Intended Learning Outcomes: (a) Determine and interpret the static-error constants of a system; (b) Identify the system type and evaluate the steady-state error based on the static-error constants of the system; (c) Design the gain of the system to meet steady-state error specification objective.
In this section, the parameters specifying steady-state error performance of unity negative feedback systems will be defined. These steady-state error performance specifications are called static error constants.
The static error constants are
• The position constant, K୮
K୮= limୱ→Gሺsሻ (5.7)
• The velocity constant, K୴
K୴ = limୱ→sGሺsሻ (5.8)
• The acceleration constant, Kୟ
Stability and Steady-State Errors Page 26
Kୟ = limୱ→sଶGሺsሻ (5.9)
These quantities can assume the value of zero, finite constant or infinity depending on the form of Gሺsሻ.
Also, the steady state error decreases when the static error constant increases.
Example 5.18
For each of the system shown below, evaluate the static error constants and find the expected error for the standard step, ramp and parabolic inputs.
Answers:
For system (a) the static error constants are K୮ = 5.208, K୴ = 0 and Kୟ = 0. The expected errors are eୱ୲ୣ୮ሺ∞ሻ = 0.161, e୰ୟ୫୮ሺ∞ሻ = ∞ and e୮ୟ୰ୟሺ∞ሻ = ∞.
For system (b) the static error constants are K୮ = ∞, K୴ = 31.25 and Kୟ = 0. The expected errors are eୱ୲ୣ୮ሺ∞ሻ = 0, e୰ୟ୫୮ሺ∞ሻ = 0.032 and e୮ୟ୰ୟሺ∞ሻ = ∞.
For system (c) the static error constants are K୮ = ∞, K୴ = ∞ and Kୟ = 875. The expected errors are eୱ୲ୣ୮ሺ∞ሻ = 0, e୰ୟ୫୮ሺ∞ሻ = 0 and e୮ୟ୰ୟሺ∞ሻ = 1.14 × 10ିଷ.
Stability and Steady-State Errors Page 27 System Type. The following table summarizes the system type of systems, which depends on the number of integrations of the forward transfer function.
Steady-State Error Specifications. Static error constants can be used to specify the steady-state error characteristics of control systems, such as what has been done with the damping ratio, natural frequency, rise, peak and settling time and percent overshoot specifying transient response performance of systems.
Example 5.19
What information can be deduced from the system whose static error constant is K୴ = 1000?
Answer:
What information is obtained in the specification K୮= 1000?
Answer:
1. The system is stable.
2. The system is type 0.
3. The input test signal is a step.
4. The steady-state error is eሺ∞ሻ =ଵଵଵ .
Stability and Steady-State Errors Page 28 Static-error constants can be used to design the gain of a system to meet steady-state error specifications.
Example 5.21
Given the control system below, find the value of K so that there is 10% error in the steady-state.
Answer:
K = 672
Example 5.22
A unity feedback system has the following forward transfer function Gሺsሻ = Kሺs + 12ሻ
ሺs + 14ሻሺs + 18ሻ
Find the value of K to yield a 10% error in the steady-state.
Answer:
Find the steady-state error in position.
Stability and Steady-State Errors Page 29 2. A system has K୮= 4. What steady-state error can be expected for inputs of 70 uሺtሻ and 70t uሺtሻ?
3. A type 3 unity feedback system has rሺtሻ = 10tଷ applied to its input. Find the steady-state position error for this input if the forward transfer function is Gሺsሻ =ଵଷ൫ୱమୱା଼ୱାଶଷ൯൫ୱయሺୱାሻሺୱାଵଷሻమାଶଵୱାଵ.଼൯.
4. A unity feedback system as shown in item (1) has an open-loop transfer function Gሺsሻ = ൫ୱሺୱାହሻమାୱା൯మሺୱାଷሻ.
a. Find the system type.
b. What error can be expected for an input of 12 uሺtሻ?
c. What error can be expected for an input of 12t uሺtሻ?
5. Find the system type for the system shown below.
6. For the system shown below
a. Find K୮, K୴ and Kୟ.
b. Find the steady-state error for an input of 50 uሺtሻ, 50t uሺtሻ and 50tଶ uሺtሻ.
c. State the system type.
Stability and Steady-State Errors Page 30 7. For the systems shown below, find
a. The closed-loop transfer function.
b. The system type.
c. The steady-state error for an input of 5 uሺtሻ.
d. The steady-state error for an input of 5t uሺtሻ.
8. For the system shown below,
a. What value of K will yield a steady-state error in position of 0.01 for an input of ଵଵ t?
b. What is the K୴ for the value of K found in (a)?
c. What is the minimum possible steady-state position error for the input given in (a)?
9. Given the system shown below, design the value of K so that for an input of 100t uሺtሻ, there will be a 0.01 error in the steady-state.
Stability and Steady-State Errors Page 31 10. The unity feedback system shown below, where Gሺsሻ = ൫ୱୱమାଷୱାଷ൯ሺୱାହሻ , is to have 1/6000 error
between an input of 10t uሺtሻ and the output in the steady-state.
a. Find K and n to meet the specifications.
b. What are K୮, K୴ and Kୟ?
11. Given the unity feedback control system of item 10 where Gሺsሻ =ୱሺୱାୟሻ , find the following:
a. K and a to yield K୴ = 1000 and a 20% overshoot.
b. K and a to yield a 1% error in the steady-state and a 10% overshoot.
12. For the unity feedback system of item 10, where
Gሺsሻ = K
sሺs + 4ሻሺs + 8ሻሺs + 10ሻ
find the minimum possible steady-state position error if a unit ramp is applied. What places the constraint upon the error?
13. The unity feedback system of item 10 where Gሺsሻ =ሺୱାሻୱሺୱାஒሻ is to be designed to meet the following requirements: The steady-state position error for a unit ramp input equals 1/10; the closed-loop poles will be located at −1 ± j1. Find K, α and β in order to meet the specifications.
14. Given the unity feedback control system of item 10 where Gሺsሻ =ୱሺୱାୟሻ , find the values of n, K and a to meet specifications of 12% overshoot and K୴ = 100.
15. The system shown below is to have the following specifications: K୴ = 10; d୰ = 0.5. Find the values of Kଵ and K required for the specifications of the system to be met.
Stability and Steady-State Errors Page 32 5.6 Steady-State Error for Disturbances and Non-unity Feedback Systems
Intended Learning Outcome: Evaluate the steady-state error or steady-state actuating signal and design components for systems with disturbances and non-unity feedback.
One advantage of feedback systems is that it can compensate for disturbances or unwanted inputs that enter the system. Thus, systems can be designed to follow the input with small or zero error. Figure 5.9 shows a feedback control system with a disturbance Dሺsሻ, injected between the controller and the plant.
Figure 5. 9. Feedback control system with disturbance.
It can be shown that the steady-state error for this system will be eሺ∞ሻ = limୱ→sEሺsሻ = limୱ→ s
1 + GଵሺsሻGଶሺsሻ Rሺsሻ − limୱ→ sGଶሺsሻ
1 + GଵሺsሻGଶሺsሻ Dሺsሻ (5.10)
Equation 5.10 shows that the error of the system of figure 5.9 has two components: error due to the input Rሺsሻ, eୖሺ∞ሻ,
eୖሺ∞ሻ = limୱ→ s
1 + GଵሺsሻGଶሺsሻ Rሺsሻ (5.11a)
Stability and Steady-State Errors Page 33 which have been derived earlier; and error due to the disturbance Dሺsሻ, eୈሺ∞ሻ,
eୈሺ∞ሻ = − limୱ→ sGሺsሻ
1 + GଵሺsሻGଶሺsሻ Dሺsሻ (5.11b)
If the disturbance is of step form, the steady-state error component due to the disturbance is
eୈሺ∞ሻ = − 1
limୱ→ୋమଵሺୱሻ+ limୱ→Gଵሺsሻ (5.12)
This equation shows that the steady-state error produced by a step disturbance can be reduced by increasing the dc gain of Gଵሺsሻ or decreasing the dc gain of Gଶሺsሻ.
The following examples demonstrate the analysis of steady-state error in the presence of step disturbance.
Example 5.23
Find the steady-state error component due to a step disturbance for the system shown below.
Answer:
eୈሺ∞ሻ = − 1 1000
Example 5.24
Evaluate the steady-state error component due to a step disturbance for the system shown below.
Stability and Steady-State Errors Page 34 Answer:
eୈሺ∞ሻ = −9.98 × 10ିସ
Control systems often do not have unity feedback because of the compensation used to improve performance or because of the physical model for the system. The feedback path can be a pure gain other than unity or have some dynamic representation. The analysis of these type of systems will first require a reduction of the general feedback control system shown in figure 5.10 into the one in figure 5.11.
Figure 5. 10. The general non-unity feedback systems.
Figure 5. 11. The reduced form of figure 5.10.
Note that in the system of 5.11, the signal Eሺsሻ is the steady-state error provided the units of the input and the output of system of 5.10 are the same. If they are not, the difference at the summing junction of 5.10 is called the actuating signal Eୟଵሺsሻ and its steady-state value eୟଵሺ∞ሻ from the figure, it is given as
Stability and Steady-State Errors Page 35 eୟଵሺ∞ሻ = limୱ→ sRሺsሻGଵሺsሻ
1 + GଶሺsሻHଵሺsሻ (5.13)
For non-unity feedback systems, it will always be assumed that the input and output have the same units unless otherwise stated.
The following example demonstrates the analysis of steady-state error and steady-state actuating signal.
Example 5.25
For the system shown below, find the system type, 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 the same.
Answer:
The system is Type 0, the position constant is K୮ = −5/4 and the steady-state error is eሺ∞ሻ = −4.
Example 5.26
Find the steady-state actuating signal for the system shown in Example 5.25 for a unit step input and unit ramp input.
Answer:
For the unit step, eୟሺ∞ሻ = 0 while for the ramp input, eୟሺ∞ሻ = 1/2.
Example 5.27
(a) Find the steady-state error, eሺ∞ሻ = rሺ∞ሻ − cሺ∞ሻ, for a unit step input given the non-unity feedback system shown below. Repeat for a unit ramp input. Assume input and output units are the same.
(b) Find the steady-state actuating signal, eୟሺ∞ሻ, for a unit step given the non-unity feedback system below. Repeat for a unit ramp input.
Stability and Steady-State Errors Page 36 Answers:
(a) eୱ୲ୣ୮ሺ∞ሻ = 3.846 × 10ିଶ; e୰ୟ୫୮ሺ∞ሻ = ∞.
(b) For unit step input eୟሺ∞ሻ = 3.846 × 10ିଶ; for unit ramp input eୟሺ∞ሻ = ∞.
For the general system shown in figure 5.12, the restrictions for Gଵሺsሻ, Gଶሺsሻ, and Hሺsሻ can be obtained.
Figure 5. 12. A non-unity feedback system with disturbance.
The steady-state error for this system, eሺ∞ሻ = rሺ∞ሻ − cሺ∞ሻ is eሺ∞ሻ = limୱ→sEሺsሻ = limୱ→s ቊቈ1 − GଵሺsሻGଶሺsሻ
1 + GଵሺsሻGଶሺsሻHሺsሻRሺsሻ − ቈ Gଶሺsሻ
1 + GଵሺsሻGଶሺsሻHሺsሻ Dሺsሻቋ (5.14)
If both the input and the disturbance are of step form, this equation becomes
eሺ∞ሻ = limୱ→sEሺsሻ = ൝1 − limୱ→[GଵሺsሻGଶሺsሻ]
limୱ→[1 + GଵሺsሻGଶሺsሻHሺsሻ]൩ −
limୱ→Gଶሺsሻ
limୱ→[1 + GଵሺsሻGଶሺsሻHሺsሻ]൩ൡ (5.15)
From this, for zero error,
Stability and Steady-State Errors Page 37
1. Find the total steady-state error due to a unit step input and a unit step disturbance for the system shown below.
2. Design values of Kଵ and Kଶ in the system shown below to meet the following specifications: steady-state error component due to a unit step disturbance is −0.000012; steady-state error component due to a unit ramp input is 0.003.
3. For each of the systems shown below, find the following:
a. The system type
b. The appropriate static error constant
c. The input waveform to yield a constant error
d. The steady-state error for a unit input of the waveform found in Part (c).
Stability and Steady-State Errors Page 38 e. The steady-state value of the actuating signal
4. For each of the system shown below, find the appropriate static error constant as well as the steady-state error, rሺ∞ሻ − cሺ∞ሻ for unit step, ramp and parabolic inputs.
Stability and Steady-State Errors Page 39 5. Given the system shown below, find the following:
a. The system type
b. The value of K to yield 0.1% error in the steady-state.