chapter 10 time-domain analysis and design of control systems · me 413 systems dynamics &...

ME 413 Systems Dynamics & Control Chapter 10: Time-Domain Analysis and Design of Control Systems

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Chapter 10

Time-Domain Analysis and

Design of Control Systems

A. Bazoune

10.1 INTRODUCTION

Block Diagram: Pictorial representation of functions performed by each component of a system and that of flow of signals.

( )C s( )R s

( ) ( ) ( )=C s G s R s

Figure 10-1. Single block diagram representation.

Terminology:

( )C s( )R s ( )G s1 ( )G s

2

( )H s

( )Disturbance U s

± ( ) ( ) ( )E s R s b s= ± ( )M s

( )B s

Figure 10-2. Block Diagram Components.

1. Plant: A physical object to be controlled. The Plant ( )G s2

, is the controlled system, of which

a particular quantity or condition is to be controlled.


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2. Feedback Control System: A system which compares output to some reference input and keeps output as close as possible to this reference.

3. Closed-loop Control System: same as feedback control system.

4. Open-loop Control System: Output of the system is not feedback to the system.

5. The Control Elements ( )G s1

, also called the controller, are the components required to

generate the appropriate control signal ( )M s applied to the plant.

6. The Feedback Elements ( )H s is the components required to establish the functional

relationship between the primary feedback signal ( )B s and the controlled output ( )C s .

7. The Reference Input ( )R s is an external signal applied to a feedback control system in order

to command a specified action of the plant. It often represents ideal plant output behavior.

8. The Controlled Output ( )C s is that quantity or condition of the plant which is controlled.

9. The Primary Feedback signal ( )B s is a signal which is a function of the controlled output

( )C s , and which is algebraically summed with the reference input ( )R s to obtain the

actuating signal ( )E s .

10. The Actuating Signal ( )E s , also called the error or control action, is the algebraic sum

consisting of the reference input ( )R s plus or minus (usually minus) the primary feedback

( )B s .

11. The Manipulated Variable ( )M s (control signal) is that quantity or condition which the

control elements ( )G s1

apply to the plant ( )G s2

.

12. A Disturbance ( )U s is an undesired input signal which affects the value of the controlled

output ( )C s . It may enter the plant by summation with ( )M s , or via an intermediate

point, as shown in the block diagram of the figure above.

13. The Forward Path is the transmission path from the actuating signal e to the controlled output

( )C s .

14. The Feedback Path is the transmission path from the controlled output ( )C s to the primary

feedback signal ( )B s .

15. Summing Point: A circle with a cross is the symbol that indicates a summing point. The ( )+

or ( )− sign at each arrowhead indicates whether that signal is to be added or subtracted.

16. Branch Point: A branch point is a point from which the signal from a block goes concurrently

to other blocks or summing points.


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10.2 BLOCK DIAGRAMS AND THEIR SIMPLIFICATION

Definitions:

( )C s( )R s ( )G s

( )H s

( )B s

( )E s

Figure 10-3 Block diagram of a closed-loop system with a feedback element.

1. ( )G s ≡Direct transfer function = Forward transfer function

2. ( )H s ≡Feedback transfer function

3. ( ) ( )G s H s ≡ Open-loop transfer function

4. ( ) ( )C s R s ≡ Closed-loop transfer function = Control ratio

5. ( ) ( )C s E s ≡Feedforward transfer function

Closed Loop Transfer Function:

For the system shown in Figure 10-3, the output ( )C s and input ( )R s are related as follows:

( ) ( ) ( )=C s G s E s

where

( ) ( ) ( ) ( ) ( ) ( )= − = −E s R s B s R s H s C s

Eliminating ( )E s from these equations gives

( ) ( ) ( ) ( ) ( )[ ]= −C s G s R s H s C s

This can be written in the form

( ) ( )[ ] ( ) ( ) ( )+ =G s H s C s G s R s1

or

( )

( )

( )

( ) ( )=

+

C s G s

R s G s H s1

The Characteristic equation of the system is defined as an equation obtained by setting the denominator polynomial of the transfer function to zero. The Characteristic equation for the above

system is ( ) ( )G s H s+ =1 0 .


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Block Diagram Reduction Rules

In many practical situations, the block diagram of a Single Input-Single Output (SISO), feedback control system may involve several feedback loops and summing points. In principle, the block diagram of (SISO) closed loop system, no matter how complicated it is, it can be reduced to the standard single loop form shown in Figure 10-3. The basic approach to simplify a block diagram can be summarized in Table 1:

TABLE 10-1 Block Diagram Reduction Rules

1 Combine all cascade blocks

2 Combine all parallel blocks

3 Eliminate all minor (interior) feedback loops

4 Shift summing points to left

5 Shift takeoff points to the right

6 Repeat Steps 1 to 5 until the canonical form is obtained

TABLE 10-2. Some Basic Rules with Block Diagram Transformation

G1u

2u

y

1/ G

1u

1

y G u

u yG

=

=

y Gu=

Gu

u

y

( )2 1 2e G u u= −

Gu

y

y

G

G1u

2u

yG1

u

2u

y

G

uy

y

G

u

y

1/ G

Gu

G

2u

y1 2

y Gu u= −

u ( )1 2y G G u= −1

G y21/G2

G

( )Y GG X=1 21

G Y2

GX

( )Y G G X= ±1 2

1G

2G

XY±±±±

1 2G GX Y

1 2±G GX Y


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█ Example 1

A feedback system is transformed into a unity feedback system


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=±

⋅=±

=GH

GH

HGH

G

R

C

1

1

1Closed-loop Transfer function

█ Example 2

Let us reduce the following block diagram to canonical form.

⇒⇒⇒⇒

⇒⇒⇒⇒

⇒⇒⇒⇒

Step 4:

⇒⇒⇒⇒

█ Example 3


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█ Example 4

Use block diagram reduction to simplify the block diagram below into a single block relating

( )Y s to ( )R s .

█ Solution


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█ Example 5

Use block diagram algebra to solve the previous example.

█ Solution

Multiple-Inputs cases

In feedback control system, we often encounter multiple inputs (or even multiple output cases). For a linear system, we can apply the superposition principle to solve this type of problems, i.e. to treat each input one at a time while setting all other inputs to zeros, and then algebraically add all the outputs as follows:

TABLE 10-3: Procedure For reducing Multiple Input Blocks

1 Set all inputs except one equal to zero

2 Transform the block diagram to solvable form.

3 Find the output response due to the chosen input action alone

4 Repeat Steps 1 to 3 for each of the remaining inputs.

5 Algebraically sum all the output responses found in Steps 1 to 5


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█ Example 6

We shall determine the output C of the following system:

█ Solution

Step1: Put 0≡U

Step2: The system reduces to

Step 3: The output RC due to input R is RGG

GGCR ⋅

+=

21

21

1

Step 4a: Put R = 0. Step 4b: Put -1 into a block, representing the negative feedback effect:

Rearrange the block diagram:

Let the -1 block be absorbed into the, summing point:

Step 4c: By Equation (7.3), the output UC due to input U is UGG

GCU ⋅

+=

21

2

1

Step 5: The total output is C:


11/11

[ ]URGGG

GU

GG

GR

GG

GGCCC UR +⋅

+=⋅

++⋅

+=+=

1

21

2

21

2

21

21

111

█ Example 7


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Chapter 10

Time-Domain Analysis and Design of

Control Systems

A. Bazoune

10.5 TRANSIENT RESPONSE SPECIFICATIONS

Because systems that stores energy cannot respond instantaneously, they exhibit a transient response when they are subjected to inputs or disturbances. Consequently, the transient response characteristics constitute one of the most important factors in system design.

In many practical cases, the desired performance characteristics of control systems can

be given in terms of transient-response specifications. Frequently, such performance characteristics are specified in terms of the transient response to unit-step input, since such an input is easy to generate and is sufficiently drastic. (If the response of a linear system to a step input is known, it is mathematically possible to compute the system’s response to any input).

The transient response of a system to a unit step-input depends on initial conditions.

For convenience in comparing the transient responses of various systems, it is common practice to use standard initial conditions: The system is at rest initially, with its output and all time derivatives thereof zero. Then the response characteristics can be easily compared.

Transient-Response Specifications. The transient response of a

practical control system often exhibits damped oscillations before reaching a steady state. In specifying the transient-response characteristics of a control system to a unit-step input, it is common to name the following:


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1. Delay time, dT

2. Rise time, rT

3. Peak time, pT

4. Maximum overshoot, pM

5. Settling time, sT

These specifications are defined next and are shown in graphically in Figure 10-21.

Delay Time. The delay time dT is the time needed for the response to reach half of

its final value the very first time.

Rise Time. The rise time rT is the time required for the response to rise from 10%

to 90%, 5% to 95%, or 0% to 100% of its final value. For underdamped second order systems, the 0% to 100% rise time is normally used. For overdamped systems, the 10% to 90% rise time is common.

Peak Time. The peak time pT is the time required for the response to reach the first

peak of the overshoot.

Maximum (percent Overshoot). The maximum percent overshoot pM is the

maximum peak value of the response curve [the curve of ( )c t versus t ], measured from

( )c ∞ . If ( ) 1c ∞ = , the maximum percent overshoot is 100%pM × . If the final steady state

value ( )c ∞ of the response differs from unity, then it is common practice to use the

following definition of the maximum percent overshoot:

( ) ( )

( )100Maximum percent overshoot %

pC t C

C

− ∞= ×

∞

Settling Time. The settling time sT is the time required for the response curve to reach

and stay within 2% of the final value. In some cases, 5% instead of 2% , is used as the percentage of the final value. The settling time is the largest time constant of the system.

Comments. If we specify the values of dT , rT , pT , sT and pM , the shape of the

response curve is virtually fixed as shown in Figure 10.22.

Figure 10-22 Specifications of transient-response curve.


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A Few Comments on Transient Response-Specifications.

In addition of requiring a dynamic system to be stable, i.e., its response does not increase unbounded with time (a condition that is satisfied for a second order system provided that

0ζ ≥ , we also require the response:

• to be fast

• does not excessively overshoot the desired value (i.e., relatively stable) and

• to reach and remain close to the desired reference value in the minimum time possible.

Second-Order Systems and Transient-Response-Specifications. The response for a unit step input of an underdamped second order system ( )0 1ζ< < is

given by

( )2

2

1 sin cos

1

1 sin cos

1

ζω ζω

ζω

ζω ω

ζ

ζω ω

ζ

− −

−

= − −

−

= − +

−

n n

n

t t

d d

t

d d

c t e t e t

e t t

(10-13)

or

( )2

1

2

11 sin tan

1

ζωζ

ωζζ

−

− −= − +

−

nt

d

ec t t (10-14)

A family of curves ( )c t plotted against t with various values of ζ is shown in Figure 10-24.

2

2 22

n

n ns s

ω

ζ ω ω+ +− − − − −��

In p u t

t

( )u t

1

0 2 4 6 8 10 12 14 16 18 200

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6Step Response

Time (sec)

Am

plit

ude

ζ = 0.2

0.5

0.7

1

2

5

Output

_______________��

Figure 10-24 Unit step response curves for a second order system.


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Delay Time. We define the delay time by the following approximate

formula:

1 0 7.d

n

Tζ

ω

+=

Rise Time. We find the rise time rT by letting ( ) 1rc T = in Equation (10-13), or

( )2

1 1 sin cos

1

ζω ζω ω

ζ

−= = − +

−

n rT

d dr r rc T e T T (10-15)

Since 0ζω−

≠nt

e , Equation (10-15) yields

2

sin cos 0

1

ζω ω

ζ

+ =

−d dr rT T

or

21

tanζ

ωζ

−= −

d rT

Thus, the rise rT is

2

1 11tan

ζ

ω ζ ω

π β− −= −

−=

d d

rT (10-16)

where β is defined in Figure 10-25. Clearly to obtain a large value of rT we must have a

large value of β .

Oσ

S-plane

jω

nω

σ−

nζω

21nω ζ−

djω

β

( )

( )

1

1 2

21

cos

or sin 1

1or tan

β ζ

β ζ

ζβ

ζ

−

−

−

=

= −

−=

Figure 10-25 Definition of angle β

Peak Time. We obtain the peak time pT by differentiating ( )c t in Equation (10-13),

with respect to time and letting this derivative equal zero. That is,


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( )2

sin 01

ζωωω

ζ

−= =

−

ntn

de

dc tt

dt

It follows that

sin 0ω =dt

or

0, , 2 , 3 ,... , 0,1,2.....ω π π π π= = =dt n n

Since the peak time pT corresponds to the first peak overshoot ( )1n = , we have ω π=d pT .

Then

21

π π

ω ω ζ= =

−p

d n

T (10-17)

The peak time pT corresponds to one half-cycle of the frequency damped oscillations.

Maximum Overshoot pM The maximum overshoot pM occurs at the peak

π ω=p dT . Thus, from Equation (10-13),

( ) ( )�2

1

1

0

1sin cos

n dp pM c T e

ζω π ω ζπ π

ζ

−

= −

− −

=

+ −

= =

��

or

21pM e πζ ζ− −= (10-18)

Since ( ) 1c ∞ = , the maximum percent overshoot is

21100% %pM e πζ ζ− −

×=

The relationship between the damping ratio ζ and the maximum percent overshoot is

shown in Figure 10-26. Notice that no overshoot for 1ζ ≥ and overshoot becomes negligible

for 0 7.ζ > .


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Figure 10-26 Relationship between the maximum percent overshoot %pM and damping ratio ζ

Settling Time sT Based on 2%criterion the settling time sT is defined as:

( )( ) 4

0 02

0 020 02

lnln

.

..n s s

n n

n sT

T T

e ζω

ζωζω ζω

−

− = = ≈−

=

⇒

( )4

2%Criterionn

sTζω

= (10-19)

Similarly for 5%we can get

( )3

5%Criterionn

sTζω

= (10-20)


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REVIEW AND SUMMARY

TRANSIENT RESPONSE SPECIFICATIONS OF A SECOND ORDER SYSTEM

TABLE 1. Useful Formulas and Step Response Specifications for the Linear

Second-Order Model )(tfxkxcxm =++ �� where m, c, k constants

1. Roots m

mkccs

2

42

2,1

−±−=

2. Damping ratio or mkc 2/=ζ

3. Undamped natural frequency m

kn

=ω

4. Damped natural frequency 21 ζωω −=

nd

5. Time constant n

cm ζωτ /1/2 == if 1≤ζ

6. Logarithmic decrement 2

1

2

ζ

πζδ

−= or

224 δπ

δζ

+=

7. Stability Property Stable if, and only if, both roots have negative real parts, this

occurs if and only if , m, c, and k have the same sign.

8. Maximum Percent Overshoot: The maximum % overshoot pM is the maximum peak

value of the response curve. 2

1/100

ζπζ −−= eM

p

9. Peak time: Time needed for the response to reach the first peak of the overshoot

21/p nT π ω ζ= −

10. Delay time: Time needed for the response to reach 50% of its final value the first time

1 0 7.

d

n

Tζ

ω

+≈

11. Settling time: Time needed for the response curve to reach and stay within 2% of the final

value 4

s

n

Tζω

=

12. Rise time: Time needed for the response to rise from (10% to 90%) or (0% to 100%) or (5% to

95%) of its final value r

d

Tπ β

ω

−= (See Figure 10-25)


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SOLVED PROBLEMS

█ Example 1

Figure 4-20 (for Example 1)

█ Example 2


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Figure 4-21 (for Example 2)

█ Solution

First The transfer function of the system is

█ Example 3 (Example 10-2in the Textbook Page 520-521)

Determine the values of dT , rT , pT , sT when the control system shown in Figure 10-28 is

subject to a unit step input


10/10

( )

1

1+s s( )C s( )R s

Figure 10-28 Control System

█ Solution

The closed-loop transfer function of the system is

( )

( )

( )

( )

2

1

1 1

1 11

1

C s s s

R s s ss s

+= =

+ ++

+

Notice that 1nω = rad/s and 0 5.ζ = for this system. So 2 2

1 1 0 5 0 866. .d n

ω ω ζ= − = − =

Rise Time. ω

π β−=

d

rT

where

( ) ( )1 10 866 1 1 05sin sin . .d nβ ω ω− −

= = = rad

or

( ) ( ) ( )1 1 10 5

1 05

cos cos cos

d

.

ra.

n nβ ζω ω ζ− − −= = =

=

Therfore,

1.052.41

0.866

π −= =

rT s

Peak Time. 3 630 866

..

p

d

Tπ π

ω= = = s

Delay Time.

( )1 0 7 0 51 0 7

1 351

. ...d

n

Tζ

ω

++= = = s

σ

jω

nω

σ−

nζω

21nω ζ−

djω

β

Maximum Overshoot : 1 81

2 21 0 5 1 0 50 163 16 3

.. . . . %pM e e eπζ ζ π −− − − × −= = = ==

Settling time: 4 4

80 5 1.

s

n

Tζω

= = =×

s

ME 413 Systems Dynamics & Control Section 10-4: Stability

1/20

9 Chapter 10 Time Domain Analysis and Design of

Control System

A. Bazoune

10.7 STABILITY ANALYSIS

Stability Conditions Using Rolling Ball

The base of the hollow is in equilibrium point. If the ball is displaced a small distance

from this position and released, it oscillates but ultimately returns to its rest position

at the base as it loses energy as a result of friction. This is therefore a stable

equilibrium point. (With no energy dissipation the ball would roll back and forth forever

and exhibit neutral stability).

The ball is in equilibrium if placed exactly at the top of the surface, but if it is displaced

an infinitesimal distance to either side, the net gravitational force acting on it will cause

it to roll down the surface and never return to the equilibrium point. This equilibrium is

therefore unstable.

The ball neither moves away nor returns to its equilibrium position. The flat portion

represents a neutrally stable region.


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Stability Analysis in the Complex Plane

The stability of a linear closed-loop system can be determined from the location of the

closed-loop poles in the s-plane. If any of these poles lie in the Right-Half of the s-plane

(RHS), then with increasing time, they give rise to the dominant mode, and the transient

response increases monotonically or oscillates with increasing amplitude. Either of these

motions represents an unstable motion.

Location of roots of characteristic equations and the corresponding impulse response

For such a system, as soon as the power is turned on, the output may increase with time.

If no saturation takes place in the system and no mechanical stop is provided, then the

system may eventually be damaged and fail, since the response of a real physical system

cannot increase indefinitely.

Stability may be defined as the ability of a system to restore its equilibrium position

when disturbed or a system which has a bounded response for a bounded output.

Consider a simple feedback system

( )C s( )E s( )R s( )G s

( )H s

The overall transfer function is given by

( )

( )

( )

( ) ( )1

C s G s

R s G s H s=

+

The denominator of the transfer function ( ) ( )1 G s H s+ is known as the characteristic

polynomial. The characteristic equation is of the above system is

( ) ( )1 0G s H s+ =

What are the roots of the characteristic equation?


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Let us write the characteristic equation in the form:

( )( )( )( ) ( )

( )( )( ) ( )1 2 3

1 2 3

...

...n

m

s z s z s z s zT s

s p s p s p s p

− − − −=

− − − −

where 1 1, , ..., nz z z are the zeros of ( )T s and

1 1, , ..., mp p p are the poles of ( )T s . If the

input ( )R s is given then

( ) ( ) ( )C s T s R s=

or

( )( )( )( ) ( )

( )( )( ) ( )

( )( )( ) ( )( )( )( ) ( )

1 2 3

1 2 3

......

... ...

i ii iii jn

m i ii iii k

s z s z s z s zs z s z s z s zC s

s p s p s p s p s p s p s p s p

− − − − − − − − =

− − − − − − − −

In order to compute the output, we resort the above equation into partial fraction

expansion

( )( ) ( ) ( ) ( ) ( ) ( )

1 21 2

1 2

... ...ijn i i

m i ii ik

KK K KK KC s

s p s p s p s p s p s p

= + + + + + + +

− − − − − −

The first bracket contains time that originates from the system itself. Terms in the

first bracket will give rise to time in the form

1 2

1 2, , ..., np t p t p t

nK e K e K e

where p j tσ ω= + . The above expression can be rewritten as

1 21 2

1 2, , ..., nnj t j t j t

nK e e K e e K e eω ω ωσσ σ

Terms of the imaginary exponent are basically sinusoidal functions(which will never die

away). Terms with real exponent must have negative values in order to force the

sinusoidal functions to approach a very small magnitude (die away). The location of the

roots of the characteristic equation on the s-plane will indicate the condition of

stability.

Stable region

Unstable region

Im

Re

Margin Stability

s-plane

s jσ ω= +


4/20

Routh Stability criterion

The characteristic equation of the simple feedback system can be written as

( ) ( ) ( ) 1 2

1 2 1 01 0

n n nn n nF s G s H s a s a s a s a s a− −

− −= + = + + + + + =�

Two Necessary But Insufficient Conditions

There are two necessary but insufficient conditions for the roots of the characteristic

equation to lie in Left Hand Side (LHS)-plane (i.e., stable system)

1. All the coefficients 1 2 1

, , , ... ,n n na a a a− −

and 0a should have the same sign.

2. None of the coefficients vanish (All coefficients of the polynomial should exist).

█ Example 1 Given the characteristic equation,

( ) 6 5 4 3 224 3 4 4a s s s s s s s= + + + + +−

Is the system described by this characteristic equation stable?


5/20

█ Solution

One coefficient (-2) is negative. Therefore, the system does not satisfy the necessary

condition for stability.

█ Example 2 Given the characteristic equation,

( ) 6 5 4 24 3 4 4a s s s s s s= + + + + +

Is the system described by this characteristic equation stable?

█ Solution

The term 3s is missing. Therefore, the system does not satisfy the necessary condition for stability.

Necessary and Sufficient Condition

The necessary and sufficient condition for the roots of the characteristic equation to

lie in Left Hand Side (LHS)-plane (i.e., stable system) is

0 1 2 3Polynomial Hurwitz Determinant , , , ...kD k> = (4)

This criterion is an algebraic method that:

1. provides information on the absolute stability of a Linear Time Invariant (LTI)

system that has a characteristic equation with constant coefficients.

2. indicates whether any of the roots of the characteristic equation lie in (RHS)-

plane.

3. indicates the number of roots that lie on the jω -axis and gives the range of

some system parameters over which the system can be stable.

How to Apply the Test

Let us write the characteristic equation in the form:

( ) ( ) ( ) 1 2

1 2 1 01 0

n n nn n nF s G s H s a s a s a s a s a− −

− −= + = + + + + + =�

Step 1: Arrange the coefficients of the above equation into two rows. The

first row consists of the first, third, fifth, …, coefficients. The

second row consists of the second, fourth, sixth,…, coefficients, all

counting from the highest term as shown in the tabulation below

na 2na −

4na −

6na − �

1na −

3na −

5na −

7na − �


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Step 2: Form the following array of numbers by the indicated operations,

illustrated here for a sixth-order equation:

6 5 4 3 2

6 5 4 3 2 1 00a s a s a s a s a s a s a+ + + + + + =

6s

5s

4s

3s

2s

1s

0s

6a

4a

2a

0a

5a

3a

1a 0

5 4 6 3

5

=−

Aa a a a

a5 2 6 1

5

=−

Ba a a a

a 0

5 0 6

5

0=

− ×a

a a a

a

53 =−

CAa a B

A051 =

−D

Aa a a

A5 0

0 0=

× − ×A a

A

=−

EBC AD

C0

0

0=

− ×a

Ca A

C0

0 0=

× − ×C A

C

0 =−

FED Ca

E0 0

0

0

0

0

0

0

0=

− ×a

Fa E

F0 0 0

The previous table can also be obtained as follows:


7/20

6s

5s

4s

3s

2s

1s

0s

6a

4a

2a

0a

5a

3a 1

a 0

6 4

5 3

5

=

−

A

a a

a a

a

6 2

5 1

5

=

−

B

a a

a a

a0

6 0

5

5

0=

−

a

a a

a

a

5 3

=

−

C

a a

A B

A0

5 1

=

−

D

a a

A a

A

5

0

0

0=

−a

A

A

=

−

E

A C

B D

C0

0

0=

−

a

A C

a

C0

0 0=

−A C

C

0=

−

F

C E

D a

E

0 0

0

0

0

0

0

0

0=

−

a

E F

a

F

0 0 0

█ Example 1 Given

( )( )2

5 4 3 2

2 2 25

3 9 16 10

s sH s

s s s s s

+ +=

+ + + + +

Check whether this system is stable or not.

█ Solution

The characteristic equation is: 5 4 3 23 9 16 10 0s s s s s+ + + + + =

Construct the Routh array


8/20

5s 1 3 16 0

4s 1 9 10 0

3s 1 3 9 1

61

× − ×= −

1 16 10 16

1

× − ×= 0 0

2s 6 9 6 1

106

− × − ×=

−

6 10 0 110

6

− × − ×=

− 0 0

1s ( )10 6 10 6

1210

× − × −=

( )10 0 0 60

10

× − × −= 0 0

0s ( )12 10 0 10

1012

× − ×= 0 0 0

There are 2 sign changes. There are 2 poles on the right half of the “s” plane.

Therefore, the system is unstable.

Pole-Zero Map

Real Axis

Imagin

ary

Axis

-1.5 -1 -0.5 0 0.5 1-5

-4

-3

-2

-1

0

1

2

3

4

5

pzmap plot

█ Example 2 In the figure below, determine the range of K for the system to be

stable

( )C s( )E s( )R s

( )( )1 2

K

s s s+ +

█ Solution: The characteristic equation is:

( )( )1 ( ) ( ) 1 0

1 2+ = + =

+ +

KG s H s

s s s

or ( )( ) 3 21 2 0 3 2 0+ + + = ⇒ + + + =s s s K s s s K


9/20


3s 1 2 0

2s 3 K 0

1s 2 3 1 6

3 3

K K× − × −= 0 0

0s

63 0

3

6

3

KK

KK

− × − ×

=

−

0 0

For the system to be stable we must have the following two conditions satisfied

simultaneously:

i) 0>K

and

ii) 6

0 6 0 63

−> ⇒ − > ⇒ <

KK K

The above two conditions can be shown graphically on the real line

0 1 2 3 4 5 6− ∞ +∞

0>K

6<K

0 6< <K

The intersection of the two conditions above gives the condition for the stability

of the above system, that is 0 6< <K . What will happen if 0=K or 6=K ?

For 0=K , the above characteristic equation becomes 3 2

3 2 0 0+ + + =s s s or

( )

+ + = ⇒ = − = −

=

⇒2

0 The system is marginally stable

3 2 0 1

2

s

s

s s s

s

For 6=K , the above characteristic equation becomes

3 23 2 6 0+ + + =s s s or ( ) ( )2

2 3 03 =+ + +s ss

or

( )( )

= −

+ = + ⇒

= − ⇒

+ = ⇒

2 2 The system is marginally stable

2 The system is marginally stable

3

3 2 0 s j

s

s j

s s


10/20

The frequency of oscillations in the previous case is 2 rad/s.

Special Cases

i) Zero Coefficient in the First Column of the Routh’s

Array

When the first term in a row is zero, the following element of (power-1) s will be

infinite. One of the following methods may be used.

1 Substitute for =1

sx in the original equation The same

criterion applies for x . Then solve for x .

2 Multiply the original equation by ( )+s a , 1, 2,3,...,=a This

introduces an additional negative root. Then follow the same procedure to check for

stability.

3 Substitute a small parameter ε for the 0 . Complete the

array and take the limit as 0ε → .

█ Example 3

Check whether the following system given by its characteristic equation is stable

or not

( )Γ = + = + + + + =4 3 21 ( ) ( ) 2 2 5 0s G s H s s s s s

█ Solution


4s 1 2 5

3s 1 2 0

2s 1 2 1 2

01

× − ×=

1 5 1 05

1

× − ×=

1s

0s

Method 1: Substitute for 1

=sx in the original equation.

The above characteristic equation is:

Zero element in the first column,

we can’t continue as usual.


11/20

( ) 4 3 22 2 5 0Γ = + + + + =s s s s s

Substitute 1

=sx in the above

( ) ( ) ( ) ( ) ( )4 3 2

1 1 2 1 2 1 5 0= + + + + =B x x x x x

or ( ) 4 3 25 2 2 1 0= + + + + =B x x x x x

4x 5 2 1

3x 2 1 0

2x 2 2 1 5 1

2 2

× − × −= 1

1x ( )

( )

1 2 1 1 25

1 2

− × − ×=

− 0

0x 1

There are two sign changes. Therefore, there are 2 roots that lie on the right half of the

s-plane. The system is unstable.

Method 2: Multiply by ( )+s a in the original equation.

Let 1=a an arbitrary value. Notice that if we get again a zero in the first column, this

means that 1=a is a root of the polynomial. In this case change the value ofa . For 1=a , the above polynomial becomes

( ) ( ) ( )= + + ++ +��

W

4 3 2

e are multiplyingby

Previousthis quantit

Polyny

omial

2 2 51B s s s s ss

or

( ) = + + + + +5 4 3 22 3 4 7 5B s s s s s s

5s 1 3 7 0

4s 2 4 5 0

3s × − ×

= +2 3 1 4

12

× − ×

=2 7 1 5 9

2 2 0 0

2s ( )× − ×= −

1 4 9 2 25

1

× − ×=

1 5 2 05

1 0 0

1s ( )− × − ×

=−

5 9 2 5 111 2

5 0

0s ( )

( )

× − ×=

11 2 5 0 55

11 2

( )Γ s and

( )B x have

same stability

characteristic

First Sign Change.

Second Sign Change.

First Sign Change.

Second Sign Change.


12/20



Method 3: Substitute ε for the zero element

Remember the expression for ( )Γ = + + + +4 3 22 2 5s s s s s . The Routh array is

4s 1 2 5

3s 1 2 0

2s 1 2 1 20

1ε

× − ×=

1 5 1 05

1

× − ×=

1s ε

ε ε

× − × = − = −∞

2 5 1 52

0

0s

εε

ε

× − − ×

=

−

55 2 0

55

2



The roots of ( )Γ s can be solved by MATLAB

MATLAB PROGRAM:

>> Q=[1 1 2 2 5];

>> roots (Q)

ans = 0.5753 + 1.3544i 0.5753 - 1.3544i -1.0753 + 1.0737i -1.0753 - 1.0737i

Pole-Zero Map

Real Axis

Imagin

ary

Axis

-1.2 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6-1.5

-1

-0.5

0

0.5

1

1.5

Tends to −∞ as ε → 0

0 is substituted by ε

First Sign Change.

Second Sign Change.

2 roots with positive real parts 0.5753. They will lie in the right half of the s-plane as shown in the pzmap


13/20

ii) A Row of Zeros

This situation occurs when the characteristic equation has a pair of real roots with

opposite sign ( )± r , complex conjugate roots on the imaginary axis ( )ω± j , or a pair of

complex conjugate roots with opposite real parts ( )− ± ±,a jb a jb as shown in the

figure below.

The procedure to overcome this situation is as follows:

1. Form the “auxiliary equation” as shown in the example below from the preceding

row.

2. Complete Routh array by replacing the zero row with the coefficients obtained

by differentiating the auxiliary equation.

3. Roots of the auxiliary equation are also roots of the characteristic equation (i.e.,

( ) ( ) ( )= + =1 0Q s G s H s ). The roots of the auxiliary equation occur in pairs

and are of opposite sign of each other. In addition the auxiliary equation is always

“even” order.

█ Example 3

Check whether the following system given by its characteristic equation is stable

or not

( ) = + = + + + + + =5 4 3 21 ( ) ( ) 5 11 23 28 12 0A s G s H s s s s s s

█ Solution

Step 1 Form the auxiliary equation ( ) = 0a s by use of the coefficients

from the row just preceding the row of zeros.


14/20

Step 2 Take the derivative of the auxiliary equation ( )[ ]d

d

a s

s with

respect to s .

Step 3 Replace the row of zeros with the coefficients of ( )[ ]d

d

a s

s.

Step 4 Continue with Routh’s tabulation in the usual manner with the

newly formed row of coefficients replacing the row of zeros.

Step 5 Interpret the change of signs, if any, of the coefficients in the

first column of Routh’s tabulation in the usual manner.

5s 1 11 28

4s 5 23 12

3s 5 11 1 23

6 45

× − ×= .

5 28 12 125 6

5

× − ×= . 0

2s × − ×

=6.4 23 25.6 5

6.43

× − ×=

6.4 12 0 5

6.412 0

1s × − ×

=3 25.6 6.4 12

30 0

Row

of zeros

0s

The auxiliary equation is ( ) = +23 12a s s and

( )[ ]=

d6

d

a ss

s

The zeros in the row of 1s are replaced by the coefficient of the derivative

of the auxiliary equation ( )[ ]

=d

6d

a s

ss

1s 6 0

0s × − ×

=6 12 0 3

66

There are no sign changes in the first column of the Routh array. All roots have

negative real parts except for a pair on the imaginary axis.

Roots of the auxiliary equation are

( ) ( )= = ⇒= = ±+ +2 2

11 ,23 12 3 4 0 2a s s s js

Roots of the characteristic equation are


15/20

( )

= − = −

= + + + + + = ⇒ = +

= − = −

1

25 4 3 2

3

4

5

3 (real)

25 11 23 28 12 0 complex conjugates

2

1real repeated

1

s

s jA s s s s s s

s j

s

s

Pole-Zero Map

Real Axis

Imagin

ary

Axis

-3.5 -3 -2.5 -2 -1.5 -1 -0.5 0-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

Limitations

It should be reiterated that the Routh-Hurwitz criterion is valid only if the

characteristic equation is algebraic with real coefficients. If any one of the coefficients

is complex, or if the equation is not algebraic, such as containing exponential functions

or sinusoidal functions of s , the Routh-Hurwitz criterion simply cannot be applied. Another limitation of the Routh-Hurwitz criterion is that it is valid only for the

determination of the roots of the characteristic equation with respect to the left half

or the right half of the s-plane. The stability boundary is just the ω −j axis of the

−s plane. The criterion cannot be applied to any other stability boundaries in a complex

plane, such as the unit circle in the −z plane, which is the stability of discrete data

system.


16/20

Solved Problems


17/20


18/20


19/20


20/20

ME 413 Systems Dynamics & Control Section 10-5: Steady State Errors and System Types

1/15

Chapter 10

Time-Domain Analysis and Design of

Control Systems

A. Bazoune

10.5 STEADY STATE ERRORS AND SYSTEM TYPES

Steady-state errors constitute an extremely important aspect of the system performance, for it would be meaningless to design for dynamic accuracy if the steady output differed substantially from the desired value for one reason or another.

The steady state error is a measure of system accuracy. These errors arise from the nature of the inputs, system type and from nonlinearities of system components such as static friction, backlash, etc. These are generally aggravated by amplifiers drifts, aging or deterioration. The steady-state performance of a stable control system is generally judged by its steady state error to step, ramp and parabolic inputs.

Consider a unity feedback system as

shown in the Figure. The input is ( )R s ,

the output is ( )C s , the feedback signal

( )H s and the difference between input

and output is the error signal ( )E s .

( )C s

( )H s

( )E s( )R s( )G s

From the above Figure

( )

( )

( )

( )1

C s G s

R s G s=

+ (1)

On the other hand

( ) ( ) ( )C s E s G s= (2)

Substitution of Equation (2) into (1) yields

( )( )

( )1

1E s R s

G s=

+ (3)

The steady-state error sse may be found by use of the Final Value Theorem (FVT) as follows:

( ) ( )( )

( )0 0 1lim lim limsst s s

sR se e t SE s

G s→∞ → →= = =

+ (4)

Equation (4) shows that the steady state error depends upon the input ( )R s and the forward

transfer function ( )G s . The expression for steady-state errors for various types of standard

test signals are derived next.


2/15

1. Unit Step (Positional) Input.

Input ( ) ( )1r t t=

or ( ) ( )1

R s L r ts

= =

From Equation (4)

( )

( )

( )

( ) ( ) ( )0 0 0

1 1 1 1

1 1 1 1 0 1lim lim limsss s s

p

s ssR se

G s G s G s G K→ → →= = = = =

+ + + + +

( )r t

t

1

where ( )0pK G= is defined as the position error constant.

2. Unit Ramp (Velocity) Input.

Input ( ) ( ) 1orr t t r t= =�

or ( ) ( )2

1R s L r t

s= =

From Equation (4)

( )

( )

( )( ) ( ) ( )

2

0 0 0 0

1 1 1 1

1 1lim lim lim limsss s s s

v

s ssR se

G s G s s sG s sG s K→ → → →= = = = =

+ + +

( )r t

t

1

1

where ( )0

limvs

K sG s→

= is defined as the velocity error constant.

3. Unit Parabolic (Acceleration) Input.

Input ( ) ( )211

2orr t t r t= =��

or ( ) ( )3

1R s L r t

s= =

From Equation (4)

( )

( )

( )( ) ( ) ( )

3

2 2 20 0 0 0

1 1 1 1

1 1lim lim lim limsss s s s

a

s ssR se

G s G s s s G s s G s K→ → → →= = = = =

+ + +

( )r t

t

where ( )2

0limas

K s G s→

= is defined as the acceleration error constant.


3/15

10.6 TYPES OF FEEDBACK CONTROL SYSTEMS

The open-loop transfer function of a unity feedback system can be written in two standard forms:

• The time constant form

( )( )( ) ( )

( )( ) ( )

1 2

1 2

1 1 1

1 1 1

z z zj

np z zk

K T s T s T sG s

s T s T s T s

+ + +=

+ + +

�

� (8)

where K and T are constants. The system type refers to the order of the pole of ( )G s at 0s = .

Equation (8) is of type n .

• The pole-zero form

( )( )( ) ( )( )( ) ( )

1 2

1 2

' j

nk

K s z s z s zG s

s s p s p s p

+ + +=

+ + +

�

� (9)

The gains in the two forms are related by

'j

j

kk

z

K Kp

=

∏

∏ (10)

with the gain relation of Equation (10) for the two forms of ( )G s , it is sufficient to obtain

steady state errors in terms of the gains of any one of the forms. We shall use the time constant form in the discussion below.

Equation (8) involves the term ns in the denominator which corresponds to number of

integrations in the system. As 0s→ , this term dominates in determining the steady-state

error. Control systems are therefore classified in accordance with the number of integration in

the open loop transfer function ( )G s as described below.

1. Type-0 System.

If 0,n = ( )0

KG s K

s= = the steady-state errors to various standard inputs, obtained from

Equations (5), (6), (7) and (8) are

( )( )

( )( )

( )( )

0 0

2 20 0

1 1 1

1 1 1

1 1

1 1

0

lim l

Position

im

li

Velocity

Acceleration m lim

ss

p

sss s

sss s

G K

K

eK

esG s s

es G s s K

→ →

→ →

= = =+ + +

= = = ∞

= = = ∞

(11)


4/15

Thus a system with 0,n = or no integration in ( )G s has

• a constant position error,

• infinite velocity error and

• infinite acceleration error

2. Type-1 System.

If 1,n = ( )1

KG s

s= , the steady-state errors to various standard inputs, obtained from


( )( )

( )( )

( )( )

0

0 0

20 0 2

1 1 10

1 0 11

1 1 1 1

1 1 1

0

lim

lim li

Positi

m

lim

on

Velocity

Acceleration lim

sss

sss s

v

sss s

eG

esG s K Ks

es G s s

K

s

K

s

K

s

→

→ →

→ →

= = = =+ + ∞+

= = = =

= = = = ∞

(12)

Thus a system with 1,n = or with one integration in ( )G s has

• a zero position error,

• a constant velocity error and

• infinite acceleration error

3. Type-2 System.

If 1,n = ( )2

KG s

s= , the steady-state errors to various standard inputs, obtained from


( )( )

( )( )

( )( )

0

0 0

20

2

2

2

0 2

1 1 10

1 0 11

1 1 10

1 1 1 1

lim

lim li

Posit

m

lim

ion

Velocity

Acceleration lim

sss

sss s

sss s

a

K

s

K

s

K K

eG

esG s

s

s

es G s Ks

→

→ →

→ →

= = = =+ + ∞+

= = = =∞

= = = =

(13)

Thus a system with 2,n = or with one integration in ( )G s has

• a zero position error,

• a zero velocity error and

• a constant acceleration error


5/15

TABLE 1. Steady-state errors in closed loop systems

Typ

e-0 sy

stem

Typ

e-1 sy

stem

Typ

e-2 sy

stem

0lim ( )ps

K G s→

=0

lim ( )vs

K sG s→

= 2

0lim ( )as

K s G s→

=

sse =∞

ss

v

Ae

K=

( )r t At=

0sse =

sse =∞

sse = ∞

ss

a

Ae

K=

21

2( )r t At=

1ss

p

Ae

K=

+

( )r t A=

0sse =

0sse =

where ( )0pK G= is defined as the position error constant.

where ( )0

limvs

K sG s→

= is defined as the velocity error constant.

where ( )2

0limas

K s G s→

= is defined as the acceleration error constant.


6/15

10.7 STEADY STATE ERROR FOR NON-UNITY FEEDBACK SYSTEMS

( )C s( )E s( )R s( )G s

( )H s

Add to the previous block two feedback blocks ( )11H s = − and ( )1

1H s =

( )C s( )E s( )R s( )G s

( )H s

1−

Parallel blocks. Can be combined in one

( )C s( )E s( )R s( )G s

( ) 1H s −

( )C s( )E s( )R s ( )

( ) ( ) ( )1

G s

G s H s G s+ −

( )eG s


7/15

█ Example 1

For the system shown below, find

• The system type

• Appropriate error constant associated with the system type, and

• The steady state error for unit step input

( )C s( )E s( )R s

( )

100

10s s +

1

5s+

█ Solution


8/15


9/15


10/15


11/15


12/15


13/15


14/15


15/15

MCEN 467 – Control Systems

Chapter 4: Chapter 4:

Basic Properties of FeedbackBasic Properties of Feedback

Part D: The Classical Three-

Term Controllers


Basic Operations of a Feedback Control

Think of what goes on in domestic hot water thermostat:

• The temperature of the water is measured.

• Comparison of the measured and the required values

provides an error, e.g. “too hot’ or ‘too cold’.

• On the basis of error, a control algorithm decides what to do.

→ Such an algorithm might be:

– If the temperature is too high then turn the heater off.

– If it is too low then turn the heater on

• The adjustment chosen by the control algorithm is applied to

some adjustable variable, such as the power input to the

water heater.


Feedback Control Properties

• A feedback control system seeks to bring the measured quantity to its required value or set-point.

• The control system does not need to know why the measured value is not currently what is required, only that is so.

• There are two possible causes of such a disparity:

– The system has been disturbed.

– The set point has changed. In the absence of external

disturbance, a change in set point will introduce an error.

The control system will act until the measured quantity

reach its new set point.


The PID Algorithm

• The PID algorithm is the most popular feedback controller

algorithm used. It is a robust easily understood algorithm

that can provide excellent control performance despite the

varied dynamic characteristics of processes.

• As the name suggests, the PID algorithm consists of three

basic modes:

the Proportional mode,

the Integral mode

& the Derivative mode.


P, PI or PID Controller

• When utilizing the PID algorithm, it is necessary to decide which modes are to be used (P, I or D) and then specify the parameters (or settings) for each mode used.

• Generally, three basic algorithms are used: P, PI or PID.

• Controllers are designed to eliminate the need for continuous operator attention.

→ Cruise control in a car and a house thermostat

are common examples of how controllers are used to

automatically adjust some variable to hold a measurement

(or process variable) to a desired variable (or set-point)


Controller Output

• The variable being controlled is the output of the controller

(and the input of the plant):

• The output of the controller will change in response to a change

in measurement or set-point (that said a change in the tracking

error)

provides excitation to the plant system to be controlled


PID Controller

• In the s-domain, the PID controller may be represented as:

• In the time domain:

dt

tdeKdtteKteKtu d

t

ip

)()()()(

0++= ∫

)()( sEsKs

KKsU d

ip

++=

proportional gain integral gain derivative gain


PID Controller


• The signal u(t) will be sent to the plant, and a new output y(t)

will be obtained. This new output y(t) will be sent back to

the sensor again to find the new error signal e(t). The

controllers takes this new error signal and computes its

derivative and its integral gain. This process goes on and on.

dt

tdeKdtteKteKtu d

t

ip

)()()()(

0++= ∫


Definitions


++=

++=

∫

∫

dt

tdeTdtte

TteK

dt

tdeKdtteKteKtu

d

t

i

p

d

t

ip

)()(

1)(

)()()()(

0

0

i

dd

i

p

iK

KT

K

KTwhere == ,

proportional gain integral gain

derivative gain

derivative time constantintegral time constant


Controller Effects

• A proportional controller (P) reduces error responses to

disturbances, but still allows a steady-state error.

• When the controller includes a term proportional to the

integral of the error (I), then the steady state error to a

constant input is eliminated, although typically at the cost

of deterioration in the dynamic response.

• A derivative control typically makes the system better

damped and more stable.


Closed-loop Response

Small

change

DecreaseDecreaseSmall

change

D

EliminateIncreaseIncreaseDecreaseI

DecreaseSmall

change

IncreaseDecreaseP

Steady-

state error

Settling

time

Maximum

overshoot

Rise time

• Note that these correlations may not be exactly accurate,

because P, I and D gains are dependent of each other.


Example problem of PID

• Suppose we have a simple mass, spring, damper problem.

• The dynamic model is such as:

• Taking the Laplace Transform, we obtain:

• The Transfer function is then given by:

fkxxbxm =++ ��

)()()()(2

sFskXsbsXsXms =++

kbsmssF

sX

++=

2

1

)(

)(


Example problem (cont’d)

• Let

• By plugging these values in the transfer function:

• The goal of this problem is to show you how each of

contribute to obtain:

fast rise time,

minimum overshoot,

no steady-state error.

Nf,m/Nk,m/s.Nb,kgm 120101 ====

2010

1

)(

)(2

++=

sssF

sX

dip KandKK ,


Ex (cont’d): No controller

• The (open) loop transfer function is given by:

• The steady-state value for the output is:

2010

1

)(

)(2

++=

sssF

sX

20

1

)(

)()(lim)(lim)(lim

00====

→→∞→ sF

sXssFssXtxx

sstss


Ex (cont’d): Open-loop step response

• 1/20=0.05 is the final value

of the output to an unit step

input.

• This corresponds to a

steady-state error of 95%,

quite large!

• The settling time is about

1.5 sec.


Ex (cont’d): Proportional Controller

• The closed loop transfer function is given by:

)20(10

20101

2010

)(

)(2

2

2

p

p

p

p

Kss

K

ss

Kss

K

sF

sX

+++=

+++

++=


Ex (cont’d): Proportional control

• Let

• The above plot shows that

the proportional controller

reduced both the rise time

and the steady-state error,

increased the overshoot, and

decreased the settling time

by small amount.

300=pK


Ex (cont’d): PD Controller


)20()10(

20101

2010

)(

)(2

2

2

pd

dp

dp

dp

KsKs

sKK

ss

sKKss

sKK

sF

sX

++++

+=

++

++

++

+

=


Ex (cont’d): PD control

• Let

• This plot shows that the

proportional derivative

controller reduced both

the overshoot and the

settling time, and had

small effect on the rise

time and the steady-state

error.

10,300 == dp KK


Ex (cont’d): PI Controller


ip

ip

ip

ip

KsKss

KsK

ss

sKKss

sKK

sF

sX

++++

+=

++

++

++

+

=)20(10

2010

/1

2010

/

)(

)(23

2

2


Ex (cont’d): PI Controller

• Let

• We have reduced the proportional

gain because the integral controller

also reduces the rise time and

increases the overshoot as the

proportional controller does

(double effect).

• The above response shows that the

integral controller eliminated the

steady-state error.

70,30 == ip KK


Ex (cont’d): PID Controller


ipd

ipd

idp

idp

KsKsKs

KsKsK

ss

sKsKKss

sKsKK

sF

sX

+++++

++=

++

+++

++

++

=)20()10(

2010

/1

2010

/

)(

)(23

2

2

2


Ex (cont’d): PID Controller

• Let

• Now, we have obtained

the system with no

overshoot, fast rise time,

and no steady-state

error.

5500

,300,350

=

==

d

ip

K

KK


Ex (cont’d): Summary

PDP

PI PID


PID Controller Functions

• Output feedback

→ from Proportional action

compare output with set-point

• Eliminate steady-state offset (=error)

→ from Integral action

apply constant control even when error is zero

• Anticipation

→ From Derivative action

react to rapid rate of change before errors grows too big


Effect of Proportional,

Integral & Derivative Gains on the

Dynamic Response


Proportional Controller

• Pure gain (or attenuation) since:

the controller input is error

the controller output is a proportional gain

)()()()( teKtusUKsE pp =⇒=


Change in gain in P controller

• Increase in gain:

→ Upgrade both steady-

state and transient

responses

→ Reduce steady-state

error

→ Reduce stability!


P Controller with high gain


Integral Controller

• Integral of error with a constant gain

→ increase the system type by 1

→ eliminate steady-state error for a unit step input

→ amplify overshoot and oscillations

dtteKtusUs

KsE

t

ii ∫=⇒=

0

)()()()(


Change in gain for PI controller


→ Do not upgrade steady-

state responses

→ Increase slightly

settling time

→ Increase oscillations

and overshoot!


Derivative Controller

• Differentiation of error with a constant gain

→ detect rapid change in output

→ reduce overshoot and oscillation

→ do not affect the steady-state response

dt

tdeKtusUsKsE dd

)()()()( =⇒=


Effect of change for gain PD controller


→ Upgrade transient

response

→ Decrease the peak and

rise time

→ Increase overshoot

and settling time!


Changes in gains for PID Controller


Conclusions

• Increasing the proportional feedback gain reduces steady-state errors, but high gains almost always destabilize the system.

• Integral control provides robust reduction in steady-state errors, but often makes the system less stable.

• Derivative control usually increases damping and improves stability, but has almost no effect on the steady state error

• These 3 kinds of control combined from the classical PIDcontroller


Conclusion - PID

• The standard PID controller is described by the

equation:

)(1

1)(

)()(

sEsTsT

KsUor

sEsKs

KKsU

d

i

p

di

p

++=

++=


Application of PID Control

• PID regulators provide reasonable control of most

industrial processes, provided that the performance

demands is not too high.

• PI control are generally adequate when plant/process

dynamics are essentially of 1st-order.

• PID control are generally ok if dominant plant dynamics

are of 2nd-order.

• More elaborate control strategies needed if process has long

time delays, or lightly-damped vibrational modes

chapter 10 time-domain analysis and design of control systems · me 413 systems dynamics &...

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