自动控制原理 英文版共61页
自控原理课件3(英文版)
r(t) 1 t 2 1(t)
1 t2 2
t 0
2
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t 0
r (t )
L{1 2
t2
1(t)}
1 s3
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3-1. Introduction
Function imagine
t
t
0
Original function of Time-domain
image
relation
Image function
Unit-impulse response of first-order system
For the unit-impulse input, R(s)=1, so
(s) C(s) 1 R(s) Ts 1
C(s) 1 Ts 1
c(t) 1 et /T , for t 0 T
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3-2 Time response of first-order system
Figure 5-4 Unit-impulse response of the system shown
in Figure 5–1(a).
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3-2 Time response of first-order system
r(t)§R3(s).2C.3(s)=一F(阶s) R系(s)统的典c型(t) 响应
Given an absolute stable system, we will discuss further the relative stability of the system.
If the output of a system at steady state does not exactly agree with the input, the system is said to have steady-state error. This error is indicative of accuracy of the system.
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AUTOMATIC CONTROL THEOREM (1)⒈ Derive the transfer function and the differential equation of the electric network⒉ Consider the system shown in Fig.2. Obtain the closed-loop transfer function)()(S R S C , )()(S R S E . （12%） ⒊ The characteristic equation is given 010)6(5)(123=++++=+K S K S S S GH . Discuss the distribution of the closed-loop poles. (16%)① There are 3 roots on the LHP ② There are 2 roots on the LHP② There are 1 roots on the LHP ④ There are no roots on the LHP . K=?⒋ Consider a unity-feedback control system whose open-loop transfer function is )6.0(14.0)(++=S S S S G . Obtain the response to a unit-step input. What is the rise time for this system? What is the maximum overshoot? （10%）Fig.15. Sketch the root-locus plot for the system )1()(+=S S K S GH . ( The gain K is assumed to be positive.)① Determine the breakaway point and K value.② Determine the value of K at which root loci cross the imaginary axis.③ Discuss the stability. (12%)6. The system block diagram is shown Fig.3. Suppose )2(t r +=, 1=n . Determine the value of K to ensure 1≤e . (12%)Fig.37. Consider the system with the following open-loop transfer function:)1)(1()(21++=S T S T S K S GH . ① Draw Nyquist diagrams. ② Determine the 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%)⒈212121121212)()()(C C S C C R R C S C C R S V S V ++++=⒉ 2423241321121413211)()(H G H G G G G G G G H G G G G G G G S R S C ++++++=⒊ ① 0<K<6 ② K ≤0 ③ K ≥6 ④ no answer⒋⒌①the breakaway point is –1 and –1/3; k=4/27 ② The imaginary axis S=±j; K=2③⒍5.75.3≤≤K⒎ )154.82)(181.34)(1481.3)(1316.0()11.0(62.31)(+++++=S S S S S S GHAUTOMATIC CONTROL THEOREM (2)⒈Derive the transfer function and the differential equation of the electric network⒉ Consider the equation group shown in Equation.1. Draw block diagram and obtain the closed-loop transfer function )()(S R S C . （16% ） Equation.1 ⎪⎪⎩⎪⎪⎨⎧=-=-=--=)()()()()]()()([)()]()()()[()()()]()()[()()()(3435233612287111S X S G S C S G S G S C S X S X S X S G S X S G S X S C S G S G S G S R S G S X⒊ Use Routh ’s criterion to determine the number of roots in the right-half S plane for the equation 0400600226283)(12345=+++++=+S S S S S S GH . Analyze stability.（12% ）⒋ Determine the range of K value ,when )1(2t t r ++=, 5.0≤SS e . （12% ）Fig.1⒌Fig.3 shows a unity-feedback control system. By sketching the Nyquist diagram of the system, determine the maximum value of K consistent with stability, and check the result using Routh ’s criterion. Sketch the root-locus for the system （20%）（18% ）⒎ Determine the transfer function. Assume a minimum-phase transfer function.（10% ）⒈1)(1)()(2122112221112++++=S C R C R C R S C R C R S V S V⒉ )(1)()(8743215436324321G G G G G G G G G G G G G G G G S R S C -+++=⒊ There are 4 roots in the left-half S plane, 2 roots on the imaginary axes, 0 root in the RSP. The system is unstable.⒋ 208<≤K⒌ K=20⒍⒎ )154.82)(181.34)(1481.3)(1316.0()11.0(62.31)(+++++=S S S S S S GHAUTOMATIC CONTROL THEOREM (3)⒈List the major advantages and disadvantages of open-loop control systems. (12% )⒉Derive the transfer function and the differential equation of the electric network⒊ Consider the system shown in Fig.2. Obtain the closed-loop transfer function)()(S R S C , )()(S R S E , )()(S P S C . （12%）⒋ The characteristic equation is given 02023)(123=+++=+S S S S GH . Discuss the distribution of the closed-loop poles. (16%)5. Sketch the root-locus plot for the system )1()(+=S S K S GH . (The gain K is assumed to be positive.)④ Determine the breakaway point and K value.⑤ Determine the value of K at which root loci cross the imaginary axis. ⑥ Discuss the stability. (14%)6. The system block diagram is shown Fig.3. 21+=S K G , )3(42+=S S G . Suppose )2(t r +=, 1=n . Determine the value of K to ensure 1≤SS e . (15%)7. Consider the system with the following open-loop transfer function:)1)(1()(21++=S T S T S K S GH . ① Draw Nyquist diagrams. ② Determine the stability of the system for two cases, ⑴ the gain K is small, ⑵ K is large. (15%)⒈ Solution: The advantages of open-loop control systems are as follows: ① Simple construction and ease of maintenance② Less expensive than a corresponding closed-loop system③ There is no stability problem④ 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.⒉ 1)(1)()()(2122112221122112221112+++++++=S C R C R C R S C R C R S C R C R S C R C R S U S U ⒊351343212321215143211)()(H G G H G G G G H G G H G G G G G G G G S R S C +++++= 35134321232121253121431)1()()(H G G H G G G G H G G H G G H G G H G G G G S P S C ++++-+=⒋ R=2, L=1⒌ S:①the breakaway point is –1 and –1/3; k=4/27 ② The imaginary axis S=±j; K=2⒍5.75.3≤≤KAUTOMATIC CONTROL THEOREM (4)⒈ Find the poles of the following )(s F :se s F --=11)( (12%)⒉Consider the system shown in Fig.1,where 6.0=ξ and 5=n ωrad/sec. Obtain the rise time r t , peak time p t , maximum overshoot P M , and settling time s t when the system is subjected to a unit-step input. （10%）⒊ Consider the system shown in Fig.2. Obtain the closed-loop transfer function)()(S R S C , )()(S R S E , )()(S P S C . （12%）⒋ The characteristic equation is given 02023)(123=+++=+S S S S GH . Discuss the distribution of the closed-loop poles. (16%)5. Sketch the root-locus plot for the system )1()(+=S S K S GH . (The gain K is 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. 21+=S K G , )3(42+=S S G . Suppose )2(t r +=, 1=n . Determine the value of K to ensure 1≤SS e . (12%)7. Consider the system with the following open-loop transfer function:)1)(1()(21++=S T S T S K S GH . ① Draw Nyquist diagrams. ② Determine the 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%)⒈ Solution: The poles are found from 1=-s e or 1)sin (cos )(=-=-+-ωωσωσj e e j From this it follows that πωσn 2,0±== ),2,1,0(K =n . Thus, the poles are located at πn j s 2±=⒉Solution: rise time sec 55.0=r t , peak time sec 785.0=p t ,maximum overshoot 095.0=P M ,and settling time sec 33.1=s t for the %2 criterion, settling time sec 1=s t for the %5 criterion.⒊ 351343212321215143211)()(H G G H G G G G H G G H G G G G G G G G S R S C +++++= 35134321232121253121431)1()()(H G G H G G G G H G G H G G H G G H G G G G S P S C ++++-+=⒋R=2, L=15. S:①the breakaway point is –1 and –1/3; k=4/27 ② The imaginary axis S=±j; K=2⒍5.75.3≤≤KAUTOMATIC CONTROL THEOREM (5)⒈ Consider the system shown in Fig.1. Obtain the closed-loop transfer function )()(S R S C , )()(S R S E . （18%）⒉ The characteristic equation is given 0483224123)(12345=+++++=+S S S S S S GH . Discuss the distribution of the closed-loop poles. (16%)⒊ Sketch the root-locus plot for the system )15.0)(1()(++=S S S K S GH . (The gain K is assumed to be positive.)① Determine the breakaway point and K value.② Determine the value of K at which root loci cross the imaginary axis. ③ Discuss the stability. (18%)⒋ The system block diagram is shown Fig.2. 1111+=S T K G , 1222+=S T K G . ①Suppose 0=r , 1=n . Determine the value of SS e . ②Suppose 1=r , 1=n . Determine the value of SS e . (14%)⒌ Sketch the Bode diagram for the following transfer function. )1()(Ts s K s GH +=, 7=K , 087.0=T . (10%)⒍ A system with the open-loop transfer function )1()(2+=TS s K S GH is inherently unstable. This system can be stabilized by adding derivative control. Sketch the polar plots for the open-loop transfer function with and without derivative control. (14%)⒎ Draw the block diagram and determine the transfer function. (10%)⒈∆=321)()(G G G S R S C ⒉R=0, L=3,I=2⒋①2121K K K e ss +-=②21211K K K e ss +-= ⒎11)()(12+=RCs s U s UAUTOMATIC CONTROL THEOREM (6)⒈ Consider the system shown in Fig.1. Obtain the closed-loop transfer function )()(S R S C , )()(S R S E . （18%）⒉The characteristic equation is given 012012212010525)(12345=+++++=+S S S S S S GH . Discuss thedistribution of the closed-loop poles. (12%)⒊ Sketch the root-locus plot for the system )3()1()(-+=S S S K S GH . (The gain K is assumed to be positive.)① Determine the breakaway point and K value.② Determine the value of K at which root loci cross the imaginary axis. ③ Discuss the stability. (15%)⒋ The system block diagram is shown Fig.2. SG 11=, )125.0(102+=S S G . Suppose t r +=1, 1.0=n . Determine the value of SS e . (12%)⒌ Calculate the transfer function for the following Bode diagram of the minimum phase. (15%)⒍ For the system show as follows, )5(4)(+=s s s G ,1)(=s H , (16%) ① Determine the system output )(t c to a unit step, ramp input.② Determine the coefficient P K , V K and the steady state error to t t r 2)(=.⒎ Plot the Bode diagram of the system described by the open-loop transfer function elements )5.01()1(10)(s s s s G ++=, 1)(=s H . (12%)w⒈32221212321221122211)1()()(H H G H H G G H H G G H G H G H G G G S R S C +-++-+-+= ⒉R=0, L=5 ⒌)1611()14)(1)(110(05.0)(2s s s s s s G ++++= ⒍t t e e t c 431341)(--+-= t t e e t t c 41213445)(---+-= ∞=P K , 8.0=V K , 5.2=ss eAUTOMATIC CONTROL THEOREM (7)⒈ Consider the system shown in Fig.1. Obtain the closed-loop transfer function)()(S R S C , )()(S R S E . （16%）⒉ The characteristic equation is given 01087444)(123456=+--+-+=+S S S S S S S GH . Discuss the distribution of the closed-loop poles. (10%)⒊ Sketch the root-locus plot for the system 3)1()(S S K S GH +=. (The gain K is assumed to be positive.)① Determine the breakaway point and K value.② Determine the value of K at which root loci cross the imaginary axis. ③ Discuss the stability. (15%)⒋ Show that the steady-state error in the response to ramp inputs can be made zero, if the closed-loop transfer function is given by:nn n n n n a s a s a s a s a s R s C +++++=---1111)()(Λ ;1)(=s H （12%）⒌ Calculate the transfer function for the following Bode diagram of the minimum phase.（15%）w⒍ Sketch the Nyquist diagram (Polar plot) for the system described by the open-loop transfer function )12.0(11.0)(++=s s s S GH , and find the frequency and phase such that magnitude is unity. （16%）⒎ The stability of a closed-loop system with the following open-loop transfer function )1()1()(122++=s T s s T K S GH depends on the relative magnitudes of 1T and 2T . Draw Nyquist diagram and determine the stability of the system.（16%） ( 00021>>>T T K )⒈3213221132112)()(G G G G G G G G G G G G S R S C ++-++=⒉R=2, I=2,L=2 ⒌)1()1()(32122++=ωωωs s s s G⒍o s rad 5.95/986.0-=Φ=ωAUTOMATIC CONTROL THEOREM (8)⒈ Consider the system shown in Fig.1. Obtain the closed-loop transfer function)()(S R S C , )()(S R S E . （16%）⒉ The characteristic equation is given 04)2(3)(123=++++=+S K KS S S GH . Discuss the condition of stability. (12%)⒊ Draw the root-locus plot for the system 22)4()1()(++=S S KS GH ;1)(=s H .Observe that values of K the system is overdamped and values of K it is underdamped. (16%)⒋ The system transfer function is )1)(21()5.01()(s s s s K s G +++=,1)(=s H . Determine thesteady-state error SS e when input is unit impulse )(t δ、unit step )(1t 、unit ramp t and unit parabolic function221t . (16%)⒌ ① Calculate the transfer function (minimum phase);② Draw the phase-angle versus ω (12%) w⒍ Draw the root locus for the system with open-loop transfer function.)3)(2()1()(+++=s s s s K s GH (14%)⒎ )1()(3+=Ts s Ks GH Draw the polar plot and determine the stability of system. (14%)⒈43214321432143211)()(G G G G G G G G G G G G G G G G S R S C -+--+= ⒉∞ππK 528.0⒊S:0<K<0.0718 or K>14 overdamped ;0.0718<K<14 underdamped⒋S: )(t δ 0=ss e ; )(1t 0=ss e ; t K e ss 1=; 221t ∞=ss e⒌S:21ωω=K ; )1()1()(32121++=ωωωωs s ss GAUTOMATIC CONTROL THEOREM (9)⒈ Consider the system shown in Fig.1. Obtain the closed-loop transfer function)(S C , )(S E . （12%）⒉ The characteristic equation is given0750075005.34)(123=+++=+K S S S S GH . Discuss the condition of stability. (16%)⒊ Sketch the root-locus plot for the system )1(4)()(2++=s s a s S GH . (The gain a isassumed to be positive.)① Determine the breakaway point and a value.② Determine the value of a at which root loci cross the imaginary axis. ③ Discuss the stability. (12%)⒋ Consider the system shown in Fig.2. 1)(1+=s K s G i , )1()(2+=Ts s Ks G . Assumethat the input is a ramp input, or at t r =)( where a is an arbitrary constant. Show that by properly adjusting the value of i K , the steady-state error SS e in the response to ramp inputs can be made zero. (15%)⒌ Consider the closed-loop system having the following open-loop transfer function:)1()(-=TS S KS GH . ① Sketch the polar plot ( Nyquist diagram). ② Determine thestability of the closed-loop system. (12%)⒍Sketch the root-locus plot. (18%)⒎Obtain the closed-loop transfer function )()(S R S C . （15%）⒈354211335421243212321313542143211)1()()(H G G G G H G H G G G G H G G G G H G G H G H G G G G G G G G G S R S C --++++-= 354211335421243212321335422341)()(H G G G G H G H G G G G H G G G G H G G H G H G G G H H G S N S E --+++--= ⒉45.30ππK⒌S: N=1 P=1 Z=0; the closed-loop system is stable ⒎2423241321121413211)()(H G H G G G G G G G H G G G G G G G S R S C ++++++=AUTOMATIC CONTROL THEOREM (10)⒈ Consider the system shown in Fig.1. Obtain the closed-loop transfer function)()(S R S C ,⒉ The characteristic equation is given01510520)(1234=++++=+S S KS S S GH . Discuss the condition of stability. (14%)⒊ Consider a unity-feedback control system whose open-loop transfer function is)6.0(14.0)(++=S S S S G . Obtain the response to a unit-step input. What is the rise time forthis system? What is the maximum overshoot? （10%）⒋ Sketch the root-locus plot for the system )25.01()5.01()(s S s K S GH +-=. (The gain K isassumed to be positive.)③ Determine the breakaway point and K value.④ Determine the value of K at which root loci cross the imaginary axis. Discuss the stability. (15%)⒌ The system transfer function is )5(4)(+=s s s G ,1)(=s H . ①Determine thesteady-state output )(t c when input is unit step )(1t 、unit ramp t . ②Determine theP K 、V K and a K , obtain the steady-state error SS e when input is t t r 2)(=. （12%）⒍ Consider the closed-loop system whose open-loop transfer function is given by:①TS K S GH +=1)(; ②TS K S GH -=1)(; ③1)(-=TS KS GH . Examine the stabilityof the system. （15%）⒎ Sketch the root-locus plot 。
自控原理英文版分析
Desired temperature
thermometer
operator
gas
valve
air
furnace
mixer
1-3 What is an automatic control system?
Definition of automatic control system
A control system without people involved is called automatic control system.
2. According to whether the system is satisfied the superposition principle
Linear Systems versus Nonlinear System
A linear system satisfies the properties of superposition and Homogeneity: (Principle of Superposition).
1-3 What is an automatic control system?
water level automatic control system
1-4 types and Basic components of control systems.
1. According to the form of given signal Regulator Systems versus Servo Systems (恒值系 统和随动系统）
Open-loop system, closed-loop system (feedback control systems), Compound Control systems
完整word版,《自动控制原理》试卷及答案(英文10套)
AUTOMATIC CONTROL THEOREM(1)⒈Derive the transfer function and the differential equation of the electric network shown in Fig.1. (12% )R1C1V1(S)C2V2(S) R2⒉Consider the system shown in Fig.2. Obtain the closed-loop transferfunction C (S),E( S).（12%）R(S) R(S)G4R ECG1G2G3H2H1⒊ The characteristic equation is given 1 GH (S) S35S2(6 K )S 10K 0. Discuss the distribution of the closed-loop poles. (16%)①There are 3 roots on the LHP ② There are 2 roots on the LHP②There are 1 roots on the LHP ④ There are no roots on the LHP . K=?⒋Consider a unity-feedback control system whose open-loop transfer function is1G(S). Obtain the response to a unit-step input. What is the rise time for S( S0.6)this system? What is the maximum overshoot?（ 10%）5. Sketch the root-locus plot for the systemK. ( The gain K is GH (S)S(S1)assumed to be positive.)①Determine the breakaway point and K value.②Determine the value of K at which root loci cross the imaginary axis.③ Discuss the stability.(12%)6. The system block diagram is shown Fig.3. Supposer( 2 t),n. Determine1the value of K to ensure e SS1 .(12%)NR E4CKS 2S(S 3)7. Consider the system with the following open-loop transfer function:GH (S)K① Draw Nyquist diagrams. ② Determine the.S(T1S1)(T2 S 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)S 2S3C(S) (S2)(S 5)(S10)⒈V2(S)R2C1C2 S C1V1( S) ( R1R2 )C1C2 S C1 C2⒉C(S)G1G2 G3 G1G4R(S) 1 G1G2H1G1G2G3 G1G4 G2G3H 2 G4H 2⒊ ① 0<K<6 ② K ≤0 ③ K ≥6 ④ no answer⒋⒌① the breakaway point is –1 and –1/3; k=4/27 ② The imaginary axis S=±j; K=2 ③⒍31.62( S1)⒎ GH (S)S S SS1)(1)((1)(1)AUTOMATIC CONTROL THEOREM(2)⒈Derive the transfer function and the differential equation of the electric network shown in Fig.1. (12% )R1R1V1(S)C1C2V2(S)⒉Consider the equation group shown in Equation.1. Draw block diagram and obtainthe closed-loop transfer function C (S). （16%）R(S)X1 (S) G1 (S)R(S)G1(S)[G7 (S) G8 (S)]C(S)X2 (S)G2(S)[ X1( S)G6 (S)X 3(S)][ X2(S)C(S)G5 (S)]G3(S)X3(S)C(S)G4(S)X3 (S)⒊Use Routh’s criterion to determine the number of roots in the right-half S plane forthe equation 1GH (S) S53S428S3226S2600S 400 0 .Analyze stability.（12% ）⒋ Determine the range of K value ,when r(1 t t 2 ) , e SS0.5 .（12%）RE3S2 4S KS2(S 2)C⒌Fig.3 shows a unity-feedback control system. By sketching the Nyquist diagram of the system, determine the maximum value of K consistent with stability, and check the result using Routh’s criterion. Sketch the root-locus for the system（20%）R K CS(S24S 5)R E⒍Sketch root-locus diagram（.18% ）Im Im ImRe Re ReIm Im ImRe Re Re⒎Determine the transfer function. Assume a minimum-phase transfer function. （10% ）L(dB)0200–20–40 -604030205ωω1ω2ω3ω4⒈V2(S)1V1 ( S) R1C1R2 C2 S2(R1C1 R2 C2 R1C2 )S 1⒉C(S)G1 G2G3G 4R( S) 1 G2G3G6G3G4 G5 G1G2G3G4 (G7 G8 )⒊There are 4 roots in the left-half S plane, 2 roots on the imaginary axes, 0 root in the RSP. The system is unstable.⒋8K20⒌K=20⒍31.62( S1)⒎GH (S)S S SS1)(1)((1)(1)AUTOMATIC CONTROL THEOREM(3)⒈List the major advantages and disadvantages of open-loop control systems. (12% )⒉Derive the transfer function and the differential equation of the electric network shown in Fig.1.（ 16% ）C1R1U2U1R2C2⒊Consider the system shown in Fig.2. Obtain the closed-loop transferC(S)E(S)C(S)function,,.（12%）E G5R P CG1G2G3G4H2H1H3⒋ The characteristic equation is given 1GH (S) S33S22S 200 . Discuss the distribution of the closed-loop poles. (16%)5. Sketch the root-locus plot for the systemK. (The gain K is GH (S)S( S1)assumed to be positive.)④Determine the breakaway point and K value.⑤Determine the value of K at which root loci cross the imaginary axis.⑥ Discuss the stability.(14%)6. The system block diagram is shown Fig.3.G1K4. Suppose ,G2S2S(S3)r (2t ) , n 1 . Determine the value of K to ensure e SS 1 .(15%)RNEG2C G17. Consider the system with the following open-loop transfer function:GH (S)K① Draw Nyquist diagrams. ② Determine the.S(T1S1)(T2 S 1)stability of the system for two cases, ⑴ the gain K is small, ⑵ K is large. (15%)⒈Solution: The advantages of open-loop control systems are as follows:①Simple construction and ease of maintenance②Less expensive than a corresponding closed-loop system③ There is no stability problem④ Convenient when output is hard to measure or economically not feasible. (Forexample, 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:①Disturbances and changes in calibration cause errors, and the output may bedifferent from what is desired.②To maintain the required quality in the output, recalibration is necessaryfrom time to time.U2(S)R1C1 R2 C2 S2(R1C1 R2C2 )S 1⒉U1(S)R1C1 R2 C2 S2(R1C1 R2 C2 R1C2 )S 1⒊ C(S)G1G2 G3G4 G1G5R(S) 1 G1G2H 1 G2G3H 2G1G2G3G4 H 3G1G5H 3C( S)G3G4 (1 G1G2 H 1) G3G5 H 2P( S) 1 G1G2H 1 G2G3H 2G1G2G3G4H 3G1G 5H 3⒋R=2, L=1⒌S:① the breakaway point is –1 and –1/3; k=4/27 ② The imaginary axis S=±j; K=2⒍ 3.5 KAUTOMATIC CONTROL THEOREM(4)⒈Find the poles of the following F ( s) :F ( s)1(12%)e s1⒉ Consider the system shown in Fig.1,where0.6 and n 5 rad/sec. Obtain the rise time t r , peak time t p , maximum overshoot M P , and settling time t s when the system is subjected to a unit-step input. （ 10%）2ns(s2n )C(s)R(s)⒊Consider the system shown in Fig.2. Obtain the closed-loop transferC(S)E(S)C(S)function,,.（12%）G5ERP CG1G2G3G4H2H1H3⒋ The characteristic equation is given 1GH (S) S33S22S 200 . Discuss the distribution of the closed-loop poles. (16%)5. Sketch the root-locus plot for the systemK. (The gain K is GH (S)S( S1)assumed to be positive.)⑦Determine the breakaway point and K value.⑧Determine 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.G1K4. Suppose ,G2S2S(S3)r (2t ) , n 1 . Determine the value of K to ensure e SS 1 .(12%)RNEG2C G17. Consider the system with the following open-loop transfer function:GH (S)K① Draw Nyquist diagrams. ② Determine the.S(T1S1)(T2 S 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)S 2S3C(S) (S2)(S 5)(S10)⒈ Solution: The poles are found from e s1or e(j ) e (cosj sin ) 1From this it follows that0,2n(n0,1,2, ) . Thus, the poles are located at s j 2n⒉ Solution: rise timet rt0.785 sec, , peak time pmaximum overshoot M P0.095 ,and settling time t s 1.33 sec for the 2%criterion, settling time t s1sec for the 5% criterion.⒊C(S)G1G2 G3G4G1G5R(S) 1 G1G2H 1G2G3H 2G1G2G3G4 H 3G1G5H 3C( S)G3G4 (1 G1G2 H 1) G3G5 H 2P(S) 1 G1G2H 1G2G3H 2G1G2G3G4H 3G1G 5H 3⒋R=2, L=15.S:①the breakaway point is –1 and –1/3; k=4/27 ② The imaginary axis S=±j; K=2⒍ 3.5 KAUTOMATIC CONTROL THEOREM(5)⒈Consider the system shown in Fig.1. Obtain the closed-loop transferfunction C (S),E( S).（18%）R(S) R(S)H2R E CG1G2G3H1H3H4⒉The characteristic equation is given1GH (S) S5 3S4 12S3 24S2 32S 48 0 . Discuss the distribution of the closed-loop poles. (16%)⒊ Sketch the root-locus plot for the system GH (S)K. (The gain S( S1)(0.5S 1)K is assumed to be positive.)①Determine the breakaway point and K value.②Determine the value of K at which root loci cross the imaginary axis.③ Discuss the stability.(18%)⒋ The system block diagram is shown Fig.2. G1K 1, G2K 2.①T1S 1T2S 1Suppose r 0 , n 1 . Determine the value of e SS.②Suppose r 1 , n 1. Determine the value of e SS . (14%)RNEG2C G1⒌ Sketch the Bode diagram for the following transfer function. GH (s)K, s(1Ts)K 7 , T 0.087. (10%)⒍ A system with the open-loop transfer function GH (S)K is inherently2 (TSs1)unstable. This system can be stabilized by adding derivative control. Sketch the polar plots for the open-loop transfer function with and without derivative control. (14%)⒎Draw the block diagram and determine the transfer function. (10%)U1(s)R C U2(s)⒈C (S) G 1G 2 G 3 R(S)⒉ R=0, L=3,I=2⒋① e ssK 21 K2 K 1K 2② e ssK 1K 21 1 ⒎U 2 ( s) 1U 1 ( s)RCs 1AUTOMATIC CONTROL THEOREM(6)⒈Consider the system shown in Fig.1. Obtain the closed-loop transferfunction C (S),E( S).（18%）R(S) R(S)R E CG1G2H1H2H3⒉The characteristic equation is given1 GH (S) 25S5105S4120S3122S220S 1 0 . Discuss the distribution of the closed-loop poles. (12%)⒊ Sketch the root-locus plot for the system GH ( S)K (S 1). (The gain K isS( S3) assumed to be positive.)①Determine the breakaway point and K value.②Determine the value of K at which root loci cross the imaginary axis.③ Discuss the stability.(15%)⒋ The system block diagram is shown Fig.2. G11, G210. Suppose S1)r 1t , n 0.1 . Determine the value of e SS .(12%)R E N CG1G2⒌ Calculate the transfer function for the following Bode diagram of the minimum phase. (15%)dB14816w-40-200dB/dec 200⒍ For the system show as follows, G(s)4,(16%), H (s) 1s(s5)①Determine the system output c(t ) to a unit step, ramp input.② Determine the coefficient K P , K V and the steady state error to r (t )2t .⒎Plot the Bode diagram of the system described by the open-loop transfer functionelements G(s)10(1s), H (s) 1.(12%)s(10.5s)⒈C(S)G 1G 2 (1 G 2H 2)R(S) 1 G 1H 1G 2H 2 G 1G 2H 3H 2 G 1G 2H 1H 2 G 2H 2H 3⒉ R=0, L=50.05(10s 1)( s 1)(s1)⒌ G(s)41s)s 2 (116⒍c(t ) 1 4e t 1 e 4 tc(t) t5 4 e t 1 e 4t K P, K V ,334 312essAUTOMATIC CONTROL THEOREM(7)⒈Consider the system shown in Fig.1. Obtain the closed-loop transferfunction C (S),E( S).（16%）R(S) R(S)R E CG1G2G3⒉The characteristic equation is given1 GH (S) S64S54S44S37S28S 10 0 . Discuss the distribution ofthe closed-loop poles. (10%)⒊Sketch the root-locus plot for the system GH ( S)K(S 1). (The gain K is assumed to be positive.)S3①Determine the breakaway point and K value.②Determine the value of K at which root loci cross the imaginary axis.③ Discuss the stability.(15%)⒋Show that the steady-state error in the response to ramp inputs can be made zero, if the closed-loop transfer function is given by:C( s)a n 1s a n; H ( s) 1（12%）R( s)s n a1s n 1a n 1 s a n⒌ Calculate the transfer function for the following Bode diagram of the minimum phase.dB-40-20dB/decw（15%）w1 w2w3-40⒍Sketch the Nyquist diagram (Polar plot) for the system described by the open-looptransfer function GH (S)1, and find the frequency and phase such that 1)magnitude is unity.（16%）⒎ The stability of a closed-loop system with the following open-loop transferfunction GH (S)K (T2 s1)s2 (T1s depends on the relative magnitudes of T1 and T2 .1)Draw Nyquist diagram and determine the stability of the system.（ 16%）( K 0T1 0T20 )⒈C (S)G1G1G2G 3R(S) 2 G1G1G2G2G3 G1G2 G3⒉R=2, I=2,L=222 (s1)⒌ G(s)1ss2 ()31⒍0.986rad / s oAUTOMATIC CONTROL THEOREM (8)⒈ Consider the system shown in Fig.1. Obtain the closed-loop transfer functionC (S) , E(S) . （16%）R( S) R(S)G1G2RECG3G4⒉ The characteristic equation is given 1 GH ( S) S 33KS 2 (2 K)S 4 0.Discuss the condition of stability. (12%)⒊ Draw the root-locus plot for the system GH ( S)K2 ; H (s) 1.1)2( S 4)( SObserve that values of K the system is overdamped and values of K it is underdamped. (16%)K (1 0.5s) . Determine the⒋ The system transfer function is G (s), H ( s) 1s(1 2s)(1 s)steady-state error e SS when input is unit impulse (t) 、unit step 1(t ) 、unit ramp t and unit parabolic function1 t2 . (16%)2⒌ ① Calculate the transfer function (minimum phase);② Draw the phase-angle versusdB-40-20dB/decw(12%)w1 w2 w3-40⒍Draw the root locus for the system with open-loop transfer function.K (1 s)(14%)GH (s)s( s 2)(s3)⒎ GH ( s)K Draw the polar plot and determine the stability of system.3 (Tss1)(14%)⒈ C(S)G 1G 2 G 3G 4 1 G 1G 2G 3G 4 R(S)G 1G 2 G 3G 4 G 1G 2 G 3G 4⒉ 0.528 K⒊ S:0<K<0.0718 or K>14 overdamped ;0.0718<K<14 underdamped ⒋ S:(t)ess0 ; 1(t ) e ss0 ; t e ss1 ; 1 t2 e ssK 21 2 (s1)⒌S:K12 ;G( s)1ss 2 ( )31AUTOMATIC CONTROL THEOREM(9)⒈Consider the system shown in Fig.1. Obtain the closed-loop transferfunction C (S),E(S) .（12%）R(S)N(S)G5RE N C G1G2G3G4H1H2H3⒉The characteristic equation is given1 GH (S) S327500S 7500K 0 . Discuss the condition of stability. (16%)⒊Sketch the root-locus plot for the system(s a)4a is GH (S)2 (s. (The gains1)assumed to be positive.)①Determine the breakaway point and a value.②Determine the value of a at which root loci cross the imaginary axis.③ Discuss the stability.(12%)⒋ Consider the system shown in Fig.2. G1(s) K i s1, G 2 ( s)K. Assume s(Ts 1)that the input is a ramp input, or r (t)at where a is an arbitrary constant. Show that by properly adjusting the value of K i, the steady-state error e SS in the response to ramp inputs can be made zero.(15%)R(s)E(s)C(s)G1(s)G2(s)⒌ Consider the closed-loop system having the following open-loop transfer function:K . ① Sketch the polar plot ( Nyquist diagram). ② Determine theGH (S)S(TS1)stability of the closed-loop system. (12%)⒍ Sketch the root-locus plot. (18%)Im Im ImRe Re ReIm Im ImRe Re Re⒎ Obtain the closed-loop transfer functionC (S). （15%）R( S)G4RCG1G2G3H2H1⒈C(S)G1G2G3G4G1G2 G4G5 (1 G3 H 1 )R(S) 1 G3H1G2G3H 2G1G2G3 G4 H 2G1G2 G4G5 H 3G3 H 1G1G2 G4 G5 H 3 E (S)G4H 3H 2G2G4G5 H 3N(S) 1 G3H1G2G3H 2G1G2G3G4H 2G1G2 G4G5 H 3G3 H 1G1G2 G4G5 H 3⒉⒌S: N=1 P=1 Z=0; the closed-loop system is stable⒎C(S)G1G2 G3 G1G4R(S) 1 G1G2H 1G1G2G3 G1G4 G2G3H 2 G4H 2AUTOMATIC CONTROL THEOREM(10)⒈Consider the system shown in Fig.1. Obtain the closed-loop transfer functionC(S) ,C(S). （16%）R( S)N (S)G3NR CG1G2G4G5H⒉The characteristic equation is given1 GH (S) S420KS 35S210S 15 0 . Discuss the condition of stability. (14%)⒊Consider a unity-feedback control system whose open-loop transfer function is1G(S). Obtain the response to a unit-step input. What is the rise time for S( S0.6)this system? What is the maximum overshoot?（ 10%）⒋ Sketch the root-locus plot for the system GH (S)K (10.5s). (The gain K isS(10.25s) assumed to be positive.)③Determine the breakaway point and K value.④Determine the value of K at which root loci cross the imaginary axis. Discuss the stability.(15%)⒌4, H (s) 1 .① Determine the The system transfer function is G ( s)s(s5)steady-state output c(t ) when input is unit step1(t )、unit ramp t .②Determine the K P、K V and K a, obtain the steady-state error e SS when input is r (t ) 2t .（12%）⒍Consider the closed-loop system whose open-loop transfer function is given by:K K K. Examine the stability ① GH (S); ② GH(S); ③GH(S)1 TS 1 TS TS1of the system.（15%）⒎Sketch the root-locus plot。
自动控制原理第一部分自动控制的一般概念汇总
减速器
电动机
SM
if
当电位器电刷位于中点位置时，电动机不动，控制阀门 有一定的开度，使水箱中流入水量与流出水量相等，从而液 面保持在希望高度h0上。一旦流入水量或流出水量发生变化， 水箱液面高度便相应变化。如当液面升高时，浮子位置亦相 应升高，通过杠杆作用使电位器电刷从中点位置下移，从而 给电动机提供一定的控制电压，驱动电动机通过减速器减小 阀门开度，使进入水箱的流量减少。此时，水箱液面下降， 浮子位置相应下降，直到电位器电刷回到中点位置，系统重 新处于平衡状态，液面恢复给定高度。反之，若水箱液位下 降，则系统会自动增大阀门开度，加大流入水量，使液位升 到给定高度h0 。
3 反馈控制原理(feedback control theory)
自动控制系统：为了实现各种复杂的控制任务，首 先要将被控对象和控制装置按照一定的方式连接起 来，组成一个有机整体，这就是自动控制系统。
反馈控制原理：在自动控制系统中，控制装置(控 制器)对被控对象施加的控制作用，是取自被控量 的反馈信息，用来不断修正被控量与输入量之间的 偏差，从而实现对被控对象进行控制的任务，这就 是反馈控制的原理。
20世纪60年代，数字计算机技术的迅速发展为复杂系统的 基于时域分析的现代控制理论提供了可能。
从1960年至今，确定性系统、随机系统的最佳控制，及复 杂系统的自适应和智能控制技术，都得到充分的研究。
在现代控制理论发展中，特别应该提到三位学者的重 大贡献：
➢1956年，前苏联科学家Pontryagin 提出的极大值原理；
1884年，E. J. Routh提出了有关线性系统稳定性的判据。
1932年，Nyquist研制出电子管振荡器，同时提出了著名的 Nyquist稳定性判据；此后Bode总结出了负反馈放大器。
自动控制原理_英文论文
Automatic Control Applications In the social life班级：学号：：Programmable controller to control watersupply systemConstant pressure water supply system for a certain industry or a particular user is very important, for example in certain production processes, if the tap water supply or short-time shortages due to insufficient water, which may affect product quality, serious product scrap and damage to the equipment. When a fire occurs, if the water pressure is insufficient or no water supply, no rapid fire, can lead to significant economic losses and casualties. So some of the water area with constant voltage water supply system, has great economic and social significance.Mechanical technologyOld pressurized equipment, General starts or stops usingconstant-pressure water supply pressurizing station exit of pumps and regulating valve opening is to be achieved. Control system is the use of relay-contactor control circuits, this line complex, difficult to maintain, and operate trouble, workers to guards on duty 24 hours, labor intensive. It is necessary to reform, improve the level of automation.Electrical technologyPresented to a tap water pressure station quick starting of constant pressure water supply control system, the FP3 produced by Matsushita programmable logic controller (PLC) controls with Advantech industrial computer monitor, high degree of automation, the whole program to work automatically, clearly shows the real-time status of each device, and automatically adjust the water pressure. The system also has a wide range of protection, such as water pressure alarm fault alarm, water alarm, valves, pump motor current flow and processing alarm processing and so on.System structure and control requirements of mechanical and electrical engineering technology network，Constant pressure water supply system consists of the main loop, the alternate loop of water supply,Composed of 2 water tank and pump house, as shown in Figure 1.Pumping station equipped with a 1# ~ 6# a total of 6 sets of150kW pumps. There is more than one (V1 ~ V23) electric valvecontrols the water circuit and the flow of water.Requires the constant pressure water-supply system has the following basic operating functions.Electrical technical machinery technolog.When municipal water pressure is higher than the setting pressure 21.56x104Pa, directly by the municipal water supply in electrical technology. When city water is lower than the set pressure, but under the pressure of not less than 7.84x104Pa when using direct pumping pressurized water supply solutions. Of progressively starting 2 pumps to pipe network pressure. When city water higher than the set pressure is detected, then converted to city water supply directly. When the tap water pressure is lower than 2.94x104Pa, or when there is a negative pressure signal exactly, should immediately convert water pressure, but should ensure that the pool water level above the minimum water level conditions. Mechanical technology.When pumping or pumping water pressure water supply is used, should be able to automatically adjust the water pressure for a given value of its total exports, control deviation is less than or equal to 10%. Electrical technologyCAD/CAM technologyDesign of PLC control system CAD/CAM technologyConstant pressure water supply system for detection and control of more is a large control system. According to its characteristics, we have chosen Panasonic FP3 programmable controller is a controller.The controller with it can programming sequence controller comparedto, has some obviously of advantages, as FP3 used has module ofdesign, can according to actual needs flexible assembled, usingconvenient, I/O distribution used free programming way; capacity big, program volume only by scan cycle limit, and scan cycle can in mustrange within itself change; has A/D, and D/A, and pulse output, andlocation control, senior unit, can achieved "shared memory"; additionalso some special of function. Mechanical technologyConstant pressure water supply system of PLC system structure isshown in Figure 1-dashed border. Industrial computer to monitor theentire system, the display shows the total pressurized systemstructure, read the real-time status of each valve and pump, waterpressure and flow rate, the valve opening, the pool water level andother parameters, and real-time display alarm and fault record.Electrical technologyBoth analog input and switch input. Analog by a/d module input,total 27 channels. There are 96 I/O points.CAD/CAM technologyMechanical technologyPLC software designElectrical technologyAccording to constant pressure water supply system operational requirements, PLC control system to monitor City tap water, and waterhas to decide whether to start the water pump, or direct waterpumping programme or by pumping pressurized water full solution. Control system the procedure is more complex. Electrical/mechanical engineering technology networkIn the control process, water supply and pressure regulation isan important and one of the more distinctive design, focusing onsoftware design of automatic constant-pressure function.Due to the large water system piping length and diameter, the opening and closing of the valve, pipe network pressure is slow, so the system is a system with large time delay. And because it is based on the old equipment transformation, to make use of existing equipment, it does not use speed regulator, instead of using the various methods to adjust the water pressure. First used segment regulation method, put hydraulic deviation is divided into four segment, that 10%, and 20%, and 30%, and 40%, dang detection to deviation smaller Shi, output of control volume (butterfly valve of incremental) smaller, and operation cycle also larger; In addition, when the deviation is less than or equal to ± 10%, coupled with the fuzzy control, according to d EK=EK-EK-1 to determine whether regulating butterfly valve opening, to further reduce the errors to ensure its error less than or equal to ± 10% requirements. Pressure is regula ted by multiple methods combination of outlet pressure can be satisfied with the effect. Electrical technical machinery technologyConclusionThe design of constant pressure water supply control system with PLC have successfully applied to an industrial zone, results showed that the system satisfies its design requirements, with convenient operation, high reliability, data integrity and monitor timely advantage and significantly reduce the labor intensity of workers, shorten the operating time, operators, maintainers, managers at home. The successful design of the monitoring system, as well as similar systems of old equipment modification to provide a good experience.。
自动控制原理英文版课后全部_答案
Module3Problem 3.1(a) When the input variable is the force F. The input variable F and the output variable y are related by the equation obtained by equating the moment on the stick:2.233y dylF lk c l dt=+Taking Laplace transforms, assuming initial conditions to be zero,433k F Y csY =+leading to the transfer function31(4)Y k F c k s=+ where the time constant τ is given by4c kτ=(b) When F = 0The input variable is x, the displacement of the top point of the upper spring. The input variable x and the output variable y are related by the equation obtained by the moment on the stick:2().2333y y dy k x l kl c l dt-=+Taking Laplace transforms, assuming initial conditions to be zero,3(24)kX k cs Y =+leading to the transfer function321(2)Y X c k s=+ where the time constant τ is given by2c kτ=Problem 3.2 P 54Determine the output of the open-loop systemG(s) = 1asT+to the inputr(t) = tSketch both input and output as functions of time, and determine the steady-state error between the input and output. Compare the result with that given by Fig3.7 . Solution ：While the input r(t) = t , use Laplace transforms, Input r(s)=21sOutput c(s) = r(s) G(s) = 2(1)aTs s ⋅+ = 211T T a s s Ts ⎛⎫ ⎪-+ ⎪ ⎪+⎝⎭the time-domain response becomes c(t) = ()1t Tat aT e ---Problem 3.33.3 The massless bar shown in Fig.P3.3 has been displaced a distance 0x and is subjected to a unit impulse δ in the direction shown. Find the response of the system for t>0 and sketch the result as a function of time. Confirm the steady-state response using the final-value theorem. Solution ：The equation obtained by equating the force:00()kx cxt δ+=Taking Laplace transforms, assuming initial condition to be zero,K 0X +Cs 0X =1leading to the transfer function()XF s =1K Cs +=1C1K s C+The time-domain response becomesx(t)=1CC tK e -The steady-state response using the final-value theorem:lim ()t x t →∞=0lim s →s 1K Cs +1s =1K00000()()()1;11111()K t CK x x Cx t Kx X K Cs Kx Kx X C Cs K K s KKx x t eCδ-++=⇒++=--∴==⋅++-=⋅According to the final-value theorem:0001lim ()lim lim 01t s s Kx sx t s X C K s K→∞→→-=⋅=⋅=+ Problem 3.4 Solution:1.If the input is a unit step, then1()R s s=()()11R s C s sτ−−−→−−−→+ leading to,1()(1)C s s sτ=+taking the inverse Laplace transform gives,()1tc t e τ-=-as the steady-state output is said to have been achieved once it is within 1% of the final value, we can solute ―t‖ like this,()199%1tc t e τ-=-=⨯ (the final value is 1) hence,0.014.60546.05te t sττ-==⨯=(the time constant τ=10s)2.the numerical value of the numerator of the transfer function doesn’t affect the answer. See this equation, If ()()()1C s AG s R s sτ==+ then()(1)A C s s sτ=+giving the time-domain response()(1)tc t A e τ-=-as the final value is A, the steady-state output is achieved when,()(1)99%tc t A e A τ-=-=⨯solute the equation, t=4.605τ=46.05sthe result make no different from that above, so we said that the numerical value of the numerator of the transfer function doesn’t affect the answer.If a<1, as the time increase, the two lines won`t cross. In the steady state the output lags the input by a time by more than the time constant T. The steady error will be negative infinite.R(t)C(t)Fig 3.7 tR(t)C(t)tIf a=1, as the time increase, the two lines will be parallel. It is as same as Fig 3.7.R(t)C(t)tIf a>1, as the time increase, the two lines will cross. In the steady state the output lags the input by a time by less than the time constant T.The steady error will be positive infinite.Problem 3.5 Solution: R(s)=261s s+, Y(s)=26(51)s s s +⋅+=229614551s s s -+++ /5()62929t y t t e -∴=-+so the steady-state error is 29(-30). To conform the result:5lim ()lim(62929);tt t y t t -→∞→∞=-+=∞6lim ()lim ()lim ()lim(51)t s s s s y t y s Y s s s →∞→→→+====∞+.20lim ()lim ()lim [()()]161lim [()1]()lim (1)()5130ss t s s s s e e t S E S S Y S R S S G S R S S S S S→∞→→→→==⋅=⋅-=⋅-=⋅-⋅++=- Therefore, the solution is basically correct.Problem 3.623yy x += since input is of constant amplitude and variable frequency , it can be represented as:j tX eA ω=as we know ,the output should be a sinusoidal signal with the same frequency of the input ,it can also be represented as:R(t)C(t)t0j t y y e ω=hence23j tj tj tj yyeeeA ωωωω+=00132j y Aω=+ 0294Ayω=+ 2tan3w ϕ=- Its DC(w→0) value is 003Ay ω==Requirement 01122w yy==21123294AA ω=⨯+ →32w = while phase lag of the input:1tan 14πϕ-=-=-Problem 3.7One definition of the bandwidth of a system is the frequency range over which the amplitude of the output signal is greater than 70% of the input signal amplitude when a system is subjected to a harmonic input. Find a relationship between the bandwidth and the time constant of a first-order system. What is the phase angle at the bandwidth frequency ？ Solution ：From the equation 3.41000.71r A r ωτ22=≥+ (1)and ω≥0 (2) so 1.020ωτ≤≤so the bandwidth 1.02B ωτ=from the equation 3.43the phase angle 110tan tan 1.024c πωτ--∠=-=-=Problem 3.8 3.8 SolutionAccording to generalized transfer function of First-Order Feedback Systems11C KG K RKGHK sτ==+++the steady state of the output of this system is 2.5V .∴if s →0, 2.51104C R→=. From this ,we can get the value of K, that is 13K =.Since we know that the step input is 10V , taking Laplace transforms,the input is 10S.Then the output is followed1103()113C s S s τ=⨯++Taking reverse Laplace transforms,4/4332.5 2.5 2.5(1)t t C e eττ--=-=-From the figure, we can see that when the time reached 3s,the value of output is 86% of the steady state. So we can know34823(2)*4393τττ-=-⇒-=-⇒=, 4/3310.8642t t e ττ-=-=⇒=The transfer function is3128s +146s+Let 12+8s=0, we can get the pole, that is 1.5s =-2/3- Problem 3.9 Page 55 Solution:The transfer function can be represented,()()()()()()()o o m i m i v s v s v s G s v s v s v s ==⋅While,()1()111//()()11//o m m i v s v s sRCR v s sC sC v s R R sC sC =+⎛⎫+ ⎪⎝⎭=⎡⎤⎛⎫++ ⎪⎢⎥⎝⎭⎣⎦Leading to the final transfer function,21()13()G s sRC sRC =++ And the reason:the second simple lag compensation network can be regarded as the load of the first one, and according to Load Effect , the load affects the primary relationship; so the transfer function of the comb ination doesn’t equal the product of the two individual lag transfer functio nModule4Problem4.14.1The closed-loop transfer function is10(6)102(6)101610S S S S C RS s +++++==Comparing with the generalized second-order system,we getProblem4.34.3Considering the spring rise x and the mass rise y. Using Newton ’s second law of motion..()()d x y m y K x y c dt-=-+Taking Laplace transforms, assuming zero initial conditions2mYs KX KY csX csY =-+-resulting in the transfer funcition where2Y cs K X ms cs K +=++ And521.26*10cmkc ζ== Problem4.4 Solution:The closed-loop transfer function is210263101011n n d n W EW E W W E ====-=2121212K C K S S K R S S K S S ∙+==+++∙+Comparing the closed-loop transfer function with the generalized form,2222n n nCR s s ωξωω=++ it is seen that2n K ω= And that22n ξω= ; 1Kξ=The percentage overshoot is therefore21100PO eξπξ--=11100k keπ-∙-=Where 10%PO ≤When solved, gives 1.2K ≤（2.86）When K takes the value 1.2, the poles of the system are given by22 1.20s s ++=Which gives10.45s j =-±±s=-1 1.36jProblem4.5ReIm0.45-0.45-14.5 A unity-feedback control system has the forward-path transfer functionG (s) =10)S(s K+Find the closed-loop transfer function, and develop expressions for the damping ratio And damped natural frequency in term of K Plot the closed-loop poles on the complex Plane for K = 0,10,25,50,100.For each value of K calculate the corresponding damping ratio and damped natural frequency. What conclusions can you draw from the plot?Solution: Substitute G(s)=(10)K s s + into the feedback formula : Φ(s)=()1()G S HG S +.And in unitfeedback system H=1. Result in: Φ(s)=210Ks s K++ So the damped natural frequencyn ω=K ,damping ratio ζ=102k =5k.The characteristic equation is 2s +10S+K=0. When K ≤25,s=525K -±-; While K>25,s=525i K -±-; The value ofn ω and ζ corresponding to K are listed as follows.K 0 10 25 50 100 Pole 1 1S 0 515-+ -5 -5+5i 553i -+Pole 2 2S -10 515-- -5 -5-5i553i --n ω 010 5 52 10 ζ ∞2.51 0.5 0.5Plot the complex plane for each value of K:We can conclude from the plot.When k ≤25,poles distribute on the real axis. The smaller value of K is, the farther poles is away from point –5. The larger value of K is, the nearer poles is away from point –5.When k>25,poles distribute away from the real axis. The smaller value of K is, the further (nearer) poles is away from point –5. The larger value of K is, the nearer (farther) poles is away from point –5.And all the poles distribute on a line parallels imaginary axis, intersect real axis on the pole –5.Problem4.61tb b R L C b o v dv i i i i v dt C R L dt=++=++⎰Taking Laplace transforms, assuming zero initial conditions, reduces this equation to011b I Cs V R Ls ⎛⎫=++ ⎪⎝⎭20b V RLs I Ls R RLCs =++ Since the input is a constant current i 0, so01I s=then,()2b RLC s V Ls R RLCs==++ Applying the final-value theorem yields ()()0lim lim 0t s c t sC s →∞→==indicating that the steady-state voltage across the capacitor C eventually reaches the zero ,resulting in full error.Problem4.74.7 Prove that for an underdamped second-order system subject to a step input, thepercentage overshoot above the steady-state output is a function only of the damping ratio .Fig .4.7SolutionThe output can be given by222222()(2)21()(1)n n n n n n C s s s s s s s ωζωωζωζωωζ=+++=-++- (1)the damped natural frequencyd ω can be defined asd ω=21n ωζ- (2)substituting above results in22221()()()n n n d n d s C s s s s ζωζωζωωζωω+=--++++ (3) taking the inverse transform yields22()1sin()11tan n t d e c t t where ζωωφζζφζ-=-+--=(4)the maximum output is22()1sin()11n t p d p p d n e c t t t ζωωφζππωωζ-=-+-==-(5)so the maximum is2/1()1p c t eπζζ--=+the percentage overshoot is therefore2/1100PO eπζζ--=Problem4.8 Solution to 4.8:Considering the mass m displaced a distance x from its equilibrium position, the free-body diagram of the mass will be as shown as follows.aP cdx kxkxmUsing Newton ’s second law of motion,22p k x c x mx m x c x k x p--=++=Taking Laplace transforms, assuming zero initial conditions,2(2)X ms cs k P ++= results in the transfer function2/(1/)/((/)2/)X P m s c m s k m =++ 2(2/)(2/)((/)2/)k k m s c m s k m =++As we see2(2)X m s c s k P++= As P is constantSo X ∝212ms cs k ++ . When 56.25102cs m-=-=-⨯ ()25min210mscs k ++=4max5100.110X == This is a second-order transfer function where 22/n k m ω= and/2/22n c w m c k m ζ== The damped natural frequency is given by 2212/1/8d n k m c km ωωζ=-=-22/(/2)k m c m =- Using the given data,462510/2100.050.2236n ω=⨯⨯⨯== 462502.79501022100.05ζ-==⨯⨯⨯⨯ ()240.22361 2.7950100.2236d ω-=⨯-⨯= With these data we can draw a picture14.0501160004.673600p de s e T T πωτζωτ======222222112/1222()22,,,428sin (sin cos )0tan 7.030.02n n pp dd n dd n ntd d t t t n d p d d p ddd p p p nX k m c k P ms cs k k m s s s m m k c k c cm m m m km p x e tm p xe t t m t t x m ζωζωωωζωωωωζζωωωζωωωωωωωζω--===⋅=⋅++++++=-===∴==-+=∴=⇒=⇒= 其中Problem4.10 4.10 solution:The system is similar to the one in the book on PAGE 58 to PAGE 63. The difference is the connection of the spring. So the transfer function is2222l n d n n w s w s w θθζ=++222(),;p a m ld a m p m l m l l m mm l lk k k N RJs RCs R k k N k J N J J C N c c N N N θθωθωθ=+++=+=+===p a mn K K K w NJ R='damping ratio 2p a m c NRK K K J ζ='But the value of J is different, because there is a spring connected.122s m J J J J N N '=++Because of final-value theorem,2l nd w θθζ=Module5Problem5.45．4 The closed-loop transfer function of the system may be written as2221010(1)610101*********CR K K K S S K K S S K S S +++==+++++++ The closed-loop poles are the solutions of the characteristic equation6364(1010)3110210(1)n K S K JW K -±-+==-±+=+ 210(1)6310(1)E K E K +==+In order to study the stability of the system, the behavior of the closed-loop poles when the gain K increases from zero to infinte will be observed. So when12K = 3010E =321S J =-± 210K = 3110110E =3101S J =-± 320K = 21070E =3201S J =-±双击下面可以看到原图ReProblem5.5SolutionThe closed-loop transfer function is2222(1)1(1)KC K KsKR s K as s aKs Kass===+++++∙+Comparing the closed-loop transfer function with the generalized form, 2222nn nCR s sωξωω=++Leading to2nKa Kωξ==The percentage overshoot is therefore2110040%PO eξπξ--==Producing the result0.869ξ=（0.28）And the peak time241PnT sπωξ==-Leading to1.586nω=（0.82）Problem5.75.7 Prove that the rise time T r of a second-order system with a unit step input is given byT r = d ω1 tan -1n dζωω = d ω1 tan -1d ωζ21--Plot the rise against the damping ratio.Solution:According to (4.33):c(t)=1-2(cos sin )1n t d d e t t ζωζωωζ-+-. 4.33When t=r T ,c(t)=1.substitue c(t)= 1 into (4.33) Producing the resultr T =d ω1 tan -1n dζωω = d ω1 tan -112ζζ--Plot the rise time against the damping ratio:Problem5.9Solution to 5.9:As we know that the system is the open-loop transfer function of a unity-feedback control system.So ()()GH S G S = Given as()()()425KGH s s s =-+The close-loop transfer function of the system may be written as()()()()()41254G s C Ks R GH s s s K ==+-++ The characteristic equation is()()2254034100s s K s s K -++=⇒++-=According to the Routh ’s method, the Routh ’s array must be formed as follow20141030410s K s s K -- For there is no closed-loop poles to the right of the imaginary axis4100 2.5K K -≥⇒≥ Given that 0.5ζ=4103 4.752410n K K K ωζ=-=⇒=- When K=0, the root are s=+2,-5According to the characteristic equation, the solutions are349424s K =-±-while 3.0625K ≤, we have one or two solutions, all are integral number.Or we will have solutions with imaginary number. So we can drawK=102 -5 K=0K=3.0625K=2.5 K=10Open-loop polesClosed-loop polesProblem5.10 5.10 solution:0.62/n w rad sζ==according to()211sin()21n w t d e c w t ζφζ-=-+=- 1.2sin(1.6)0.4t e t φ-⋅+= 4t a n3φ= finally, t is delay time:1.23t s ≈(0.67)Module6Problem 6.3First we assume the disturbance D to be zero:e R C =-1011C K e s s =⋅⋅⋅+Hence:(1)10(1)e s s R K s s +=++ Then we set the input R to be zero:10()(1)C K e D e s s =⋅+⋅=-+ ⇒ 1010(1)e D K s s =-++Adding these two results together:(1)1010(1)10(1)s s e R D K s s K s s +=⋅-⋅++++21()R s s =; 1()D s s= ∴222110910(1)10(1)100(1)s s e Ks s s Ks s s s s s +-=-=++++++ the steady-state error:232200099lim lim lim 0.09100100ss s s s s s s e s e s s s s s →→→--=⋅===-++++Problem 6.4Determine the disturbance rejection ratio(DRR) for the system shown in Fig P.6.4+fig.P.6.4 solution ：from the diagram we can know :0.210.05mv K RK c === so we can get that()0.21115()0.05v m m OL n CL K K DRR cR ωω∆⨯==+=+=∆210.10.050.050.025s s =++, so c=0.025, DRR=9Problem 6.5 6.5 SolutionFor the purposes of determining the steady-state error of the system, we should get to know the effect of the input and the disturbance along when the other will be assumed to be zero.First to simplify the block diagram to the following patter:110s +2021Js Tddθoθ0.220.10.05s ++__+d T—Allowing the transfer function from the input to the output position to be written as01220220d Js s θθ=++ 012222020240*220220(220)dJs s Js s s Js s sθθ===++++++ According to the equation E=R-C:022*******(2)()lim[()()]lim[(1)]lim 0.2220220ssr d s s s Js e s s s s Js s Js s δδδθθ→→→+=-=-==++++问题;1. 系统型为2，对于阶跃输入，稳态误差为0.2. 终值定理写的不对。
自动控制原理2008(双语)chapter5
The Principle of Automatic Control 2008
5-1 Frequency characteristics
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The Principle of Automatic Control 2008
Obtaining steady-state outputs to sinusoidal inputs
a AG( j) a AG( j)
2 j
2j
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r(t) Asint
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frequency characteristic / frequency response
The Principle of Automatic Control 2008
RC网络只允许低频信号通过,具 有低通滤波器(low-pass filter) 的性质。
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The Principle of Automatic Control 2008
A few conclusions from the example
• Physical meaning of FC:表示系统或环节对不同频率正弦 信号的跟踪能力或复现能力。确定了在不同频率正弦信号作 用下,系统或环节输入与输出的单值对应关系。 • Properties of FC: G(jw)只与系统或环节本身的结构参数 有关,是系统或环节本身的属性,与输入信号和初始条件无关。 • 频率法就是利用稳态的频率响应，间接地研究系统的动态特 性，从而避免了求解高阶微分方程的困难。
自动控制原理2021双语cha幻灯片PPT
控制系统简介
2
The Principle of Automatic Control 2008
Course Description
• This course is concerned with the analysis and design of control system
Manipulated variable 〔操纵量〕:按照指定规律变化使得被控 量到达指定目标的变量(varied by controller so as to affect the controlled variable)
e.g. Temperature control system (空调〕
Speed control system 〔汽车〕
转速0~1000转 扰动
P
电位器 ur
电压 uk 放大器
可控硅 ua 直流 n
功放
电动机
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The Principle of Automatic Control 2008
Comments on open-loop control system
• The output is not measured and compared with the input, no feedback
comparing and using the difference (error) as a means of control –feedback control system
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The Principle of Automatic Control 2008
Block diagram
r
自动控制原理中英文对照
自动控制原理中英文对照Automatic Control Principles 自动控制原理Introduction 简介Automatic control principles refer to the principles and theories that govern the design, development, and implementation of automated control systems. These systems are used in a variety of fields, including manufacturing, transportation, aerospace, and more. The goal of automatic control principles is to create systems that can operate independently and make decisions based on the input they receive.自动控制原理是指掌握设计、开发和实现自动控制系统的原理和理论。
这些系统应用于各种领域，包括制造业、交通运输、航空航天等。
自动控制原理的目标是创建能够独立运作并根据所接收的输入做出决策的系统。
Types of Control Systems 控制系统的类型There are two main types of control systems: open-loop and closed-loop. Open-loop systems are those that operate without any feedback, meaningthat they do not adjust their output based on the input they receive. Closed-loop systems, on the other hand, use feedback to adjust their output based on the input they receive.控制系统主要有两种类型：开环和闭环。
自动控制原理第八版英文版课程设计
Automatic Control Principles 8th Edition - CourseDesign (English Version)AbstractThe course design of Automatic Control Principles 8th Editionfocuses on the application of theoretical knowledge and practical skills. It ms to enhance students’ understanding of automatic controlprinciples and their ability to analyze and design control systems. This document contns a summary of the course design, including the course objectives, content, teaching methods, materials, assessments, and expected outcomes.Course ObjectivesUpon completion of this course, students should be able to:•Understand the fundamental principles of automatic control and the mathematical models of dynamic systems.•Analyze the stability, performance, and robustness of control systems using frequency response and time domn methods.•Design control systems using classical and modern control techniques.•Implement and simulate feedback control systems using computer tools (e.g., MATLAB, Simulink).•Evaluate the effectiveness and limitations of different control strategies in specific applications (e.g., robotics,automotive systems, industrial processes).Course ContentThe course is structured into 14 lectures and 10 lab sessions. The lectures cover the following topics:1.Introduction to automatic control principles2.Mathematical models of dynamic systems3.Time domn analysis of control systems4.Stability analysis of control systems5.Root locus method for control system design6.Frequency domn analysis of control systems7.Bode and Nyquist plots for control system design8.Feedback compensation techniques9.State space analysis of control systems10.Controllability and observability11.Pole placement and observer design12.Nonlinear control systems13.Digital control systems14.Advanced topics in control systems (e.g., adaptivecontrol, fuzzy control, neural networks)The lab sessions provide hands-on experience in designing andtesting control systems using MATLAB and Simulink. The lab topics include:1.Modeling and simulation of dynamic systems2.Time and frequency response analysis of control systems3.Design of proportional-integral-derivative (PID) controllers4.Root locus and frequency response design of control systems5.State space design of control systems6.Robust control design using H-infinity and µ-synthesis7.Nonlinear control system design using feedback linearizationand sliding mode control8.Digital control system design using discretization anddigital filtering9.Experimental validation of control system designs10.Final project on control system design andimplementationTeaching MethodsThe course adopts a blended learning approach, combining face-to-face lectures with online resources and activities. The lectures provide the theoretical foundations of automatic control principles, while the lab sessions offer hands-on practice and experimentation. The online resources include recorded lectures, slides, notes, quizzes, and forums. Students are expected to read the course materials before each lecture and participate in class discussions and group projects. The teaching methods also emphasize the integration of real-world applications and examples into the course content.MaterialsThe required course materials include the textbook。
自动控制原理 英文版
y
y 0 y
y0
df dx
x0
f ( x)
y k x
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x
x 0 x0 x
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F ( x ) kx
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F ( x ) kx
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operating point x0 0.25
1) Identify the output and input variables of the control systems.
2) Write the differential equations of each system’s component in terms of the physical laws of the components. * necessary assumption and neglect. * proper approximation. 3) Dispel the intermediate(across) variables to get the input-output description which only contains the output and input variables. 4) Formalize the input-output equation to be the “standard” form: Input variable —— on the right of the input-output equation . Output variable —— on the left of the input-output equation. Writing the equation—according to the falling-power order.
自控原理课件(英文版)分析
1-4 types and Basic components of control systems.
Features of open-loop control
优点：简单，相对来说成本低 缺点：对可能出现的被控量偏离给定 值的偏差没有任何修正能力，抗干扰 能力差，控制精度不高。
1-4 types and Basic components of control systems.
Open-loop Control System
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Controller
被Pl控ant对象
Controller
Measurement U
n
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Reference input
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Feedback signal
Measurement
Plant
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1-4 types and Basic components of control systems.
Features of feedback control systems
Example: Furnace temperature control system
Switch
Heater
Furnace
Power
Command input Switch & Heater Controller
自动控制原理课件 CH3_4_Stability Analysis
FIGURE 6.3 Tacoma
(a)
(b)
Narrows Bridge (a)
as oscillation begins
and (b) at
catastrophic failure.
CEPE
CH3 Time Domain Analysis
4
2. IFF Conditions for LSS
The bridge was found to oscillate whenever the wind blew. After four months on November 7, 1940, a wind produced an oscillation that grew in amplitude until the bridge broke apart. Figure 6.3(a) shows the condition of beginning oscillation; Fig. 6.3(b) shows the catastrophic failure.
FIGURE 6.1 The
stability of a cone.
CEPE
CH3 Time Domain Analysis
3
Example of an unstable system is shown in Fig. 6.3. The first bridge across the Tacoma Narrows at Puget Sound, Washington, was opened to traffic on July 1, 1940. 塔科马峡谷
The concept of stability can be illustrated by considering a right circular cone placed on a plane horizontal surface. If the cone is resting on its base and is tipped slightly, it returns to its original equilibrium position. This position and response are said to be stable. If the cone rests on its side and is displaced slightly, it rolls with no tendency to leave the position on its side. This position is designated as the neutral stability. On the other hand, if the cone is placed on its tip and released, it falls onto its side. This posi-tion is said to be unstable. These three positions are illustrated in Fig. 6.1.