《自动控制原理》试卷及答案(英文10套)

⒈ Solution: The advantages of open-loop control systems are as follows: ① Simple construction and ease of maintenance

② Less expensive than a corresponding closed-loop system ③ There is no stability problem

④ Convenient when output is hard to measure or economically not feasible. (For example, it would be quite expensive to provide a device to measure the quality of the output of a toaster.)

The disadvantages of open-loop control systems are as follows:

① Disturbances and changes in calibration cause errors, and the output may be different from what is desired.

② To maintain the required quality in the output, recalibration is necessary from time to time.

U2(S)R1C1R2C2S2?(R1C1?R2C2)S?1⒉ ?2U1(S)R1C1R2C2S?(R1C1?R2C2?R1C2)S?1 ⒊

G1G2G3G4?G1G5C(S)? R(S)1?G1G2H1?G2G3H2?G1G2G3G4H3?G1G5H3G3G4(1?G1G2H1)?G3G5H2C(S)? P(S)1?G1G2H1?G2G3H2?G1G2G3G4H3?G1G5H3

⒋ R=2, L=1

⒌ S:①the breakaway point is –1 and –1/3; k=4/27 ② The imaginary axis S=±j; K=2

⒍3.5?K?7.5

AUTOMATIC CONTROL THEOREM (4)

⒈ Find the poles of the following F(s):

F(s)?

1 (12%) 1?e?s⒉Consider the system shown in Fig.1,where??0.6 and ?n?5rad/sec. Obtain the rise timetr, peak timetp, maximum overshootMP, and settling timets when the system is subjected to a unit-step input. （10%）

2 ?n C(s) s(s?2??n)R(s) Fig.1

⒊ Consider the system shown in Fig.2. Obtain the closed-loop transfer function

C(S)E(S)C(S), , . （12%） R(S)R(S)P(S)G5 E R P C G1 G2 G3 G4 H2 Fig.2 H1 H3

⒋ The characteristic equation is given 1?GH(S)?S3?3S2?2S?20?0. Discuss the distribution of the closed-loop poles. (16%)

5. Sketch the root-locus plot for the system GH(S)?K. (The gain K is

S(S?1)assumed to be positive.)

⑦ Determine the breakaway point and K value.

⑧ Determine the value of K at which root loci cross the imaginary axis. ⑨ Discuss the stability. (12%)

6. The system block diagram is shown Fig.3. G1?4K, G2?. Suppose

S(S?3)S?2r?(2?t), n?1. Determine the value of K to ensure eSS?1. (12%)

N

R E C G1 G2

Fig.3

7. Consider the system with the following open-loop transfer function:

GH(S)?K. ① Draw Nyquist diagrams. ② Determine the

S(T1S?1)(T2S?1)stability of the system for two cases, ⑴ the gain K is small, ⑵ K is large. (12%)

8. Sketch the Bode diagram of the system shown in Fig.4. (14%)

R(S) C(S) (S?2)S?2

S3(S?5)(S?10)

Fig.4

⒈ Solution: The poles are found from e?s?1 or e?(???j)?e??(cos??jsin?)?1 From this it follows that ??0,???2n? (n?0,1,2,?). Thus, the poles are located at s??j2n?

⒉Solution: rise timetr?0.55sec, peak timetp?0.785sec, maximum overshootMP?0.095,

and settling timets?1.33sec for the2% criterion, settling timets?1sec for the5% criterion. ⒊

G1G2G3G4?G1G5C(S)? R(S)1?G1G2H1?G2G3H2?G1G2G3G4H3?G1G5H3G3G4(1?G1G2H1)?G3G5H2C(S)? P(S)1?G1G2H1?G2G3H2?G1G2G3G4H3?G1G5H3

⒋R=2, L=1

5. S:①the breakaway point is –1 and –1/3; k=4/27 ② The imaginary axis S=±j; K=2

⒍3.5?K?7.5