is called the steady-state response and K is called the steady-state
value. The second term is called the transient response. If τ > 0 the transient
response tends to 0 as t → ∞. The step response of a first-order system as given
in (3.1) is shown in Figure 3.1. Note that the exponentially decaying term has
an initial slope of K/τ ; that is,
d   K K
K − Ke−t/τ = e−t/τ =
dt t=0 τ t=0 τ
 52 CHAPTER 3. TIME-DOMAIN ANALYSIS OF CONTROL SYSTEMS

y(t)

K(1 − e−t/τ )
0.632K

Figure 3.1: Step response of first -order systems.

Mathematically, the exponential term does not decay to zero in a finite length
of time. However, if the term continued to decay at its initial rate, it would
reach a value of zero in τ seconds. The parameter τ is called the time constant
and has the units of seconds. The exponential function decays to about 2% of
its initial value within 4 time constants. The output y(t) reaches about 63% of
its final vaue when t = τ .

Example 3.1 An example of a first order system is provided by the following RC circuit.

Figure 3.2: A simple resistor-capacitor circuit.

Vout (s) 1
= (K = 1, τ = RC) 
Vin (s) 1 + sRC

3.4.3 Steady-state response

The concept of finding the steady-state response to a unit step for a system of
any order is now developed. Suppose that

Y (s) = G(s)R(s)

where G(s) is a given transfer function. From Section 2.2.3, the final-value
theorem of the Laplace transform is

lim y(t) = lim sY (s) = lim sG(s)R(s)

t→∞ s→0 s→0

provided that the limit

lim y(t)
t→∞
3.5. RESPONSE OF SECOND ORDER SYSTEMS 53

exists, i.e., y(t) has a final value. For the case that the input is a unit step, R(s)
is equal to 1/s and
1
lim y(t) = lim sG(s) = lim G(s) (3.2)
t→∞ s→0 s s→0
= G(0)
G(0) is often called the dc gain of the system and is defined as the ratio of the
output of a system to a constant input after all transients has decayed. Care
must be taken to apply the Final value Theorem only to stable systems (i.e.,
y(t) is bounded) and with at most a single pole at s = 0.

Find the dc gain of the system whose transfer function is Example 3.2
3(s + 2)
G(s) =
(s2 + 2s + 10)
 Solution Applying (3.2), we get
(3)(2)
dc gain = G(s) = = 0.6 
s=0 (10)

3.5 Response of second order systems

3.5.1 Standard form
Consider a second-order differential equation
d2 y(t) dy(t)
a 2
+b + cy(t) = er(t)
dt dt
Take Laplace transforms, with zero initial conditions
as2 Y (s) + bsY (s) + CY (s) = eR(s)
(as2 + bs + c)Y (s) = eR(s)
The transfer function is
Y (s) e
G(s) = = 2
R(s) as + bs + c
To obtain the standard form, divide by c
e/c
G(s) =
(a/c)s2 + (b/c)s + 1
which is written as
K
G(s) =
(1/ωn2 )s2 + (2ζ/ωn )s + 1
and with the s2 coefficient normalized to unity
Kωn2
G(s) =
s2 + 2ζωn s + ωn2
where
54 CHAPTER 3. TIME-DOMAIN ANALYSIS OF CONTROL SYSTEMS

ζ : damping ratio

ωn : undamped natural frequency

K : dc gain

Since K only determines the dc magnitude of the response, we will set K = 1

and so
ωn2
G(s) = (3.3)
s2 + 2ζωn s + ωn2

where p
s1 , s2 = −ζωn ± ωn ζ2 − 1
are the roots of the characteristic equation and are the poles of G(s).

3.5.2 Pole locations

All system characteristics of the standard second-order system are functions
of only ζ and ωn , since ζ and ωn are the only parameters that appear in the
transfer function (3.3). There are four cases of practical interest:

Case 1: ζ > 1 (Overdamped system)

p
When ζ > 1 the roots s1 , s2 = −ζωn ±ωn ζ 2 − 1 are real, negative and unequal.

ζ>1

Figure 3.3: Pole locations in the s-plane when ζ > 1.

Case 2: ζ = 1 (Critically damped system)

When ζ = 1 the roots s1 , s2 = −ζωn are real, negative and equal.

ζ=1

Figure 3.4: Pole locations in the s-plane when ζ = 1.

3.5. RESPONSE OF SECOND ORDER SYSTEMS 55

Case 3: 0 ≤ ζ < 1 (Underdamped system)

When 0 ≤ ζ < 1 the roots
p
s1 , s2 = −ζωn ± jωn 1 − ζ2

are complex conjugate and have negative real parts. The real parts are zero if
ζ = 0 and the system is called undamped.

0≤ζ<1

Figure 3.5: Pole locations in the s-plane when 0 ≤ ζ < 1.

Case 4: ζ < 0 (Unstable system)

When ζ < 0 the roots s1 , s2 have positive real parts and will be covered later.

Figure 3.6: Pole locations in the s-plane when ζ < 0.

Figure 3.7 summarizes all four cases.

Figure 3.7: Pole locations in the s-plane and the corresponding transient response
type.
56 CHAPTER 3. TIME-DOMAIN ANALYSIS OF CONTROL SYSTEMS

3.5.3 Step response

Consider the second-order system described by the transfer function given in
(3.3), the step response is

ωn2
Y (s) = G(s)R(s) =
s(s2 + 2ζωn s + ωn2 )
Expanding in partial fraction
K1 K2 s + K3
Y (s) = + 2
s s + 2ζωn s + ωn2
Using the cover-up method, we find

K1 = sY (s) =1
s=0

Equating the numerators,

(s2 + 2ζωn s + ωn2 ) + K2 s2 + K3 s = ωn2

After equating the powers of s on the two sides of the above equation, we find
that K2 = −1 and K3 = −2ζωn , hence
1 s + 2ζωn
Y (s) = −
s s2 + 2ζωn s + ωn2
Completing the square
1 s + 2ζωn
Y (s) = −
s (s + ζωn )2 + ωn2 − ζ 2 ωn2
1 s + 2ζωn
= − 2
s
 p
(s + ζωn )2 + ωn 1 − ζ 2

and writing it in the standard forms in Table 2.1 we have

 
1 s + ζωn ζωn ωd
Y (s) = − − (3.4)
s (s + ζωn )2 + ωd2 ωd (s + ζωn )2 + ωd2
where p
ωd = ωn 1 − ζ2
is called the damped natural frequency. Taking the inverse Laplace transform
ζ
y(t) = 1 − e−ζωn t cos ωd t − p
 −ζωn t 
e sin ωd t
1−ζ 2
!
−ζωn t ζ
=1−e cos ωd t + p sin ωd t (3.5)
1 − ζ2

Using the trigonometric identity, sin(a + b) = sin a cos b + cos a sin b, (3.5) can
be written as
1
y(t) = 1 − p e−ζωn t sin(ωd t + α) (3.6)
1 − ζ2
3.5. RESPONSE OF SECOND ORDER SYSTEMS 57

where
1
p
−1 1 − ζ2
α = tan (3.7) p
ζ 1 − ζ2
p α
is as shown (note that cos α = ζ and sin α = 1 − ζ 2 ). The step responses for a
ζ
second-order system are shown in Figure 3.8 for several values of ζ as a function
of ωn t.

y(t)

ωn t
Figure 3.8: Step response for second-order system.

With reference to Figure 3.8:

• If ζ = 0, from (3.5), noting that ωd = ωn

y(t) = 1 − cos ωn t

the frequency of the sinusoid is ωn , called the undamped natural frequency.

The step input will cause the system to oscillate continuously at ωn .
p
• If 0 < ζ < 1, ωd = ωn 1 − ζ 2 is the frequency of the sinusoid, hence,
damped natural frequency. The response is a damped sinusoid, and is
called damped (or underdamped) system. The time constant of the ex-
ponential envelope is τ = ζω1n . As ζ increases from 0 to 1, the response
becomes less oscillatory, and hence the name damping ratio.

• If ζ ≥ 1, the oscillations have ceased and hence the system is called an

overdamped system when ζ > 1 and critically damped when ζ = 1. For
the case ζ = 1 it is not clear from (3.5) how the oscillations in y(t) have
ceased. However, before taking the inverse Laplace to Y (s) in (3.4) let
ζ = 1 to obtain
 
1 s + ωn ωn ωd
Y (s) = − −
s (s + ωn )2 + 0 ωd (s + ωn )2 + 0
1 1 ωn
= − −
s s + ωn (s + ωn )2
58 CHAPTER 3. TIME-DOMAIN ANALYSIS OF CONTROL SYSTEMS

Taking inverse Laplace transform

y(t) = 1 − e−ωn t − ωn te−ωn t

= 1 − e−ωn t (1 + ωn t)

which clearly shows that no sinusoidal term exist. For the overdamped
case, we have two real poles at −ζωn ± ωd . The corresponding response
is easily obtained from

1 K2 K3
Y (s) = + +
s s + ζωn + ωd s + ζωn − ωd
as
y(t) = 1 + K2 e−(ζωn +ωd )t + K3 e−(ζωn −ωd )t

As an exercise find K2 and K3 .

The effect of the characteristic equation roots on the damping of the second-
order system is further illustrated by Figure 3.9

Figure 3.9: Step response comparison for various pole locations in the s-plane.
3.6. SPECIFICATIONS OF A SECOND ORDER SYSTEM 59

3.6 Specifications of a second order system

In designing control systems, specifications must be developed that describes
the characteristics the system should posses. Usually system design specifica-
tions or standard performance measures can be described in terms of the step
response of a system as shown in Figure 3.10.

Figure 3.10: Step response of a control system.

For underdamped systems, we define the following specifications:

Tr : rise time (0%-100%)

Tr1 : rise time (10%-90%)
Mp : peak value
Tp : time to first peak
yss : steady state value
Mp − yss
% overshoot : × 100%
yss
Ts : settling time

The speed of the response is measured by the rise time, Tr , and the peak
time, Tp . For underdamped systems with an overshoot, the 0 − 100% rise time
is a useful index. If the system is overdamped, then the peak time is not defined,
and the 10 − 90% rise time, Tr1 , is normally used.

The tracking properties, i.e., the similarity with which the actual response
matches the step input is measured by the percent overshoot and settling
time, Ts . The % overshoot is defined as
Mp − yss
% overshoot = × 100% (3.8)
yss
60 CHAPTER 3. TIME-DOMAIN ANALYSIS OF CONTROL SYSTEMS

Figure 3.11: Step response of a control system.

where Mp is the peak value of the time response, and yss is the steady state
value or the final value of the response.

The settling time, Ts , is defined as the time required for the system to set-
tle within a certain percentage, δ, of the input amplitude. This band of ±δ is
shown in Figure 3.10. In other words it is the time required for the transient
to decay to a small value so that y(t) is almost in the steady state. For second
order systems we usually determine the time, Ts , for which the response remains
within 2% of the final value. From Figure 3.11 we can determine an approximate
to Ts by computing the time when the decaying exponential e−ζωn t reaches 2%:
e−ζωn Ts = 0.02
or
ζωn Ts ≈ 4
Therefore, we have
4
Ts ≈ 4τ =
ζωn
The steady-state error of the system may be measured on the step response of
the system as shown in Figure 3.10. To obtain analytic expressions for Tp we
first differentiate (3.6) to obtain
ωn p
ẏ(t) = p e−ζωn t sin 1 − ζ 2 ωn t
1 − ζ2
and equating ẏ(t) = 0 gives
p
1 − ζ 2 ωn t = nπ, n = 1, 2, 3, · · ·
from which we get
nπ
t= p
ωn 1 − ζ 2
3.6. SPECIFICATIONS OF A SECOND ORDER SYSTEM 61

Tp occurs when n = 1, therefore

π π
Tp = p = (3.9)
ωn 1 − ζ 2 ωd

To determine Mp we substitute (3.9) back in (3.6)

√ ! !
1 π
 
−ζωn π/ωn 1−ζ 2
y(t) =1− p e sin ωd p +α
t=Tp 1 − ζ2 ωn 1 − ζ 2
1 √ 2
=1− p e−ζπ/ 1−ζ sin(π + α)
1 − ζ2
1 √ 2 p
=1+ p e−ζπ/ 1−ζ sin(α) (noting that sin α = 1 − ζ 2 )
1−ζ 2
√ 2
= 1 + e−ζπ/ 1−ζ = Mp

Therefore, since yss = 1, the percent overshoot is, from (3.8)

√ 2
% overshoot = 100e−ζπ/ 1−ζ

The percent overshoot is thus a function only of ζ and is plotted versus ζ in

Figure 3.12. We can express (3.9) as
π
ωn Tp = p
1 − ζ2

Thus the product ωn Tp is also a function of only ζ, and this is also plotted in
Figure 3.12. Since Tp is an approximate indication of the rise time, Figure 3.12
also roughly indicates rise time. As ζ increases from 0 to 1, Mp , Ts , and the
percent overshoot decreases, while Tp and Tr increase.

Figure 3.12: Percent overshoot and normalized peak time versus damping ratio ζ.
 62 CHAPTER 3. TIME-DOMAIN ANALYSIS OF CONTROL SYSTEMS

Comment
The transient response of the system may be described in terms of two factors:
1. The speed of response, measured by the rise time, Tr , and the peak
time, Tp .
2. The tracking properties, i.e., the closeness of the response to the desired
response, is measured by the % overshoot and the settling time, Ts .
It is important to realize that the two factors are contradictory requirements;
thus, a compromise must be obtained.

Example 3.3 Consider the following transfer function

4
G(s) =
s2 + 2s + 4
Determine, Ts , Tp and the % overshoot.

 Solution We have

ωn2 = 4 =⇒ ωn = 2 rad/s, 2ζωn = 2 =⇒ ζ = 0.5

p √
ωd = ωn 1 − ζ 2 = 3 rad/s
π π
The peak time is obtained as Tp = = √ = 1.82s.
ωd 3
4
The settling time is found to be Ts = = 4s.
ζωn
√ 2
The maximum % overshoot = 100e−ζπ/ 1−ζ = 16.3%

3.7 Specifications vs. pole locations

Figure 3.13 illustrates the relationships between the location of the char-
acteristic equation roots and ζ, ωn , and ωd . For the complex conjugate roots
shown,

Figure 3.13: Relationship between poles of second-order system and ζ, ωn , and ωd .

3.7. SPECIFICATIONS VS. POLE LOCATIONS 63

• ωn is the radial distance from the roots to the origin of the s-plane.
• ζωn is the real part of the roots.
• ωd is the imaginary part of the roots.
• ζ is the cosine of the angle between the radial line to the roots and the
negative axis when the roots are in the left-halp s-plane, or ζ = cos α.
The effect of increasing ωn and ζ on pole locations in the s-plane is shown in
Figure 3.14.

Figure 3.14: Pole locations tin the s-plane as ωn and ζ increases respectively.

Settling time and pole locations

The settling time Ts is related to the roots in Figure 3.14 by the relation
4 4
Ts = =
ζωn σ
The settling time is then inversely related to the real parts of the poles. If in

Faster response

Figure 3.15: Pole locations to achieve desired settling time.

design the settling is specified to be less than or equal to some value Ts,desired ,
64 CHAPTER 3. TIME-DOMAIN ANALYSIS OF CONTROL SYSTEMS

ζωn ≥ 4/Ts,desired , and the pole locations are then restricted to the region of
the s-plane indicated in Figure 3.15. Hence the speed of response is increased
by moving the poles to the left in the s-plane.

% Overshoot and pole locations

The angle α in Figure 3.13 satisfies the relationship
p
−1 1 − ζ2
α = tan = cos−1 ζ (3.10)
ζ
The percent overshoot is given by
√ 2
% overshoot = e−ζπ/ 1−ζ × 100

Hence this equation can be expressed as

% overshoot = e−π/ tan α (3.11)

Decreasing the angle α reduces the percent overshoot. Hence, specifying the
percent overshoot to be less than a particular value restricts the pole locations
to the region of the s-planre, as shown in Figure 3.16

Figure 3.16: Pole locations and their relation to % overshoot.

Peak time and pole locations

From (3.9) the peak time Tp is related to the roots in Figure 3.14 by the relation
π
Tp =
ωd
The peak time is inversely proportional to the imaginary part of the pole.
3.7. SPECIFICATIONS VS. POLE LOCATIONS 65

Lines of constant Tp , % overshoot, and Ts are shown in Figure 3.17. Note that
Ts2 < Ts1 ; Tp2 < Tp1 ; and %OS1 < %OS2 .

Figure 3.17: Lines of constant Tp , Ts , and % overshoot.

Suppose that, in the design of a second-order system, the %overshoot in a step Example 3.4
response is limited to 4.32%. A maximum settling time of 2s is also required.
On an s-plane show the region to which the pole locations are limited to.

 Solution To find ζ, from (3.11)

4.32 = e−π/ tan α × 100
=⇒ 0.0432 = e−π/ tan α
π
=⇒ 3.14 = after taking natural log of both sides
tan α
∴ tan α = 1
Thus, α = 45o , hence ζ = cos 45o = 0.707. A settling time of 2s implies ζωn ≥ 2.
Hence the pole locations are limited to the regions of the s-plane shown in Figure
3.18. The pole locations that exactly satisfy the limits of the specifications are
s = −2 ± j2. 

Figure 3.18: Pole plot for Example 3.4.

Consider the pole plot shown in Figure 3.19. Find ζ, ωn , Tp , and Ts . Example 3.5
66 CHAPTER 3. TIME-DOMAIN ANALYSIS OF CONTROL SYSTEMS

Figure 3.19: Pole plot for Example 3.5.

 Solution The damping ratio is given by

ζ = cos α = cos tan−1 (7/3) = 0.394

 

√
The natural frequency ωn is given by ωn = 72 + 32 = 7.616 rad/s
π π
The peak time Tp is Tp = = = 0.449s
ωd 7
4 4
The settling time Ts is Ts = = = 1.33s 
ζωn 3

3.7.1 Step response vs. pole location

In Figure 3.20 the step responses are shown as the poles are moved in vertical
direction, keeping the real part the same. We see that the frequency changes, but
the envelope remains the same. Since all curves fit under the same exponential
decay curve, the settling time is virtually the same for all waveforms.

Figure 3.20: Step response as poles move with constant real part.

In Figure 3.21 the step responses are shown as the poles are moved in horizontal
3.8. STEADY-STATE ACCURACY 67

direction, keeping the imaginary part the same. As the poles move to the left,
the response damps out rapidly, while the frequency remains the same. Notice
that the peak time is the same for all waveforms.

Figure 3.21: Step response as poles move with constant imaginary part.

In Figure 3.22 the poles are moved along a constant radial line. We see that the
% overshoot remains the same. The farther the poles are from origin, the more
rapid the response.

Figure 3.22: Step response as poles move with constant damping ratio.

3.8 Steady-state accuracy

It has been observed, in the previous sections, that much information about
a system can be obtained from the analysis of its response to test inputs. In
control system design, one of the major objectives is to be able to track reference
inputs precisely, and to maintain this precision in the face of disturbances. In
many cases an error between the desired and resulted final value does occur. In
this section, we consider how such precision can be achieved.

3.8.1 Error signal

Consider the feedback system in Figure 3.23. Assume the feedback loop is sta-
ble. We have seen in Section 1.3 how a feedback control system can reduce
68 CHAPTER 3. TIME-DOMAIN ANALYSIS OF CONTROL SYSTEMS

Figure 3.23: Closed loop control system.

sensitivity of the system. Furthermore, the effect of disturbances can also be

reduced significantly. However, as a further requirement, we must examine and
compare the final steady-state error of the closed loop system.

The output Y (s) is required to track the reference input R(s). The input into
plant G(s) is the tracking error

E(s) = R(s) − Y (s)

R(s)
=
1 + G(s)

Let e(t) denote its inverse Laplace transform. The limit

ess = lim e(t)

t→∞

is referred to as the steady-state error. The value of ess characterises the

final value of error as a difference between the value of the input r(t) and the
final value of the output y(t). In other words, ess is the value of error after
transients have died out.

To calculate the steady-state error, we utilize the final value theorem, as long
as E(s) does not have any poles in the right half of the s-plane, except maybe,
at s = 0, then
sR(s)
ess = lim e(t) = lim sE(s) = lim
t→∞ s→0 s→0 1 + G(s)
Step Input The steady-state error for a step input is
sR(s) 1
ess = lim =
s→0 1 + G(s) 1 + G(0)
Note that in this case, the steady-state error is determined by the dc gain of
G(s). The larger is dc gain, the smaller is the steady-state error. Furthermore,
if G(s) has one or more poles at s = 0, then lims→0 G(s) = ∞. In this case,
ess = 0. A pole at s = 0 implies G(s) includes an integrator. The number of
poles at the origin of the loop gain (i.e. the number of integrators)
defines the system’s type number, N.

Type zero systems (N = 0) do not include an integrator and therefore have

a finite dc gain G(0). The constant G(0) is denoted by Kp , the position error
constant, and is given by
Kp = lim G(s)
s→0
3.8. STEADY-STATE ACCURACY 69

The steady-state error for a step input is thus given by

1
ess =
1 + Kp

Note that a large position error constant corresponds to a small steady-state

error. Furthermore, if N ≥ 1 the steady-state error ess = 0.

1
1

Figure 3.24: Steady-state error for step input.

Ramp Input Now consider the steady -state error in response to a ramp input
r(t) = t. The Laplace transform of the ramp input is R(s) = 1/s2 and

s(1/s2 ) 1 1
ess = lim = lim = lim
s→0 1 + G(s) s→0 s + sG(s) s→0 sG(s)

Again, the steady-state error depends upon the number of integrators, N . For
a type zero system, N = 0, the steady-state error is infinite. For a type one
system, N = 1, the error is
1
ess =
Kv
where Kv the velocity error constant and is computed as

Kv = lim sG(s)
s→0

For N ≥ 2, Kv is infinite and the steady-state error for a ramp input is zero.

Parabolic Input for a parabolic input r(t) = t2 /2, we take R(s) = 1/s3 ,
the steady-state error is

s(1/s3 ) 1
ess = lim = lim 2
s→0 1 + G(s) s→0 s G(s)

The steady-sate error is infinite for type zero and type one systems. For type
two, N = 2, and we obtain
1
ess =
Ka
 70 CHAPTER 3. TIME-DOMAIN ANALYSIS OF CONTROL SYSTEMS

Figure 3.25: Steady-state error for ramp input.

where Ka is called the acceleration error constant. The acceleration error

constant is
Ka = lim s2 G(s)
s→0

The error constants and the steady-state error for the three inputs are summa-
rized in Table 3.1.

Table 3.1: Summary of steady-state errors.

1 1 1
Type R(s) = R(s) = R(s) = Error constants
s s2 s3

1
0 ess = ∞ ∞ Kp = lim G(s)
1 + Kp s→0

1
1 ess = 0 ∞ Kv = lim sG(s)
Kv s→0

1
2 ess = 0 0 Ka = lim s2 G(s)
Ka s→0

Example 3.6 Consider the control system in Figure 3.23 with

200(s + 1)2
G(s) =
(s + 2)(s + 3)(s + 4)

Find the steady-state error when (a) r(t) is a unit step, (b) r(t) is a unit ramp.

 Solution G(s) is a type zero system, therefore for a step input, the posi-
tion error constant, Kp = G(0) = 200/24 = 8.333. Then

1 1
ess = = = 0.1071
1 + Kp 9.333
3.8. STEADY-STATE ACCURACY 71

For a ramp input Kv = lim sG(s) = 0. Then,

s→0

1
ess = =∞ 
Kv