P3.4 The open-loop transfer function of a unity negative feedback system is)1(1)(+=s s s GDetermine the rise time, peak time, percent overshoot and setting time (using a 5% setting criterion).Solution: Writing he closed-loop transfer function 2222211)(nn ns s s s s ωςωωΦ++=++=we get 1=n ω, 5.0=ς. Since this is an underdamped second-order system with 5.0=ς, thesystem performance can be estimated as follows.Rising time.sec 42.25.0115.0arccos 1arccos 22≈-⋅-=--=πςωςπn r tPeak time.sec 62.35.011122≈-⋅=-=πςωπn p tPercent overshoot %3.16% 100% 100225.015.01≈⨯=⨯=--πςπςσee pSetting time.sec 615.033=⨯=≈ns t ςω(using a 5% setting criterion)P3.5 A second-order system gives a unit step response shown in Fig. P3.5. Find the open-loop transfer function if the system is a unit negative-feedback system.Solution: By inspection we have %30% 100113.1=⨯-=pσSolving the formula for calculating the overshoot,3.021==-ςπςσep, we have362.0ln ln 22≈+-=pp σπσςSince .sec 1=p t , solving the formula for calculating the peak time, 21ςωπ-=n p t , we gets e c / 7.33rad n =ωHence, the open-loop transfer function is )4.24(7.1135)2()(2+=+=s s s s s G n nςωωP3.6 A feedback system is shown in Fig. P3.6(a), and its unit step response curve is shown in Fig. P3.6(b). Determine the values of 1k , 2k , and a ..1.1Figure P3.5Solution: The transfer function between the input and output is given by2221)()(k as sk k s R s C ++=The system is stable and we have, from the response curve,21lim )(lim 122210==⋅++⋅=→∞→k sk as sk k s t c s tBy inspection we have %9% 10000.211.218.2=⨯-=pσSolving the formula for calculating the overshoot, 09.021==-ςπςσep, we have608.0ln ln 22≈+-=pp σπσςSince .sec 8.0=p t , solving the formula for calculating the peak time,21ςωπ-=n p t , we gets e c / 95.4rad n =ωThen, comparing the characteristic polynomial of the system with its standard form, we have22222n n s s k as s ωςω++=++5.2495.4222===n k ω02.695.4608.022=⨯⨯==n a ςωP3.8 For the servomechanism system shown in Fig. P3.8, determine the values of k and a that satisfy the following closed-loop system design requirements. (a) Maximum of 40% overshoot. (b) Peak time of 4s.Solution: For the closed-loop transfer function we have 22222)(nn ns sks k sk s ωςωωαΦ++=++=hence, by inspection, we getk n=2ω, αςωk n =2, and nnkωςςωα22==Taking consideration of %40% 10021=⨯=-ςπςσepresults in280.0=ς.In this case, to satisfy the requirement of peak time, 412=-=ςωπn p t , we have.s e c / 818.0r a d n =ω.2.2(a)(b)Figure P3.6Figure P3.8Hence, the values ofkandaare determined as67.02==n k ω, 68.02==nωςαP3.10 A control system is represented by the transfer function)13.04.0)(56.2(33.0)()(2+++=s ss s R s CEstimate the peak time, percent overshoot, and setting time (%5=∆), using the dominant polemethod, if it is possible.Solution: Rewriting the transfer function as]3.0)2.0)[(56.2(33.0)()(22+++=s s s R s Cwe get the poles of the system: 3.02.02 1j s ±-=，, 56.23-=s . Then, 2 1，s can be considered as a pair of dominant poles, because )Re()Re(32 1s s <<，.Method 1. After reducing to a second-order system, the transfer function becomes13.04.013.0)()(2++=s ss R s C (Note:1)()(lim==→s R s C k s Φ)which results in sec / 36.0rad n =ω and 55.0=ς. The specifications can be determined ass e c 0.42112ςωπ-=n p t , %6.12% 10021=⨯=-ςπςσeps e c 67.2011ln 12=⎪⎪⎪⎭⎫⎝⎛-=ς∆ςωns t Method 2. Taking consideration of the effect of non-dominant pole on the transient components cause by the dominant poles, we haves e c 0.8411)(231=--∠-=ςωπn p s s t%6.13% 10021313=⨯-=-ςπςσes s s ps e c 6.232ln 1313=⎪⎪⎭⎫⎝⎛-⋅=ss s t ns ∆ςωP3.13 The characteristic equations for certain systems are given below. In each case, determine the value of k so that the corresponding system is stable. It is assumed that k is positive number.(a) 02102234=++++k s s s s (b) 0504)5.0(23=++++ks s k sSolution: (a) 02102234=++++k s s s s .The system is stable if and only if⎪⎪⎩⎪⎪⎨⎧<⇒>=>9 022010102203k k D ki.e. the system is stable when 90<<k .(b) 0504)5.0(23=++++ks s k s . The system is stable if and only if⎪⎩⎪⎨⎧>-+⇒>-+⇒>+=>>+0)3.3)(8.34( 05024 041505.00 ,05.022k k k k k k D k ki.e. the system is stable when 3.3>k .P3.14 The open-loop transfer function of a negative feedback system is given by)12.001.0()(2++=s ss Ks G ςDetermine the range of K and ς in which the closed-loop system is stable. Solution: The characteristic equation is02.001.023=+++K s s s ς The system is stable if and only if⎪⎩⎪⎨⎧<⇒>-⇒>=>>ςςς20 001020 0101.02.002.0 ,02K K .ς.K D kThe required range is20>>K ς.P3.17 A unity negative feedback system has an open-loop transfer function )16)(13()(++=s s s K s GDetermine the range ofkrequired so that there are no closed-loop poles to the right of the line1-=s . Solution: The closed-loop characteristic equation is18)6)(3( 0)16)(13(=+++⇒=+++K s s s K s s si.e. 01818923=+++K s s sLetting 1~-=s s resulting in 0)1018(~3~6~ 018)5~)(2~)(1~(23=-+++⇒=+++-K s s s K s s sUsing Lienard-Chipart criterion, all closed-loop poles locate in the right-half s~-plane, i.e. to theright of the line 1-=s , if and only if⎪⎩⎪⎨⎧<⇒>-⇒>-=>⇒>-14 08.182 0311018695 ,010182K K K D K KThe required range is 91495 <<K , or56.10.56 <<KP3.18 A system has the characteristic equation0291023=+++k s s sDetermine the value of k so that the real part of complex roots is 2-, using the algebraic criterion.Solution: Substituting 2~-=s s into the characteristic equation yields 02~292~102~ 23=+-+-+-k s s s ）（）（）（ 0)26(~~4~ 23=-+++k s s sThe Routh array is established as shown.If there is a pair of complex roots with real part of 2-, then026=-ki.e. 30=k . In the case of 30=k , we have the solution of the auxiliary equation j s ±=~, i.e. j s ±-=2.3s 1 12s 4 26-k1s 0sP3.22 The open-loop transfer function of a unity negative feedback system is given by)1)(1()(21++=s T s T s Ks GDetermine the values of K , 1T , and 2T so that the steady-state error for the input, bt a t r +=)(, is less than 0ε. It is assumed that K , 1T , and 2T are positive, a and b are constants. Solution: The characteristic polynomial is K s s T T s T T s ++++=221321)()(∆Using L-C criterion, the system is stable if and only if2121212121212 0 01T T T T K T KT T T T T K T T D +<⇒>-+⇒>+=Considering that this is a 1-type system with a open-loop gain K , in the case of 2121T T T T K +<,we have 00.. εεεεεbK Kb v ss r ss ss>⇒<=+=Hence, the required range for K is21210T T T T K b+<<εP3.24 The block diagram of a control system is shown in Fig. P3.24, where )()()(s C s R s E -=. Select the values of τ and b so that the steady-state error for a ramp input is zero.Solution: Assuming that all parameters are positive, the system must be stable. Then, the error response is)()1)(1()(1)()()(21s R K s T s T b s K s C s R s E ⎥⎦⎤⎢⎣⎡++++-=-=τ)()1)(1()1()(2121221s R Ks T s T Kb s K T T sT T ⋅+++-+-++=τLetting the steady-state error for a ramp input to be zero, we get 221212210.)1)(1()1()(lim )(lim sv K s T s T Kb s K T T sT T s s sE s s r ss ⋅+++-+-++⋅==→→τεwhich results in ⎩⎨⎧=-+=-0121τK T T Kb I.e. KT T 21+=τ,Kb 1=.P3.26 The block diagram of a system is shown in Fig. P3.26. In each case, determine the steady-state error for a unit step disturbance and a unit ramp disturbance, respectively. (a) 11)(K s G =,)1()(222+=s T s K s GFigure P3.24Figure P3.26(b)ss T K s G )1()(111+=,)1()(222+=s T s K s G , 21T T >Solution: (a) In this case the system is of second-order and must be stable. The transfer function from disturbance to error is given by 212212.)1(1)(K K Ts s K G G G s d e ++-=+-=ΦThe corresponding steady-state errors are 1212.11)1(lim K s K K Ts s K s s p ss -=⋅++-⋅=→ε∞→⋅++-⋅=→2212.1)1(lim sK K Ts s K s s ass ε(b) Now, the transfer function from disturbance to error is given by )1()1()(121222.+++-=s T K K s T s sK s d e Φand the characteristic polynomial is21121232)(K K s T K K s s T s +++=∆ Using L-C criterion,0)(121211212212>-==T T K K T K K T K K Dthe system is stable. The corresponding steady-state errors are 01)1()1(lim 1212220.=⋅+++-⋅=→ss T K K s T s sK s s p ss ε121212220.11)1()1(lim K ss T K K s T s sK s s a ss -=⋅+++-⋅=→ε。