P3.4 The open-loop transfer function of a unity negative feedback system is
G(s)?1s(s?1)
Determine the rise time, peak time, percent overshoot and setting time (using a 5% setting criterion).
Solution: Writing he closed-loop transfer function ?(s)?1s?s?12??n222s?2??ns??n
?0.5we get ?n?1, ??0.5. Since this is an underdamped second-order system with ?system performance can be estimated as follows.
Rising time
tr?, the
??arccos??n1??2???arccos0.51?1?0.52?2.42 sec.
Peak time
tp???n1??2??1?1?0.52?3.62 sec.
Percent overshoot Setting time
ts??3p?e???1??2?100 %?e0.5?1?0.52?100 %?16.3 %
30.5?1???6 sec. (using a 5% setting criterion)
c(t)1.3nP3.5 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 ?p1.0?1.3?112?100 %?30 %
00.1t(s)Figure P3.5
Solving the formula for calculating the overshoot,
?p
?e??1???0.3, we have
????ln?2p2?0.362
??n1??2?ln?pSince tp?1 sec., solving the formula for calculating the peak time, ?n?33.7 rad/sec
Hence, the open-loop transfer function is G(s)??n2tp?, we get
s(s?2??n)?1135.7s(s?24.4)
P3.6 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 k1, k2, and a.
c(t)R(s)k1?k2s(s?a)C(s)2.182.00t0(a)Figure P3.6
0.8(b)
Solution: The transfer function between the input and output is given by
C(s)R(s)?sk1k22
?as?k2k1k2s2The system is stable and we have, from the response curve,
limc(t)?lims?t??s?0?1s?as?k2?k1?2
By inspection we have ?p?2.18?2.112.00?100 %?9 %
pSolving the formula for calculating the overshoot, ? ????ln?2p2?e??1??2?0.09, we have
?0.608
??n1??2?ln?pSince tp?0.8 sec., solving the formula for calculating the peak time,
tp?, we get
?n?4.95 rad/sec
Then, comparing the characteristic polynomial of the system with its standard form, we have
2 s2?as?k2?s2?2??ns??n
22 k2??n?4.95?24.5
a?2??n?2?0.608?4.95?6.02
P3.8 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 ?(s)?ks22?nR(s)?ks2C(s)1?asFigure P3.8
?k?s?k?s2?2??ns??2??k22n
hence, by inspection, we get
2?k, 2??n?k?, and ? ?n?n?2??n
results in
??n1??2Taking consideration of ??0.280.
?p?e??1???100 %?40 %In this case, to satisfy the requirement of peak time, ?n?0.818 rad/sec.
tp??4, we have
Hence, the values of
k and
?a are determined as
?0.682 k??n?0.67, ?2??n
P3.10 A control system is represented by the transfer function
C(s)R(s)?(s?2.56)(s0.332
?5%?0.4s?0.13)Estimate the peak time, percent overshoot, and setting time (?method, if it is possible.
Solution: Rewriting the transfer function as
C(s)R(s)?0.33(s?2.56)[s(?0.2)2), using the dominant pole
?0.3]2
we get the poles of the system: s1，, s3??2.56. Then, s1， can be considered as 2 2??0.2?j0.3a pair of dominant poles, because Re(s1， 2)??Re(s3).
Method 1. After reducing to a second-order system, the transfer function becomes
C(s)R(s)?s0.132?0.4s?0.13 (Note:
k??limC(s)R(s)s?0?1)
which results in ?n?0.36 rad/sec and ?
tp??0.551??2. The specifications can be determined as
?100 %?12.6 %??n1??210.4 2sec, ?p?e??
ts?1??n?1?ln???1???2???20.67 sec ???Method 2. Taking consideration of the effect of non-dominant pole on the transient components cause by the dominant poles, we have
tp????(s1?s3)?n1??2?10.8 4sec
?
p?s3s1?s3e??1??2?100 %?13.6 %
ts?1??n?2s3ln????s?s13????23.6 sec ??P3.13 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) s4?2s3?10s2?2s?k?0 (b) s3?(k?0.5)s2?4ks?50?0
Solution: (a) s4?2s3?10s2?2s?k?0.The system is stable if and only if
?k?0?2???D3?1?0?21020k?0 ? k ?920?k?9
i.e. the system is stable when .
(b) s3?(k?0.5)s2?4ks?50?0. The system is stable if and only if
?k?0.5?0, k?0?k?0.550?2D??0 ? 4k?2k?50?0 ? 4(k?3.8)(k?3.3)?02?14k?k?3.3.
i.e. the system is stable when
Ks(0.01s2P3.14 The open-loop transfer function of a negative feedback system is given by
G(s)??0.2?s?1)
Determine the range of K and ? in which the closed-loop system is stable. Solution: The characteristic equation is
0.01s3?0.2?s2?s?K?0 The system is stable if and only if
?k?0, 0.2??0?0.2?K?D??0 ? 0.2??0.01K?0 ? K?20?2?0.011?20??K?0
The required range is
Ks(s3?1)(.
P3.17 A unity negative feedback system has an open-loop transfer function G(s)?s6
?1)kDetermine the range of
s??1.
s3s6 required so that there are no closed-loop poles to the right of the line
Solution: The closed-loop characteristic equation is
s(?1)(?1)?K?0 ? s(s?3)(s?6)?18K?0
i.e. s3?9s2?18s?18K?0
s?1 resulting in Letting s?~~~~3~2~ (~s?1)(s?2)(s?5)?18K?0 ? s?6s?3s?(18K?10)?0
Using Lienard-Chipart criterion, all closed-loop poles locate in the right-half right of the line s??1, if and only if
~s-plane, i.e. to the
?18K?10?0, ? K?59?618K?10?D??0 ? 28?1.8K?0 ? K?149?213? 59?K?149
The required range is , or
0.56?K?1.56
P3.18 A system has the characteristic equation
s3?10s2?29s?k?0
Determine the value of k so that the real part of complex roots is ?2, using the algebraic criterion.
s?2 into the characteristic equation yields Solution: Substituting s?~32 （ ~s?2）?10（~s?2）?29（~s?2）?k?0 3s ~3~2~ s?4s?s?(k?26)?0
2s The Routh array is established as shown.
1 4
1
k?26
If there is a pair of complex roots with real part of ?2, then
k?26?0
i.e. k?30. In the case of k?30, we have the solution of the auxiliary equation ~s??j, i.e. s??2?j.
ss10
P3.22 The open-loop transfer function of a unity negative feedback system is given by
G(s)?Ks(T1s?1)(T2s?1)
Determine the values of K, T1, and T2 so that the steady-state error for the input, r(t)?a?bt, is less than ?0. It is assumed that K, T1, and T2 are positive, a and b are constants. Solution: The characteristic polynomial is ?(s)?T1T2s3?(T1?T2)s2?s?K
Using L-C criterion, the system is stable if and only if
D2?T1?T2T1T2K1?0 ? T1?T2?KT1T2?0 ? K?T1?T2T1T2
K?T1?T2T1T2Considering that this is a 1-type system with a open-loop gain K, in the case of we have ?ssb??ss.r??ss.v?bK??0 ? K?b,
?0
Hence, the required range for K is
?0?K?T1?T2T1T2
P3.24 The block diagram of a control system is shown in Fig. P3.24, where E(s)?R(s)?C(s). Select the values of ? and b so that the steady-state error for a ramp input is zero.
R(s)? s?b-K(T1s?1)(T2s?1)C(s)Figure P3.24
Solution: Assuming that all parameters are positive, the system must be stable. Then, the error response is
??K(?s?b)E(s)?R(s)?C(s)??1??R(s)
(T1s?1)(T2s?1)?K?? ?T1T2s2?(T1?T2?K?)s?(1?Kb)(T1s?1)(T2s?1)?KT1T2s2?R(s)
Letting the steady-state error for a ramp input to be zero, we get ?ss.r?limsE(s)?lims?s?0s?0?(T1?T2?K?)s?(1?Kb)v0?2(T1s?1)(T2s?1)?Ks
which results in I.e. ???1?Kb?0??T1?T2?K??0 .
T1?T2K,
b?1KP3.26 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) G1(s)?K1,
G2(s)?K2s(T2s?1)
D(s)R(s)G1(s)G2(s)C(s)-Figure P3.26
(b)
G1(s)?K1(T1s?1)s,
G2(s)?K2s(T2s?1), T1?T2
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 ?e.d(s)??G21?G1G2??K2s(Ts?1)?K1K21s1s2
The corresponding steady-state errors are ?ss.p ?ss.a?lims?s?0?K2s(Ts?1)?K1K2?K2s(Ts?1)?K1K2???1K1
?lims?s?0???
(b) Now, the transfer function from disturbance to error is given by ?e.d(s)??K2ss(T2s?1)?K1K2(T1s?1)2
and the characteristic polynomial is
?(s)?T2s3?s2?K1K2T1s?K1K2 Using L-C criterion,
D2?1T2K1K2K1K2T1?K1K2(T1?T2)?0
the system is stable. The corresponding steady-state errors are ?ss.p ?ss.a
?lims?s?0?K2s2s(T2s?1)?K1K2(T1s?1)s?1?0
1K1?lims?s?0?K2s2?12??s(T2s?1)?K1K2(T1s?1)s