Digital control system analysis and design 4th edition phillips solutions manual

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Digital Control System Analysis & Design 4e Instructor Manual

CHAPTER 6 6.2-1. Consider the closed-loop system of Fig. P6.2-1.

R(s)

+ -

T= 2s

D(z)

1 - ε-Ts s

0.5 s + 0.5

C(s)

Figure P6.2-1 System for Problem 6.2-1. (a) Calculate and plot the unit-step response at the sampling instants, for the case that D (z) = 1 . (b) Calculate the system unit-step response of the analog system, that is, with the sampler, digital controller, and data hold removed. Plot the response on the same graph with the results of part (a). (c) For the system of Fig. P6.2-1, let D (z) = 1 and T = 0.4 s . Calculate the unit-step response and plot these results on the same graph used for parts (a) and (b). (d) Use the system dc gains to calculate the steady-state responses for each of the systems of parts (a), (b), and (c). Why are these gains equal?

Solution: (a)

0.6321 z − 1 ⎡ 0.5 ⎤ z − 1 (1 − ε −1 ) = = ⎢ ⎥ z ⎣ s(s + 0.5) ⎦ z (z − 1)(z − ε −1 ) z − 0.3679

G(z) =

G( z ) 0.6321 = 1 + G( z ) z − 0.2642 C( z) 0.6321 0.5 −0.5 = = + z ( z − 1)( z + 0.2642) z − 1 z + 0.2642

∴ c(kT ) = 0.5[1 − (−0.2642)k ]

(b)

(c)

G p (s) 1 + G p (s)

=

0.5 s +1

C ( s) =

0.5 0.5 0.5 = − ⇒ c(t ) = 0.5(1 − ε −t ) s(s + 1) s s +1

G( z ) =

0.1813 0.1813 G( z ) , = z − 0.8187 1 + G( z) z − 0.6374

∴

C( z) 0.1813 0.5 0.5 = = + z ( z − 1)( z − 0.6374) z − 1 z − 0.6374

c(kT ) = 0.5[1 − (0.6374)k ] In (b), t = 0.4k, and c (t) t=0.4k = 0.5[1 − ε −0.4k ] = 0.5[1 − (0.6703) k ]

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C (a) (b) (c)

0.5

0.25

0

(d)

1

(a) T (1) = 0.5

2

3

(c)

T (1) = 0.5

4

t, s

(b) T (0) = 0.5 dc gain is a function of only Gp(0) and is independent of T. 6.2-2. Consider the system of Fig. P6.2-1, with D (z) = 1 . Use the results of Problem 6.2-1 if available. (a) Find the system time constant τ for T = 2 s . (b) With the input a step function, find the time required for the system output c(kT ) to reach 98 percent of its final value, for T = 2 s . Recall that four time constants (4τ) are required. (c) Repeat parts (a) and (b) for T = 0.4 s . (d) Repeat parts (a) and (b) for the analog system, that is, for the system with the sampler-datahold removed.

Solution: −T −2 = = 1.503s ln r ln(0.2642)

(a)

τ=

(b)

ε−t τ = 0.02 ∴ t = −(1.503) ln(0.02) = 5.88s ≅ 4τ

(c)

(a) τ =

(d)

(a) τ = 1 s;

−0.4 = 0.888 s; ln (0.6374)

(b)

t = −τ ln (0.02) = 3.47s ≈ 4τ

(b) t = −τ ln(0.02) = 3.91s ≈ 4τ

6.2-3. In Example 6.1 the response of a sampled-data system between sample instants was expressed as a sum of delayed step responses. (a) Use this procedure to find the system output y (t) of Fig. P6.2-1 at t = 1 s . (b) Repeat part (a) at t = 3 s . (c) The equation for c(t) in part (a) and that for c(t) in part (b) should give the same value at t = 2 s . Why? Hint: Consider the step response of the analog plant. (d) Show that the statement in part (c) is true.

Solution:

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(a) From Problem 6.2-1(a): G( z ) =

0.6321 z − 0.3678

From Example 6.2-1: C ( s ) = G ( s )

R( z ) 1 + G ( z ) z =εTs

z − 0.3678 1 = = 1 − 0.632z −1 + !

1 + G z + 0.2642

C1 ( s) =

0.5 1 1 = − s(s + 0.5) s s + 0.5

c1 (t) = (1 − ε −0.5t )

∴ c(t) = (1 − ε −0.5t )u(t) − 0.632(1 − ε −0.5(t−2) )u(t − 2) + ! ∴ c(t) t=1 = 1 − ε −0.5 = 0.3935

(b)

c(t) t=3 = 1 − ε −1.5 − 0.632 (1 − ε −0.5 ) = 0.5281

(c)

The output c(t) cannot change instantaneously for step inputs.

(d)

(a) c1 (2) = 1 − ε −1;

(b) c(2) = 1 − ε −1 − (0) = 1 − ε −1

6.2-4. Shown in Fig. P6.2-4 is the block diagram of a temperature control system for a large test chamber. This system is described in Problem 1.6-1. Ignore the disturbance input for this problem.

Chamber Disturbance Rd(s)

2.5 s + 0.5

Controller R(s)

+

D(z) -

T

1 - ε-Ts s

M(s) Volts

2 s + 0.5

+

-

C(s) 5C

Sensor Volts

0.04

Figure P6.2-4 Chamber temperature control system. (a) With D ( z ) = 1 and T = 0.6 s , evaluate and plot the system response if the input is to command a 10°C step in the output. Note that the system input must be a step function with an amplitude of 0.4 V. Why? (b) Use the results of part (a) to plot the output of the zero-order hold. (c) Solve for the steady-state output for part (a). (d) Suppose that the gain of 2 in the plant is replaced with a variable gain K. What value does the output approach in the steady state as K becomes very large? Assume that the system remains stable as K is increased (an unrealistic assumption).

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Solution: From Problem 4-21: G( z ) =

(a)

T ( z) =

1.037 z − 0.7408

1.037 1.037 G( z ) = = 1 + G( z ) H z − 0.7408 + 1.037(0.04) z − 0.700

C ( z ) = R( z )T ( z ) =

0.4 z 1.037 ⎡1.383 −1.383 ⎤ = z⎢ + ⎥ z − 1 z − 0.7 ⎣ z − 1 z − 0.7 ⎦

∴ c(kT ) = 1.383[1 − (0.7)k ]

r(t) = 0.4 V, since the sensor output = 0.4 with c(t) = 10°C. m(kT ) = 0.4[1 − (0.1) c(kT )]

(b)

k

c(kT )

m(kT )

0

0

0.4

1

0.416

0.384

2

0.704

0.372

4

1.05

0.360

8

1.30

0.348

16

1.38

0.345

∞

1.383

0.345

m

0.4

0.3

0

(c) css (kT ) = lim (z − 1)C(z) = z→1

(d) T ( z ) = T ( z) =

8T

12T

16T

t

0.415 = 1.383 z − 0.7 z=1

1.037 ( K2 ) G( z ) ; G( z ) 1 + G( z ) H z − 0.7408

0.519 K 0.519 K = z − 0.7408 + 0.519 K (0.04) z − 0.7408 + 0.0207 K

dc gain = T (1) = lim T (1) =

K →∞

4T

0.519K 0.259 + 0.0207 K

0.519 K = 25 0.0207 K

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∴ css (kT ) = (0.4)(25) = 10, ∴ no steady-state error.

6.2-5. Consider the temperature control system of Problem 6.2-4 and Fig. P6.2-4. (a) Let T = 6 s , and solve for the response to the input R(s) = 0.4 s−1 . Plot this response on the same graph with the response found in Problem 6.2-4. Note the effects of increasing T from 0.6 s to 6 s, when the plant has a time constant of 2 s. (b) Find c(t ) for 0 ≤ t ≤ 6 s . This response can be calculated without the use of the z-transform.

Solution: ⎡ ⎤ z −1 ⎛ z − 1⎞ 1 2z(1 − ε −3 ) 3.801 (2) (2) = = ⎢ ⎥ ⎟ −3 z ⎝ z ⎠ (z − 1)(z − ε ) z − 0.04979 ⎣ s(s + 0.5) ⎦

(a) G(z) = ⎜ ∴ T ( z) =

C ( z) =

3.801 3.801 G( z ) = = 1 + G( z ) H z − 0.04979 + 3.801(0.04) z + 0.1022

0.4 z 3.801 1.379 z −1.379 z = + z − 1 z + 0.1022 z − 1 z + 0.1022

∴ c(kT ) = 1.379[1 − (−0.1022)k ]

0

0

1

1.52

2

1.365 …

c(kT )

…

k

∞

1.379

c(t) 6-5

1.5 1.3 1.0 6-4 0.5

0

6T

12T

t

(b) at t = 0, m(t) = 1, and doesn’t change until t = 6. ∴ C (s) =

0.4 2 0.8 1.6 1.6 × = = − s s + 0.5 s( s + 0.5) s s + 0.5

∴ c(t) = 1.6(1 − ε −0.5t ), 0 ≤ t ≤ 6

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6.2-6. Consider the system of Fig. P6.2-4, with D ( z ) = 1 . Use the results of Problems 6.2-4 and 6.2-5 if available. (a) Find the system time constant τ for T = 0.6 s . (b) With the input a step function, find the time required for the system output c( kT ) to reach 98 percent of its final value. Note that this time is approximately four time constants (4τ) . (c) Repeat parts (a) and (b) for T = 6 s . (d) Repeat parts (a) and (b) for the analog system, that is, for the system with the sampler/data hold removed.

Solution: (a)

From Problem 6.2-4, T ( z ) =

∴τ =

(b)

1.037 z − 0.700

−T −0.6 = = 1.68 s ln r ln (0.7)

ε−t τ = 0.02 ⇒ t = −τ ln (0.02) = −(1.68) ln (0.02) = 6.57s ≈ 4τ

(c) From Problem 6.2-5, (a) T ( z ) = ∴τ =

1.90 z + 0.1022

−T −6 = = 2.63 s ln r ln (0.1022)

(b) t = −τ ln(0.02) = 10.3 s ≈ 4τ (d) T ( s ) = ∴τ =

G p ( s) 1 + G p (s) H

=

2 2 = s + 0.5 + 2(0.04) s + 0.58

1 = 1.72s ⇒ ε −t τ = 0.02, ∴ t = −(1.72) ln (0.02) = 6.73s ≈ 4τ 0.58

6.2-7. The block diagram of a control system of a joint in a robot arm is shown in Fig. P6.2-7. This system is discussed in Section 1.6. Let T = 0.1 s and D ( z ) = 1 .

Power amplifier

Controller

wc +

Zero-order hold

D (z) -

T

K

Servomotor Ea

200 s (0.5 s + 1)

wm

Gears 1 100

wa

Volts Sensor Volts

0.07

Figure P6.2-7 Robot arm joint control system.

153

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(a) Evaluate C ( z ) if the input is to command a 20°C step in the output and K = 10 . Note that the system input must be a step function with an amplitude 1.4 V. Why? (b) Assuming the system to be stable, find the steady-state system output. (c) Find the (approximate) time required for the system response to reach steady state. (d) Simulate the system to verify the results in parts (b) and (c).

Solution: (a) r(t) = 1.4 V, since the sensor output = 1.4 with c(t) = 20° ⎤ 4 (z − 1) ⎡ z(0.2 − 1 + ε −0.2 ) + (1 − ε −0.2 − 0.2ε −1.2 ) ⎤ ⎛ z − 1⎞ ⎡ 4 G(z) = ⎜ ⎢ ⎥ ⎥= ⎢ 2 ⎟ ⎝ z ⎠ ⎣ s (s + 2) ⎦ 2 z ⎣ 2(z − 1)2 (z − ε −0.2 ) ⎦

= T ( z) =

0.01873z + 0.01752 ( z − 1)( z − 0.8187)

10(0.01873z + 0.01752) KG( z ) = 2 1 + KG( z ) H z − 1.8187 z + 0.8187 + 0.01311z + 0.01226

=

0.1873z + 0.1752 z − 1.8056 z + 0.8310 2

1.4 z(0.1873z + 0.1752) ⎛ 1.4 z ⎞ C ( z) = ⎜ ⎟ T ( z) = ( z − 1)( z 2 − 1.8056 z + 0.8310) ⎝ z −1⎠

(b) css (kT ) = lim (z − 1)C(z) = z→1

1.4(0.1873 + 0.1752) = 20 1 − 1.8056 + 0.8310

(c) poles: z = 0.9028 ± j0.1262 = 0.9116 ∠ ± 7.96° ∴τ = −

T −0.1 = = 1.08 s; t ≈ 4τ = 4.32 s ln r ln (0.9116)

6.4-1. The block diagram of a control system of a joint in a robot arm is shown in Fig. P6.2-7. Let T = 0.1 s , K = 10 , and D ( z ) = 1 . The results of Problem 6.2-7 are useful in this problem if these results are available. (a) Find the damping ratio ζ , the natural frequency ω n , and the time constant τ of the open-loop system. If the system characteristic equation has two real zeros, find the two time constants.

These values can be solved by inspection. Why? (b) Repeat part (a) for the closed-loop system. (c) Repeat parts (a) and (b) for the system with the sampler, digital controller, and data hold removed, that is, for the analog system. (d) Use the results in parts (b) and (c) to find the percent overshoot in the step responses for the sampled-data closed-loop system and for the analog closed-loop system.

Solution: (a) The open-loop system has the characteristic of Gp(s). ∴ poles at: s1 = 0,

τ1 = ∞

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s = −2,

τ2 = 0.5 s

(b) From Problem 6.2-7, char. eq.: z2 –1.8056z + 0.8310 = 0 zeros: z = 0.9116 ∠± 7.96° = 0.9116 ∠± 0.139 rad −T −0.1 = = 1.08 s ln r ln (0.9116)

τ=

ωn = ζ=

1 1 1 ln 2 r + θ2 = [ln 2 (0.9116) + (0.139)2 ] 2 = 1.670 T 0.1

− ln r ln 2 r + θ2

= 0.554

(c) (a) same as (a) 4 (10) 2 s + 2s (b) T ( s) = = 1 + KG p ( s) H 1 + 0.07 40 s 2 + 2s KG p ( s)

=

40 40 = 2 s + 2s + 2.8 s + 2ζωn s + ωn 2 2

−2 ± 4 − 11.2 = −1 ± j1.3416 = 1.673∠ ± 126.7° 2

poles: s = ∴ τ = 1s 1

ωn = (2.8) 2 = 1.67

(d) (b) ζ = 0.554, − ζ π

ζ=

2 = 0.5988 2(1.67)

1 − ζ 2 = −(0.554) π 0.8325 = −2.091

∴ % overshoot = (ε−2.091 )100 = 12.4% 1 − ζ 2 = −2.349, % overshoot = (ε −2.349 )100 = 9.5%

(c) −ζπ

6.4-2. The block diagram of an attitude control system of a satellite is shown in Fig. P6.4-2. Let T = 1 s , K = 100 , J = 0.1 , H k = 0.02 , and D ( z ) = 1 .

Digital Controller R(s) r(t)

+

A/D

D(z)

Amplifier and thrusters D/A

-

M(s)

K

Satellite T(s)

Torque

1 Js2

®(s)

w(t)

Sensor Hk

Figure P6.4-2 Block diagram for a satellite control system.

155

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(a) Find the damping ratio ζ the natural frequency ω n , and the time constant τ of the open-loop system. If the system characteristic equation has two real zeros, find the two time constants.

These values can be solved by inspection. Why? (b) Repeat part (a) for the closed-loop system. (c) Repeat parts (a) and (b) for the system with the sampler, digital controller, and data hold removed, that is, for the analog system. (d) The closed-loop sampled-data system is seen to be unstable and the closed-loop analog system is seen to be marginally stable. If the satellite is operated with each of these control systems, describe the resulting movement for both the sampled-data system and the analog system.

Solution: (a)

⎛ z − 1⎞ (0.1)2 z(z + 1) 0.05(z + 1) z − 1 ⎡ 10 ⎤ = 10 = ⎜⎝ z ⎟⎠ z ⎢⎣ s3 ⎥⎦ (z − 1)2 2(z − 1)2

G(z) =

(s + a) has the time constant τ = 1 a . ∴ τ = ∞ at both poles.

The characteristics of the plant are not changed by sampling. (b)

T ( z) =

KG 5( z + 1) 5( z + 1) = 2 = 2 1 + KG( z ) H k z − 2 z + 1 + 0.1( z + 1) z − 1.9 z + 1.1

poles: z = 0.95 ± j 0.4444 = 1.049∠ ± 25.07 ∴ unstable (c)

(a) GP ( s) =

(b)

T (s) =

10 s2

∴ τ = ∞ at both poles

KG p ( s ) 1 + KG p ( s ) H k

=

1000 , poles: s = ± j 4.472 s 2 + 20

∴ marginally stable − τ = ∞ for both poles. (d)

sampled-data: u(t)

t

analog: u(t)

t

6.4-3. It is shown in Example 6.6 that the system of Fig. 6.2-2 has the parameters ζ = 0.250 , ω n = 0.9191 , and τ = 4.36 s .

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(a) Find ζ , ω n , and τ for the sample period T = 0.5 . (b) Repeat part (a) for T = 0.1 . (c) Repeat part (a) for the analog system, that is, the system with the sampler and data hold removed. Note that this case can be considered to be the limit as the sample period approaches zero. (d) Give a table listing the three parameters as a function of sampling frequency fs = 1 T . State the result of decreasing the sampling frequency on the parameters.

Solution: (a)

From Example 6.2: G ( z ) = T = 0.5, G( z ) =

(T − 1 + ε−T ) z + (1 − ε −T − T ε −T ) ( z − 1)( z − ε−T )

0.1065z + 0.09020 ( z − 1)( z − 0.6065)

∴ Char. eq.: z 2 − 1.6065z + 0.6065 + 0.1065z + 0.09020 = z 2 − 1.500z + 0.6967 = 0

zeros: z = 0.75 ± j0.3663 = 0.8347∠ ± 26.03° = 0.8347 ∠ ± 0.454 rad τ=−

T = 2.767 s ln r

ωn =

1 2 [ln r + θ2 ] = 0.9773 T

ζ=

(b)

− ln r 1

2

[ln r + θ2 ] 2

T = 0.1, G( z ) =

= 0.3698

0.004837 z + 0.004679 ( z − 1)( z − 0.9048)

∴ Char. eq.: z 2 − 1.9048z + 0.9048 + 0.004837z + 0.004679 = z 2 − 1.900z + 0.9095

zeros: z = 0.950 ± j 0.0837 = 0.9537 ∠ ± 5.033° = 0.9537 ∠ ± 0.0878 rad τ=

−T = 2.109s ln r

ωn = ζ=

(c)

1 2 [ln r + θ2 ] = 0.9978 T − ln r 1

[ln r + θ2 ] 2

T (s) =

2

G p ( s) 1 + G p ( s)

= 0.4751

=

1 1 = 2 s + s + 1 s + ζωn s + ωn 2 2

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=

∴τ =

1 ( s + 0.5)2 + 0.75

1 = 2s; ωn 2 = 1 ⇒ ωn = 1 0.5

2 ζωn = 1 ⇒ ζ =

1 = 0.5 (2)(1)

(d) I

τ

ωn

ζ

1

4.36s

0.9191

0.25

1

0.5

2.77

0.977

0.369

2

0.1

2.11

0.998

0.475

10

analog

2

1

0.5

→∞

fs = 1 T

as the sampling frequency decreases: τ increases, ωn increases, and ζ decreases 6.4-4. Consider the system of Fig. P6.4-4. This system is called a regulator control system, in which it is desired to maintain the output, c(t ) , at a value of zero in the presence of a disturbance, f (t ) . In this problem the disturbance is a unit step. f(t) = u(t)

r(t) = 0

+ -

T = 0.1 s

D(z)

1 - ε-Ts s

+

+

6 s+2

c(t)

Figure P6.4-4 System for Problem 6.4-4. (a) With D ( z ) = 1 (i.e., no compensation) find the steady-state value of c(t ) . (b) For f(t) to have no effect on the steady-state value of c( kT ) , D ( z ) should have a pole at z = 1 .

Let D( z) = 1 +

0.1z z −1

Determine the steady-state value of c( kT ) . (c) Repeat parts (a) and (b) with T = 1 s . (d) Why does the value of sample period T have no effect on the steady-state response for a constant input?

Solution: (a)

Can ignore sampling if all signals are constant.

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O

+

+

1 F(s) = s +

-

C (s) =

G p ( s) 1 + G p ( s)

∴ css (t) =

(b)

G(z) =

F (s) =

6 S+Z

C(s)

6 1 6 8 −6 8 · = + s+8 s s s+8

6 = 0.75 8

⎤ 0.5438 z −1 ⎡ 6 ⎤ z −1⎡ 6z(1 − ε −0.2 ) = = ⎢ ⎢ ⎥ −0.2 ⎥ z ⎣ s(s + 2) ⎦ z ⎣ (2)(z − 1)(z − ε ) ⎦ z − 0.8187

⎡ 6 ⎤ 0.5438z ⎢ s(s + 2) ⎥ = (z − 1)(z − 0.8187) = FG p (z) ⎣ ⎦

0.5438 z ( z − 1)( z − 0.8187) = C ( z) = 1 + D( z )G( z ) ⎛ 1.1z − 1 ⎞ ⎛ 0.5438 ⎞ 1+ ⎜ ⎟ ·⎜ ⎟ ⎝ z − 1 ⎠ ⎝ z − 0.8187 ⎠ FG p ( z )

=

0.5438z z − 1.221z + 0.2749 2

poles: z = 0.924, 0.298

∴ no poles at z = 1.

∴ css (kT ) = lim(z − 1) C(z) = 0 z→1

6.4-5. Consider the sampled-data systems with the following characteristic equations: (i) z − 0.999 = 0 (v) z 2 − 1.85z + 0.854 = 0 (ii) z − 0.99 = 0 (vi) z 2 − 1 = 0 (iii) z − 0.9 = 0 (vii) z 2 − 2z + 0.99 = 0 (viii) z 2 − 1.2z + 0.7 = 0 (iv) z + 0.9 = 0 (a) What can you determine about the system natural-response characteristics for each system, for T = 0.1 s ? (b) Repeat part (a) for T = 1 s . (c) For a given characteristic equation as a function of z, which parameters of the transient response vary with the sample period T, and which are independent of T? Solution: (a) (i) τ = (ii) τ =

−T −0.1 = = 99.95s, exponential ln r ln (0.999) −0.1 = 9.95s, exponential ln (0.99)

159

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(iii) τ =

−0.1 = 0.95s, exponential ln (0.9)

(iv) τ =

ω −0.1 π = 0.95s; ω = s = = 31.42 rad s ln (0.9) 2 0.1

damped sinusoid (v) zeros: z = 0.965, 0.885 exponential, τ1 =

0.1 0.1 = 2.81s; τ2 = = 0.819s ln (0.965) ln (0.885)

(vi) zeros: z = 1, –1

z = 1 − constant output

z = −1 − sinusoidal output, ω = ωs 2 = 31.42 rad s

(vii) zeros: z = 1.01, 0.99

∴ unstable

(viii) zeros: z = 0.837 ∠± 44.2°, 0.837 ∠ ± 0.771 rad ∴ damped sinusoid τ=

−T = 0.56 s ln r

ωn =

− ln r 2

1

[ln r + θ2 ] 2

= 0.156

1 1 2 [ln r + θ2 ] 2 = 11.38 rad s T

(b) (i) τ = 999.5 s

(c)

ζ=

(ii) τ = 99.5 s

(iii) τ = 9.5 s

(iv) τ = 9.5 s, ω = 3.142 rad/s

(v) τ1 = 28.1s, τ2 = 8.19s

(vi) ω = 3.142 rad/s

(vii) unstable

(viii) τ = 5.6s , ωn = 1.138 rad/s, ζ = 0.156

τ , ωn vary with T, ζ independent of T.

6.4-6. Consider the system of Fig. P6.2-1. Suppose that an ideal time delay of 0.2 s is added to the plant, such that the plant transfer function is now given by G p ( s) =

0.5ε−0.2T s + 0.5

(a) Find the time constant τ of the system if the time delay is omitted. (b) Find the time constant τ of the system if the time delay is included. (c) Repeat part (b) for a time delay of 1 s. (d) What is the effect on the speed of response of the closed-loop system of adding time delay to the plant?

Solution: (a) From Problem 6.2-1, T ( z ) =

0.6321 z + 0.2642

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−T −1 = = 0.751s ln r ln (0.2642)

∴τ =

(b)

G(z) =

=

z − 1 ⎡ 0.5ε −0.2Ts ⎤ z − 1 ⎡ 0.5 ⎤ ⎢ ⎥= z m ⎢⎣ s(s + 0.5) ⎥⎦ m=0.8 z ⎣ s(s + 0.5) ⎦

ε−0.8 ⎤ 0.5507 z + 0.0814 z −1 ⎡ 1 − ⎢ ⎥= z ⎣ z − 1 z − ε−1 ⎦ z ( z − 0.3679)

∴ char. eq.: 1 + G(z) = 0 = z2 – 0.3679z + 0.5507z + 0.0814 = z2 + 0.1828z + 0.0814 = 0 zeros: z = 0.2853 ∠ ± 108.7° ∴τ =

−T −1 = = 0.7973 ln r ln (0.2853)

(c) From Problem 6.2-1, G( z ) =

0.6321 z ( z − 0.3679)

char. eq.: z2 – 0.3679z + 0.6321 = 0 zeros: z = 0.7950 ∠ ± 76.6° τ=

−T = 4.36 s ln r

(d) Delay slows the system response.

6.5-1. (a) Give the system type for the following systems, with D ( z ) = 1 . It is not necessary to find the pulse transfer functions to find the system type. Why? (i) Fig. P6.2-1. (ii) Fig. P6.2-4. (iii) Fig. P6.2-7. (iv) Fig. P6.4-2. (v) Fig. P6.4-4. (b) It is desired that the systems of part (a) have zero steady-state error for a constant input. Give the required characteristics for each digital controller. (c) It is desired that the systems of part (a) have zero steady-state error for a ramp input. Give the required characteristics for each digital controller.

Solution: (a) (i) type 0

(ii) type 0

(iv) type 2

(v) type 0

(iii) type 1

(b) (i), (ii), (v) one pole at z = 1, with the system stable. (iii), (iv) system stable only (c) (i), (ii), (v) 2 poles at z = 1, with the system stable. (iii) 1 pole at z = 1, with the system stable.

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(iv) system stable only.

6.5-2. Consider the system of Fig. P6.5-2. The digital filter is described by m( kT ) = e( kT ) − 0.9e ⎡⎢( k − 1)T ⎤⎥ + m ⎡⎢( k − 1)T ⎤⎥ ⎣ ⎦ ⎣ ⎦ Gp(s) R(s)

+

D(z)

T= 1s

-

1-ε s

-Ts

M(s)

1 s+1

C(s)

Figure P6.5-2 System for Problem 6.5-2. (a) Find the system type. (b) Find the steady-state response for a unit-step input, without finding C ( z ) . (c) Find the approximate time for the system to reach steady state. (d) Find the unit-step response for the system and verify parts (b) and (c).

Solution: (a)

(1 − z −1 )M ( z) = (1 − 0.9z −1 ) E( z) D( z ) =

M ( z ) z − 0.9 = E( z) z −1

∴ system is type 1

(b) Type 1 system, ∴ no steady-state error, ∴ css (kT ) = 1 (c)

G(z) =

0.6321 z −1 ⎡ 1 ⎤ z −1 z(1 − ε −1 ) = = ⎥ ⎢ z (z − 1)(z − ε −1 ) z − 0.3679 z ⎣ s + 1⎦

D( z )G( z) 0.6321( z − 0.9) = 1 + D( z )G( z) z 2 − 1.368z + 0.6379 + 0.6321z − 0.5689

=

0.6321( z − 0.9) z − 0.7358z − 0.201 2

∴ poles: z = 0.9479, − 0.2121 τ1 =

−T −1 = = 18.7s, τ2 = 0.645s ln r ln (0.9479)

∴ 4τ1 = 74.8 s (d)

C ( z) 0.6321( z − 0.9) 1 −0.500 −0.500 = = + + z ( z − 1)( z − 0.9479)( z + 0.2121) z − 1 z − 0.9479 z + 0.2121

∴ c(kT ) = 1 − (0.9479)k − 0.5(−0.2121)k

(b) css (kT ) = 1

(c) c(75) = 1 − .5(0.9479)75 − 5(−0.2121)75 = 1 − 0.0090 − ( ) ×10−51 ≈ 1

162

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6.5-3. Consider the system of Fig. P6.5-2, with the digital filter as described in Problem 6.5-2. (a) Find the system type. (b) Find the steady-state response for a unit ramp input, without finding C ( z ) . (c) Find the approximate time for the system to reach steady state. (d) Calculate the steady-state unit-ramp response for the system to verify parts (b) and (c).

Solution: (a) From Problem 6.5-2, type 1 (b)

Kv = lim

1

z →1 T

∴ ess =

( z − 1) D( z) G( z ) =

0.6321(1 − 0.9) = 0.1 (1 − 0.3679)

1 = 10 Kv

∴ ess (kT ) = rss (kT ) − Css (kT ) ∴ css (kT ) = rss (kT ) − ess (kT ) = kT − 10

(c) From Problem 6.5-2, time to reach steady-state ≈ 74.8 s (d)

[t] =

Tz z = 2 (z − 1) (z − 1)2

From Problem 6.5-2: C ( z) 0.6321( z − 0.9) = 2 z ( z − 1) ( z − 0.9499)( z + 0.2121) =

1 −10.1 9.60 0.413 + + + 2 z − 1 z − 0.9479 z + 0.2121 ( z − 1)

∴ css (kT ) = kT − 10.1 ≈ kT − 10 (numerical errors)

ess (kT ) = kT − Css (kT ) ≈ 10

6.5-4. Consider the system of Problem 6.5-2. Let the plant transfer function be given by G p ( s) =

s s +1

Hence the dc gain of the plant is zero. The digital filter is as given in Problem 6.5-2. (a) Find the system type. (b) Find the steady-state response for a unit-step input, without finding C ( z ) . (c) Find the approximate time for the system to reach steady state. (d) Find c( kT ) and verify the results of parts (b) and (c).

Solution:

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G(z) =

(a)

z −1 ⎡ 1 ⎤ z −1 z z −1 = = ⎥ ⎢ −1 z z−ε z ⎣ s + 1⎦ z − 0.3679

∴ D( z )G( z ) =

(b)

D(1)G(1) = T (1) =

z − 0.9 z − 0.3679

∴ system type = 0

1 − 0.9 = 0.158 1 − 0.3679

0.158 = 0.1364 = Css (kT ) 1 + 0.158

(c) 1 + G( z) = z − 0.3679 + z − 0.9 = 2( z − 0.634) = 0 ∴τ =

−T −1 = = 2.19s, ∴ 4τ = 8.72s ln r ln (0.634)

(d)

D( z )G( z ) z − 0.9 z − 0.9 0.5 z − 0.45 = = = 1 + D( z )G( z) z − 0.3679 + z − 0.9 2 z − 1.2679 z − 0.634

∴

C( z) 0.5 z − 0.45 0.1366 0.3634 = = + z ( z − 1)( z − 0.634) z −1 z − 0.634

∴ c(kT ) = 0.1366 + 0.3634(0.634) k

css (kT ) = 0.1364

6.6-1. Consider the system of Fig. P6.5-2, with D ( z ) = 1 . (a) With the sampler and zero-order hold removed, write the system differential equation. (b) Using the rectangular rule for numerical integration with the numerical integration increment equal to 0.25 s, evaluate the unit-step response for 0 ≤ t ≤ 1.5 s . (c) Repeat part (b) with the sampler and zero-order hold in the system, and with a sample period T = 0.5 s . (d) Solve for the exact unit-step responses for parts (b) and (c), and compare results.

Solution: (a)

G p (s) 1 + G p ( s)

=

1 1 C ( s) = = s + 1 + 1 s + 2 R( s)

(s + 2)C(s) = R(s) ! + 2c(t) = r(t), r(t) = 1 c(t) ! = 1 − 2c(t), t ≥ 0 c(t)

(b) H = 0.25s, c(0) = 0 c(kH ) = c[(k − 1)H ] + H {1 − 2* c[(k − 1) H ]}

= c[(k − 1) H ](1 − 2H ) + H = 0.5c [(k − 1) H ] + 0.25 kH

c(kH )

c(t )[exact , (d )]

164

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(c)

0

0

0

0.25

0.25

0.197

0.50

0.375

0.316

0.75

0.4375

0.388

1.00

0.4688

0.432

1.25

0.4844

0.459

1.50

0.4922

0.475

C(s) 1 ! = m(t) − c(t) = ⇒ c! + c = m; c(t) M (s) s + 1

m = 1− C

c(kH ) = c[(k − 1)H ] + H {m[(k − 1)H ]} − c[(k − 1)H ]

= [1 − H ]c[(k − 1) H ] + H m[(k − 1) H ] = 0.75 c[(k − 1) H ] + 0.25m[(k − 1) H ] kH

c(kH )

m (kH )

0

0

1

0.25

0.25

1

0.50

0.4375

0.5625

0.75

0.4688

0.5625

1.00

0.4922

0.5078

1.25

0.4961

0.5078

1.50

0.4990

(d) (b) C (s) = (c) G(z) =

c (kH ) [exact (d )]

0 0.3935 0.477 0.4952

1 1 1 = 2 − 2 ⇒ c(t) = 12 (1 − ε−2t ) s(s + 2) s s + 2

z −1 ⎡ 1 ⎤ 0.3935 = ⎢ ⎥ z ⎣ s(s + 1) ⎦ z − 0.6065

G( z ) 0.3935 = 1 + G( z ) z − 0.2130 ∴ C( z) =

0.3935 z 0.50 z −0.50 z = + ( z − 1)( z − 0.2130) z − 1 z − 0.2130

∴ c(kT ) = 0.5[1 − (0.2130) k ]

6.6-2. Consider the numerical integration of the differential equation (6-24) in Section 6.6. Applying the rectangular rule results in the difference equation (6-25), which can be expressed as x ⎡⎢( k + 1) H ⎤⎥ = x ( kH ) + H ⎡⎢−x ( kH )⎤⎥ ⎣ ⎦ ⎣ ⎦

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(a) Use the z-transform to solve this equation as a function of x (0) and H. (b) Use the results of part (a) to show that as given in Section 6.6, x (1.0) = 0.3487 with x (0) = 1 and H = 0.1 . (c) Solve for x (1.0) with x (0) = 1 and H = 0.01 . (d) The exact value of x (1.0) is given in Section 6.6. Compare the errors in the results of parts (b) and (c).

Solution: (a)

zX ( z) − zx(0) = X ( z) − HX ( z) ∴[ z − (1 − H )] X ( z) = zx(0) X ( z) =

zx(0) ⇒ x(kH ) = x(0) (1 − H ) k z − (1 − H )

(b)

x(1) = (1)(0.9)10 = 0.3487

(c)

x(1) = (1)(0.99)100 = 0.3660 ⎛ 0.3678 − 0.3487 ⎞ ⎟100 = 5.19% 0.3678 ⎝ ⎠

(d) (b) % error = ⎜

⎛ 0.3678 − 0.3660 ⎞ ⎟100 = 0.489% 0.3678 ⎝ ⎠

(c) % error = ⎜

6.6-3. Consider the numerical integration of the differential equation dx (t ) dt

+ x ( t ) = r (t )

using the rectangular rule. (a) Develop the difference equation, as in (6-25), for the numerical integration of this differential equation. (b) Let r (t ) = 1 , x (0) = 0 , and H = 0.1 . Use the z-transform to solve the difference equation of part (a) for x (1.0) . (c) Repeat part (b) for H = 0.01 . (d) Solve the given differential equation, using the Laplace transform, for the exact value of x (1.0) . (e) Compare the errors in the results in parts (b) and (c).

Solution: ! = −x(t) + r(t) (a) x(t) x(kH ) = x[(k − 1)H ] + H!x[(k − 1)H ] = x[(k − 1)H ] + H[−x(k − 1)H ] + r[(k − 1)H ]

∴ X (kH ) = (1 − H ) x[(k − 1)H ] + r[(k − 1)H ]

(b) X ( z) = (1 − H ) z −1 X ( z) + H z −1R( z)

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∴ X ( z) = ∴

H H z · R( z ) = z − (1 − H ) z − (1 − H ) z − 1

X ( z) H z 1 = = + z ( z − 1)[ z − (1 − H )] z − 1 z − (1 − H )

∴ x(kH ) = 1 − (1 − H )k ∴ x(1) = 1 − (0.9)10 = 0.6513

(c) x(1) = 1 − (0.99)100 = 0.6340 1 s

(d) sX ( s) + X ( s) = X ( s) =

1 1 −1 = + ⇒ x(t ) = 1 − ε−t s(s + 1) s s + 1

∴ x(1) = 0.6321

(e) (b) % error = (c) % error =

0.6321 − 0.6513 (100) = 3.037% 0.6321

0.6321 − 0.6340 (100) = 0.301% 0.6321

6.6-4. Consider the numerical integration of the differential equation dx (t ) dt

+ x (t ) = 0

using the predictor-corrector of Section 6.6. The predictor method is the rectangular rule, and the corrector method is the trapezoidal rule. (a) Develop the difference equation, using (6-32) through (6-35), for the numerical integration of the given differential equation. The result should be one difference equation for x ⎡⎢⎣( k + 1) H ⎤⎥⎦ as

a function of x ( kH ) .

(b) Let x (0) = 1 , and H = 0.1 . Use the z-transform to solve the difference equation of part (a) for x (1.0) . This value is given as 0.3685 in Example 6.9.

(c) Repeat part (b) for H = 0.33333 . This value is given as 0.3767 in Example 6.9. (d) Solve the given differential equation, using the Laplace transform, for the exact value of x (1.0) . (e) Give the errors in the results in parts (b) and (c).

Solution: ! = −x(t) ⇒ A = −1, B = 0 x(t)

(a)

! − 1) = −x(k − 1) (6-32) x(k !

(6-33) x(k) = x(k − 1) − HX (k − 1) = (1 − H )x(k − 1) ! = −x(k) = −(1 − H )x(k − 1) (6-34) x(k)

167

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(6-35) x(k ) = x(k − 1) +

H [− x(k − 1) − (1 − H ) x(k − 1)] 2

⎡ H H H2 ⎤ ⎡ H2 ⎤ = ⎢1 − − + ⎥ x(k − 1) = ⎢1 − H + ⎥ x(k − 1) 2 2 2 ⎦ 2 ⎦ ⎣ ⎣ ⎛ H2 ⎞ x(k) = ax(k) ∴ x(k + 1) = ⎜ 1 − H + 2 ⎟⎠ ⎝

zx(0) z−a

(b) zX ( z ) − zx(0) = aX ( z ) ⇒ X ( z ) = k

⎛ H2 ⎞ ∴ x(kH ) = a x(0) = ⎜1 − H + ⎟ x(0) 2 ⎠ ⎝ k

∴ x(1) = (0.905)10 (1) = 0.3685

(c)

1 − 0.33333 +

(0.33333)2 = 0.7222 2

x(1) = (0.7222)3 (1) = 0.3767

(d)

sX (s) − x(0) + X (s) = 0 X ( s) =

x(0) ⇒ x(t ) = x(0)ε−t s +1

∴ x(1) = 0.3679

(e) (b) % error =

0.3679 − 0.3685 (100) = 0.163% 0.3679

(c) % error =

0.3679 − 0.3767 (100) = 2.39% 0.3679

6.6-5. Consider the numerical integration of the differential equation dx (t ) dt

+ x ( t ) = r (t )

using the predictor-corrector of Section 6.6. The predictor method is the rectangular rule, and the corrector method is the trapezoidal rule. (a) Develop the difference equation, using (6-32) through (6-35), for the numerical integration of the given differential equation. The result should be one difference equation for x ( kH ) as a function of x ⎡⎢⎣( k − 1) H ⎤⎥⎦ and the input function r ⎡⎢⎣( k − 1) H ⎤⎥⎦ .

(b) Let r (t ) = 1 , x (0) = 0 , and H = 0.1 . Use the z-transform to solve the difference equation of part (a) for x (1.0) . (c) Repeat part (b) for H = 0.33333 . (d) Solve the given differential equation, using the Laplace transform, for the exact value of x (1.0) . (e) Give the errors in the results in parts (b) and (c).

168

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Solution: ! = −x(t) + r(t), ∴ A = −1, B = 1 x(t)

(a)

! − 1) = −x(k − 1) + r(k − 1) (6-32) x(k

(6-33) x(k ) = x(k − 1) + H [− x(k − 1) + r (k − 1)] = (1 − H ) x(k − 1) + H r (k − 1) ! = (−1)x(k) + r(k) (6-34) x(k)

= −(1 − H ) x(k − 1) − Hr (k − 1) + r (k )

(6-35) x(k ) = x(k − 1) +

H [− x(k − 1) + r(k − 1) − (1 − H ) x(k − 1) − Hr (k − 1) + r (k )] 2

⎡ H H H2 ⎤ ⎛ H H2 ⎞ H ∴ x(k ) = ⎢1 − − + ⎟ r (k − 1) + r (k ) ⎥ x(k − 1) + ⎜ − 2 2 2 ⎦ 2 ⎠ 2 ⎣ ⎝2

= ax(k − 1) + cr (k − 1) + br (k )

(b) (1 − az −1 ) X ( z) = (b + cz −1 ) R( z) ∴ X ( z) =

bz +c R( z ) z−a

a = 1 − 0.1 + ∴

(0.1)2 0.1 0.1 (0.1)2 = 0.905; b = = 0.05, c = − = 0.045 2 2 2 2

1 X ( z ) 0.05 z + 0.045 1 −0.95 · = = + z z − 0.905 z − 1 z − 1 z − 0.905

∴ x(k ) = 1 − 0.95(0.905)k x(1) = 1 − 0.95(0.905)10 = 0.650

(c) a = 0.7222, b = 0.16667, c = 0.1111 ∴

X ( z ) 0.16667 z + 0.1111 1 −0.8332 = = + z ( z − 1)( z − 0.7222) z − 1 z − 0.7222

∴ x(k ) = 1 − 0.8332(0.7222)k x(1) = 1 − (0.8332)(0.7222)3 = 0.6862

(d) See Problem 6.6-3 (d), x(1) = 0.6321 (e) (b) % error =

0.6321 − 0.650 (100) = 2.83% 0.6321

(c) % error =

0.6321 − 0.6862 (100) = 8.56% 0.6321

6.6-6. Repeat Problem 6.6-1, but with the process transfer function given by 169

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G ( s) =

20 s(3s + 1)

Solve for the unit-step responses for 0 ≤ t ≤ 1.0s .

Solution: 1

U

6.667

X2

S-1

-0.3333

X1

S-1

-1

⎡ ⎤ ⎡ 0 ⎤ 0 1 x(t) ! =⎢ ⎥ x(t) + ⎢ ⎥ u(t) ⎣ −6.667 −0.3333 ⎦ ⎣ 6.67 ⎦

x(k) = x(k − 1) + H!x(k − 1) x(k) ! = A x(k − 1) + Bu(k − 1)

MATLAB: A = [0 1; –6.667 –0.3333]; B = [0; 6.667]; u = 1; H = 0.25; x = [0; 0]; for k = 0:4; t = k*H; [t, x(1)] xd = A*x + B*u; x = x + H*xd; end

results:

time

output

H has been chosen too large.

0

0

0.25

0

The system is stable, but the simulation is unstable.

0.5

0.4167

0.75

1.2153

1.00

2.1905

170