Analysis of Linear Continuous Systems

Transient-Response Analysis andSteady-State Error Analysis

The first step in analyzing a control system was to derive a mathematical model of thesystem. Once such model is obtained, various methods are available for the analysis ofsystem performance.

In analyzing and designing control systems we must have a basis of comparison ofperformance of various control systems. This basis may be set up by specifying particulartest input signals and by comparing the responses of various systems to these signals. Theuse of test signals can be justified because of a correlation existing between the responsecharacteristics of a system to a typical test input signal and the capability of the systemto cope with the actual input signal.

The aim of analysis is the study of system’s behavior in transient and steady-statewhen the model of the system and the input signals are known, [1], [2].

Typical test signals

The commonly used test input signals are those of step functions, ramp functions, impulsefunctions, sinusoidal functions and the like. With these test signals, mathematical andexperimental analysis of control systems can be carried out easily since the signals arevery simple functions of time.

If the input of a control system are gradually changing functions of time, then a rampfunction of time may be a good test signal. Similarly, if a system is subjected to suddendisturbances a step function may be a good test signal, and for a system subjected toshock inputs, an impulse function may be a test. Once a control system is designed onthe basis of test signals, the performance of the system in response to actual inputs isgenerally satisfactory.

The use of such test signals enables one to compare the performance of all systems onthe same basis.

1

Transient response and steady-state response

The time response of a system consists of two parts: the transient and the steady-stateresponse, as shown in Figure 1. By transient response we mean that which goes from

transient state steady-state

t

Figure 1: Transient and steady-state response

the initial to the final state. By steady-state response we mean the manner in which thesystem output behaves as t approaches infinity.

Absolute stability, relative stability and steady-state error

In designing a control system, we must be able to predict the dynamic behavior of thesystem from a knowledge of the components. The most important characteristics of thedynamic behavior of a control system is absolute stability, that is, whether the system isstable or unstable. A control system is in equilibrium if, in the absence of any disturbanceor input, the output stays in the same state. A linear time-invariant control system isstable if the output eventually comes back to its equilibrium state when the system issubjected to a disturbance. A linear time-invariant system is unstable if either oscillationof the output continues forever or the output diverges from the bound from its equilibriumstate when the system is subjected to a disturbance. Actually, the output of a physicalsystem may increase to a certain extent but may be limited by mechanical ”stops” or thesystem may breakdown or become nonlinear after the output exceeds a certain magnitudeso that the linear differential equation no longer apply. We shall discuss the stabilityproblem in another chapter.

Important system behavior to which we must give careful consideration includes rel-ative stability and steady-state error. Since a physical control system involves energystorage, the output of the system, when subjected to an input, cannot follow the inputimmediately but exhibits a transient response before a steady state can be reached. Thetransient response of a control system often exhibits damped oscillations before reachinga steady state. If the output of a system at steady state does not exactly agree with theinput, the system is said to have steady-state error. This error is indicative of the accu-racy of the system. In analyzing a control system, we must examine transient-response

2

behavior, such as the time required to reach a new steady state and the value of the errorwhile following an input signal, as well as the steady-state behavior.

Errors in control systems can be attributed to many factors. Changes in the referenceinput will cause unavoidable errors during transient periods and may also cause steady-state errors. Imperfections in the system components, such as static friction, aging ordeterioration will cause errors at steady state. In this chapter we shall not discuss errorsdue to imperfections of the system components. Rather, we shall investigate a type ofsteady-state error that is caused by the incapability of the system to follow particulartypes of inputs.

Usually the system is decomposed in simple elements of at most second order, thatsimplifies the method of analysis and also we can know the contribution of each elementto the system behavior.

If the simple elements do not accumulate energy (they don’t have inertia or timeconstants) they are called ideal elements, otherwise they are real elements.

The number of time constants at the denominator of the transfer function define theorder of the element.

The behavior of simple elements can be studied using some characteristic parameters:

• Time constant, T

• Time delay constant, Tm

• Damping factor ζ

• Natural frequency ωn

• Gain constant, K

The main steps in system analysis are:

• Write the equation or system of equations that define the relationship betweenthe input and output signals. In the case of linear elements these are differentialequations.

• Determine the transfer function or transfer matrix and identification of main pa-rameters: gain factors, damping factors, natural frequencies, time constants etc.

• Determine the output signal for a test input signal (step, ramp, impulse, sine etc.)

• Graphical or analytical study of each element behavior emphasizing the influence ofsystem parameters.

3

First-order systems

Consider a first-order system shown in Figure 2. Physically, this system may representan RC circuit, a thermal system, or the like. The input-output relationship is given by:

Figure 2: First-order system

C(s)

R(s)=

K

Ts+ 1(1)

Example. RC circuit Consider the RC circuit shown in Figure 3. The differential

Figure 3: RC circuit

equation that describe the relationship between the input and output voltage (Vin andVout) is obtained directly from the Kirchoff’s laws:

Vin = VR + Vout

where

VR = RiR = RiC , iC = CdVout

dt

and we obtain:

Vin = RCdVout

dt+ Vout

If we apply the Laplace transform to this equation, when all the initial conditions arezero, the transfer function is calculated as:

H(s) =Vout(s)

Vin(s)=

1

RCs+ 1

The gain K = 1 and the time constant is T = RC.

4

In the following we shall analyze the system responses to such inputs as the unit step,unit ramp and unit impulse functions. The initial conditions are assumed to be zero.Note that all systems having the same transfer function will exhibit the same output inresponse to the same input. For any given physical system, the mathematical responsecan be given a physical interpretation.

Unit-step response of first-order systems

Since the Laplace transform of the unit step function is 1/s, substituting R(s) = 1/s intoequation (1), we obtain:

C(s) =K

Ts+ 1

1

s

Expanding C(s) into partial fraction gives:

C(s) = K(

1

s− T

Ts+ 1

)

Taking the inverse Laplace transform, we obtain:

c(t) = K(

1− e−t/T)

, (t ≥ 0) (2)

Equation (2) states that initially the output c(t) is zero and finally approaches K.One important characteristic of such an exponential response curve c(t) is that at t = Tthe value of c(t) is 0.632K, or the response has reached 63.2% of its total change. Thismay be easily seen by substituting t = T in c(t). That is:

c(T ) = K(1− e−1) = 0.632K

Note that the smaller the time constant T , the faster the system response. Anotherimportant characteristic of the exponential response curve is that the slope of the tangentat t = 0 is K/T , since:

dc(t)

dt=

K

Te−t/T |t=0 =

K

T(3)

The output would reach the final value at t = T if it maintained its initial speed ofresponse. From equation (3) we see that the slope of the response curve c(t) decreasesmonotonically from K/T at t = 0 to zero at t = ∞.

The exponential response curve c(t) given by equation (2) is shown in Figure 4. In onetime constant, the exponential response curve has gone from 0 to 63.2% of the final value.For t ≥ 4T the response remains within 2% of the final value. As seen from equation(2), the steady-state is reached mathematically only after an infinite time. In practice,however, a reasonable estimate of the response time is the length of time the responsecurve needs to reach the 2% line of the final value, or four time constants.

5

Figure 4: First-order system step response

Unit-ramp response of first-order systems

Since the Laplace transform of the unit ramp function is 1/s2, we obtain the output ofthe system of Figure 2 as:

C(s) =K

Ts+ 1

1

s2

Expanding C(s) into partial fraction gives

C(s) = K

(

1

s2− T

s+

T 2

Ts+ 1

)

Taking the inverse Laplace transform of this equation we obtain:

c(t) = K(

t− T + Te−t/T)

, (t ≥ 0) (4)

A plot of the unit ramp response of a general first-order system is shown in Figure 5.

In case the gain factor K = 1, the error signal e(t), calculated as the differencebetween the input and output signal, is:

e(t) = r(t)− c(t) = T (1− e−t/T )

As t approaches infinity, e−t/T approaches zero, and thus the error signal e(t) approachesT or

e(∞) = T

The unit ramp input and the system output are shown in Figure 6. The error in trackingthe unit-ramp input is equal to T for sufficiently large T . The smaller the time constantT , the smaller the steady-state error in following the ramp input.

6

Figure 5: First-order system ramp response

0 2T 4T 6T t

r(t) c(t)

r(t)=t

c(t)

steady-state error

T T

Figure 6: First-order system ramp response (K = 1)

7

Unit-impulse response of first-order systems

For the unit-impulse input, R(s) = 1 and the output of the system of Figure 2 can beobtained as:

C(s) =K

Ts+ 1or,

c(t) =K

Te−t/T , (t ≥ 0)

The response curve is shown in Figure 7

Figure 7: First-order system impulse response

An important property of linear time-invariant system

In the analysis above, it has been shown that for a unit-ramp input the output c(t) is

c(t) = K(t− T + Te−t/T ), (t ≥ 0),

for the unit-step input, which is the derivative of the unit-ramp input c(t) is

c(t) = K(1− e−t), (t ≥ 0)

and for the unit-impulse, which is the derivative of the step input, the output is:

c(t) =K

Te−t/T , (t ≥ 0)

Comparison of system response to these three inputs clearly indicates that the responseto the derivative of an input signal can be obtained by differentiating the response ofthe system to the original signal. It can also be seen that the response to the integralof the original signal can be obtained by integrating the response of the system to theoriginal signal and by determining the integration constants from the zero output initialcondition. This is a property of linear time-invariant (LTI) systems. Linear time-varyingand nonlinear systems do not posses this property.

8

Second-order systems

Consider the second-order system shown in Figure 8. where the system transfer function

H(s) R(s) C(s)

Figure 8: Second-order system

is written in the generalized form:

H(s) =C(s)

R(s)=

11ω2ns2 + 2ζ

ωns+ 1

=ω2n

s2 + 2ζωns+ ω2n

The dynamic behavior of the second-order system can be described in terms of two pa-rameters: the natural frequency ωn, and the damping factor ζ .

Example. Consider the system with the transfer function:

H(s) =1

s2 + s+ 1

The system parameters are then calculated from:

1

ω2n

= 1;2ζ

ωn

= 1; ⇒ ωn = 1; ζ =1

2

If 0 < ζ < 1, the poles of the are complex conjugates and lie in the left-half s-plane.The system is then called underdamped and the transient response is oscillatory.

The roots of the characteristic equation

s2 + 2ζωns+ ω2n = 0

are:

s1,2 = −ζωn ± ωn

√

ζ2 − 1

They are complex for 0 < ζ < 1 and real for ζ ≥ 1.

If ζ = 1, the system is called critically damped. Overdamped systems correspondto ζ > 1. The transient response of critically damped and overdamped systems do notoscillate. If ζ = 0, the transient response does not die out.

Step-response of second-order systems

The transfer function of the second-order system is:

C(s)

R(s)=

ω2n

s2 + 2ζω2ns+ ω2

n

9

and consider the input r(t) = 1, (t ≥ 0) a unit step. Then R(s) = 1/s and the systemresponse is:

C(s) =ω2n

s(s2 + 2ζω2ns+ ω2

n)

1. Underdamped case (0 < ζ < 1)

The poles of the transfer function are complex conjugates and C(s) can be expandedin partial fractions:

C(s) =1

s− s+ 2ζωn

s2 + 2ζωns+ ω2n

=1

s− s+ ζωn

(s+ ζωn)2 + ω2d

− ζωn

(s+ ζωn)2 + ω2d

(5)

where ωd = ωn

√1 = ζ2 is called the damped natural frequency.

Using a Laplace transform table we obtain:

£−1

[

s+ ζωn

(s+ ζωn)2 + ω2d

]

= e−ζωntcosωdt

£−1

[

ωd

(s+ ζωn)2 + ω2d

]

= e−ζωntsinωdt

Hence, the inverse Laplace transform of equation (5) is obtained as:

£−1[C(s)] = c(t) = 1− e−ζωnt

[

cosωdt+ζ√

1− ζ2sinωdt

]

=

= 1− e−ζωnt

√1− ζ2

· sin(

ωdt+ arctan

√1− ζ2

ζ

)

, (t ≥ 0) (6)

From equation (6) it can be seen that the frequency of transient oscillation is thedamped natural frequency ωd and thus varies with the damping ration ζ .

If the damping ratio ζ is equal to zero, the response becomes undamped and oscil-lations continue indefinitely. The response c(t) for the zero damping case may beobtained by substituting ζ = 0 in equation (6), yielding:

c(t) = 1− cosωnt, (t ≥ 0) (7)

Thus, from equation (7) we see that ωn represents the undamped natural frequencyof the system. That is, ωn is that frequency at which the system would oscillate ifthe damping were decreased to zero.

2. Critically damped case, (ζ = 1)

If the two poles of C(s)/R(s) are equal the system is a critically damped one. Fora unit step input R(s) = 1/s, C(s) can be written as:

C(s) =ω2n

(s+ ωn)2s(8)

The inverse Laplace transform of equation (8) may be found as:

c(t) = 1− e−ωnt(1− ωnt), , (t ≥ 0)

10

3. Overdamped case, (ζ > 1)

In this case, the two poles of C(s)/R(s) are negative real and undamped. For a unitstep, C(s) can be written:

C(s) =ω2n

s(s+ ζωn + ωn

√ζ2 − 1)(s+ ζωn − ωn

√ζ2 − 1)

(9)

By partial fraction expansion and taking the inverse Laplace transform of equation(9) the system output c(t) is:

c(t) = 1 +1

2√ζ2 − 1(ζ +

√ζ2 − 1)

e−(ζ+√

ζ2−1)ωnt −

− 1

2√ζ2 − 1(ζ +

√ζ2 − 1)

e−(ζ−√

ζ2−1)ωnt

= 1 +ωn

2√ζ2 − 1

(

e−s1t

s1− e−s2t

s2

)

(10)

where s1 = (ζ +√ζ2 − 1)ωn and s2 = (ζ −

√ζ2 − 1)ωn are the system poles. Thus,

the response c(t) includes two decaying exponential terms.

The transient response of a second-order system for various values of the damping ratioζ is shown in Figure 9. As ζ decreases, the closed-loop roots approach the immaginary

0

0.1

2

1

0.7

0.5

0.2

t

c(t)

Figure 9: Step response of a second-order system for various values of damping ratio

axis and the response becomes increasingly oscillatory.

11

Definitions of transient-response specifications

In many practical cases, the desired performance characteristics of control systems arespecified in terms of time-domain quantities. Systems with energy storage cannot respondinstantaneously and exhibit transient responses whenever they are subjected to inputs ordisturbances.

The transient response of a practical control system often exhibits damped oscillationsbefore reaching steady-state. It is common to specify the following:

1. Rise time

2. Peak time

3. Maximum overshoot

4. Settling time

These specifications are defined in what follows and are shown graphically in Figure 10.

0 t

c(t)

±0.05 or

±0.02

Allowable tollerance

0.1

0.9 1

M p

t r1 t r t p

t s

Figure 10: Second-order system underdamped response

1. Rise time, tr : the time required for the response to rise from 10% to 90%, 5% to95% or 0% to 100% of its final value. For underdamped second-order systems the0% to 100% rise time is normally used. For overdamped systems, the 10% to 90%rise time is commonly used.

2. Peak time, tp : the time required for the response to reach the first peak of theovershoot.

12

3. Maximum (percent) overshoot Mp : the maximum peak value of the responsecurve measured from the steady-state value of the response:

Mp =c(tp)− c(∞)

c(∞)· 100%

The amount of the maximum (percent) overshoot directly indicates the relativestability of the system.

4. Settling time, ts : the time required for the response curve to reach and stay withina range about the final value, of size specified by absolute percentage of the finalvalue (usually 2% or 5%). The settling time is related to the largest time constantof the control system.

The time-domain specifications just given are quite important since most control sys-tems are time-domain systems; that is they must exhibit acceptable time responses. Thismeans, that the control system must be modified until the transient response is satisfac-tory. Note that if we specify the values of tr, tp, ts and Mp, then the shape of the responsecurve is virtually determined.

Except for certain applications where oscillations cannot be tolerated, it is desirablethat the transient response be sufficiently damped. Thus, for a desirable transient responseof a second-order system, the damping ratio must be between 0.4 and 0.8. Smaller valuesof ζ yield excessive overshoot in the transient response and a system with a large valueof ζ responds sluggishly.

Second-order systems and transient response specifications

In the following we shall obtain the rise time, peak time, maximum overshoot and settlingtime of the second-order system response. The values will be obtained in terms of ζ andωn. The system is assumed to be underdamped.

1. Rise time, tr . Referring to equation (6) we obtain the rise time tr by letting c(tr) = 1or

c(tr) = 1 = 1− e−ζωntr(cosωdtt +ζ√

1− ζ2sinωdtr)

Since e−ζωntr 6= 0, we obtain:

cosωdt+ζ√

1− ζ2sinωdtr= 0

or

tanωdtr = − ζ√1− ζ2

= −ωd

σ

Thus, the rise time tr is:

tr =1

ωd· arctan

(

ωd

−σ

)

=π − β

ωd

where β and σ are defined in Figure 11.

13

21 z-w=w nd

s-

nw

b

djw

djw-

s

nzw

wj

Figure 11: Second-order system poles

2. Peak time, tp . Referring to equation (6) we may obtain the peak time by differen-tiating c(t) with respect to time and letting the derivative equal zero, or:

dc(t)

dt|t=tp = sin(ωdtp) ·

ωn√1− ζ2

· e−ζωntp = 0

This yields the following equation:

sin(ωdtp) = 0

orωdtp = 0, π, 2π, 3π, ...

Since the peak time corresponds to the first peak overshoot, ωdtp = π. Hence

tp =π

ωd

The peak time corresponds to one half cycle of the frequency of damped oscillation.

3. Maximum (percent) overshoot, Mp occurs at the peak time or at t = tp = πωd.

Thus, from equation (6) Mp is obtained as:

Mp = c(tp)− 1 = −e−ζωnπ/ωd

(

cosπ +ζ√

1− ζ2sinπ

)

orMp = e−πζ/

√1−ζ2

4. Settling time, ts . For an underdamped second-order system, the transient responseis obtained from equation (6) as:

c(t) = 1− e−ζωnt · sin(

ωdt + arctan

√1− ζ2

ζ

)

14

0

c(t)

c 1 (t)

c 2 (t)

t

Figure 12: Step response of a second-order system and the envelope curves

The curves c1,2(t) = ±e−ζωnt/√1− ζ2 are the envelope curves of the transient re-

sponse for a unit-step input. The response curve c(t) always remains within a pairof the envelope curves as shown in Figure 12. The two envelope curves, as well asc(t) will reach 2% from the final value approximately when

e−ζωnts < 0.02

orζωnts ∼= 4

Therefore we have:

ts =4

ζωn

Example. Consider the closed-loop system with the transfer function:

H(s) =25

s2 + 6s+ 25

From the general form of a second-order system transfer function:

H(s) =ω2n

s2 + 2ζωn + ω2n

the system parameters are:ωn = 5, ζ = 0.6

Then we obtain:

1. The damped natural frequency is:

ωd = ωn

√

1− ζ2 = 5√1− 0.62 = 4

15

and the poles negative real part:

σ = −ζωn = −3.

According to Figure 11, the angle β is:

β = arctanωd

σ= 0.93

2. The rise time is:

tr =π − β

ωd

=3.14− 0.93

4= 0.55sec

3. The peak time is:

tp =π

ωd=

3.14

4= 0.78sec

4. Maximum overshoot will be:

Mp = e−πζ/√

1−ζ2 = 0.095

The maximum percent overshoot is then: Mp = 9.5%.

5. For the 2% criterion the settling time is:

ts =4

σ=

4

ζωn=

4

3= 1.33sec

In Figure 13 is shown the step response of the system. The values of the system parameterscan be seen also from the plot.

Time (sec.) 0 0.5 1 1.5

0

0.2

0.4

0.6

0.8

1

1.2

Figure 13: Step response

16

Steady-State Error

As a further requirement in analyzing a control system one must examine and comparethe final steady-state error for an open-loop and a closed-loop system. The steady-stateerror is the error after the transient response has decayed, leaving only the steady-stateresponse.

For an open-loop system having the input r(t) and the output c(t), the error signal is:

e(t) = r(t)− c(t)

and the steady-state error, ess is defined as:

ess = limt→∞

e(t) = limt→∞

(r(t)− c(t))

The Laplace transform of the error signal for the open-loop system shown in Figure14 is:

E(s) = R(s)− C(s) = R(s)−G(s)R(s) = (1−G(s))R(s)

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

Figure 14: Open-loop system

The Laplace transform of the error for unity feedback closed-loop system, shown inFigure 15, is obtained from:

Figure 15: Closed-loop system

Ec(s) = R(s)− C(s) = R(s)−G(s)C(s)

or

Ec(s) =1

1 +G(s)R(s)

To calculate the steady-state error, we utilize the final value theorem, which is:

limt→∞

e(t) = lims→0

sE(s)

17

For the open-loop system we obtain:

ess = lims→0

s(1−G(s))R(s)

and for the closed-loop system:

ess = lims→0

s

(

1

1 +G(s)R(s)

)

Therefore, using a unit step input as a comparable input, we obtain for the open-loopsystem

ess = lims→0

s(1−G(s))(1

s) = lim

s→0(1−G(s)) = 1−G(0)

For the closed-loop system, we have:

ess = lims→0

s

(

1

1 +G(s)

)

(

1

s

)

=1

1 +G(0)

Example. The advantage of the closed-loop system in reducing the steady-state errorof the system resulting from parameter changes and calibration errors may be illustratedby an example. Let us consider a system with a process transfer function

G(s) =k

Ts+ 1

which would represent a thermal control process, a voltage regulator, or a water-level con-trol process. For a specific setting of the desired input variable, which may be representedby the normalized unit step input function, we have R(s) = 1/s. Then, the steady-stateerror of the open-loop system is:

ess = 1−G(0) = 1− k

The error for the closed-loop system of Figure 15 is:

ess =1

1 +G(0)=

1

1 + k

For the open-loop system, one would calibrate the system so that k = 1 and thesteady-state error is zero. For the closed-loop system one would set a large gain k, forexample k = 100. Then the closed-loop system steady-state error is ess = 1/101.

Effect of an additional zero

Consider a system with the transfer function H(s). The system step response can beexpressed as:

y(t) = £−1 [Y (s)] = £−1

[

H(s)

s

]

18

Now suppose we add a zero at −a and divide the transfer function with a so the gain ofthe new system is unchanged. The new transfer function is:

Hz(s) =s+ a

aH(s) =

s

aH(s) +H(s)

The step response of the system Hz(s) result:

yz(t) = £−1 [Yz(s)] = £−1[

1

sHz(s)

]

= £−1[

1

s

(

s

aH(s) +H(s)

)]

=1

ay(t) + y(t)

(since £−1[sY (s)] = y(t))

If a is small, or the zero is close to the imaginary axis, 1/a is large and the stepresponse of Hz(s) will increase with the quantity 1/a · y(t). The effect of addition of azero is the increase of the overshoot.

Example. Consider the system with the transfer function:

System 1: H1(s) =1

s2 + s+ 1

We add a zero at −1 and obtain:

System 2: H2(s) =s + 1

s2 + s+ 1

The step responses for these two systems are presented in Figure 16 Now consider the

Step Response

Time (sec)

Am

plitu

de

0 2 4 6 8 10 120

0.2

0.4

0.6

0.8

1

1.2

1.4

System 1

System 2

Figure 16: Effect of addition of a zero at -1

zero was added at -10 and the gain of the new system was divided by the value 10 so thetotal gain is equal to one (the gain of system 1).

System 3: H3(s) =0.1(s+ 10)

s2 + s+ 1

The step responses were compared in Figure 17. The general effect of additional zero tothe second order system is the increase of overshoot.

19

0 2 4 6 8 10 120

0.2

0.4

0.6

0.8

1

1.2

1.4Step Response

Time (sec)

Am

plitu

de

System 2

System 1

System 3

Figure 17: Effect of addition of a zero at -1 and -10

Transient response of higher-order systems

Consider a system with unit step input R(s) = 1/s an output C(s) and the transferfunction H(s). Then C(s) can be written:

C(s) = H(s) ·R(s) =amsm + ...+ a1s+ a0s(bnsn + ...+ b1s+ b0)

, (m ≤ n) (11)

The transient response of this system to any given input can be obtained by computersimulation. If an analytical expression for the transient response is desired then it isnecessary to factor the denominator polynomial. The poles of C(s) consist of real polesand complex conjugates poles. A pair of complex-conjugates poles yields a second orderterm in s. Since the factored form of the higher-order characteristic equation consists offirst- and second-order terms, equation (11) can be rewritten:

C(s) =K∏m

i=1(s+ zi)

s∏q

j=1(s+ pj)∏r

k=1(s2 + 2ζkωks+ ω2

k)

where q + 2r = n. If the closed-loop poles are distinct C(s) can be expanded into partialfractions as follows:

C(s) =a

s+

q∑

j=1

ajs+ pj

+r∑

k=1

bk(s+ ζkωk) + ckωk

√

1− ζ2ks2 + 2ζkωks+ ω2

k

From this last equation we see that the response of a higher-order system is composedof a number of terms involving the simple functions found in the responses of first- andsecond-order systems. The unit-step response c(t), the inverse Laplace transform of C(s)is then:

c(t) = a+q∑

j=1

aje−pjt +

r∑

k=1

bke−ζkωktcosωk

√

1− ζ2kt+r∑

k=1

bke−ζkωktsinωk

√

1− ζ2kt

20

If all the closed-loop poles lie in the left-half s-plane, then the exponential terms andthe damped exponential terms will approach zero as time increases. The steady-statevalue of the output is c(∞) = a.

Let us assume that the system considered is a stable one. Then the closed-loop polesthat are located far from the jω (imaginary) axis have large negative real parts. Theexponential terms that correspond to these poles decay very rapidly to zero.

The poles located nearest the jω axis correspond to transient response terms thatdecay slowly. Those poles that have dominant effects on the transient-response behaviorare called dominant poles. They are most important among all closed-loop poles.

Note. Consider a first-order system with the transfer function

Hk(s) =k

Ts+ 1= k ·H(s)

or a second-order system with the transfer function:

Hk(s) = k · ω2n

s2 + 2ζωn + ω2n

= k ·H(s)

The system response when the input is a unit step R(s) = 1/s is:

c(t) = £−1[Hk(s) · R(s)] = £−1[k ·H(s)

s] = k ·£−1[

H(s)

s]

That is, the time response is k · c(t) where c(t) is the response of the system with a unitygain k. The final value of the response, in case of a stable system will be c(∞) = k.

System approximation using the concept of dominant poles

Consider a system with a transfer function:

H(s) =k(s+ a)

( 1ω2ns2 + 2ζ

ωns+ 1)(Ts+ 1)

It has a zero at z1 = −a and three poles: two complex poles p1,2 = −ζωn ± j√1− ζ2 and

one real pole at p3 = −1/T . If, for example, the real pole is far from the jω axis andthe complex poles are dominant, the system order can be reduced by neglecting the realpole. Because the steady-state value of the system response to test inputs must remainthe same, the gain factor must be multiplied by the absolute value of the time constantor 1/pole (or divided by the absolute value of the pole).

Example 1. Consider a third-order system with the transfer function:

H1(s) =s+ 2

(s2 + 2s+ 2)(s+ 10)

The system poles are: p1, 2 = −1 ± j and p3 = −10. Because the real pole is located10 times far from the imaginary axis than the complex poles, p3 can be neglected. Thesystem gain will be divided by |p3| = 10 and obtain:

H2(s) =0.1(s+ 2)

s2 + 2s+ 2

21

The step responses obtained for the two systems are almost the same, as shown in Figure18.

0 1 2 3 4 5 0

0.02

0.04

0.06

0.08

0.1

0.02

0.04

0.06

0.08

0.1

0 1 2 3 4 5 0

Figure 18: Comparation of two step responses

Example 2. Consider a system with the transfer function:

H1(s) =62.5(s+ 2.5)

(s2 + 6s+ 25)(s+ 6.25)

The system poles are: p1, 2 = −3 ± 4 · j and p3 = −6.25. The gain is equal to 1 and thesteady state value for a step input must be maintained. As an approximation we neglectthe real pole and obtain:

H2(s) =10(s+ 2.5)

s2 + 6s+ 25

Using a computer simulation the step responses of the two systems are shown in Figure19. The effect of neglecting the real pole was to increase the overshoot and reduce thesettling time. This happened because the real pole was not very far from the complexpoles and cannot be neglected for a good approximation.

Systems with transport lag

Figure 20 shows a thermal system in which hot water is circulated to keep the temperatureof a chamber constant. In this system, the measuring element is placed downstream adistance L from the furnace, the air velocity is v and T = L/v sec would elapse beforeany change in the furnace temperature was sensed by the thermometer. Such a delay inmeasuring, delay in controller action, or delay in actuator operation, and the like is calledtransport lag or dead time. Dead time is present in most process control systems.

22

0 0.3 0.6 0.9 1.2 1.5 0

0.2

0.4

0.6

0.8

1

1.2

0 0.5 1 1.5 0

0.2

0.4

0.6

0.8

1

1.2

1.4

Figure 19: Comparation of two step responses

Fur

nace

L

v thermometer

blower

fuel

Figure 20: Thermal process

A dead time is the time interval between the start of an event at one point in asystem and its resulting action at another point in the system.

The input x(t) and the output y(t) of a transport lag or dead time element are relatedby

y(t) = x(t− T )

where T is the dead time. The transfer function of transport lag or dead time is given by:

H(s) =£[x(t − T )]

£[x(t)]=

X(s)e−sT

X(s)= e−sT

A linear system which exhibits dead time, is defined by the differential equation:

m∑

j=0

ajdjx(t− T )

dtj=

n∑

j=0

bjdjy(t)

dtj

23

where x(t) is the input signal, y(t) is the output, and T the dead time. The Laplacetransform of the differential equation which describes the system will give:

e−sT ·m∑

j=0

ajsj£[x(t)] =

n∑

j=0

bjsj£[y(t)]

and the transfer function is:

H(s) = e−sT

∑mj=0 ajs

j

∑nj=0 bjs

j=

Y (s)

X(s)

If we use a Taylor series expansion for e−sT :

e−sT = 1− Ts+1

2!T 2s2 − 1

3!T 3s3 + ...

and the truncated Taylor series expansion of a ratio of two polynomials:

1 + αTs

1 + βTs= 1 + (α + β)Ts− β(α− β)T 2s2 + β2(α− β)T 3s3,

we can approximate the dead time transfer function as a ratio of two polynomials, usuallycalled a Pade approximation:

a) e−sT =1− 1

2Ts

1 + 12Ts

b) e−sT =1− 1

2Ts+ 1

12T 2s2

1 + 12Ts+ 1

12T 2s2

24

Stability of linear systems

Introduction

The idea of stability is familiar to us. Knowing that an unstable system will exhibit anerratic and destructive response, we seek to ensure that a system is stable and exhibits abounded transient response.

The stability of a feedback system is related to the location of the roots of the charac-teristic equation of the system transfer function. Thus, we wish to develop a few methodsfor determining whether a system is stable, and if so, how stable it is.

Whether a linear system is stable or unstable is a property of the system itself anddoes not depend on the input or driving function of the system. The input contributeonly to the steady-state response terms in the solution.

The transient response of a system is of primary interest and must be investigated. Avery important factor regarding the transient performance of a system is the stability ofthe system.

A stable system is defined as a dynamic system with a bounded (limited)system response to a bounded input.

That is, if the system is subjected to a bounded input or disturbance and the responseis bounded in magnitude, the system is said to be stable.

Example. The concept of stability can be illustrated by considering a right circularcone placed on a plane horizontal surface. If the cone is resting on its base and is tippedslightly, it returns to its original equilibrium position. This position and response is saidto be stable. If the cone rests on its side and is displaced slightly, it rolls with no tendencyto leave the position on its side. This position is designated as the neutral stability. Onthe other hand, if the cone is placed on its tip and released, it falls onto its side. Thisposition is said to be unstable.

The stability of a dynamic system is defined in a similar manner. The response to adisplacement, or initial condition, will result in either a decreasing, neutral or increasingresponse. Specifically, it follows from the definition of stability that a linear system isstable if and only if the absolute value of its impulse response g(t), integrated

25

over an infinite range, is finite. The location in the s-plane of the poles of a systemindicated the resulting transient response. The poles in the left-hand portion of the s-planeresult in a decreasing response for disturbance inputs. Similarly, poles on the jω axis andin the right-hand plane result in a neutral and an increasing response, respectively, for adisturbance input. This division of the s-plane is shown in Figure 21. Clearly, the polesof desirable dynamic systems must lie in the left-hand portion of the s-plane.

Figure 21: Stability in the s-plane

The system transfer function is written as:

C(s) =C(s)

R(s)=

k∏m

i=1(s+ zi)

sn∏q

j=1(s+ σj)∏r

k=1(s2 + 2αks+ (α2

k + ω2k))

The system poles are either real: pj = −σj or complex conjugates pk1,2 = αk ± jωk. Theoutput response for an impulse input is then:

c(t) =q∑

j=1

Aje−σjt +

r∑

k=1

Bk(1

ωk)e−αktsinωkt

Clearly, to obtain a bounded response, the poles of the closed-loop system must be in theleft-hand portion of the s-plane. Thus a necessary and sufficient condition that a system

be stable is that all the poles of the system transfer function have negative real parts.

We will call a system not stable if not all the roots are in the left-hand s-plane. If thecharacteristic equation has roots on the imaginary axis, with all the other roots in theleft-hand plane, the steady-state output will be sustained oscillations for a bounded input.Such a system is called marginally stable or critically stable. For an unstable system, thecharacteristic equation has at least one root in the right-hand half plane or repeated jωroots; for this case, the output is unbounded for any input.

To ascertain the stability of a feedback control system, one could determine the rootsof the characteristic equation.

The Routh-Hurwitz stability criterion

The discussion and determination of stability has occupied the interest of many engineers.Maxell and Vishnegradski first considered the question of stability of dynamic systems.

26

In the late 1800’s, A. Hurwitz and E.J.Routh published independently a method of in-vestigating the stability of linear systems. The Routh-Hurwitz stability method providesan answer to the question of stability by considering the characteristic equation of thesystem. The characteristic equation in the Laplace variable is written as:

∆(s) = q(s) = ansn + an−1s

n−1 + ... + a1s+ a0 = 0 (12)

To ascertain the stability of the system, it is necessary to determine whether any oneof the roots of q(s) lies in the right-hand half of the s-plane.

The procedure in the Routh’s stability criterion is as follows:

1. Write the polynomial in s in the form (12) where the coefficients are real quantities.We assume that a0 6= 0 that is, any zero root has been removed.

2. Necessary condition. All the coefficients of the polynomial must have the samesign if all the roots are in the left-hand half plane. Also, it is necessary that allthe coefficients for a stable system be nonzero. However, although necessary, theserequirements are not sufficient. That is, we immediately know the system is unstableif they are not satisfied, yet if they are satisfied, we must proceed to ascertain thestability of the system.

3. Sufficient condition. If all the coefficients are positive, arrange the coefficients ofthe polynomial in rows according to the following pattern:

sn : an an−2 an−4 an−6 . .sn−1 : an−1 an−3 an−5 an−7 . .sn−2 : b1 b2 b3 b4 . .sn−3 : c1 c2 c3 c4 . .. : . . . . . .. : . . . . . .s2 : e1 e2s1 : f1s0 : g1

The coefficients b1, b2, and so on are evaluated as follows:

b1 =an−1an−2 − anan−3

an−1

b2 =an−1an−4 − anan−5

an−1

b3 =an−1an−6 − anan−7

an−1

The evaluation of b’s is continued until the remaining ones are all zero. The samepattern of cross-multiplying the coefficients of the two previous rows is folowed inevaluating the c’s, ..e’s, f, g. That is:

c1 =b1an−3 − an−1b2

b1

27

c2 =b1an−5 − an−1b3

b1

The process is continued until the n-th row is completed. The complete array ofcoefficients is triangular. Note that in developing the array, an entire row maybe multiplied or divided by a positive number in order to simplify the subsequentnumerical calculation without altering the stability conclusion.

Routh’s stability criterion states that the number of roots of the characteristic equa-

tion q(s) = 0 with positive real parts is equal to the number of changes in sign of

the coefficients of the first column of the array.

The necessary and sufficient condition that all roots of the characteristic equationlie in the left-half s-plane is that all the coefficients of the characteristic equation bepositive and all terms in the first column of the array have positive signs.

Example. Consider the following polynomial:

s4 + 2s3 + 3s2 + 4s+ 5 = 0

Let us follow the procedure just presented and construct the array of coefficients. The firsttwo rows are obtained directly from the given polynomial. The remaining are obtainedfrom these. If any coefficients are missing they may be replaced by zeros in the array.

s4 : 1 3 5s3 : 2 4 0s2 : 2·3−1·4

2= 1 2·5−1·0

2= 5

s1 : 1·4−2·51

= −6s0 : −6·5−1·0

−6= 5

In this example, the number of changes in sign of the coefficients in the first column istwo. This means that there are two roots with positive real parts. Indeed, if we calculatethe roots of the polynomial we obtain: p1 = 0.2878 + 1.4161i, p2 = 0.2878− 1.4161i, p3 =−1.2878 + 0.8579i, p4 = −1.2878− 0.8579i.

Example. Special cases. If a first-column term in any row is zero, but the remainingterms are not zero or there is no remaining term, then the zero term is replaced by a verysmall positive number ǫ and the rest of the array is evaluated. For example consider thecharacteristic equation:

s3 + 2s2 + s+ 2 = 0

The array of coefficients is:s3 : 1 1s2 : 2 2s1 : 0 ≈ ǫs0 : 2

If the sign of the coefficient above the zero (ǫ) is the same as the one below it, it indicatesthat there are a pair of imaginary roots. Actually, this equation has two roots at ±j.

If however, the sign of the coefficient above zero is opposite that below it, it indicatesthat there is one sign change.

28

Application of Routh’s stability criterion to control system anal-ysis.

Routh stability criterion is of limited usefulness in linear control system analysis because itdoes not suggest how to improve relative stability or how to stabilize an unstable system.It is possible, however, to determine the effects of changing one or two parameters byexamining the values that cause instability. In the following, we shall consider the problemof determining the stability range of a parameter value.

Consider the system shown in Figure 22. Let us determine the range of k for stability.The closed-loop transfer function is:

k R(s) C(s)

s(s+2)(s 2 +s+1)

Figure 22: Stability in the s-plane

C(s)

R(s)=

k

s(s+ 2)(s2 + s+ 1) + k

The characteristic equation is:

s4 + 3s3 + 3s2 + 2s+ k = 0

The array of coefficients becomes:

s4 : 1 3 ks3 : 3 2 0s2 : 7/3 ks1 : 2− 9k/7s0 : k

For stability, k must be positive, and all coefficients in the first column be positive.Therefore

2− 9 · k7

> 0

or14

9> k > 0

When k = 14/9, the system becomes oscillatory and mathematically, the oscillation issustained at constant amplitude.

29

Bibliography

[1] Richard C. Dorf and Robert H. Bishop, Modern Control Systems, Addison-Wesley,1995.

[2] Katsuhiko Ogata, Modern Control Engineering, Prentice Hall International Editions,1990.

30