feedcon[unit 5].pdf

Feedback and Control Systems

Stability and Steady-State Errors Page 1

STABILITY AND STEADY-STATE ERRORS

At the end of this chapter, the students shall be able to:

5.1. (a) Define stability and cite the necessary conditions for a system to become stable; (b) Relate

stability to the location of the closed-loop poles of the system.

5.2. (a) Generate and interpret a Routh tableto determine how many closed-loop poles are in the left-

half plane, in the right-half plane and on the j-axis of the complex s-plane does a higher-ordered system have; (b) Use the Routh-Hurwitz criterion to determine the stability of a higher-ordered

system; (c) Use Routh-Hurwitz criterion for stability design.

5.3. Define steady-state error and discuss the errors arising from the system configuration and the

type of input.

5.4. Evaluate steady-state errors for unity feedback systems using the open-loop or closed-loop

transfer functions.

5.5. (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.

5.6. Evaluate the steady-state error or steady-state actuating signal and design components for

systems with disturbances and non-unity feedback.

5.7. Evaluate the sensitivity of transfer function and steady-state error of a system due to a change in

parameter.

5.1. Stability

Intended Learning Outcomes: (a) Define stability and cite the necessary conditions for a system to become

stable; (b) Relate stability to the location of the closed-loop poles of the system.

Stability is the most important system specification. If a system is unstable, transient response and steady-

state errors are moot points. An unstable system cannot be designed for specific transient response

performance or steady-state error specifications. Stability can be defined in many ways, depending on the

kind of system being analyzed or the point of view of the definition.

For linear, time-invariant systems, stability can be defined in terms of the natural response. Recall that the

total response of the system ct is the sum of two responses: the forced response and the natural



response. Also, it is said that if a system is controlled the steady-state output consists only of the forced

response. Thus, it can be concluded that:

An LTI system is stable if the natural response approaches zero as time approaches infinity.

An LTI system is unstable if the natural response grows without bound as time approaches infinity.

An LTI system is marginally stable if the natural response neither decays nor grows but remains

constant or oscillates as time approaches infinity.

These definitions rely on natural response, but this perspective is difficult to look into when considering the

total response. For the total response, the following points can be raised:

If the input is bounded and the total response does not approach infinity as time approaches

infinity, then the natural response is not approaching infinity; thus the system is stable.

If the input is unbounded, the total response becomes unbounded also, but it cannot be determined

whether it is the natural response that becomes unbounded, or it is the forced response that

becomes unbounded.

With this, alternative definitions, one that looks into the input and the total response, are in order. Thus,

A system is stable if every bounded input yields a bounded output (called the BIBO stability

requirement).

A system is unstable if any bounded input yields an unbounded output.

A system is marginally stable if it is stable for some inputs and unstable for others.

How can a system be concluded as stable? Looking into the natural response, which is required to

approach zero as time approaches infinity, it can be seen that the system poles that yield the natural

response should be negative if they are real or have negative real parts if they are complex. Thus,

Stable systems have closed-loop transfer functions with poles only in the left half-plane of the

complex s-plane. Those systems which have closed-loop poles that are on the right half-plane of the complex s-

plane are unstable.



Systems whose closed-loop poles that are on the j-axis and multiplicity of one are marginally stable systems. When at least one of the closed-loop poles on the j-axis have a multiplicity of two or more, the system is considered unstable.

Step responses of unity feedback system and their closed-loop poles plotted on the complex s-plane are shown in Figure 5.1 below. The first system is considered to be stable, the second one unstable.

Determining the stability of feedback control systems is not a simple task however. Some systems will have

higher order poles when reduced to an equivalent closed-loop transfer function, which requires factoring

and solving for the roots of the denominator of the transfer function. But without actually solving for the

poles, the following observations can be used to determine whether the system is stable or not:

Since all of the poles must be negative if real or have negative real part if complex, the

denominator of the transfer function can be factored into s + a where a real and positive, or complex with positive real part. The product of all such terms is a polynomial with all coefficients

having the same sign.

No term in the polynomial can be missing, since this will imply cancellation between positive and

negative coefficients or imaginary axis roots in the factor, which is not the case.

Thus a system is unstable (or at best, marginally stable) if at least one of the coefficients of the

denominator of the closed-loop transfer function has a sign that is different from the others, and/or at

least one term in the polynomial is missing. These observations can be used to determine outright

whether the system is stable or not. However, when both conditions exist (all coefficients have the

same sign and no power of s missing) other methods, like computer-aided methods, of solving for the roots are employed.



Figure 5. 1. Step response and pole plot of stable and unstable systems.

5.2. Routh-Hurwitz Criterion for Stability

Intended Learning Outcomes: (a) Generate and interpret a Routh table to determine how many closed-loop

poles are in the left-half plane, in the right-half plane and on the -axis of the complex -plane does a higher-ordered system have; (b) Use the Routh-Hurwitz criterion to determine the stability of a higher-

ordered system; (c) Use Routh-Hurwitz criterion for stability design.

There is a method by which the stability of feedback control systems can be determined without actually

solving for the roots. Using the Routh-Hurwitz criterion for stability, one can tell how many poles are in the

left half-plane (LHP), right half-plane (RHP) and on the j-axis.



The method requires two steps:

Generate a table called a Routh table.

Interpret the Routh table to tell how many closed-loop system poles are in the LHP, RHP and j-axis.

Generating a Basic Routh Table. For the closed-loop transfer function shown in figure 5.2, the initial

layout for the Routh table is shown in figure 5.3, and a completedtable in figure 5.4.

Figure 5. 2. Closed-loop transfer function to be used in the generation of Routh table.

Figure 5. 3. Initial layout of the Routh table.

Figure 5. 4. Completed Routh table.



Example 5.1

Make a Routh table for the system shown below.

Answer:

Interpreting a Basic Routh Table. The basic Routh table applies to systems with poles in LHP and RHP.

Systems with imaginary poles and the kind of the Routh table that results will be discussed next. Simply

stated, the Routh-Hurwitz criterion declares that the number of roots of the polynomial that are in the right-

half-plane is equal to the number of sign changes in the first column.

Example 5.2

Based on the completed basic Routh table in Example 5.1, determine whether the system is stable.

Answer:

The system is unstable since there are two poles in the RHP of the pole map of the system, as shown by

two sign changes in the first column of the Routh table.

Example 5.3

Make a Routh table for the following polynomial and tell how many roots are in the RHP and LHP.

Ps = 3s + 9s + 6s + 4s + 7s + 8s + 2s + 6



Answer:

Since there are four sign changes in the first column, then the polynomial has four poles in the RHP and three poles

in the LHP.

Routh-Hurwitz Special Case Zero Only in the First Column. If the first element of a row is zero,

division by zero would be required to form the next row. To avoid this phenomenon, an epsilon, , is assigned to replace zero in the first column. The value is then allowed to approach zero from either the positive or the negative side, after which the signs of the entries in the first column can be determined.

Example 5.4

Determine the stability of the closed-loop transfer function

Ts = 10s + 2s + 3s + 6s + 5s + 3

Answer:

The Routh table is completed as follows



If is taken to approach 0# and 0$, the following sign changes results:

Thus, with two sign changes in either case, there are two poles in the RHP and the system is unstable.

Another method that can be used when a zero appears only in the first column of a row is by obtaining a

polynomial that has roots reciprocal that of the original polynomial which, in this case will not arrive at a

Routh table with a zero in the first column.

Example 5.5

Redo Example 5.4 by obtaining a polynomial whose roots are the reciprocal of the original polynomial.

Answer:

If the reciprocal polynomial is used, the following Routh table results.

Thus, just like in the previous example, there are two sign changes and therefore there are two poles in the

RHP. The system is unstable.



Routh-Hurwitz Special Case Entire Row Zero.Proceeding first with an example, the procedure for the

case when an entire row becomes zero is demonstrated.

Example 5.6

Determine the number of right half-plane poles in the closed-loop transfer function

Ts = 10s + 7s + 6s + 42s + 8s + 56

Answer:

There will be an entire row of zeros in this case. However, using the procedure described when handling

this case, the resulting Routh table will be:

The table shows no sign changes in the first column; hence the system has no poles in the RHP.

An entire row of zeros will appear in the Routh table when a purely even or a purely odd polynomial is a

factor of the original polynomial. Even polynomials only have roots that are symmetrical about the origin.

They can be, as shown in figure 5.5,

symmetrical and real (A).

symmetrical and imaginary (B).

quadrantal (C).



Figure 5. 5. Roots of an even polynomial.

Another characteristic of the Routh table for the case in question is that the row previous to the row of zeros

contains the even polynomial that is a factor of the original polynomial. Finally, everything from the row

containing the even polynomial down to the end of the Routh table is a test of only the even polynomial.

Example 5.7

For the transfer function

Ts = 20s% + s + 12s + 22s + 39s + 59s + 48s + 38s + 20

tell how many poles are in the right half-plane, in the left half-plane and on the j-axis.

Answer:

Example 5.8

Use the Routh-Hurwitz criterion to find how many poles of the following closed-loop system Ts are in the RHP, LHP and on the j-axis.



Ts = s + 7s 21s + 10s + s 6s + 0s s s + 6

Answer:

Two RHP poles, two LHP and two j.

Example 5.9

Find the number of poles in the LHP, RHP and on the j-axis for the system shown below.

Answer:

Two in RHP, two in LHP

Example 5.10

Find the number of poles in LHP, RHP and on the j-axis for the system shown

Answer:

Three poles in the LHP and two poles in RHP.



Example 5.11

Find the number of poles in LHP, RHP and on the j-axis for the system shown

Answer:

Two in RHP, 4 in LHP and 2 in j-axis

The Routh-Hurwitz criterion gives vivid proof that changes in the gain of a feedback control system results

in differences in transient response because of the changes in closed-loop pole locations. It will be seen

that, aside from variations in the transient response, gain variations can also cause the closed-loop poles to

move from stable regions of the s-plane onto the j-axis and then into the right-half plane. Example 5.12

Find the range of gain K for the system shown below that will cause the system to be stable, unstable and marginally stable.

Answer:

When K < 1386, the system is stable. When K > 1386, the system is unstable. When K = 1386, the system is marginally stable.

Example 5.13

For a unity feedback system with the forward transfer function

Gs = Ks + 20ss + 2s + 3



find the range of K to make the system stable.

Answer:

The gain K should be between, but not equal to 0 and 2.

Drill Problems 5.1

1. Tell how many roots of the following polynomial are in the right half-plane, left half-plane and on the j-axis:

a. Ps = s + 3s + 5s + 4s + s + 3 b. Ps = s + 6s + 5s + 8s + 20 c.

2. Using the Routh table tell how many poles of the following closed-loop transfer functions are in the

RHP, LHP, and on the j-axis. a. Ts = +#%+,$+-#+.$+/#+$ b. Ts = +.#+/#+#0+,$+-#+.$+/#+$

3. How many poles are in the right half-plane. In the left half-plane and on the j-axis for the open-loop systems shown below:

4. For the system below with a positive feedback, tell how many closed-loop poles are in the right half-

plane, in the left half-plane, and on the j-axis.



5. Using the Routh-Hurwitz criterion, tell how many closed-loop poles of the system shown below lie in the

left-half plane, in the right half-plane, and on the j-axis.

6. Determine if the unity feedback system of the figure below when Gs = 12+/#03+#0+#

can be unstable.

7. In the system shown below, let Gs = 1+$4++$5 . Find the range of K for closed-loop stability when a. a < 0, b < 0 b. a < 0, b > 0 c. a > 0, b < 0 d. a > 0, b > 0

8. For the unity feedback system shown below, find the range of the gain K for which the system will become stable for the following open-loop transfer functions:



a. Gs = 1+#+#+$+$ b. Gs = 1+#+$+/# c. Gs = 1+#0+-+# d. Gs = 1+$+#+#+/#0 e. Gs = 1+#+/#0+#+$0

9. Given the unity feedback system shown below with (i) Gs = 1+#++#0.+# , (ii) Gs = 1+$0+$8#+/#+# , (iii) Gs = 1++#0+#+# , (iv) Gs = 1+#0.+# , and (v) Gs = 1+#9+/#+#, find the following:

a. The range of K that keeps the system stable. b. The value of K that makes the system oscillate. c. The frequency of oscillation when K is set to the value that makes the system oscillate.

10. For each of the systems shown below, find the range of K for which the system is stable.



11. For the system shown below, find the value of gain K that will make the system oscillate and find the frequency of oscillation.

12. Find the value of K for the system shown that will place the closed-loop poles as shown.



13. The closed-loop transfer function of a system is Ts = +/#1:+#1/+-#1:+.#1/+/#+#0. Determine the range of K0 for the system to be stable. What is the relationship between K0 and K for stability?

14. For the closed-loop transfer function Ts = 1:+#1/+-#1:+.#+/#1/+#0, find the constraints on K0 and K such that the function will have only two j poles.

15. The transfer function relating the output engine fan speed (rpm) to the input main burner fuel flow rate

(lb/h) in a short takeoff and landing (STOL) fighter aircraft, ignoring the coupling between engine fan

speed and the pitch control command, is

a. Find how many poles are in the right half-plane, in the left-half plane, and on the j-axis. b. Is this open-loop system stable?



5.3. Steady-State Errors

Intended Learning Outcome: Define steady-state error and discuss the errors arising from the system

configuration and the type of input.

Steady-state error is the difference between the input and the output for a prescribed test input as t .Test inputs used for analysis and design are summarized in the table below.

Bear in mind always that only systems that are stable can be analyzed for steady-state errors or transient response.

Thus, it is a practice that systems must be checked for stability first before proceeding with the analysis of errors and

transient response. All the derived expressions here assume that the system under consideration is stable, but for

the exercises, stability must be checked first before proceeding with the evaluation of the steady-errors.

Figure 5.6 will help explain how steady-state errors are evaluated.

For the step input in figure 5.6a, output 1 has no steady-state error (e0 = 0) while output 2 has a finite steady-state error (e < ) as measured vertically between the input and output 2 after the transients have died down. For the ramp input in figure 5.6b, again, output 1 has no steady-state error while that of

output 2 has finite steady-state error. For this input, another possibility exists, such as when an output 3

has a different slope than that of the ramp input; which in that case, output 3 has an infinite steady-state

error (e = .



Figure 5. 6. Steady-state error for step and ramp inputs.

Note that the error is the difference between the input and the output, and to measure the error, a closed-

loop system is formed as shown in figure 5.7a. Note that this system uses the closed-loop transfer function

Ts in the forward path. For figure 5.7b, this feedback system uses the open-loop transfer function Gs.

Figure 5. 7. Measurement of steady-state error.

Many steady-state errors in control systems arise from nonlinear sources. However, in the discussions,the

steady-state errors being referred to are errors that arise from the configuration of the system itself and the

type of applied input.



Figure 5. 8. System with a finite and zero steady-state error for a step input.

For a system such that in figure 5.8a, where Rs is the input and Cs is the output, Es = Cs Rs is the error signal. Note that in this case when there is only a pure gain K, for ct to be of finite value and nonzero, an error signal et must always be present. From this system, it can be said that

e++ = 1K c++ (5.1) where e++ is the steady-state error and c++ is the steady-state output. Based on 5.1, it can be generalized that although the steady-state error does not become zero, it diminishes as K, the value of the gain is increased.

If the forward path gain is replaced by an integrator, such as the one shown in figure 5.8b, there will be zero

error in the steady-state for a step input. As ct increases, et decreases, because ct = rt ct. This decrease will continue until et is zero, in which the integrator continues to have an output. Thus, systems like that of figure 5.8b have zero steady-state output for a step input.

5.4. Steady-State Error for Unity Feedback Systems

Intended Learning Outcome: Evaluate steady-state errors for unity feedback systems using the closed-loop

or open-loop transfer functions.

Steady-state error can be calculated from a systems closed-loop transfer function, Ts or the open-loop transfer function Gs for unity feedback systems. Steady-State Error in Terms of BC. Consider figure 5.7a. Using the final value theorem of Laplace transform, the steady-state error in terms of the closed-loop transfer function is

e = lim+G s Rs[1 Ts] (5.2)



Example 5.13

Find the steady-state error for the system of figure 5.7a if Ts = +/#+#0G and the input is a unit step.

Answer:

e = 12

Steady-State Error in Terms of JC. More insights can be deduced when the steady-state error is evaluated using the open-loop transfer function Gs. Using the final value theorem, and ensuring that the closed-loop system is stable, the steady-state error is

e = lim+G s Rs1 + Gs (5.3)

For the step input, 5.3 becomes

e+KLM = 11 + lim+G Gs (5.4)

The term lim+G Gs is called the dc gain of the forward transfer function. In order to have zero steady-state error, the limit lim+G Gs must approach infinity. For it to happen, the open-loop transfer function must have at least one pole in the origin, or, there should be at least a pure integration of the open-loop

transfer function. If there are no integrations, the lim+G Gs attains a finite value and therefore a finite steady-state error will result.

For the ramp input, 5.3 becomes

eN4OM = 1lim+G sGs (5.5)



For 5.5 to become zero, lim+G sGs must approach infinity. Therefore, there should be at least two integrations of the open-loop transfer function. If there is only one integration, the steady-state error

becomes a finite value. If there are no integrations in the forward path, this will result to an infinite steady-

state error.

For the parabolic input, 5.3 becomes

eM4N4 = 1lim+G sGs (5.6)

If there are three integrations, the steady-state error is zero. If there are only two, a finite steady-state error

results. Finally, if there is at most one integration, the steady-state error is infinite.

Example 5.14

Find the steady-state errors for inputs 5 ut, 5t ut and 5t ut to the system shown below.

Answer:

e = e+KLM = 521

e = eN4OM =

e = eM4N4 =



Example 5.15

Find the steady-state errors for inputs 5 ut, 5t ut and 5t ut to the system shown below.

Answer:

e = e+KLM = 0

e = eN4OM = 120

e = eM4N4 =

Example 5.16

A unity feedback system has the following forward transfer function:

Gs = 10s + 20s + 30ss + 25s + 35

Find the steady-state error for the following inputs: 15 ut, 15t ut and 15t ut.

Answer:

e = e+KLM = 0

e = eN4OM = 2.1875

e = eM4N4 =



Example 5.17

A unity feedback system has the following forward transfer function:

Gs = 10s + 20s + 30ss + 25s + 35s + 50

Find the steady-state error for the following inputs: 15 ut, 15t ut and 15t ut.

Answer:

The closed-loop system is unstable. Calculations cannot be made.

Drill Problems 5.2

1. For the unity feedback system shown below, where Gs = %G+#%+#0+#0++#%+/#+#% , find the steady-state errors for the following test inputs: 25 ut, 37t ut and 47tut.

2. For the unity feedback system shown in item 1, where Gs = G+#+#+#+/+#+#0 , find the steady-state error if the input is 80t ut.

3. For the unity feedback system shown in item 1, where Gs = GG+#+/#%+#0, find the steady-state error for inputs of 30 ut, 70t ut, and 81t ut.

4. The steady-state error in velocity of a system is defined as

QRdrdt dcdtTUKV where r is the system input and c is the system output. Find the steady-state error in velocity for an input of tut to a unity feedback system with a forward transfer function of Gs = 0GG+#0+#+/+#+#0G .



5. For the system shown below, what steady-state error can be expected for the following test inputs:

15 ut, 15t ut, 15t ut.

5.5. Static Error Constants, System Type and Steady-State Error Specifications

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, KM KM = lim+G Gs (5.7)

The velocity constant, KW KW = lim+G sGs (5.8)

The acceleration constant, K4



K4 = lim+G sGs (5.9)

These quantities can assume the value of zero, finite constant or infinity depending on the form of Gs. 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 KM = 5.208, KW = 0 and K4 = 0. The expected errors are e+KLM = 0.161, eN4OM = and eM4N4 = .

For system (b) the static error constants are KM = , KW = 31.25 and K4 = 0. The expected errors are e+KLM = 0, eN4OM = 0.032 and eM4N4 = .

For system (c) the static error constants are KM = , KW = and K4 = 875. The expected errors are e+KLM = 0, eN4OM = 0 and eM4N4 = 1.14 10$.



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 KW = 1000?

Answer:

1. The system is stable.

2. The system is Type 1.

3. A ramp is the test input signal.

4. The steady-state error is e = 00GGG = 0.001

Example 5.20

What information is obtained in the specification KM = 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 e = 00GG0.



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

Gs = Ks + 12s + 14s + 18

Find the value of K to yield a 10% error in the steady-state.

Answer:

K = 189

Drill Problems 5.3

1. An input of 25t ut is applied to the input of a Type 3 unity feedback system shown below, where Gs = 210s + 4s + 6s + 11s + 13ss + 7s + 14s + 19

Find the steady-state error in position.



2. A system has KM = 4. What steady-state error can be expected for inputs of 70 ut and 70t ut?

3. A type 3 unity feedback system has rt = 10t applied to its input. Find the steady-state position error for this input if the forward transfer function is Gs = 0GG2+/#%+#32+/#0+#0.%3+.+#+#0 .

4. A unity feedback system as shown in item (1) has an open-loop transfer function Gs = 12+/#+#3+#/+#.

a. Find the system type.

b. What error can be expected for an input of 12 ut? c. What error can be expected for an input of 12t ut?

5. Find the system type for the system shown below.

6. For the system shown below

a. Find KM, KW and K4. b. Find the steady-state error for an input of 50 ut, 50t ut and 50t ut. c. State the system type.



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 ut. d. The steady-state error for an input of 5t ut.

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 00G t? b. What is the KW 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 ut, there will be a 0.01 error in the steady-state.



10. The unity feedback system shown below, where Gs = 12+/#+#G3+Z+# , is to have 1/6000 error between an input of 10t ut and the output in the steady-state.

a. Find K and n to meet the specifications. b. What are KM, KW and K4?

11. Given the unity feedback control system of item 10 where Gs = 1++#4, find the following: a. K and a to yield KW = 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

Gs = Kss + 4s + 8s + 10 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 Gs = 1+#]++#^ 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, and in order to meet the specifications.

14. Given the unity feedback control system of item 10 where Gs = 1+Z+#4, find the values of n, K and a to meet specifications of 12% overshoot and KW = 100.

15. The system shown below is to have the following specifications: KW = 10; dN = 0.5. Find the values of K0 and Kb required for the specifications of the system to be met.



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 Ds, 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

e = lim+G sEs = lim+G s1 + G0sGs Rs lim+GsGs1 + G0sGs Ds (5.10)

Equation 5.10 shows that the error of the system of figure 5.9 has two components: error due to the input

Rs, ed, ed = lim+G s1 + G0sGs Rs (5.11a)



which have been derived earlier; and error due to the disturbance Ds, ee, ee = lim+G sGs1 + G0sGs Ds (5.11b)

If the disturbance is of step form, the steady-state error component due to the disturbance is

ee = 1lim+G 0f/+ + lim+G G0s (5.12)

This equation shows that the steady-state error produced by a step disturbance can be reduced by

increasing the dc gain of G0s or decreasing the dc gain of Gs. 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:

ee = 11000

Example 5.24

Evaluate the steady-state error component due to a step disturbance for the system shown below.



Answer:

ee = 9.98 10$

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 Es 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 E40s and its steady-state value e40 from the figure, it is given as



e40 = lim+G sRsG0s1 + GsH0s (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 KM = 5/4 and the steady-state error is e = 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, e4 = 0 while for the ramp input, e4 = 1/2.

Example 5.27

(a) Find the steady-state error, e = r c, 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, e4, for a unit step given the non-unity feedback system below. Repeat for a unit ramp input.



Answers:

(a) e+KLM = 3.846 10$; eN4OM = . (b) For unit step input e4 = 3.846 10$; for unit ramp input e4 = .

For the general system shown in figure 5.12, the restrictions for G0s, Gs, and Hs can be obtained.

Figure 5. 12. A non-unity feedback system with disturbance.

The steady-state error for this system, e = r c is e = lim+G sEs = lim+G s hi1 G0sGs1 + G0sGsHsj Rs i

Gs1 + G0sGsHsj Dsk (5.14)

If both the input and the disturbance are of step form, this equation becomes

e = lim+G sEs = lm1 lim+G[G0sGs]lim+G[1 + G0sGsHs]n m

lim+G Gslim+G[1 + G0sGsHs]no (5.15)

From this, for zero error,



lim+G[G0sGs]lim+G[1 + G0sGsHs] = 1

and

lim+G Gslim+G[1 + G0sGsHs] = 0

These equations will be satisfied if (1) the system is stable; (2) G0s is a Type 1 system; (3) Gs is a Type 2 system; and (4) Hs is a Type 0 system. Drill Problems 5.4

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 K0 and K 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).



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, r c for unit step, ramp and parabolic inputs.



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.

5.7. Sensitivity Analysis

Intended Learning Outcome: Evaluate the sensitivity of transfer function and steady-state error of a system

due to a change in parameter.

In designing systems, ideally, parameter changes of components should not appreciably affect a systems

performance. The degree by which changes in a system parameter affect system transfer function, and

hence system performance, is called sensitivity. A system with changes in system parameters produces no

effect on transfer functions has zero sensitivity. The greater the sensitivity, the less desirable the effect of

the parameter change.

More formally defined, sensitivity is the ratio of the fractional change in the function to a fractional change in

the parameter as the fractional change of the parameter approaches zero, or

Sq:s = limtsG Fractional change in the function, FFractional change in the parameter, P (5.16)

or

Sq:s = PF |F|P (5.17) This definition will now be applied to transfer functions and steady-state error through the following

examples.



Example 5.28

Given the system shown below, calculate the sensitivity of the closed-loop transfer function to changes in

parameter a. How can sensitivity be reduced?

Answer:

The sensitivity of the transfer function to changes in a is given as S}:4 = ass + as + K

The sensitivity of the transfer function to changes in a decreses as K is increased for a given value of s.

Example 5.29

For the system shown below, find the sensitivity of the steady-state error to changes in parameter K and parameter a with ramp inputs.

Answer:

The sensitivity of e to changes in parameter a is SL:4 = 1 while that of e to changes in parameter K is SL:1 = 1.

Example 5.30

Find the sensitivity of the steady-state error to changes in parameter K and parameter a for the system shown below with a step input.



Answer:

The sensitivity of e to changes in parameter a is SL:4 = Kab + K

while that in parameter K SL:1 = Kab + K

Example 5.31

Find the sensitivity of the steady-state error to changes in K for the system shown below.

Answer:

SL:1 = 7K10 + 7K

The examples show that there are cases that feedback reduces the sensitivity of a systems steady-state

error to changes in system parameters. The concept of sensitivity can be applied to other measures of

control system performance, as well; it is not limited to the sensitivity of the steady-state error performance.

Drill Problems 5.5

1. Given the system shown below, find the sensitivity of the steady-state error to parameter a. Assume a step input. Plot the sensitivity as a function of a.



2. For the system shown below, show that the sensitivity to plant changes is

S}:s = 11 + Ls where Ls = GsPsHs and Ts = q++0#d+ .

3. For the system in item (2), if Ps = +, Ts = 01+#0+#+/#+#0 and S}:s = +/#++/#+#0. a. Find Fs and Gs. b. Find the value of K that will result in zero steady-state error for a unit step input.

4. For the system shown below, find the sensitivity of the steady-state error for changes in K0 and K when K0 = 100 and K = 0.1. Assume step inputs for both the input and the disturbance.



REFERENCES:

N. Nise. (2008). Control Systems Engineering (6th ed.). United States of America: John Wiley & Sons.

R. Dorf& R. Bishop. (2008). Modern Control Systems (12th ed). New Jersey: Prentice Hall.

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