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## (线性控制系统工程)_英文原版答案

s:61012:" Solutions Manual to Accompany Homework For Linear Control System Engineering Liu Sijiu HIT 651 Copyright 2006 P1-2 Redesign the turbine speed control system discussed in Sample Problem 1.1, but replace the fly-ball governor with the tachometer shown in Fig.1.2. A tachometer consists basically of a small DC motor operated in reverse as a generator, the shaft being rotated continuously, producing a DC voltage proportional to shaft speed. Solution 2 P1-4. Shown in Fig.P1.4 is a water-level control system comprising a tank, inlet pipe, slide valve, and float. Details of the operation of the flow valve are also shown in the figure. Draw a block diagram of the feedback control system and identify the main elements, writing down the mathematical transfer functions where appropriate. Solution P1-5 A method of producing a displacement proportional to an input displacement but with a much larger output force is shown in Fig.P1.5. The input displacement x causes the movement of the spool valve to produce a differential flow to the hydraulic actuator. Draw a block diagram of the system, label the principal parts of the control loop, and identify as many of the transfer functions as possible. State any assumptions made in the analysis. Solution 3 P2-1 A system unknown transfer function is shown in Fig.P2.1. If a unit impulse applied at the input produces at the output a signal described by the time function = e2)t(c − t3 , determine the unknown transfer function. Solution C 2 = e2)t(c − t3 => = R + 3s P2-2. Find the solution of the differential equation 2 xd dx dz x8 +=++ z3 dt 2 dt dt When = e)t(z − t2 and all other initial conditions are zero. ⎛ ⎞ ⎜ ⎟ 1 + 3s ⎛ 1 + Cs 2 ⎞ ⎜ 1 −11s ⎟ 1 Solution )s(Z = ⋅ = ⎜ − ⎟ C1 =⋅ − ⋅ + 2s 2 ++ 8ss + 2s 2 ++ 8ss ⎜ + 2s 1 2 31 ⎟ 10 ⎝ ⎠ ⎜ s( ) ++ ⎟ ⎝ 2 4 ⎠ 1 − t ⎛ 31 23 31 ⎞ − t2 −= 2 ⎜cose1.0e1.0)t(y t − sin t⎟ ⎜ ⎟ ⎝ 2 31 2 ⎠ P2-3 For the system shown in Fig.P2.3., determine the relationship between voltage and current, express this relationship in the form of a transfer function and determine the current as a function of time when the voltage is a step change from zero to 10V. 1 Cs Solution = , ++ Cs/1sLR 2 ++ 1RCsLCs 10 Cs ⋅1010 −6 10 )s(Y = = = s 2 ++ 1RCsLCs − −326 ++ 1s10s10 ++ 10s10s 632 2 3 = 10)t(y 2 −− t500 sin(e ⋅= − t500 )t866sin(e01154.0)t1000 3 2 P2-8 For the system shown in Fig2.8 determine the closed –loop transfer function C/R. C G1 ⎛ G 2 ⎞ + GG 21 Solution = ⎜1+⋅ ⎟ = R + 1HG1 ⎝ G1 ⎠ + 1HG1 P2-9 For the single input system shown in Fig2.9, find the transfer function of output to input C/R. GG GGG Solution 21 => 321 + HGG1 232 + + HGGHGG1 121232 C GGG = 321 R + + + HGGGHGGHGG1 3321121232 4 P3.2 Determine the output of the open-loop system G(s)=a/(1+sT) to the input r(t)=t. Sketch both input and output as function of time, and determine the steady-state error between the input and output. Compare the result with that given by Fig.3.7. a a T/t Solution: )s(R)s(c ⋅= = +−⋅= − )TeTt(a)t(c + )sT1( 2 +⋅ )sT1(s −+−=−= − T/t )TeTt(atcre ⎧ = 1aforT As t ∞→ , ss aTatte =−−= ⎨ ⎩∞ ≠ 1afor 1 P3.5 An open-loop first-order system is characterized b the transfer function )S(G = , where the time τ+ s1 constant is =τ s5 . Calculate the steady-state error when the system input is r(t)=1+6t. Confirm the result by using the final-value theorem. Solution: By the superposition theorem, the system output c(t) could be considered as a sum of a step response and a ramp response. That is − 5/t − 5/t −=−−+−⋅−+=−= e530)e1()e5t(6)1t6(cr)t(e − 5/t , yields =∞ 30)(e By the final value theorem for Laplace’s transform, 6 1 ⎛ 1 ⎞ ⋅+ 5)s6( []()s(G1)s(R)s(C)s(R)s(E ⎜1) −⋅+=−⋅=−= ⎟ = s2 s ⎝ +1s5 ⎠ +⋅ )1s5(s ⋅+ 5)s6( sLim)S(EsLim)t(eLim ⋅=⋅= = 30 t ∞→ →0s s ∞→ +⋅ )1s5(s P3.7 One 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 time constant of a first-order system. What is the phase angle at the bandwidth frequency? 1 1 2 22 Solution: = 707.0 ≈= or1 b =τω+ 2 τ+ s1 22 2 1 bτω+ /1 τ=ω −1 −1 i.e. b 45)1(tan)(tan)j(G °−=−=ωτ−=ω∠ , phase lag is 45° at the bandwidth frequency. P3.8 Figure P3.8 shows the experimentally obtained voltage output of an unknown system subjected to a step input of +10V. Determine the transfer function of the system and locate its pole on the complex plane. Solution: System appears to be first order. So suppose the transfer function is as follow: )s(G C == K , then −⋅= /t τ− )e1(K10)t(c R τ+ s1 From final value theory 10 K sLim)s(CsLim)t(cLim ⋅=⋅= =⋅= V5.2K10 , hence K=0.25 t ∞→ →0s →0s s +τ 1s Now consider time constant， t ⋅=τ 1 =1.4983 ⎛ 1 ⎜1ln −− )t(c ⎟⎞ ⎝ K10 ⎠ Matlab: K=0.25, c=[1.22, 1.84, 2.16, 2.33, 2.41], t=1:5, tao=t./(-log(ones(1,5)-C/10/K))*ones(5,1)/5 5 P4.1 Figure P4.1 shows a closed-loop feedback system with a second-order plant. Determine the damped natural frequency and damping ratio of the closed-loop response. )s(C ω2 = G = 10 = n )s(R + GH1 2 2 2 ++ s2s10s6s ω+ζω+ nn Hence n ==ω 16.310 rad/s, ζωn == 32/6 => ==ζ 103.010/3 2 nd 1 =−⋅=ζ−ω=ω 19.0110 rad/s P4.4 Calculate the required value of gain K shown in follow such that the closed-loop response of the system to a step input is limited to no more than 10% overshoot. Plot the closed-loop poles on the complex plane for the same value of gain. )s(C ω2 = G = K = n ζω = = 12/2 , 2 =ω K )s(R + GH1 2 2 2 n n ++ s2sKs2s ω+ζω+ nn By textbook fig.4.17, 10% overshoot is responding to ζ = 6.0 , 5 4 =ω 1 5 == 67.1 , K 25 == 78.2 , 1 2 6.01 2 ==−=ς−ω=ω 33.1 n ζ 3 9 nd 3 3 2 25 4 The closed-loop poles will be s2s =++ 0 => ±−= j1s or s = −ςω ± dn ±−=ω 33.1j1j 9 3 P4.5 A unity-feedback control system has the forward-path transfer function G(s)=K/s(s+10). Find the closed –loop transfer function, and develop expression for the damping ratio and damped natural frequency in terms of K. Plot the close-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? )s(C ω2 = K = n )s(R 2 2 2 ++ s2sKs10s ω+ζω+ nn 5 Hence n =ω K and ζωn = 102 or =ζ K Closed loop poles located at 2 s n 1j n −±−=ως−±ζω−= 25Kj5 P4.7 Prove that for an under damped second-order system subject to a step input, the percentage overshoot above the steady-state output (as shown in textbook Fig.P4.7) is a function only of the damping ratio ζ. 1 ζω− t By step response 1)t(c −= n d φ+ω )tsin(e 1 ζ− 2 ζω− t ζω− t At the peak time, ζω− n n d ω+φ+ω d n d =φ+ω 0)tcos(e)tsin(e π d +ω φ /)t(tg ςωω= nd = φ => pd ππ=ω ...2,,0T => Tp = ωd π ζπ− ζπ− ζω− 1 n ω π 1 1 ζ− 2 1 ζ− 2 PO = e d sin(ωd ) =φ+ =φ esine ω 1 ζ− 2 d 1 ζ− 2 6 P5.2 A second-order system is given by the transfer function )s(C 10 10 9 = = )s(R 2 ++ 9s2s 9 2 ++ 9s2s Determine: a. damping ratio b. The damped and undamped natural frequency. c. Maximum peak modulus Mp d. The frequency at which Mp occurs. e. The bandwidth, defined as the frequency range over which the modulus does not fall more then 3db below the low-frequency (DC) value. f. The steady-state output for a unit step input. g. The location of the closed-loop poles on the complex plane. h. The rise time. i. The 2% settling time Solution: 2 2 n =ω s/rad3 and ζωn = 1, so ζ = 3/1 , nd n ==−=ζω−ω=ω s/rad83.22219)( 9/10 10 5 2 M p = === 77.12 , np =ζ−ω=ω 721 =2.65 4 12 ζ−⋅ς 2 82 1 From 707.0M == , where r=ω/ωn ζ+− r4)r1( 2222 Squaring, and setting x = r 2 , gives 2 =++− 2x)9/4(1x2x => 2 =−− 01x)9/14(x 7 + 8149 ± 1307 1 29.4 x= ± = , r 07.77 =+= , =ω⋅=ω 29.4r 9 9 9 3 3 b n 1 10 10 f. R(s)=1/s, ss = slim)t(clime ⋅⋅= = t ∞→ →0s s 2 ++ 9s2s 9 g. closed-loop poles for 2 ++ 9s2s , ±−= 22j1s ⎛ 1 ζ− 2 ⎞ h. Rise time 1 ⎜ −1 ⎟ 1 Tr = ⎜ −π tan ⎟ =⋅= s68.0rad91.1 ωd ⎜ ζ ⎟ 22 ⎝ ⎠ 4 i. 2% settling time Ts = = sec4 ζωn P5.3 Consider each of the following closed-loop transfer functions. By considering the location of the poles on the complex plane, sketch the unit step response, explaining the results obtained: Solution: )s(C a. = 20 )s(R ++ )10s)(2s( 1 20 ss= slim)t(clime ⋅⋅= = 1 t ∞→ →0s s ++ )10s)(2s( )s(C b. = 6 )s(R +++ )3s)(2s)(1s( 1 6 Dominant time constant=1s, ss slime ⋅⋅= = 1 →0s s +++ )3s)(2s)(1s( 7 )s(C c. = 9 )s(R 2 ++ 3ss Poles: ±−= 115.0j5.0s n =ζ=ω = 289.0732.1/5.0,3 from Fig4.7, = Δ = 05.4c,%35PO 1 9 ss slime ⋅⋅= = 3 →0s s 2 ++ 3ss )s(C d. = 10 )s(R 2 +++ )5.0s)(5s4s( Solving 2 =++ 05s4s gives = − ± j2s sin 2 ===θ=ζ 89.054.0 5 No overshoot, however, real axis pole will dominate with a time constant at 2s.ess=10/2.5=4 )s(C e. = 1 )s(R 2 +++ )5s2s)(5s( Solving 2 =++ 05s2s gives ±−= 2j1s sin 1 ==θ=ζ 45.0 5 The complex pair will dominate the real axis pole, ess=1/25=0.04 )s(C f. = 100 )s(R 2 2 ++++ )40s12s)(13s6s( Solving 2 ++ 13s6s , gives −= ± j23s Solving 2 ++ 40s12s ,gives −= ± j26s Complex pair closer to im aginary axis will dominate those further away. 3 dom ==ζ 83.0 High damping, no or little overshoot, Steady state output=100/(13*40)=0.192 13 P5.6 Determine the magnitude of the first overshoot of an under damped, second-order system subjected to a unit impulse input. Plot a graph of this magnitude as a function of damping ratio ζ. )s(C ω2 ω2 Solution: = n If R(s)=1 then C )s( = n )s(R 2 2 2 2 2s ω+ζω+ nn n )s( ω+ζω+ d ω2 n ζω− n t 2 = ωd tsine)t(c where nd 1 ζ−ω=ω ωd )t(dc ζω− n t ζω− n t π let = 0 , ζω− n ω+ω dd d =ω 0tcosetsine hence Tp = dt φ φ 2 ςω− n ⎛ 2 ⎞ ωn ω φ ⎜ ς −1 1− ζ ⎟ at this time ,output = eM d sin ωd n exp⎜−ω= tg ⎟ ωd ωd ⎜ 2 ς ⎟ ⎝ 1 ζ− ⎠ This is a function of ωn and ζ ,so plot M/ωn against ζ . 8 P6.1 Figure P6.1 shows a unity-feedback control system with one input R and two disturbance T1 and T2. Find the transfer function relating the output to each of the three inputs and determine the characteristic equation. What is the output if all inputs are present? Solution: )s(C KK KK = 21 = 21 )s(R +++ )10s)(1s(KK 2 21 +++ 21 )KK10(s11s )s(C 1 ⋅ + )10s(K − + )10s(K ⋅−= 2 = 2 )s(T s +++ )10s)(1s(KK 2 1 21 +++⋅ 21 )KK10s11s(s )s(C + + )10s)(1s( ++ )10s)(1s( = = )s(T +++ )10s)(1s(KK 2 212 +++ 21 )KK10(s11s 2 2 The characteristic equation is 21 =+++ 0)KK10(s11s or 21 =+++ 0)]KK10(s11s[s . KK ⋅ + )10s(K + + )10s)(1s( )s(C = 21 )s(R − 2 )s(T + )s(T 2 2 1 2 2 +++ KK10s11s 21 +⋅ 11s(s ++ 21 +++ KK10s11s)KK10s 21 P6.3 A unity-feedback control system with a single disturbance input is shown in Fig.P6.3. Calculate the transfer function relating the error e to the two inputs R and D. Calculate the steady-state error ess when r(t) is a unit ramp. d(t) is a unit step, and K=10. Solution: )s(E + )1s(s + )1s(s = = , )s(R ++ K10)1s(s 2 ++ K10ss )s(E 10 10 −= −= )s(D ++ K10)1s(s 2 ++ K10ss 1 + )1s(s 1 10 −101 −9 ss = slim)t(elime ⋅⋅= slim ⋅⋅− = = =-0.09 t ∞→ →0s s 22 ++ K10ss →0s s 2 ++ K10ss K10 100 P6.5 A radar-tracking antenna is shown in a simplified schematic form in Fig.P6.5. If the commanded input is a step, θd = s/2)s( . While the disturbance Td is a unit step, determine the steady-state error e in the antenna position, assuming J=0.5 N-m s2/rad. Solution: )s(E 2 +⋅ )s2Js(20 +⋅⋅ )2Js(s20 = = , 2 2 θd )s( ++ 20)s2Js( ++ 20s2Js )s(E 1 1 −= −= 2 2 τd )s( +⋅+ )10s(2Js ++ 20s2Js 2 ⋅ + )2s5.0(s20 1 1 ss slime ⋅⋅= slim ⋅⋅− −= 05.0 →0s s 2 ++ 20s2s5.0 →0s s 2 ++ 20s2s5.0 9 P6.10 Consider the system discussed in SP6.3. Represent graphically, in as compact a form as possible, the relationship between ζ , ω n and the system parameters Ka and Kl. What conclusions may be drawn from the graph regarding the possible range of ζ. If the sign of the rate feedback signal is changed from a minus to a plus, what effect does this have on the graph? Solution: θ ⎡ 0ap N/KKK ⎤ K J/K l = ⎢ ⎥ = = 2 2 θd ⎣⎢ cJs(s ++⋅ + 0ap0v1 N/KKK)KKK ⎦⎥ ++ KBsJs s ++ )J/K(s)J/B( = 0ap N/KKKK , = cB + KKK 0v1 n ==ω 0ap NJ/KKKJ/K ⋅⋅= a = K7.14K)0125.030/(5.136 a J/B + KKKc + ⋅ ⋅ K5.1302.005.0 + K27.005.0 =ς = 0v1 = 1 = 1 ωn J22 ω⋅⋅ n ⋅ 02 ⋅ K7.140125. a 3675.0 ⋅ Ka Clearly, only Ka affects ωn. As for ζ , it is likely that K1 will affect it the most, so we will plot graph for ζ − K1 with fixed values of Ka. Suppose Ka=1, then ζ = + K75.0139.0 1 . So that if 1 << 1K0 then < ζ < 89.0139.0 . Now if Ka=2, then ζ = + K53.0098.0 1 , and so on. A conclusion is B to be only dependent K1 and independent of Ka , butζwill be varied with either Ka or K1 . If the rate feedback signs changes, it is the same as making K1 < 0. That is say ζ −= K75.0139.0 1 ( as Ka=1) . It means the value of ζ could be decreased by K1. In order to represent the relationship betweenζ, ωn and the system parameters Ka and Kl graphically in as compact a form as possible, several curves could be drawn in one graph. ωn ζ(Kl1) ζ(Kl2) Ka 10 For problems 7.1-7.4 reduce the transfer function to an approximate first of second-order system. Plot the closed-loop poles on the complex plane, and sketch the unit step response. C 100 P7.2 = R 2 +++ )20s12s)(1s( C 100 5 5 Solution: = = ≅ = 5C R +++ )10s)(2s)(1s( 1 1 1 ss 1)(s1( ++ 1)(s + )s 1)(s1( ++ )s 2 10 2 That is a second-order, over damped system. Further reduction might be C ≅ 5 R + )s1( first order , with dominant time constant τ=1. P7.6 For the system shown in C = 10 R + s)(1s( 2 + 2 + )6s Determine the maximum value of the output when the input takes the form R(s)=4/s Solution: Since ςωn = 1 and τ=1, then =β 1, hence the percentage overshoot could be negligible. So the maximum output should be the steady state output. 10 If R=4/s, then max ∞ 4)(cc =⋅== .676 6 P7.8 Calculate the maximum value of output when a unit step input is applied to the system described by the closed-loop transfer function + )3s(4 C = R 2 ++ 10s3s2 2/3 τ 3/1 Solution: n ==ςω 75.0 , τ = 3/1 => =γ == 4 2 ςωn 75.0 By n ==ω 52/10 , yields =ς = 34.0575.0 From textbook Fig7.3 , when ς = 34.0 and γ = 4 , the percentage overshoot should be 45%. The maximum value of output with unit step input is 12 )45.01( =+⋅ 1.74 10 11 P8.1 For unity-feedback systems described by the following open-loop transfer functions, determine the system type, any finite, nonzero error coefficients, and the corresponding steady-state error. = 1)s(H 10 2 (a) G )s( ∴= Type 2 system, a =⋅= 10)s(GslimK , s 2 →0s 1 steady- state error to a unit acceleration input is ss )(e ==∞ 1.0 K a +1s (b) )s(G = type 1 system, v =⋅= 2.0)s(GslimK 23 +++ )5s7s2s(s →0s 1 steady state error to a unit ramp input is ss )(e ==∞ 5 K v 7 (c) )s(G = type 0 system, p == 7)s(GlimK 2345 +++++ 1s2s3s3s2s →0s steady state error to a unit step input is )(e =∞ 1 = 125.0 ss + Kp1 5 (d) )s(G = type 0 syst em, p == 5)s(GlimK + )1s( 3 →0s steady state error to a unit step input is )(e 1 ==∞ 16.0 7 ss + Kp1 2 +++ )4s)(3s)(2s( (e) )s(G = type 1 system, v =⋅= 24)s(GslimK s →0s steady state error to a unit ramp input is )(e 1 ==∞ 0417.0 ss Kv P8.8 Figure P8.8 shows a speed control system subject to a disturbance torque T. If the system transfer functions are given by Gc(s)=K/(s+1), H(s)=1. Determine the steady-state er ror due to a unit step change in input R (assuming T constant) and a unit step change in the disturbance torque T(assuming R constant), assuming K=1. Determine K such that the result ing chan ge in speed is less than 10% of the disturbance value. Since System is type I, so the steady-state error due to a unit step change in input R is zero. E +1s As R=0, −= T ++⋅ K)1s(Js 1 +− )1s( 1 so ss= slim)t(elime ⋅⋅= −= → →0s0t s ++⋅ K)1s(Js K when K=1, ess=1 Obviously, as ess=10% of disturbance change value, the gain K should be increased to K=10 12 P9.6 For each of the open-loop poles and zeros shown in Fig.P9.6, sketch the corresponding root locus. ψ=180-90-135=-45 P9.7 The open-loop pole-zero map of a satellite attitude control system is shown in Fig.P9.7. Draw the root locus and determine, as accurately as possible, the point at which the locus leaves the real axis. 1 Solution: )s(GH = s +⋅+⋅ )5s()4s( Asymptotes: a +=σ∠ = 300,180,603/k3603/180 , a =σ − − = −33/)54( ω ω ω tg−1 + tg−1 −π= tg−1 +σ 4 +σ 5 σ ω ω + ω +σ 92 1 => +σ +σ 44 −= => −= ω ω 2 2 1− ⋅ σ 209 ω−+σ+σ σ +σ +σ 44 2 2 2 2 ω 20 183 =+ω−σ+σ 020 => )3( 9 −=−+σ (Hyperbola) 3 3 =ω 0 => −=−=σ 47.133/7 leaving point on the real axis =σ 0 , 20 ==ω ±4.47 intersect with imaginary axis 13 P10.8 Shown in Fig.P10.8 is the pole-zero map of a feedback control system in which the location of the real-axis pole is given by the variableφ. Determine the angle of emergence and entry into the complex poles and zeros when (a)φ=0 and (b)φ=-3 .Sketch the resulting root locus for each case. Solution: (a) φ=0 emergence from pole −−−=ψ + = −18724590135180 entry into zero ++=ψ + − = 10890108135135180 (b)φ=-3 emergence from pole −−−=ψ + = 9072459027180 entry into zero +++=ψ − = 369010813563180 P10.9 Plot the root locus for the feedback control system shown in Fig.P10.9, and determine the location of all roots for K=20. e u 1 c r K + s - 1 2 ++⋅+ )2s2s()2s( Solution: Conjugate poles locates on −±− 12j1 Asymptotes: a =σ∠ 180 + = 315,225,135,454/k3604/ a −=σ 4/4 −= 1 Leaving point on the real axis S=-1 since 2 ) = 11)(1)(1(1K ε−ε−ε+ε−>= 4 Consider the location of root for gain K=20, x=y+1 and with [ ] [ ] [ ] [ ++⋅−+=+⋅−+−= y)1y()1y(yyx)1y()1x(K 22222222 ]. = [2 2 ] [ 2 +⋅+− 1]=+ y4y4)1y2(y2y21y2y 422 +=−+ 1 . 4 −⋅= 1K707.0y =1.48 It results in 4 roots of charaterstic equation for closed-loo p: -2.48-j1.48, -2.48+j1.48, 0.48-j1.48, 0.48+j1.48, 14 P11.5 A robot take television tubes from a convayer belt and places then in a test fixture. The vetical axis of the robot is controlled by a system with open-loop transfer function 2 +⋅ )16s(K s)(Gh = 2 ++⋅ )120s10s(s Determine the “best” value of K to accomplish this task, stating reasons for the choice. Solution: There two possibilities for the locus go to imaginary zeros. It depends the leaving points on the real axis. So from check gain on the real axis s=0 to -∞ 2 ++⋅ )120s10s(s K −= 2 + )16s( dK 2 2 2 ++⋅⋅−+++ )120s10s(ss2)16s)(120s20s3( = ds + )16s( 22 24 =++− 01920s320s78s s=-3.5, s=-9.5 For robot conveyer, the best way is to avoid the overshoot. − ⋅ − + )1203525.12(5.3 So the damping ratio zeta=1, and take s=-3.5.The gain is about 12, K −= = 12 + )1625.12( P11.8 Consider the control system shown in Fig.P11.8. Sketch the root locus and estimate the range of K for which the closed-loop poles satisfies the following design constraints: a. Not more than 15% overshoot. b. B. Dominant time constraint less than 2s. r e K u +⋅ )1s(2 c c. C. Damped natural frequency of oscillation + + 5.1s +⋅ )2s(s less than 2 rad/s. - Solution: Asymptotes: ∠σa = 180 − + − = 270,90)13/(k360)13/( . σa = − − + −= 25.12/)125.1( +⋅+⋅ )2s()5.1s(s dK 2 2 ++⋅−+++ )5.2s5.3s(s)1s)(5.2s7s3( K −= = +⋅ )1s(2 ds + )1s( 2 23 =+++ 05.2s7s5.6s2 Leaving point on the real axis is s=-1.25 Consider the dominant effect of single pole on the real axis, is no any overshoot and damped natu ral frequency. Only condition to choice is time constant less than 2s. 1 So <=τ 2 , then the dominant pole should be = ςωn > 5.0p . ςωn −⋅−⋅ 25.05.15.05.0 => K > = 5.1 − )15.0 15 P12.1 For the following open-loop transfer functions, sketch the nyquist diagrams: K 10 5 4 a. )s(GH = b. )s(GH = c. )s(GH = d. )s(GH = + s21 s +⋅ )s41( +⋅ )2s(s 2 2 ++ 1ss 5 e. )s(GH = ++⋅ )2s)(1s(s Solution: P12.9 Figure P12.9 shows some experimentally determined open-loop frequency response plots. Match each plot with one of the transfer functions listed below: 1 a. G )s(h = +⋅+⋅+ 321 )sT1()sT1()sT1( K b. )s(Gh = +⋅ s)(as(s + )b 5 c. )s(Gh = 3 +⋅ )1s(s 1 d. )s(Gh = s 10 e. )s(Gh = + ⋅ +⋅+⋅+ )ds()cs()bs()as( f. += sT1)s(Gh Solution: a - (a), b – (d), c - (e), d –(c), e –(f), f - (b), 16 P13.1 a. In Fig. P13.3a, match each s-plane contour with the corresponding GH(s) con tour. b. In Fig. P13.3b, how many zeros are enclosed by the s plane contour? c. In Fig. P13.3c, are any of the points A-E encircled by the contour? If so, how many and in which direction do the encirclements occur? Solution: a. As arrowhead on the figure shown. b. 3 c. A- 1 with clockwise B- 0 C- 2 with anti-clockwise D- 0 with clockwise E- 2 with anti-clockwise P13.4 For the following open-loop transfer functions, sketch the Nyquist diagrams for both positive and negative frequency and comment on the system stability: K a. )s(Gh = + s21 10 b. )s(Gh = +⋅ )s41(s 10 c. )s(Gh = +⋅ )2s(s 2 4 d. )s(Gh = 2 ++ 1ss 5 e. )s(Gh = +⋅+⋅ )2s()1s(s Solution: a stable b stable c Unstable 5 5 d stable e stable: =ω 2 )j(GH =ω <= 1 +⋅+⋅ 42122 6 17 P13.7 Fig. P13.7 shows the open-loop frequency response of a system. Estimate the transfer function, and determine the stability of the closed-loop system. Solution: +⋅+⋅ )bs()as(K )s(Gh = s3 It is stable, since Z=P+N = 2+ (-2) =0 P13.8 The open-loop frequency response shown in Fig.P13.8 was obtained experimentally. If the critical point (-1,0j) is located first at A, then at B, and finally at C, comment on the system stability. Reconsider stability of the modified contour shown in Fig. P13.6. Solution: A-unstable, B- stable, C-unstable 18 P14.1 From the Nyquist diagram, find the maximum value of K for which the following systems are stable: K K5 K K a. )s(Gh = b. )s(Gh = c. )s(Gh = d. )s(Gh = +⋅ )as(s 2 +⋅ )2s(s + )1s( 3 +⋅+ )2s()1s( 2 Solution: a. Stable for any value of K b. Unstable for any value of K. c. Stable for K=8: tg3)j(GH −1ω⋅−=ω∠ = 180 => =°=ω 360tg => K= += )13(K 3 =8 d. Stable for K=36 ω ω + ∠ −1 −1ω⋅−ω−=ω 2/tg2tg)j(GH => =ω− 22 => =ω 8 ω2 1− 4 K => )j(GH =ω => K=36 ()+⋅+ 4818 P14.2 A unity-feedback control system has the forward-path transfer function K )s(G = 2 ++⋅ )9ss(s Find the maximum value of K consistent with stability and check the result using Routh’s method and by drawing the root locus. Solution: 1 K Obviously as ω=3, ∠GH(jω)=-180, K)j(GH ⋅=ω = => K<9 ⋅33 9 Gheck with Routh’s: 23 +++= Ks9ss)s(G so K < 9 P14.5 Fo the systems described by the Nyquistg diagrams shown in Fi. P14.5 estimate the gain margin and phase magin. Solution: a. PM>0 , GM= infinity b. PM>0, GM>0 c. PM<0 , GM<0 d. PM>0 , infinity e. PM>0 , GM= infinity f. PM<0, >0 , GM<0, >0 19 P15.1 Using straight-line approximations, plot the Bode diagram of the systems described by the following open-loop transfer function: 10 25 s( + )1 s16 a. )s(Gh = b. )s(Gh = c. )s(Gh = 1 1 +⋅ )5s(s +⋅+ )2s()1s( 1( 1)(s ++ )s 3 2 1 1(2 +⋅ )s ⋅ )2s(4.0 ⋅+⋅ d. )s(Gh = 10 .e. . )s(GH = 1 1 1 1 1(s +⋅ 1()s 1()s +⋅+⋅ )s + )2.0s( 1()1s( +⋅+⋅ )s 100 1000 2000 10 Solution: 10 -20db/de a. )s(GH = ++ )2/s1)(3/s1( 10 -40db/de ω -20db/de 2 3 20 60 +⋅ )1s(5 -20db/dec b )S(GH = +⋅ )5/s1(s -20db/dec 5 ω 1 5 25 s8 c.. )s(GH = +⋅+ )2/s1()1s( +20db/dec -20db/dec 8 ω 0.125 1 2 16 )10/s1(2 ⋅+⋅ d. )s(GH = . +⋅+⋅+⋅ )2000/s1()1000/s1()100/s1(s ω 10 2 100 1000 2000 -20db/dec -40db/dec )2/s1(4 ⋅+⋅ -60db/dec .e. . )s(GH = . 1( + )2.0/s +⋅+⋅ /s1()s1( )10 -20db/dec 4 ω 0.2 1 2 4 10 -20db/dec -40db/dec 20 P15.2 The following systems contain second-order components. By estimating the magnitude and phase contributions from those elements, plot the Bode diagram for each system: 15 + s2.01 +⋅ )3s(100 a. )s(GH = b. GH )s( = , c. )s(GH = , 2 ++ 64ss 2 ++⋅ )1.0s5.0s(s 2 ++⋅ 2 )s01.0s2.01(s 20 2 ++⋅ )4s8.0s(625.0 d. )s(GH = , e. )s(GH = , 2 2 ++⋅++ )50s3s()17s2s( 2 ++ 25s4s Solution: 64/15 a. )s(GH = 1 1 s2 ++ 1s 64 64 15 ω = −∠ arctg 2 )64( ω+ω− 222 64 ω− +⋅ )s2.01(10 b. )s(GH = 2 ++⋅ )10s5s10(s + 04.01 ω2 5.0 ω⋅ ω = −−∠ tg90 −1 + tg −1 2 22 25.0)1( ω+ω−⋅ω 2 1 ω− 5 +⋅ )1s333.0(300 c. )s(GH = 2 ++⋅ 0.0s2.01(s 2 )s1 9100 ω+ 2 2.0 ω⋅ ω = −−∠ tg180 −1 + tg −1 2 ω+⋅ω 2 )01.01( 01.01 ω− 2 3 85/2 d. )s(GH = , 1 2 1 3 ( s2 ()1s s2 ++⋅++ )1s 17 17 50 50 2.0 = 222 222 9)50(4)17( ω+ω−⋅ω+ω−⋅ω 2ω 3ω −∠ tg −1 − tg −1 17 ω− 2 50 ω− 2 2 ++⋅ )1s2.0s25.0(1.0 e. GH )s( = 2 ++ 1s016s04.0 22 ω+ω−⋅ )2.0()5.01(1.0 2 = 22 2 ω+ω− )16.0()04.01( 16.0 ω 2.0 ω −∠ tg −1 + tg −1 04.01 ω− 2 5.01 ω− 2 21 P16.1 For following systems, sketch the Bode diagram, and from the straight-line approximations to the gain and phase plots, estimate the maximum value of K for which the system is stable: K K ⋅sK a. )s(Gh = b. )s(GH = c. )s(Gh = +⋅+⋅ )4s()1s(s 2 +⋅ )s1(s + )2s( 4 K )s1(K5 ⋅+⋅ d. )s(Gh = .e. . s(GH ) = 2 ++⋅ )16s2s(s s2 +⋅ )3/s1( 2 Solution: 4/K a. )s(Gh = 4/s()1s(s +⋅+⋅ )1 −1 −1 Let c ω−ω−−= c 4/tgtg90180PM =0 ω + ωcc 4/ => 90tg =° ∞= => ωc = 2 ω⋅ω− cc 4/1 4/K From c =ω 1)j(GH , =1 => K=20 2 2 +⋅+⋅ 15.0122 K b. )s(GH = , Always unstable. 2 +⋅ )s1(s ⋅sK c. )s(GH = + )12/s( 4 In order to let PM=0, ω When ⋅−+= tg490180PM − c1 = 0 , 2 270° c ⋅=ω tg2 = 85.4 4 + )1426.2( 22 From c =ω 1)j(GH => K = = 7.9 85.4 16/K d. )s(GH = . 2 ++⋅ )18/s16/s(s PM=0 => ωc = 4 16/K From c =ω 1)j(GH = => K=32 ⋅ )2/1(4 )s1(K5 ⋅+⋅ e. . )s(GH = 2 +⋅ )3/s1(s 2 −1 −1 Let c ω⋅−ω+−= c 3/tg2tg180180PM =0 By strait line approximations to the phase plot, suppose ωc + ωc 3/3/ => c =ω => c =ω 3 ω⋅ω− cc 9/1 +⋅ 31K5 From c =ω 1)j(GH , = 1 => K=2/5=0.4 +⋅ )3/11(3 22 P16.2 Estimate for each system given below the gain margin and phase margin: +⋅ )2s(15 20 2 ++⋅ )9s2s(1.0 a. )s(GH = , b. )s(GH = , c. )s(GH = , 2 +⋅ )9s(s 2 ⋅+⋅ s()5s(s 2 ++ 4s2 ) 2 +⋅ )10/s1(s 2 +⋅ )1s(2000 +⋅+⋅ )20s()10s(5.0 d. )s(GH = , e. )s(GH = , +⋅+⋅ )5s()2s(s +⋅ 10s( +⋅ )20s() 2 +⋅+⋅ )100s()1s(s Solution: +⋅ )12/s(3/10 a. )s(GH = , 2 +⋅ )19/s(s c ==ω 83.13/10 , 83.1 83.1 = tgPM −1 − tg −1 = 42.5-11.5=31 2 9 1 b. G )s(H = , 2 2 ++⋅+⋅ )12/s4/s()15/s(s 1 −− 11 5.0 c =ω 1, PM −= tg − tg =-45 5 − 25.01 ⋅ 2 9/s(9.0 + s2 + )19/ c. )s(GH = , 2 +⋅ )10/s1(s 2 c =ω 9.0 =0.95, ⋅ 9/95.02 .0 95 PM = tg −1 ⋅− tg2 −1 =2.3 − 2 9/95.01 10 +1s d. GH )s( = , )12/s(s 5/s( + +⋅+⋅⋅+⋅ )120/s()110/s()1 c =ω 5.1 , 1 1 1 1 tg1tgPM −− 11 −−= tg −1 − tg − − tg −11 2 5 10 20 =45 – 26.6 – 11.3 – 5.7 – 2.9 = -1.6 ⋅+ s()110/s( 20/ + )1 e. G )s(H = , 2 +⋅+⋅ )1100/s()1s(s c =ω 1, +−= −− 0tg1tgPM 1. + −111 − −1 01.0tg05.0tg = -45 –5.7 –2.86 – 0.57 = -54.13 23 P17.7 Consider the system described by the open-loop transfer function element GH(s)=5/s and H(s)=1/(s+1) Draw the Bode diagram of the o pen-loop transfer function and write down an expression for the unit impulse response of the closed-loop system. Solution +⋅ )1s(5 GH(s)=5/s, and H(s)=1/(s+1), then )s(G = s +⋅ )1s(5 S+1 5 s s 1 +1s Y 5 + −+ 2025s55s5 4⋅5 By block diagram algebra, = )1s( =+⋅ = 5 −= . + 5sR + 5s + 5s + 5s For unit impulse input, −δ⋅= e4)t(5)t(y − t5 P17.8 Sketch the root locus of t he system shown in Fig.P17.8 and determine the value of K needed to give the dominant closed-loop poles a damping ratio of ζ=0.707. Estimate the closed–loop unit step response of the system by potting the Bode diagram of the open-loop transfer function for this value of K, and discuss the two results. Compare the result with direct analysis of the closed-loop transfer function. Solution . By root locus, the K should be calculated as follow: By Bode diagram, ζ=0.707 => PM=63 −1 =⋅= 21/22K by ω−−= c )1/(tg90180PM => c =ω 2 2 by K= 1)K( ω⋅= c =2 C + )s1(K + )s1(K K = = n =ω K , =ς If ζ=0.707, then the K=2. R 2 ++ KKss 2 +⋅⋅⋅+ KsK2/K2s 2 24 P18.1 Figure P18.1 shows the magnitude part of the frequency response of different systems. For each case determine the bandwidth, the size of the resonance peak, and the frequency at which the peak occurs. Solution (a) db3 =− ⋅ = .1DV707.0 414 b =ω s/rad4 p 5.2M ω= b ωr = s/rad3 (b) take –3db with Mp 1 =ω 2.28 1 =ω 3.794 12b =ω−ω=ω 766 p −= db5M ωr = s/rad105 P18.7 Consider the system shown in Fig. P 18.7. Estimate the peak time and percentage overshoot for the closed-loop system if the input is a unit step. Solution From Mr=9db => r = 8.2M r =ω s/rad15 By Fig18.9, or formulation 1 M p = 12 ς−⋅ς 2 => ζ= 0.16 From ωr = s/rad15 By Fig18.10, or formulation 2 ω nr 21 ς−⋅ω= => ωn= take –3db with M, ωb = 40 Form Fig 18.10 or formulation 2/1 ⎛ 2 42 ⎞ ω ⎜ 21 +ζ−⋅ω= 442 ζ+ζ− ⎟ b n ⎝ ⎠ n =ω 7.26 => d =ω 3.26 => /T ωπ= = s12.0dp , From Fig4.17 PO=60% 25 P19.2 Consider the unity-feedback system in Fig.P19.2. Design a phase lead compensation network that give the closed-loop system: GH(s)=K/(s*(1+0.5s)) a. A phase margin of 50º. b. A velocity error of less than 10%. -20db/dec Solution -40db/dec By requirement ess=10% for ramp input, => K=Kv=10 New 52202K ===⋅=ω 4.47 -20db/dec c ω −1 47.4 c −−=ω−= tg90180)(GH1800PM = 24.2 2 2.9 4.47 7.2 10 17.5 2 By requirement PO=10% => ζ=0.6 => PM =60 => + 45sin1 m −=φ = − = 3624600PMPM Take 45, =α = 7.5 -40db/dec − sin1 45 New cm =⋅=⋅ω=ω 15.76.147.46 mp =⋅=⋅ω=ω 5.1745.215.76 , mz =ω=ω = 92.245.2/15.76/ , 1 T =α = 3.0 4, = 057.0T ωz α+ Ts1 + s34.01 c )s(G = = , + Ts1 + s057.01 +⋅ )s34.01(10 c )s(G)s(GH = +⋅+⋅ )s057.01()s5.01(s P19.3 Design a phase lead cascade compensation filter that provides the unity-feedback system 20 )s(GH = + )s101( 2 with a percentage overshoot to a unit step of less than 15%. Solution =⋅=ω 45.01.020 c -40db/dec −1 c 355.4tg2180)(GH1800PM °=⋅−=ω−= 20 PO=15% => ζ=0.6 => PM =60° m −=φ = − = 2535600PMPM Take 30 -20db/dec + 30sin1 ω =α = 3 0.1 0.35 0.45 0.6 1.02 − 30sin1 -40db/dec New cm =⋅=⋅ω=ω 6.0316.145.03 mp =⋅=⋅ω=ω 02.1732.16.03 , mz =ω=ω = 35.0732.1/6.03/ α+ Ts1 + s9.21 c )s(G = = + Ts1 + s97.01 +⋅ )35.0/s1(20 c )s(G)s(GH = 2 +⋅+ )02.1/s1()s101( 26 P20.2 A unity feedback control system has the open-loop transfer function K )s(GH = s +⋅ )s67.01( Design a suitable phase lag compens ation filter that ensures that the closed-loop system meets the following performance requirements: a. A velocity error no more than 20%.. -20db/de b. A maximum percentage overshoot to a step of 10% Solution -40db/de By requirement ess=20% for velocity error => K=Kv=5 By requirement PO=10% => ζ=0.6 => PM =60° Draw block diagram, -20db/de the integrator forms a line with slope of -20db/dec. 6.67 ω =ω 5 => ω = =⋅ 5.767.0/15 =2.74 0 c 0.011 0.075 0.75 1.5 2.7 5 -40db/de −1 45.2 c 90180)(GH180PM −−=ω−= tg <60° 5.1 Select new ω’c, let PM=60° => ωc = ⋅ = 75.0)67.02/(1 => c =ω=β = = 67.675.0/5)75.0(GH)'(GH ω cz =ω= = 075.010/75.010/' zp =βω=ω = 011.067.6/075.0/ , ++ 075.0/s1Ts1 c )s(G = = 1 β+ Ts + 011.0/s1 ⋅ + 5(10 )07.0/s1 )s('GH = 67.01(s )s +⋅+⋅ )011.0/s1( P20.4 Design a phase lag cascade compensation filter Gc(s) for open-loop system K )s(GH = ++⋅ )s4.01)(s2.01(s such that the closed-loop system has a damping ratioζ=0.7 and a velocity error Kv=100. What is the estimate settling time of the compensat ed-system output when the input is a unit step? Sketch the root locus of the uncompensated and compen sated systems. Solution By requirement K=Kv=100, ζ=0.7 => PM =65. => new ωc = 7.0 for PM c =ω=β GH)'(GH 1( = = 143)7.0/100() z =ω = 07.010/7.0 , zp =βω=ω 0005.0/ -20db/de + Ts1 + 07.0/s1 100 + 07.0/s1 c )s(G = = c )s('GH = ⋅ β+ Ts1 + 0005.0/s1 ++⋅ )s4.01)(s2.01(s + 0005.0/s1 -40db/de 10.8 -20db/de ω 0.000 0.07 0.7 2.5 5 15. 100 -40db/de 27 P21.2 A unity feedback comtrol system has the open-loop transfer function 40 )s(GH = +⋅+ )2s()4s( Use the root locus method to design a suitable PI controller such that the system has zero steady-state position error and the dominant closed-loop poles have a damping ratio of 0.5. Estimate the percentage overshoot, peak time, and 2% settling time of the closed-loop, unit step resp onse. Solution ζ=0.5 By requirement ess=0 for position error => Type I K=8 By requirementζ=0.5 => PM >45 5 Bode form )s(GH = +⋅+ )2/s1()4/s1( ╳ ○╳ ╳ -4 -2 Using a PI controller, the transfer function become +⋅⋅ )1sT(K5 ('GH )s = 11 +⋅+⋅ )2/s1()4/s1(s Since there is a integrator in the GH’(s), then ess for position error will be zero. ω In order to meet the need ofζ=0.5, set T1=1/2=0.5, and 5K1=4 , that is K1=0.8 4 +⋅ )1s5.0(4 +⋅ )1s5.0(8.0 )s('GH = , )s(Gc = +⋅+⋅ /s1()41(s )2/s s ζ=0.5 => PO=16%, 4 4 8 ts = = = = s6.1 ςωn c ς−ω⋅ )67.01.1/(5.0 78.0/4 π 14.3 peek time t p = = = 0 s8. ωd 2 ⋅− 78.0/4)5.01( 28 P21.6 For the unity-feedback system described by the open-loop transfer function 5 )s(GH = + )10s( 2 Design a PD controller such that the closed-loop system has a damping ratio of ζ=0.7 and the 2 settling time is 32 less than 0.1s. Solution ω Using PD controller the new GH becomes 10 40 80 11 +⋅⋅ )1sT(K05.0 )s('GH = 3200 1( + )10/s 2 By requirement, ζ=0.7,ts=0.1s => ωc=8/ts=80 ζ=0.7=> PM=63°=> 1/T1 =ω c/2 =40 2 ⎛ 3200 ⎞ K05.0K =⋅= ⎜ ⎟ = 32 1 ⎜ ⎟ 10 ⎠⎝ => T1=1/40=0.025, K1=32/0.05=640 +⋅ )40/s1(32 )s('GH = Gc(s)=640 (0.025s+1) + )10/s1( 2 29";