Module-4: Time Domain Analysis of Linear Systems
Static Error Constants & Steady State Error
Akshya Swain
Department of Electrical, Computer & Software Engineering, The University of Auckland,
Auckland, New Zealand.
Akshya Swain
Module-4: Time Domain Analysis of Lin
ear Systems 1 / 31
Learning Outcome of This Module
After completing this module you would be able to the following:
1. Compute the static error constants of linear unity feedback systems.
2. Determine the steady state error due to standard test inputs such as step, ramp
and parabolic signals.
3. Could compute the steady state error due to disturbances.
4. Could compute the steady state error for non-unity feedback systems.
Akshya Swain
Module-4: Time Domain Analysis of Lin
ear Systems 2 / 31
Steady State Error in Unity Feedback Systems
Consider a unity feedback system shown below
I The closed loop transfer function is given by C(s) = G(s)
R(s) 1 + G(s)
I The error E(s) between the input R(s), ( reference signal) and the output, C)(s)
is given by
E(s) = R(s) − C(s) = R(s) − G(s)E(s); * C(s) = G(s)E(s) Solving for E(s) gives
E(s) = R(s) 1+G(s)
Akshya Swain
Module-4: Time Domain Analysis of Lin
ear Systems 3 / 31
Steady State Error in Unity Feedback Systems
I The steady state error ess of a feedback control system is defined as the error when the time reaches infinity. Thus
ess = e(∞) = lim e(t) t→∞
I By applying final value theorem, the steady state error is computed from ess = e(∞) = lim e(t)
t→∞
sR(s) s→0 s→0 1 + G(s)
= lim sE(s) = lim
1. Thus the steady state error is expressed as:
e(∞) = lim
sR(s) s→0 1+G(s)
2. This shows that the steady state error depends on the reference input R(s) as well as on the form of the transfer function G(s)
Akshya Swain
Module-4: Time Domain Analysis of Lin
ear Systems 4 / 31
Classification of Control Systems : System TYPE
I Control systems may be classified according to their ability to follow step inputs, ramp inputs, parabolic inputs, and so on.
I This is a reasonable classification scheme because actual inputs may frequently be considered combinations of such inputs.
I The magnitudes of the steady-state errors due to these individual inputs are indicative of the goodness of the system.
Consider a unity feedback control system shown in the Figure
I The open loop transfer function of the system is described by G(s) = K(s + z1)(s + z2)……
sn(s + p1)(s + p2)……
I This involves a term sn in the denominator, representing a pole of multlipicity n at the origin.
Akshya Swain
Module-4: Time Domain Analysis of Lin
ear Systems 5 / 31
Classification of Control Systems : System TYPE
I The open loop transfer function of the system shown in Figure( last slide) is G(s) = K(s + z1)(s + z2)……
sn(s + p1)(s + p2)……
which involves a term sn in the denominator, representing a pole of multiplicity n at the origin.This also implies the number of pure integrators present in the forward path.
I The present classification scheme is based on number of pure integrators present in the forward path.
1. If the number of integrators present in the forward path is zero ( i.e. n = 0), the system is called a Type-0 system.
2. Similarly if number of integrators present in the forward path is one ( i.e. n = 1), the system is called a Type-1 system.
3. In general, if number of integrators present in the forward path is n, the system is a Type-n system .
1. Note that as the type number is increased, accuracy is improved; however, increasing the type number aggravates the stability problem. A compromise between steady-state accuracy and relative stability is always necessary.
Akshya Swain
Module-4: Time Domain Analysis of Lin
ear Systems 6 / 31
Static Error Constants
I The static error constants defined in the following are figures of merit of control systems from the the perspective of steady state error.
I The higher the value of these constants, the smaller is the steady-state error.
I Note that the steady state error exhibited by a system is dependent on the nature of the input
I In the following we define three static error constants such as position error constant Kp, velocity error constant Kv and the acceleration error constant Ka when the input of the system is step, ramp or parabolic respectively.
Akshya Swain
Module-4: Time Domain Analysis of Lin
ear Systems 7 / 31
Static Position Error Constant Kp
I This is calculated when the input signal is a unit step signal
r(t) = 1(t), =⇒ R(s) = 1 s
I The steady state error due to this input is given by e(∞) = lim sR(s) = lim s
1 s→0 1+G(s) s→0 1+G(s)s
=lim 1 = 1
s→0 1+G(s) 1+limG(s)
s→0 I The static position error constant Kp, is defined by
Kp = lim G(s) = G(0) s→0
I Thus the steady state error due to unit step input is given as
e(∞)= 1 ; Kp=limG(s) 1+Kp s→0
Akshya Swain
Module-4: Time Domain Analysis of Lin
ear Systems 8 / 31
Static Position Error Constant Kp for Different System Types I For Type-0 system,
Kp = lim G(s) = lim K(s + z1)(s + z2)…… = Kz1z2…. = K1(say) s→0 s→0 (s + p1)(s + p2)…… p1p2…
I For Type-1 or higher Type system, n ≥ 1,
Kp = limG(s)= lim K(s+z1)(s+z2)…… =∞, n≥1
s→0 s→0 sn(s + p1)(s + p2)……
I Hence, for a type-0 system, the static position error constant Kp, is finite, while
for a type-1 or higher system, Kp is infinite.
I Summary: The steady state error for a unit-step input:
ess=1, fortype0system K1
ess = 0, for type 1 or higher system
Akshya Swain
Module-4: Time Domain Analysis of Lin
ear Systems 9 / 31
Static Velocity Error Constant Kv
I This is calculated when the input signal is a unit ramp signal
r(t)=t, =⇒ R(s)=1 s2
I The steady state error due to this input is given by
e(∞) = lim sR(s) = lim s 1
=lim 1 = 1
s→0 s + sG(s) lim sG(s)
s→0 I The static velocity error constant Kv, is defined by
Kv = lim sG(s) s→0
I Thus the steady state error due to unit ramp input is given as
s→0 1+G(s) s→0 1+G(s)s2
e(∞)= 1 ; Kv = limsG(s) Kv s→0
Akshya Swain
Module-4: Time Domain Analysis of Lin
ear Systems 10 / 31
Static Velocity Error Constant Kv for Different System Types I For Type-0 system,
Kv = lim sG(s) = lim sK(s + z1)(s + z2)…… = 0 s→0 s→0 s(s + p1)(s + p2)……
I For Type-1 system
Kv = lim sG(s) = lim K(s + z1)(s + z2)…… = Kz1z2…. = K1(say)
s→0 s→0 s(s + p1)(s + p2)…… p1p2… I For Type-2 or higher Type system, n ≥ 2,
Kv = limsG(s)= lim K(s+z1)(s+z2)…… =∞, n≥2 s→0 s→0 s2(s + p1)(s + p2)……
I Summary:The steady-state error ess, for unit ramp input:
ess = 1 =∞, fortype0system Kv
ess = 1 = 1 fortype1system Kv K1
ess = 0, for type 2 or higher system
Akshya Swain
Module-4: Time Domain Analysis of Lin
ear Systems 11 / 31
Static Acceleration Error Constant Ka
I This is calculated when the input signal is a unit parabolic signal
t2 1 r(t)= 2, ∀t≥0, r(t)=0 ∀t<0 =⇒ R(s)= s3
I The steady state error due to this input is given by e(∞)= lim sR(s) = lim s 1
=lim 1 = 1
s→0 s2 + s2G(s) lim s2G(s)
s→0
I The static acceleration error constant Ka, is defined by
Ka = lim s2G(s) s→0
I Summary:The steady state error due to unit accerelation input:
s→0 1+G(s) s→0 1+G(s) s3
e(∞)= 1; Ka=lims2G(s) Ka s→0
Akshya Swain
Module-4: Time Domain Analysis of Lin
ear Systems 12 / 31
Static Acceleration Error Constant Ka for Different System Types I For Type-0 system,
Ka = lim s2G(s) = lim s2K(s + z1)(s + z2)...... = 0 s→0 s→0 (s + p1)(s + p2)......
I For Type-1 system
Ka = lim s2G(s) = lim s2K(s + z1)(s + z2)...... = 0
s→0 s→0 s(s + p1)(s + p2)......
Ka = lim s2G(s) = lim s2K(s + z1)(s + z2)...... = Kz1z2.... = K1(say)
I For Type-2 system
s→0 s→0 s2(s + p1)(s + p2)...... p1p2...
I For Type-3 or higher Type system, n ≥ 3,
Ka = lims2G(s)= lim s2K(s+z1)(s+z2)...... =∞, n≥3
s→0 s→0 sn(s + p1)(s + p2)......
I Summary: the steady-state error ess, for unit acceleration input:
ess = 1 = ∞, for type 0 and type 1systems Ka
ess = 1 = 1 fortype2system Ka K1
ess = 0, for type 3 or higher system
Akshya Swain
Module-4: Time Domain Analysis of Lin
ear Systems 13 / 31
Summary of Steady State Error
Akshya Swain
Module-4: Time Domain Analysis of Lin
ear Systems 14 / 31
Summary: Steady State Error and Static Error Coefficients
1. For unit step, unit ramp and unit acceleration ( parabolic) input i.e. if
u(t)=1(t)+t+ 1t2 2
the various error constants and associated steady state error are
Kp = lim G(s), e(∞) = 1 s→0 1+Kp
2. For a general input of the form u(t) = a 1(t)+bt+ct2
the various error constants and associated steady state error are
Kp = lim G(s), e(∞) = a s→0 1+Kp
Kv = lim sG(s), s→0
Ka = lim s2G(s), s→0
e(∞) = 1 Kv
e(∞) = 1 Ka
Kv = lim sG(s), s→0
Ka = lim s2G(s), s→0
e(∞) = b Kv
e(∞) = 2c Ka
Akshya Swain
Module-4: Time Domain Analysis of Lin
ear Systems 15 / 31
Example-1: Steady State Error
I Determine the steady state error for inputs of 10u(t),15tu(t) and 20t2u(t) for the
system shown below. Note that the function u(t) is the unit step.
= 120×2 =20 3×4
Solution: Now
Kp = limG(s)= lim 120(s+2)
s→0 s→0 (s+3)(s+4) e(∞) = estep(∞) = 10 = 10
The velocity and acceleration constants are
1+20 21
Kv = limsG(s)= lims 120(s+2) =0, =⇒ eramp(∞)= 15 =∞
0
=⇒ eparabolic(∞)= 40 =∞ 0
s→0 s→0 (s+3)(s+4)
Ka = lims2G(s)= lims2 120(s+2) =0, s→0 s→0 (s+3)(s+4)
Akshya Swain
Module-4: Time Domain Analysis of Lin
ear Systems 16 / 31
Example-2: Steady State Error
I Determine the steady state error for inputs of 10u(t),15tu(t) and 50t2u(t) for the
system shown below. Note that the function u(t) is the unit step.
Solution: Now the position error constant Kp is computed as:
Kp = limG(s)= lim 500(s+2)(s+4)(s+5)(s+6)(s+7) =∞, estep(∞)=0
s→0 s→0 s2(s + 8)(s + 10)(s + 12) The velocity and acceleration constants are:
Kv = limsG(s)= lim 500(s+2)(s+4)(s+5)(s+6)(s+7) =∞, eramp(∞)=0 s→0 s→0 s2(s + 8)(s + 10)(s + 12)
Ka = lims2G(s)= lims2500(s+2)(s+4)(s+5)(s+6)(s+7) s→0 s→0 s2(s + 8)(s + 10)(s + 12)
= 500×2×4×5×6×7 =875, =⇒ eparabolic(∞)= 100 =0.1142 8×10×12 875
Akshya Swain
Module-4: Time Domain Analysis of Lin
ear Systems 17 / 31
Example-3:Steady State Error
I Consider a unity feedback system with forward path transfer function
G(s) = 100(s + 1)(s + 2) s2(s + 3)(s + 10)
For an input of t3u(t), determine the steady state error in velocity of the system, where the velocity is defined as:
dr − dc dt dt
where r is the system input and c is the system output . Solution: Now
Lde = sE(s); Given r(t) = t3u(t), =⇒ R(s) = 3! = 6 dt s4 s4
I The steady state error is computed as:
2 2R(s) 2 6/s4 9
e(∞) = lim s E(s) = lim s = lim s 100(s+1)(s+2) = s→0 s→0 1 + G(s) s→0 1 + s2(s+10)(s+3)
10
Akshya Swain
Module-4: Time Domain Analysis of Lin
ear Systems 18 / 31
Example-4: Steady State Error
Consider a unity feedback system with forward path transfer function
G(s) = 5000 s(s + 75)
a. What is the expected percent overshoot for a unit step input ? b. What is the settling time for a unit step input ?
c. What is the steady state error for an input of 5u(t)? d. What is the steady state error for an input of 50tu(t)?
e. What is the steady state error for an input of 10t2u(t)? Solution:
a. The closed loop transfer function is
C(s) = 5000 = ωn2
R(s) s2 +75s+5000 s2 +2ζωns+ωn2
Comparing with the standard second order transfer function gives
√
ωn= 5000, 2ζωn=75, =⇒ ζ=0.53
I The percentage overshoot
%Mp = exp(−πζ/1 − ζ2) × 100 = 14.01%
Akshya Swain
Module-4: Time Domain Analysis of Lin
ear Systems 19 / 31
Example-4: Solution (contd) b. The settling time is
Ts= 4 = 4 =0.017sec ζ ωn 75/2
c. Since the system is Type-1, the steady state error due to the input 5u(t) is zero. d. For a velocity input, the velocity error constant is computed as:
Kv = limsG(s)= lims 5000 = 5000 s→0 s→0 s(s+75) 75
For the input 50tu(t), the steady state error is
e(∞) = 50 = 50 × 75 = 0.75
Kv 5000
e. Since the system is Type-1, the steady state error due to parabolic input is infinity i.e. ess = ∞.
Akshya Swain
Module-4: Time Domain Analysis of Lin
ear Systems 20 / 31
Example-5: Steady State Error
Consider a unity feedback system with forward path transfer function
G(s) = K(s + α) (s+β)2
Determine the values of K, α and β such that the steady-state error for a unit √
step input = 0.1; the damping ratio = 0.5 and the natural frequency = 10. Solution: Now, the error is
R(s) 1 E(s)= 1+G(s) = K(s+α)
s 1+ (s+β)2
Given that the steady state error due to step input is 0, 1. Thus
1 e(∞)=ess = limsE(s)= lims
s→0 s→0 s 1+K(s+α) (s+β)2
(s + β)2 β2 =lim2 =2=0.1
s→0 (s+β) +K(s+α) β +Kα This implies Kα = 9β2
Akshya Swain
Module-4: Time Domain Analysis of Lin
ear Systems 21 / 31
Example-5: Solution (Contd)
The closed loop transfer function is given by
C(s) = K(s+α)
R(s) s2 +(K +2β)s+(β2 +Kα)
I Thus
ωn2 = β2 + Kα = β2 + 9β2 =⇒ 10β2 = 10, =⇒ I Now taking β = +1
β = ±1
√
K+2β=2ζωn, =⇒ K+2β= 10 =⇒ K=1.16 and α= K =7.76
9β2
Akshya Swain
Module-4: Time Domain Analysis of Lin
ear Systems 22 / 31
Steady State Error for Disturbances
I Note that one of the advantage of a feedback control system is that it can compensate for disturbances or unwanted inputs.
I It is possible to design systems which can follow the input with small or zero error.
I Consider the system shown in the Figure below where the disturbance D(s) is injected between the controller and the plant.
I The transform of the output is given by
C(s) = E(s)G1(s)G2(s) + D(s)G2(s)
But
C(s) = R(s) − E(s)
Akshya Swain
Module-4: Time Domain Analysis of Lin
ear Systems 23 / 31
Steady State Error for Disturbances
Now
or, E(s) [1 + G1(s)G2(s)] = R(s) − D(s)G2(s)
R(s) − E(s) = E(s)G1(s)G2(s) + D(s)G2(s)
or, E(s)= 1 R(s) − G2(s) D(s)
1 + G1(s)G2(s) 1 + G1(s)G2(s)
where we can think that the transfer function relating E(s) and R(s) and that
between E(s) and D(s) are respectively equals to
1 & G2(s)
1 + G1(s)G2(s) 1 + G1(s)G2(s)
I Now applying final value theorem to compute the steady state error gives
where
and
s→0 1 + G1(s)G2(s)
R(s)
e(∞) = lim sE(s) = lim s R(s) − lim sG2(s)
D(s)
s→0
s→0 1 + G1(s)G2(s)
s→0 1 + G1(s)G2(s) eR(∞) = lim s
eD(∞) = − lim sG2(s)
s→0 1 + G1(s)G2(s)
D(s)
Akshya Swain
Module-4: Time Domain Analysis of Lin
ear Systems 24 / 31
Steady State Error for Disturbances
Now the steady state error due to disturbance is given by
eD(∞) = − lim sG2(s)
s→0 1 + G1(s)G2(s)
D(s)
Let us establish the condition on eD(∞) that must exist to reduce the error due to disturbance.
I Let us assume a step disturbance such that D(s) = 1/s.
The steady state error due to step disturbance is therefore can be expressed as:
eD(∞) = − lim sG2(s) D(s) = − lim sG2(s) 1 s→0 1 + G1(s)G2(s) s→0 1 + G1(s)G2(s) s
=−lim 1 = 1
s→0 1 + G1(s) lim 1 + lim G1(s)
G2(s) s→0 G2(s) s→0
I This relation shows that the steady state error due to step disturbance can be reduced by increasing the dc gain of G1(s) or decreasing the dc gain of G2(s)
Akshya Swain
Module-4: Time Domain Analysis of Lin
ear Systems 25 / 31
Example-1: Steady State Error for Disturbances
Find the steady state error component due to step disturbance for the system shown below
The system is stable. Now
inversely proportional to the dc gain of G1(s).
I Note that the dc gain of G2(s) is infinite in this example.
eD(∞)=
s→0 G2(s)
1 =− 1 =− 1 + lim G1(s) 0 + 1000 1000
lim 1
I The result shows that the steady state error produced by the step disturbance is
s→0
Akshya Swain
Module-4: Time Domain Analysis of Lin
ear Systems 26 / 31
Example-2: Steady State Error for Disturbances
Find the steady state error component due to step disturbance for the system shown below
The system is stable. Now
eD (∞) =
s→0 G2(s)
1 = −
+ lim G1(s) s→0
1 = −9.98 × 10−4 2 + 1000
lim 1
Akshya Swain
Module-4: Time Domain Analysis of Lin
ear Systems 27 / 31
Steady State Error for Non-Unity Feedback Systems
I Often, most of the practical control systems are not unity feedback. Some of the reasons include
a. Compensation is introduced in the feedback path to improve performance of the system.
b. The feedback element may be an inherent part of the physical model of the system. I If we want to find the steady state error for the non-unity feedback system, then
Step-1: Represent the non-unity feedback system by an equivalent unity feedback system.
Step-2: Apply the methods/formula used for unity feed back system.
I Consider the general feedback (non-unity) system shown in the figure
where G1(s) is the input transducer, G2(s) is the plant/system and H1(s) is the feedback path system ( may be a compensator).
Akshya Swain
Module-4: Time Domain Analysis of Lin
ear Systems 28 / 31
Steady State Error for Non-Unity Feedback Systems
I Pushing the input transducer to the right past the summing junction gives the general non-unity feedback system shown Fig-b, where G(s) = G1(s)G2(s) and H(s) = H1(s)/G1(s)
I Note that, unlike the unity feedback systems, where H(s) = 1, the error is not the difference between the input and the output.
I For this case let us call the signal at the output of the summing junction as actuating signal, Ea(s)
I It is possible to form a unity feedback system from the non-unity feedback system shown in Fig-b by following few simple steps of block-diagram reduction which include
a. Add and subtract unity feedback paths, as shown in Fig-c ( next slide).
b. Combine H(s) with negative unity feedback,as shown in Fig-d ( next slide).
c. Finally combine the feedback system consisting of G(s) and H(s) − 1, shown in Fig-e
Akshya Swain
Module-4: Time Domain Analysis of Lin
ear Systems 29 / 31
Steady State Error for Non-Unity Feedback Systems
I It is possible to form a unity feedback system from the non-unity feedback system shown in Fig-b by following few simple steps of block-diagram reduction which include
a. Add and subtract unity feedback paths, as shown in Fig-c.
b. Combine H(s) with negative unity feedback,as shown in Fig-d.
c. Finally combine the feedback system consisting of G(s) and H(s) − 1, shown in Fig-e
Akshya Swain
Module-4: Time Domain Analysis of Lin
ear Systems 30 / 31
Example-Steady State Error for Non-Unity Feedback Systems
For the system shown in the Figure, find the system type and the appropriate error constant associated with the system type, and the steady state error for a unit step input. Assume input and output units are same.
Solution: Given
G(s) = 100 , H(s) = 1 s(s+10) s+5
I The equivalent transfer function is computed as:
Ge(s) = G(s) = 100(s + 5)
1+G(s)H(s)−G(s) s3 +15s2 −50s−400
I Thus the system is Type 0 as there are no pure integrators. The appropriate
error constant is Kp and steady state error e(∞) are calculated as:
Kp = limGe(s)= 100×5 =−5, e(∞)= 1 =−4. s→0 −400 4 1+Kp
Akshya Swain
Module-4: Time Domain Analysis of Lin
ear Systems 31 / 31