Chapter 7:

System Time Response Characteristics

1

In this chapter:

• The time response of a sampled data system is investigated and compared with the response of a similar continuous system.

• The mapping between the s-domain and the z-domain is examined, the important time response characteristics of continuous systems are revised and their equivalents in the discrete domain are discussed.

2

7.1 Time response comparison

Consider the following continuous-time closed-loop system.

Let us add a sampler (A/D converter) and a zero-order hold (D/A converter) to the previous system to form an equivalent discrete-time system.

3• We shall now derive equations for the step responses of both systems and then plot and compare them.

The continuous system• The transfer function of the closed-loop continuous system is

• Since the input is unit step, r(s)=1/s, the output becomes

• To find the inverse Laplace transform, we can write

• From the inverse Laplace transform tables we find that the time response is 4

.1

1)(1)(

)()(

2

sssGsG

srsy

.)1(

1)( 2

ssssy

.)866.0()5.0(

5.0)866.0()5.0(

5.011

11)( 22222

sss

ssss

ssy

)].866.0sin(573.0)866.0[cos(1)( 5.0 ttety t

The discrete-time systemThe transfer function of the previous discrete-time system is given by

where

And the z-transform of the plant is given by

Expanding by means of partial fractions, we obtain5

,)(1)(

)()(

zGzG

zrzy

,1

)(

zzzr

.)1(

1)( 2

ssesGTs

.1111)1()( 2

sssesG Ts

• And the z-transform is

• From z-transform tables we obtain

• Setting T=1 s and simplifying gives

• Therefore, The transfer function of the closed loop discrete-time system is given by

and the output will be

• The inverse z-transform can be found by long division: the first several terms are

6.)632.0)(1()264.0368.0()( 2

zzzzzzy

7654321 8.09.015.14.14.1368.0)( zzzzzzzzy

.1111)1()( 2

1

sssZzzG

.1)1(

)1()( 21

Tezz

zz

zTzzzG

.368.0368.1

264.0368.0)( 2

zz

zzG

,632.0264.0368.0

)(1)(

)()(

2

zzz

zGzG

zrzy

• The time response of both discrete-time system and its continuous-time equivalent are shown.

• It is clear that the discrete system has higher overshoot. This indicates that the sampling process has a destabilizing effect on the system.

7

• Let us now decrease the sampling interval to 0.5 and 0.1, respectively, and repeat the comparison.

• It is clear that the response of the discrete system approaches that of the continuous response as T becomes sufficiently small.

8

T=0.5

T=0.1

Matlab code for previous simulations:syms ss = tf('s') t=0:0.1:35; % The continuous systemG = 1/(s^2+s);Gcl = G/(1+G)y=step(Gcl,t)plot(t,y,'LineWidth',2.5)hold on % The discrete systemT = 0.1; % The sampling intervalGd = c2d(G,T);Gcld = Gd/(1+Gd);[y,t]=step(Gcld)plot(t,y,'*r','LineWidth',2.5) axis([0 12 0 1.5])

9

7.2 Time Domain Specifications

• The performance of a control system is usually measured in terms of its response to a step input.

• The step input is used because it is easy to generate and gives the system a nonzero steady-state condition, which can be measured.

• Most commonly used time domain performance measures refer to a second-order system with the transfer function:

where ωn is the undamped natural frequency of the system and ζ is the damping ratio of the system.

10

,2)(

)(22

2

nn

n

sssRsY

• When a second-order system is excited with a unit step input, the typical response is shown below. Based on this figure, the following performance parameters are usually defined: maximum overshoot; peak time; rise time; settling time; and steady-state error.

11

The maximum overshoot (Mp)

• It is the peak value of the response curve measured from unity. This parameter is usually quoted as a percentage. The amount of overshoot depends on the damping ratio and directly indicates the relative stability of the system.

• It is given by

• The lower the damping ratio, the higher the maximum overshoot as shown :

12

,21/ eM p

The peak time (Tp)

• Is the time required for the response to reach the first peak. The system is more responsive when the peak time is smaller, but this gives rise to a higher overshoot.

• The peak time It is calculated as

where

is the damped natural frequency.13

21

,

nd

dpT

The rise time (Tr)

• Is the time required for the response to go from 0% to 100% of its final value. It is a measure of the responsiveness of a system, and smaller rise times make the system more responsive.

• It is given by

where

14

.tan

,

1

n

d

drT

The settling time (Ts)

• It is the time required for the response curve to reach and stay within a range about the final value. A value of 2–5% is usually used in performance specifications.

• It is given by

(for 5% settling time),

(for 2% settling time).

15

.4

,3

ns

ns

T

T

The steady-state error (Ess)

• It is the error between the system response and the reference input value (unity) when the system reaches its steady-state value.

• A small steady state error is a requirement in most control systems. In some control systems, such as position control, it is one of the requirements to have no steady-state error.

• The steady-state error can be found by using the final value theorem, i.e. if the Laplace transform of the output response is y(s), then the final value (steady-state value) is given by

• and the steady-state error when a unit step input is applied can be found from

16).(lim10

ssYEsss

),(lim)(lim0

ssYtyystss

Example 7.1Determine the performance parameters of the system with the following closed-loop transfer function

Answer:Comparing this system with the standard second-order system transfer function

We find that ζ = 0.5 and ωn = 1 rad/s. Thus, the damped natural frequency is

17

.1

1)()(

2

sssrsy

,2)(

)(22

2

nn

n

sssrsy

866.01 2 nd

Answer (continued)

18.0

)1(1lim1)(lim1

sec,84

sec,42.2866.0047.1

,047.1tan

200

1

ssssssYE

T

T

ssss

ns

dr

n

d

The peak overshoot is

The peak time is

%.1616.021/ eM p

.627.3d

pT

Note:

Although the performance indices such as peak overshoot, rise time, peak time, and settling time are obtained using a continuous-time second order system, they can be used equally well in discrete-time systems if the sampling interval is chosen sufficiently small as we have already seen.

19

Choice of sampling interval• Whenever a digital control system is designed, a suitable

sampling interval must be chosen.

• Choosing a large sampling time has destabilizing effects on the system. In addition, information loss occurs when large sampling times are selected.

• It has been found from practical applications in the process industry that a sampling interval of 1s is generally short enough for most applications such as pressure control, temperature control and flow control. Systems with fast responses such as electromechanical systems (e.g. motors) require much shorter sampling intervals, usually of the order of milliseconds. 20

Choice of sampling interval• Various empirical rules have been suggested by many

researchers for the selection of the sampling interval. These rules are based on practical experience and simulation results. Among them are the following:

If the plant has the dominant time constant τ, then the sampling interval T for the closed-loop system should be selected such that T< τ/10.

If the closed-loop system is required to have a settling time Tss then choose the sampling interval T such that T < Tss/10.

21

7.3 Mapping the s-plane into z-plane

• It is known that the behavior and stability of a closed-loop continuous-time system can be predicted given its pole locations in the s-plane. It is desirable to do the same for sampled data systems.

• This section describes the mapping of the poles in the s-plane into z-plane. This allows us to predict the stability and behavior of discrete systems given their pole locations in the z-plane.

22

• Let s = σ + jω describes a point in the s-plane. Then, this point is mapped into z-plane as

• First, consider the mapping of the imaginary axis in the s-plane. Along the jω axis, σ = 0, so we have

• Hence, the s-plane poles on the imaginary axis are mapped onto the unit circle in the z-plane. As ω changes along the imaginary axis in the s-plane, the angle of the poles on the unit circle in the z-plane changes.

23

.TjTTs eeez

.1)sin()cos( TTjTez Tj

• For the left-half of s-plane, σ is negative, and hence, the entire left-hand s-plane is mapped into the interior of the unit circle in the z-plane.

• Similarly, the right-half s-plane is mapped into the exterior of the unit circle in the z-plane.

• Therefore a sampled data system will be stable if the closed-loop poles (or the zeros of the characteristic equation) lie within the unit circle in z-plane.

24

• Poles inside the unit circle are stable

• Poles outside the unit circle are unstable

• Poles on the unit circle are oscillatory

Lines of constant σ in the s-plane are mapped into circles in the z-plane with radius eσT:• If the line is in the left-half s-plane then the radius of the

circle in the z-plane is < 1. • If the line is in the right-half s-plane then the radius of the

circle in the z-plane is > 1.

25

ExampleFind the mapping of the s-plane poles shown below into the z-plane.

26

/T

Damping ratio and undamped natural frequency in the z-plane

27

Idea of pole placement design approach

Given some specifications of maximum overshoot, rise time or settling time

↓Find the damping ratio (ζ) and undamped natural

frequency (ωn) of the closed-loop system

↓Locate the desired poles in the z-plane

↓Design a controller (see Chapter 9) 28