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178 SYSTEM TIME RESPONSE CHARACTERISTICS

1

σ

s-plane

z-plane

Figure 7.6 Mapping the lines of constant σ

1

2

8

4

7

3

4

7

8

3

5

6

9

0

2

1

5

6

9

0

3

4

7

8

3

4

7

8

1

2

s-plane

z-plane

Figure 7.7 Poles in the s-plane and their corresponding z-plane locations

The time responses of a sampled data system based on its pole positions in the z-plane are shown in Figure 7.8. It is clear from this figure that the system is stable if all the closed-loop poles are within the unit circle.

7.4DAMPING RATIO AND UNDAMPED NATURAL FREQUENCY IN THE z-PLANE

7.4.1 Damping Ratio

As shown in Figure 7.9(a), lines of constant damping ratio in the s-plane are lines where ζ = cos α for a given damping ratio. The locus in the z-plane can then be obtained by the substitution z = es T . Remembering that we are working in the third and fourth quadrants in


DAMPING RATIO AND UNDAMPED NATURAL FREQUENCY IN THE z-PLANE

179

X

X

X

X X X X

X

X X

z-plane

Figure 7.8 Time response of z-plane pole locations

the s-plane where s is negative, we get

z

=

e−σ ωT e j ωT .

(7.3)

Since, from Figure 7.9(a),

π

σ = tan

− cos−1 ζ ,

(7.4)

2

substituting in (7.3) we have

z = exp −ωT tan

π

e j ωT .

− cos−1 ζ

(7.5)

2

Equation (7.5) describes a logarithmic spiral in the z-plane as shown in Figure 7.9(b). The spiral starts from z = 1 when ω = 0. Figure 7.10 shows the lines of constant damping ratio in the z-plane for various values of ζ .

7.4.2 Undamped Natural Frequency

As shown in Figure 7.11, the locus of constant undamped natural frequency in the s-plane is a circle with radius ωn . From this figure, we can write

ω2 + σ 2 = ωn2 or σ = ωn2 − ω2. (7.6)


180 SYSTEM TIME RESPONSE CHARACTERISTICS

const ζ

jωn 1 − ζ2

β

σ

− ζ ω

s-plane

(a)

const ζ

1.0

z-plane

(b)

Figure 7.9 (a) Line of constant damping ratio in the s-plane, and (b) the corresponding locus in the z-plane

Figure 7.10 Lines of constant damping ratio for different ζ . The vertical lines are the lines of constant ωn

DAMPING RATIO AND UNDAMPED NATURAL FREQUENCY USING FORMULAE

181

Line of

constant ζ

Line of constant ωn

ωn

Figure 7.11 Locus of constant ωn in the s-plane

Thus, remembering that s is negative, we have

z = e−s T = e−σ T e− j ωT = exp −T ( ωn2 − ω2) e− j ωT

(7.7)

The locus of constant ωn in the z-plane is given by (7.7) and is shown in Figure 7.10 as the vertical lines. Notice that the curves are given for values of ωn ranging from ωn = π/10T to

ωn = π/ T .

Notice that the loci of constant damping ratio and the loci of undamped natural frequency are usually shown on the same graph.

7.5DAMPING RATIO AND UNDAMPED NATURAL FREQUENCY USING FORMULAE

In Section 7.4 above we saw how to find the damping ratio and the undamped natural frequency of a system using a graphical technique. Here, we will derive equations for calculating the damping ratio and the undamped natural frequency.

The damping ratio and the natural frequency of a system in the z-plane can be determined if we first of all consider a second-order system in the s-plane:

ω2

G(s) = n . (7.8) s2 + 2ζ ωn s + ωn2

The poles of this system are at

s1,2 = −ζ ωn ± j ωn 1 − ζ 2. (7.9)

We can now find the equivalent z-plane poles by making the substitution z = es T , i.e.

z = es T = e−ζ ωn T ± ωn T 1 − ζ 2, (7.10)


182

SYSTEM TIME RESPONSE CHARACTERISTICS

which we can write as

where

z = r

± θ ,

(7.11)

r = e−ζ ωn T

or

ζ ωn T = − ln r

(7.12)

and

θ = ωn T

.

1 − ζ 2

(7.13)

From (7.12) and (7.13) we obtain

ζ

=

− ln r

or

1 − ζ 2

θ

ζ

− ln r

,

(7.14)

and from (7.12) and (7.14) we obtain

= (ln r )2 + θ 2

1

(ln r )2 + θ 2.

(7.15)

ωn =

T

Example 7.2

Consider the system described in Section 7.1 with closed-loop transfer function

y(z)

G(z)

0.368z + 0.264

.

r (z) =

1 + G(z) = z2 − z + 0.632

Find the damping ratio and the undamped natural frequency. Assume that T = 1 s.

Solution

We need to find the poles of the closed-loop transfer function. The system characteristic equation is 1 + G(z) = 0,

i.e.

z2 − z + 0.632 = (z − 0.5 − j 0.618)(z − 0.5 + j 0.618) = 0,

which can be written in polar form as

z1,2 = 0.5 ± j 0.618 = 0.795

± 0.890 = r

± θ

(see (7.11)). The damping ratio is then calculated using (7.14) as

ζ

=

− ln r

− ln 0.795

0.25,

(ln r )2 + θ 2 =

(ln 0.795)2 + 0.8902 =

and from (7.15) the undamped natural frequency is, taking T = 1,

1

ωn =

(ln r )2 + θ 2

= (ln 0.795)2 + 0.8902

= 0.92.

T


EXERCISES 183

Figure 7.12 Finding ζ and ωn graphically

Example 7.3

Find the damping ratio and the undamped natural frequency for Example 7.2 using the graphical method.

Solution

The characteristic equation of the system is found to be

z2 − z + 0.632 = (z − 0.5 − j 0.618)(z − 0.5 + j 0.618) = 0

and the poles of the closed-loop system are at

z1,2 = 0.5 ± j 0.618.

Figure 7.12 shows the loci of the constant damping ratio and the loci of the undamped natural frequency with the poles of the closed-loop system marked with an × on the graph. From the graph we can read the damping ratio as 0.25 and the undamped natural frequency as

ωn = 0.29π = 0.91.

T

7.6 EXERCISES

1.Find the damping ratio and the undamped natural frequency of the sampled data systems whose characteristic equations are given below

(a)z2 − z + 2 = 0

(b)z2 − 1 = 0

(c)z2 − z + 1 = 0

(d)z2 − 0.81 = 0

184 SYSTEM TIME RESPONSE CHARACTERISTICS

G(s)

r(s)

e(s)

y(s)

1 − e−Ts

1

s

s + 1

Figure 7.13 System for Exercise 2

2.Consider the closed-loop system of Figure 7.13. Assume that T = 1 s.

(a)Calculate the transfer function of the system.

(b)Calculate and plot the unit step response at the sampling instants.

(c)Calculate the damping factor and the undamped natural frequency of the system.

3.Consider the closed-loop system of Figure 7.13. Do not assume a value for T .

(a)Calculate the transfer function of the system.

(b)Calculate the damping factor and the undamped natural frequency of the system.

(c)What will be the steady state error if a unit step input is applied?

4.A unit step input is applied to the system in Figure 7.13. Calculate:

(a)the percentage overshoot;

(b)the peak time;

(c)the rise time;

(d)settling time to 5 %.

5.The closed-loop transfer functions of four sampled data systems are given below. Calculate

the percentage overshoots and peak times.

(a) G(z) =

1

z2

z

+

2

(b) G(z) =

+ 1

z2

2z

+

1

(c) G(z) =

+1

z2

z

+

1

(d) G(z) =

−2

z2

+ z

+ 4

6.The s-plane poles of a continuous-time system are at s = −1 and s = −2. Assuming T = 1 s, calculate the pole locations in the z-plane.

7.The s-plane poles of a continuous-time system are at s1,2 = −0.5 ± j 0.9. Assuming T = 1 s, calculate the pole locations in the z-plane. Calculate the damping ratio and the undamped natural frequency of the system using a graphical technique.

FURTHER READING

[D’Azzo and Houpis, 1966] D’Azzo, J.J. and Houpis, C.H. Feedback Control System Analysis and Synthesis, 2nd

edn., McGraw-Hill, New York, 1966.

[Dorf, 1992]

Dorf, R.C. Modern Control Systems, 6th edn. Addison-Wesley, Reading, MA, 1992.