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Transient and steady state response analysis
CHAPTER 4
Transient and steady state response analysis
4.1 INTRODUCTION Many applications of control theory are to servomechanisms (or servos for short) which are systems using the feedback principle designed so that the output will follow the input. For example, an anti-aircraft gun must follow accurately and compensate for the rapidly varying signals received from an incoming plane or missile. In general, of course, the input signals into a system are not known, and even if they were, it would be difficult to express them analytically in a form suitable for system analysis. Because of this it is a frequent practice to apply a reference input into the system and to study its response in the time domain. The type of reference input applied does depend on the purpose of the system-for example, if the system may be subjected to sudden disturbances, a step reference input may be best; on the other hand, if the system is used to follow a gradually changing situation, a ramp reference input may be better. The time response of a system may be considered in two parts. One part of the response, known as the transient response (corresponding to the comple mentary function), which reduces to zero as time increases, the other part, known as steady-state (corresponding to the particular integral), is the response of the system when time tends to infinity. As already pointed out in Chapter 1, because of friction, inertia etc., the output of a system cannot follow an instantaneous change in the input (such as a reference step signal), so that the system response will always involve a transient part. To a control engineer, the time required for a system to reach a new steady state is of great importance, but there are various other factors he must consider. Apart from stability, which is, of course, the most important factor of all, he may have to consider factors such as steady-state error which is the difference between the steady-state response and the input. We begin by considering first and second order systems, and it will then be shown that the time responses of higher order systems are a collection of functions of the same type as those for first and second order ones.
84
Transient and steady state response analysis
[Ch.4
4.2 RESPONSE OF FIRST ORDER SYSTEMS We have already noted -(see equation (2.13)) that the Laplace transform Y(s) of the output of a linear system can be written as Y(s) = G(s)U(s)
(4.1)
where G(s) is its transfer function and U(s) the Laplace transform of the input. A first order system has an associated differential equation of the form aDy + y = u, Y(s)
hence (from Section 2.4) with y(0) = 0
1 fls + l
U(s).
For transient response analysis it is customary to use a reference unit step function u(t) for which U(s) = l/s. It then follows that Y(s)
1
l/s
(as + \)s
(s + l/a)s
s
s + \/a
(4.2)
On taking the inverse Laplace of equation (4.2), we obtain-, y(t) = 1 - e- , / a
(t>0).
(4.3)
Both the input and the output to the system are shown in Fig. 4.1. The response has an exponential form. The constant a is called the time constant of the system.
Fig. 4.1
Sec. 4.2]
Response of first order systems
85
Notice that when t = a, then y(t) = y(a) = 1 — e"1 = 0.63. The response is in two parts, the transient part e~^a, which approaches 0 as t * °° and the steady-state part 1, which means that eventually (when t = °°) the output is 1 (that is, equal to the input). If the derivatives of the input are involved in the differential equation of the system, that is, if aDy + y = bDu + u then (bs + l)
Y(s) =
(as + 1)
U(s) = K
(s + z)
(» + />)
U(s)
(4.4)
b 1 1 where K = —,z = — and p = —. a b a When i/(s) = 1/s, equation (4.4) can be written as K2 s
s+p
z z—p where ^i = K— and K2 = K . P P Hence, ^(0 = K, - K2
(4.5)
With the the assumption that z > p > 0, this response is shown in Fig. 4.2.
^ >
^Ky-K2e'pt
K2
-K2 '
^ ^ ^ _ _
K2e-pt
Fig. 4.2
We note that the responses to the two systems (Figs. 4.1,4.2) have the same form, except for the constant terms Ki and K2. It appears that the role of the numerator of the transfer function is to determine these constants, that is, the size of y(t), but its form is determined by the demoninator.
86
Transient and steady state response analysis
[Ch. 4
4.3 RESPONSE OF SECOND ORDER SYSTEMS An example of a second order system is a spring-dashpot arrangement shown in Fig. 4.3. Applying Newton's law, we find d2y dy M—-2 = — μ— dt dt
— ky + u(t) w
where k = spring constant and μ = damping coefficient, y is the distance of the system from its position of equilibrium, and it is assumed that y(0) = y'(0) = 0. V-
M
μ
S
Fig. 4.3
Hence
d2v dv «(f) = M—'-41 ++ μ— μ — + ky. dt dt
On taking Laplace transforms, we obtain 1 Y(s) =
K
2
Ms + μs + k
U(s) =
s +aiS + a2
U(s)
where K = ^ , a, = j£, and a2 = ^ . Applying a unit step input, we obtain A:
(4.6)
Y(s) =
where P\,p2 —
ai ± V(«i - 4e 2 )
. Pi and pi are the poles of the transfer function
K G{s) = — —, s +als + a2 that is, the zeros of the denominator of G(s).
Sec. 4.3]
Response of second order systems
87
There are three cases to be considered: Case 1. a\ > 4a2 (overdamped system) In this case pi and p 2 are both real and unequal. Equation (4.6) can be written as Ai
A2
Ai
S
S + Pi
S + p2
Y(s) = - ! + — 2 - + —*where A, = — - (= * ) , A2 = —~
(4.7) . and K3 = —r-*-
,
(notice that Aj + A2 + A3 = 0). On taking the inverse Laplace transform of equation (4.7), we obtain y(t) = A, + Κ&-** + Κ36-ρ>*.
(4.8)
The transient part of the solution is seen to be A 2 e _Plf + A3e~p»f. Case 2. a] = 4a2 (critically damped system) In this case, the poles p, and p 2 are real and equal: Pi = Pi = -
= P
A
(say) At
A2
A9
and
Y(s) =
Hence
y(t) = A", + A 2 e-"' + Α 3 Γβ -ρ ',
so that
r = — + —- + —,. s(s + p)2 s s+p (s + p)2
A A"i = —, A2 = P
A ; P
where A
and
K3 = P
A0 = -M - t-" - ptt-'*].
Case 3. 4e2 > a\ (underdamped system) In this case, the poles pi and p 2 are complex conjugate having the form Pu Pi = a ± iß,
(4.9)
(4.10)
88
Transient and steady state response analysis
[Ch.4
where a = °j
and
β = Κ2
Ai
Hence
Y(s) = - +
where
Kt =
—
S
\^{Λα2-α\). K-i
+
— ,
S + Pi
K
S+p2
K(-ß-ia) 2;. * 2
α2 + β
=
2ß(a2 + ß2)
£ ( - / ? + ig)
and ΛΓ3
2β(α2 + β2)'
(Notice that K2 and K3 are complex conjugates.) It follows that y(t) = K1 + / : 2 e - ( a + / « ' +
K3e-{a-i0)t
= Ki + e-°"[(K2 + K3) cos ßt + (K3 - K2)i sin ßt] (using the relation e'*" = cos ßt + /' sin ßt) K K = -n2 ; + - 2· : 2: e _ c r f [ - c o s 0 f 2 a +ß a +ß K 2
α +β where
tan e =
K 2
α + β* 1 + 2
2
^J
a ß
sin ßt].
„-at e-°"sm(ßt
+ e)
ß
Notice that when t = 0, .y(r) = 0 (see equation (4.11). The three cases discussed above are plotted in Fig. 4.4.
underdamped system
critically damped system overdamped system
Fig. 4.4
(4.11)
(4.12)
Response of second order systems
Sec. 4.3]
89
From Fig. 4.4, we see the importance of damping (note that α^ = μ/Μ, μ being the damping factor). We would expect that when the damping is 0 (that is, ax = 0) the system should oscillate indefinitely. Indeed when a, = 0, then
a = 0, and ß = \fä2 and since sin e = 1 and cos e = 0, then e = π/2, equation (4.12) becomes K y{t) =
1 — s i n j v ^ t H—
= — [1 — cos\Ja2 t]. a2
This response of the undamped system is shown in Fig. 4.5. ylrl
Fig. 4.5
Notice that this is the response of the system when the roots of s1 +fljS+ a2 = 0 are pure imaginary, that is, when the poles of the transfer functions are pure imaginary. There are two important constants associated with each second order system. The undamped natural frequency ω„ of the system is the frequency of the response shown in Fig. 4.5, that is, ω
η -
y/ä2
The damping ratio ζ of the system is the ratio of the actual damping μ(= αχΜ) to the value of the damping με, which results in the system being critically damped (that is, when αγ = 2\/ä2). Hence
t=
«1
2\/ä2
90
Transient and steady state response analysis
[Ch. 4
We can write equation (4.12) in terms of these constants. We note that a{ = 2ωηξ
Hence
a2 = ω 2 ,.
and
(1 + a2lß2) i =
1+
«Î
2\[a~-L
1
4a2 — a]
V(4e2-e?)
V(i-f2)
Equation (4.12) becomes K 1
y(0 = ω„ where
ω = ω„\/{\—
-
Vd-f2)
f2)
and
e
ω
"*' sin {ωί + e)
tan e =
(4.13)
Vo-r 2 ) r
It is conventional to choose K/a2 = 1 and then plot graphs of the 'normalised' response y(t) against cof for various values of the damping ratio f. Three typical graphs are shown in Fig. 4.6.
Fig. 4.6 It is obvious from Fig. 4.4 that if the speed of response is an important criterion in the design of the system, the engineer will choose the system to function in the underdamped state, but he will then have to consider various other factors or parameters, some of which are shown in Fig. 4.7. (1) Overshoot defined as maximum overshoot X 100 final desired value (2) Time delay td, the time required for the system response to reach 50% of its final value.
Sec. 4.3]
Response of second order systems
91
(3) Rise time, the time required for the system response to rise from 10% to 90% of its final value. (4) Settling time, the time required for the eventual settling down of the system response to be within (normally) 5% of its final value. (5) Steady state error e K , the difference between the steady state response and the input. Maximum ' overshoot
Steady-state error e ss
Rjsetime Fig. 4.7
In fact, one can often improve one of the parameters but at the expense of the others. For example, the overshoot can be decreased at the expense of the time delay. In general, the quality of a system may be judged by various standards. Since the purpose of a servomechanism is to make the output follow the input signal, we may define expressions which will describe dynamic accuracy in the transient state. Such expressions are based on the system error, which in its simplest form is the difference between the system input and the output and is denoted by e(f), that is,
e(t) = y(t) - u(t) where y(i) is the actual output and u(t) is the desired output; (u(t) is the input). The expression called the performance index can take on various forms. typical among them are:
92
and
Transient and steady state response analysis
\ e 2 (r)dr, Jo
integral of error squared (IES)
1 \e(t)\dt,
integral of absolute error (IAE)
r
I t\e(t)\at, Jo
[Ch. 4
integral of time multiplied absolute error criterion (ITAE).
Having chosen an appropriate performance index, the system which minimises the integral is called optimal. The objective of modern control theory is to design a system so that it is optimal with respect to a performance index. The systems are not necessarily servomechanisms, and in fact, a common problem is to determine a signal vector so that the state vector x(f) is brought from an initial state x 0 to a final state x f in the shortest possible time. We discuss various optimisation methods in Part II of the book.
4.4 RESPONSE OF HIGHER ORDER SYSTEMS We can write the transfer function of an wth order system (see equation (2.12)) in the form K(sm + bis"1'1 + ... + bm) G(s) = — — -.
e
N
(4.14)
Systems having a transfer function of high order can arise when several com ponents, each involving a 'low' order transfer function, make up the system as illustrated by the following simple example. Example 4.1 With reference to Fig. 2.11 (see page 43), calculate the closed loop transfer function (G(s)) given the transfer functions A(s) =
1
s+3
and
2
B(s) = - . s
Solution Using6 equation (2.16), that is, G(s) = 4
G(s) -
A(s) l + A(s)B(s)
s
s
s2 + 3s + 2
(s + l)(s + 2)
, we obtain
Sec. 4.4J
Response of higher order systems
93
The response of the system having the transfer function (4.14) to a unit step input can be written in the form K(s + Zi)(s + z2) ■ ■ ■ (s + zm) Y(s) = — — — s(s + pt)(s + p2) ... (s + pn)
(4.15)
where Zi,z2,...,zm and pit p2,...,p„ are respectively the zeros of the numerator and the denominator polynomials of equation (4.14). The zeros are either real or complex numbers; in the case of a complex zero there must be another zero which is its conjugate, since the a's and b's in equation (4.14) are all real. We first assume that n > m in equation (4.14); we then have two cases to consider: Case 1 Here P\,p2,-.,pn are all distinct numbers. The partial fraction expansion of equation (4.15) has the form Ao
Ai
Y(s) = S
Ki,K2 form
A n+\
+ ——+...+
——.
S+Pi
(4.16)
S+Pn
A"„+i are called the residues of the expansions. The response has the = *i + * a e - p i f + . . . +
m
Knnt~pn'.
As we have already noticed in the previous section, if p\ is complex and p2 is its conjugate then the corresponding residues (K2 and K3) are complex conjugate. The two terms can be combined as illustrated in equation (4.12). Case 2 Here at least one of the roots, say pu is of multiplicity r, that is, K(s + Zi)(s + z 2 ) . . . (s + z m ) Y(s) = — — — . Φ + PiY(s + Pi) · · · (s + Pn-rn)
(4.17)
The partial fraction expansion has the form Λι
Ait
Since
£~l
K (s+p)l\
K2r
AM
Y(S) = — + — — + . s s + pi (s + ριγ
1 ...
' ( / - I ) ! tl-xt-Pt
1
1 .
S + Pn-r+l
(s+PiY (/ = !, 2 > . . .
(4.18)
94
Transient and steady state response analysis
[Ch. 4
the response has the form y(t) = Ki + K21e-pi'
+ K3e~p^
+ K22t
+...+
+...+
- ^ (r-l)\
i'-'e-'i'
Κη_,+ιϋ~ρ"-^.
(4.19)
We now consider that m > « in equation (4.14), which is the case (see section 2.4) when the system is improper; that is, it can happen when we consider idealised and physically non-realisable systems, such as resistanceless circuits. We then divide the numerator by the denominator until we obtain a proper fraction so that when applying a unit step input, we can write Y(s) as -m-ll + ^ + ^ - ' + . - Η J] Y(s) = K[Cl + c2s + ... + es"-™-1] + ——n „_, , · ' · , " s(s + a , s n _ 1 + . . . + a„)
(4.20)
where the cs, ds, K and K^ are all constants. The second term of the right-hand side is treated as in Case 1 or Case 2 above; the inverse Laplace transform of the first term involves the impulse function (see equation (2.7)) and various derivatives of it.
Example 4.2 Find the response of the system having a transfer function G(s) =
5(s2 + 5s + 6) s 3 + 6s 2 + 10s + 8
to a unit step input.
Solution In this case, 5(«2 + 5s + 6) Y(s) = s(s3 + 6s 2 + 10s + 8)
5(s + 2)(i + 3) s(s + 4)(s 2 + 2s + 2)
The term s2 + 2s + 2 = (s + 1 + i) (s + 1 — /'), hence K(s) can be written as the partial fraction expansion as
Y(S) = - + s
+ s+ 4
+ s + l + z'
s + 1 — /'
Sec. 4.5]
95
Steady state error
where K3 is the conjugate of K4. In fact, *i = J,K2
= -\,K3
= £(-7+i)
and
K4 = £ ( - 7 - / ) .
Hence 15 _ 1 y(t) = T4 ~ 74 e _ 4 ' 15 = -; 4
+
T e _ f I - 1 4 cos ί + 2 sin r]
1 _ 4f , V200 _ f . , , - „ . e Te + sinii + 352). 4 4
4.5 STEADY STATE ERROR It was stated in the previous section that one of the important considerations in system design is the steady state error e^ shown in Fig. 4.7. We would expect that e ss would depend on both the transfer function of the system and on the reference input. In this chapter we have used the notation u(t) to denote the input to a system - generally to denote a (unit) step input. We shall now change this notation to conform with the general usage and the fact that we consider different types of reference inputs. We shall use r(t) to denote the (reference) input and c(r) the output. A unity feedback system with an open loop transfer function A(s) will then be represented (as in Fig. 2.12) by Fig. 4.8. /?(s)
CIS)
7\ r
A[S)
Fig. 4.8 E(s),R(s) and C(s) are the Laplace transforms of e(t), r(t) and c(t) respectively.
We define the error runctionf e(t) = r(t) -
c{t).
(4.21)
Hence e ss = lim e(t). ■(This is assuming that rit) and c{t) are of the same dimensions, that is, are both voltages or speeds, etc., otherwise a non-unity feedback system must be used.
96
Transient and steady state response analysis
[Ch.4
Since E(s) = R(s) - A(s)E(s), it follows that E(s) = 1 + A(s) and by the final value theorem (equation (2.10)) e ss = lims^s) sR(s) = lim . **° 1 + A(s)
(4.22)
We now define three error coefficients which indicate the steady state error when the system is subjected to three different standard reference inputs r(t) (1) Step input, r(t) = ku(t) (k is a constant) (Fig. 4.9). e
sk/s _ ss = lim s+oi + A(s) l + \imA(s)
lim A(s) = Kp, called the position error constant, then Cec
1 + K„
(4.23)
Kp =
or
Hence if k and e ss are specified, the value of the transfer function at s = 0 is determined by equation (4.23).
Fig. 4.9
(2) Ramp input, r(t) = ktu{t) (Fig. 4.10). k k k In this case, R(s) = - j , so that e ss = — or Kv = — where Kv = lim sA(s), S
is called the velocity error constant.
Λν
é? ss
i*0
Sec. 4.5]
Steady state error
97
Fig. 4.10 (3) Parabolic input, r(t) = y kt2u(t) (Fig. 4.11). k k k R(s) = -3, so that e ss = — or Ka = — where Ka = lim s2A(s), called the acceleration error constant.
Fig. 4.11
Example 4.3 Find the (static) error coefficients for the system having the open loop transfer function A(s) =
s(4s + 2) '
98
Transient and steady state response analysis
[Ch. 4
Solution lim A(s) = oo Kv
lim sA(s) = 4 i+0
lim s2A(s) = 0. i+0
From the definitions of the error coefficients, it is seen that ess depends on the number of poles (or integrations) at s = 0 of the transfer function. This leads to the following classification. A transfer function is said to be of type N if it has N poles at the origin. Thus if A(s) =
K(s-zi)...(s-zm)
(4.24)
s'(s-Pl)...(s-p„)
where j = 0,A(s)isoitypeO, j = l,/l(s)is of type 1, / = 2, A(s) is of type 2. At s = 0,
A(s) = hm— *'0«'
where
K\ =
.
(4.25)
(ΓΡΰ-Λ-Ρη)
(Κι is called the gain of the transfer function.) Hence the steady state error eiS depends on / and r(t) as summarised in Table 4.1. Table 4.1 ^ss
/
System r(f)=ku(t)
r(t) = ktu(t)
r(f)= f'MO
0
Type 0
Finite
oo
oo
1
Type 1
0
Finite
oo
2
Type 2
0
0
Finite
It would seem that for a system to have a rapid response, capable of following various inputs with zero steady state error, the engineer must ensure that the open loop transfer function has a pole of high order at the origin. Unfortunately, such systems are prone to instability so that it is unusual to design a system having a pole at the origin of order greater than two (that is, / > 2).
Sec. 4.6]
99
Feedback control
4.6 FEEDBACK CONTROL In this section we shall discuss certain aspects of the effect of feedback on the system response. This is in fact the very basis of classical control theory, and much literature is devoted to it. We shall only consider certain general principles and not detailed system design considerations. We have already shown (see Fig. 2.12 and equation (2.16)) that the closed loop transfer function G(s) of the negative feedback system (Fig. 4.12) is related to the feed-forward transfer function A(s) and feed-back transfer function Bis) by A(s) G(s) = ——. (4.26) l+A(s)B(s)
Ris) - .
CIS)
EiS)
Als)
Ώ
Bis) Fig. 4.12
We consider a simple example of a first order system for which K as + 1 and
B(s) = c (a constant)
so that K/a
K U(S) —
as + Kc + l
—
Kc + l s+
We now take the inverse Laplace transform, that is, obtain the impulse response (see equation (2.14)) of the system, where (1 ) c = 0 (that is, response of the open loop system) K -t/a g(t) = - e-", a
and
100
Transient and steady state response analysis
[Ch.4
(2) οΦΟ K _«£±±t K _1 g(t) = - e a = - e <* a a where a =
a Kc + \
a and a are respectively the time-constants of the open loop and the closed loop systems. a is always positive (since we are considering a physical system), but since c can be made positive or negative, a may be either positive or negative. Fig. 4.13 shows how the time responses vary with different values of Kc. /Cc^-1
Fig. 4.13
If the impulse response does not decay to zero as t increases, the system is unstable. From Fig. 4.13 it is seen that the instability region is defined by Kc^-l In many applications the control system consists basically of a plant having the transfer function A(s) and a controller having a transfer function B(s), as in Fig. 4.14.
^
^
>
r
£,sl
Controller Bis)
Q{s)
l'ig. 4.14
Plant Als)
Cis)
Sec. 4.6]
Feedback control
101
The transfer function of the closed loop system can be found by application of equation (4.26) with feed-forward transfer function A(s)B(s) (see equation (2.15)) and a feed-back transfer function 1; then A(s)B(s) G(s) = —^-LL· . 1 + A(s)B(s)
(4.27)
The controllers can be of various types. (1) The on-offcontroller The action of such a controller is very simple. With reference to Fig. 4.14, the control depends on the actuating signal e(t). Thus if e(f) > 0
then
q(i) = ßi
and if e(r) < 0
then
q(t) = Q2
where q(t) is the output signal from the controller and ß x and β 2 are some constants. The on-off controller is obviously a nonlinear device and it cannot be described by a transfer function. An analysis of such systems can be accomplished by 'describing functions' which we do not consider in this book. (2) Proportional Controllers For this control action q(t) =
Kpe(t)
where Kp is a constant, called the gain of the controller. Thus the,transfer function for this controller is 2(i) Ä(s) = ~ = Kp. E(s)
(4.28)
(3) Integral Controller In this case q(t) = K \ e(t)dt,
hence
Jo B(s) = K/s.
(4.29)
102
Transient and steady state response analysis
[Ch. 4
(4) Derivative Controller In this case de q(t) = K—, dt
hence
B(s) = Ks. This type of controller is never used on its own, since its action is only effective while e(t) is changing, that is, during the transient state period, and has no effect in the steady state. (5) Proportional-plus-derivative (PD) Controller In this case de q{t) = K0e(t) + Ki—, dt
hence
B(s) = KM + - 1 s) F
*P
= Kp(l + Ks) where K =
(4.30) Kp
(6) Proportional-plus-integral (PI) Controller In this case ttf
e(t)dt,
1(0 = Kpe(t) + Kt
hence
Jo K B(s) = Kp(\ + ~) s
'where K =
(4.31) Kp
(7) Proportional-plus-derivative-plus-integral (PDI) Controller In this case
de q(t) = Kpe(t)
dt
B(s) = Kp(\ + kts + k2/s) Ki where ki = — Up
and
r
+ Ki— + K2
K2 k2 = — . Kp
e(t)dt,
Jo
hence
Sec. 4.6]
Feedback control
103
Controllers of the various types described above can be constructed for electrical, mechanical, and other systems. This was pointed out for electrical systems when discussing simulation in section 3.2. We shall not discuss the details of how such controllers are designed. For our purposes it is sufficient to appreciate that such controllers are at our disposal.
Example 4.4 Design a controller for a plant having the transfer function A(s) — l/(s + 2)) so that the resulting closed loop system has a zero steady state error to a reference ramp input. Solution From the Table 4.1 (on p. 98) we note that for zero steady state error to a ramp input, the system must De of type 2. Hence if we choose an integral controller with B(s) = K/s then the transfer function of the closed loop system including the plant and the controller (see Fig. 2.11) is A(s)B(s)
K 3
l + A(s)B(s) ~ s + 2s2 + K The response of this control system depends on the roots of the denominator polynomial s3 + 2s 2 + K = 0. It will be shown in the next chapter (example 5.3) that for all values of K at least one of the roots has a positive real part, which implies that the system is unstable. We cannot use an integral controller in this case. But if we use a PI (proportional-plus-integral) controller, say
B(s) = Kp(l + f ), the system is of type 2 and response of the system depends on the roots of s3 + 2s2 + Kps + KKP = 0. In this case, values of K and Kp can be found so that the system is stable (see example 5.3).
104
Transient and steady state response analysis
[Ch.
PROBLEMS (1) A unity feedback system has the open-loop transfer function C(s)
2 0
s + 5s2 + 6s
(i) Calculate the positional, velocity and acceleration error constants. (ii) Determine the steady-state error where the input for t > 0 is (a) r{t) = 2, (b) r(t) = 2t, (c) r(t) = \t2 and (d) r(t) = 3 - 2r. (2) Find the time constants of the system (i) whose behaviour is defined by the differential equation ay (a) 3 — + 2y = 2 dr , dy (b) i — + 3y = 3 di
given that at t = 0, y - 0. (ii) whose (open-loop) transfer function is (a) C(5) (b) G(s) =
l + 3s 1 l + 2s'
(iii) Unity feedback systems have open-loop transfer functions as in (ii) above. Determine the system time-constants. What is the effect on the response of the system on the addition of a closed loop? (3) Investigate the response to a unit step input of the system having the transfer function 4 GW = -■,
·
s J + 4s + 5 (4) Obtain the response to a unit step input of the system shown in Fig. 4.15 for K = l, 5.25, and 10.
— <
?\_ *
r
K
r,a G W
yy ( 5 )
1
-5.55*1
■
Fig. 4.15
4]
Problems
105
(5) The output of a system to a unit step input is a2 Y{s) = —1 · s(s + di s + a2) (a) Write this output (in its normalised form) as a function of the system natural frequency ωη and the damping ratio r. (b) The transfer function of the system is written as G(s) =
A:
. (s + pi)(s + p 2 ) Find the values of pt and p 2 in terms of the above two system constants. (6) A second order underdamped system has a response y(t) to a unit step input. Obtain the time of occurrence and the height of the maximum overshoot in terms of the system natural frequency ω„ and the damping ratio T.
Transient and steady state response analysis
4.1 INTRODUCTION Many applications of control theory are to servomechanisms (or servos for short) which are systems using the feedback principle designed so that the output will follow the input. For example, an anti-aircraft gun must follow accurately and compensate for the rapidly varying signals received from an incoming plane or missile. In general, of course, the input signals into a system are not known, and even if they were, it would be difficult to express them analytically in a form suitable for system analysis. Because of this it is a frequent practice to apply a reference input into the system and to study its response in the time domain. The type of reference input applied does depend on the purpose of the system-for example, if the system may be subjected to sudden disturbances, a step reference input may be best; on the other hand, if the system is used to follow a gradually changing situation, a ramp reference input may be better. The time response of a system may be considered in two parts. One part of the response, known as the transient response (corresponding to the comple mentary function), which reduces to zero as time increases, the other part, known as steady-state (corresponding to the particular integral), is the response of the system when time tends to infinity. As already pointed out in Chapter 1, because of friction, inertia etc., the output of a system cannot follow an instantaneous change in the input (such as a reference step signal), so that the system response will always involve a transient part. To a control engineer, the time required for a system to reach a new steady state is of great importance, but there are various other factors he must consider. Apart from stability, which is, of course, the most important factor of all, he may have to consider factors such as steady-state error which is the difference between the steady-state response and the input. We begin by considering first and second order systems, and it will then be shown that the time responses of higher order systems are a collection of functions of the same type as those for first and second order ones.
84
Transient and steady state response analysis
[Ch.4
4.2 RESPONSE OF FIRST ORDER SYSTEMS We have already noted -(see equation (2.13)) that the Laplace transform Y(s) of the output of a linear system can be written as Y(s) = G(s)U(s)
(4.1)
where G(s) is its transfer function and U(s) the Laplace transform of the input. A first order system has an associated differential equation of the form aDy + y = u, Y(s)
hence (from Section 2.4) with y(0) = 0
1 fls + l
U(s).
For transient response analysis it is customary to use a reference unit step function u(t) for which U(s) = l/s. It then follows that Y(s)
1
l/s
(as + \)s
(s + l/a)s
s
s + \/a
(4.2)
On taking the inverse Laplace of equation (4.2), we obtain-, y(t) = 1 - e- , / a
(t>0).
(4.3)
Both the input and the output to the system are shown in Fig. 4.1. The response has an exponential form. The constant a is called the time constant of the system.
Fig. 4.1
Sec. 4.2]
Response of first order systems
85
Notice that when t = a, then y(t) = y(a) = 1 — e"1 = 0.63. The response is in two parts, the transient part e~^a, which approaches 0 as t * °° and the steady-state part 1, which means that eventually (when t = °°) the output is 1 (that is, equal to the input). If the derivatives of the input are involved in the differential equation of the system, that is, if aDy + y = bDu + u then (bs + l)
Y(s) =
(as + 1)
U(s) = K
(s + z)
(» + />)
U(s)
(4.4)
b 1 1 where K = —,z = — and p = —. a b a When i/(s) = 1/s, equation (4.4) can be written as K2 s
s+p
z z—p where ^i = K— and K2 = K . P P Hence, ^(0 = K, - K2
(4.5)
With the the assumption that z > p > 0, this response is shown in Fig. 4.2.
^ >
^Ky-K2e'pt
K2
-K2 '
^ ^ ^ _ _
K2e-pt
Fig. 4.2
We note that the responses to the two systems (Figs. 4.1,4.2) have the same form, except for the constant terms Ki and K2. It appears that the role of the numerator of the transfer function is to determine these constants, that is, the size of y(t), but its form is determined by the demoninator.
86
Transient and steady state response analysis
[Ch. 4
4.3 RESPONSE OF SECOND ORDER SYSTEMS An example of a second order system is a spring-dashpot arrangement shown in Fig. 4.3. Applying Newton's law, we find d2y dy M—-2 = — μ— dt dt
— ky + u(t) w
where k = spring constant and μ = damping coefficient, y is the distance of the system from its position of equilibrium, and it is assumed that y(0) = y'(0) = 0. V-
M
μ
S
Fig. 4.3
Hence
d2v dv «(f) = M—'-41 ++ μ— μ — + ky. dt dt
On taking Laplace transforms, we obtain 1 Y(s) =
K
2
Ms + μs + k
U(s) =
s +aiS + a2
U(s)
where K = ^ , a, = j£, and a2 = ^ . Applying a unit step input, we obtain A:
(4.6)
Y(s) =
where P\,p2 —
ai ± V(«i - 4e 2 )
. Pi and pi are the poles of the transfer function
K G{s) = — —, s +als + a2 that is, the zeros of the denominator of G(s).
Sec. 4.3]
Response of second order systems
87
There are three cases to be considered: Case 1. a\ > 4a2 (overdamped system) In this case pi and p 2 are both real and unequal. Equation (4.6) can be written as Ai
A2
Ai
S
S + Pi
S + p2
Y(s) = - ! + — 2 - + —*where A, = — - (= * ) , A2 = —~
(4.7) . and K3 = —r-*-
,
(notice that Aj + A2 + A3 = 0). On taking the inverse Laplace transform of equation (4.7), we obtain y(t) = A, + Κ&-** + Κ36-ρ>*.
(4.8)
The transient part of the solution is seen to be A 2 e _Plf + A3e~p»f. Case 2. a] = 4a2 (critically damped system) In this case, the poles p, and p 2 are real and equal: Pi = Pi = -
= P
A
(say) At
A2
A9
and
Y(s) =
Hence
y(t) = A", + A 2 e-"' + Α 3 Γβ -ρ ',
so that
r = — + —- + —,. s(s + p)2 s s+p (s + p)2
A A"i = —, A2 = P
A ; P
where A
and
K3 = P
A0 = -M - t-" - ptt-'*].
Case 3. 4e2 > a\ (underdamped system) In this case, the poles pi and p 2 are complex conjugate having the form Pu Pi = a ± iß,
(4.9)
(4.10)
88
Transient and steady state response analysis
[Ch.4
where a = °j
and
β = Κ2
Ai
Hence
Y(s) = - +
where
Kt =
—
S
\^{Λα2-α\). K-i
+
— ,
S + Pi
K
S+p2
K(-ß-ia) 2;. * 2
α2 + β
=
2ß(a2 + ß2)
£ ( - / ? + ig)
and ΛΓ3
2β(α2 + β2)'
(Notice that K2 and K3 are complex conjugates.) It follows that y(t) = K1 + / : 2 e - ( a + / « ' +
K3e-{a-i0)t
= Ki + e-°"[(K2 + K3) cos ßt + (K3 - K2)i sin ßt] (using the relation e'*" = cos ßt + /' sin ßt) K K = -n2 ; + - 2· : 2: e _ c r f [ - c o s 0 f 2 a +ß a +ß K 2
α +β where
tan e =
K 2
α + β* 1 + 2
2
^J
a ß
sin ßt].
„-at e-°"sm(ßt
+ e)
ß
Notice that when t = 0, .y(r) = 0 (see equation (4.11). The three cases discussed above are plotted in Fig. 4.4.
underdamped system
critically damped system overdamped system
Fig. 4.4
(4.11)
(4.12)
Response of second order systems
Sec. 4.3]
89
From Fig. 4.4, we see the importance of damping (note that α^ = μ/Μ, μ being the damping factor). We would expect that when the damping is 0 (that is, ax = 0) the system should oscillate indefinitely. Indeed when a, = 0, then
a = 0, and ß = \fä2 and since sin e = 1 and cos e = 0, then e = π/2, equation (4.12) becomes K y{t) =
1 — s i n j v ^ t H—
= — [1 — cos\Ja2 t]. a2
This response of the undamped system is shown in Fig. 4.5. ylrl
Fig. 4.5
Notice that this is the response of the system when the roots of s1 +fljS+ a2 = 0 are pure imaginary, that is, when the poles of the transfer functions are pure imaginary. There are two important constants associated with each second order system. The undamped natural frequency ω„ of the system is the frequency of the response shown in Fig. 4.5, that is, ω
η -
y/ä2
The damping ratio ζ of the system is the ratio of the actual damping μ(= αχΜ) to the value of the damping με, which results in the system being critically damped (that is, when αγ = 2\/ä2). Hence
t=
«1
2\/ä2
90
Transient and steady state response analysis
[Ch. 4
We can write equation (4.12) in terms of these constants. We note that a{ = 2ωηξ
Hence
a2 = ω 2 ,.
and
(1 + a2lß2) i =
1+
«Î
2\[a~-L
1
4a2 — a]
V(4e2-e?)
V(i-f2)
Equation (4.12) becomes K 1
y(0 = ω„ where
ω = ω„\/{\—
-
Vd-f2)
f2)
and
e
ω
"*' sin {ωί + e)
tan e =
(4.13)
Vo-r 2 ) r
It is conventional to choose K/a2 = 1 and then plot graphs of the 'normalised' response y(t) against cof for various values of the damping ratio f. Three typical graphs are shown in Fig. 4.6.
Fig. 4.6 It is obvious from Fig. 4.4 that if the speed of response is an important criterion in the design of the system, the engineer will choose the system to function in the underdamped state, but he will then have to consider various other factors or parameters, some of which are shown in Fig. 4.7. (1) Overshoot defined as maximum overshoot X 100 final desired value (2) Time delay td, the time required for the system response to reach 50% of its final value.
Sec. 4.3]
Response of second order systems
91
(3) Rise time, the time required for the system response to rise from 10% to 90% of its final value. (4) Settling time, the time required for the eventual settling down of the system response to be within (normally) 5% of its final value. (5) Steady state error e K , the difference between the steady state response and the input. Maximum ' overshoot
Steady-state error e ss
Rjsetime Fig. 4.7
In fact, one can often improve one of the parameters but at the expense of the others. For example, the overshoot can be decreased at the expense of the time delay. In general, the quality of a system may be judged by various standards. Since the purpose of a servomechanism is to make the output follow the input signal, we may define expressions which will describe dynamic accuracy in the transient state. Such expressions are based on the system error, which in its simplest form is the difference between the system input and the output and is denoted by e(f), that is,
e(t) = y(t) - u(t) where y(i) is the actual output and u(t) is the desired output; (u(t) is the input). The expression called the performance index can take on various forms. typical among them are:
92
and
Transient and steady state response analysis
\ e 2 (r)dr, Jo
integral of error squared (IES)
1 \e(t)\dt,
integral of absolute error (IAE)
r
I t\e(t)\at, Jo
[Ch. 4
integral of time multiplied absolute error criterion (ITAE).
Having chosen an appropriate performance index, the system which minimises the integral is called optimal. The objective of modern control theory is to design a system so that it is optimal with respect to a performance index. The systems are not necessarily servomechanisms, and in fact, a common problem is to determine a signal vector so that the state vector x(f) is brought from an initial state x 0 to a final state x f in the shortest possible time. We discuss various optimisation methods in Part II of the book.
4.4 RESPONSE OF HIGHER ORDER SYSTEMS We can write the transfer function of an wth order system (see equation (2.12)) in the form K(sm + bis"1'1 + ... + bm) G(s) = — — -.
e
N
(4.14)
Systems having a transfer function of high order can arise when several com ponents, each involving a 'low' order transfer function, make up the system as illustrated by the following simple example. Example 4.1 With reference to Fig. 2.11 (see page 43), calculate the closed loop transfer function (G(s)) given the transfer functions A(s) =
1
s+3
and
2
B(s) = - . s
Solution Using6 equation (2.16), that is, G(s) = 4
G(s) -
A(s) l + A(s)B(s)
s
s
s2 + 3s + 2
(s + l)(s + 2)
, we obtain
Sec. 4.4J
Response of higher order systems
93
The response of the system having the transfer function (4.14) to a unit step input can be written in the form K(s + Zi)(s + z2) ■ ■ ■ (s + zm) Y(s) = — — — s(s + pt)(s + p2) ... (s + pn)
(4.15)
where Zi,z2,...,zm and pit p2,...,p„ are respectively the zeros of the numerator and the denominator polynomials of equation (4.14). The zeros are either real or complex numbers; in the case of a complex zero there must be another zero which is its conjugate, since the a's and b's in equation (4.14) are all real. We first assume that n > m in equation (4.14); we then have two cases to consider: Case 1 Here P\,p2,-.,pn are all distinct numbers. The partial fraction expansion of equation (4.15) has the form Ao
Ai
Y(s) = S
Ki,K2 form
A n+\
+ ——+...+
——.
S+Pi
(4.16)
S+Pn
A"„+i are called the residues of the expansions. The response has the = *i + * a e - p i f + . . . +
m
Knnt~pn'.
As we have already noticed in the previous section, if p\ is complex and p2 is its conjugate then the corresponding residues (K2 and K3) are complex conjugate. The two terms can be combined as illustrated in equation (4.12). Case 2 Here at least one of the roots, say pu is of multiplicity r, that is, K(s + Zi)(s + z 2 ) . . . (s + z m ) Y(s) = — — — . Φ + PiY(s + Pi) · · · (s + Pn-rn)
(4.17)
The partial fraction expansion has the form Λι
Ait
Since
£~l
K (s+p)l\
K2r
AM
Y(S) = — + — — + . s s + pi (s + ριγ
1 ...
' ( / - I ) ! tl-xt-Pt
1
1 .
S + Pn-r+l
(s+PiY (/ = !, 2 > . . .
(4.18)
94
Transient and steady state response analysis
[Ch. 4
the response has the form y(t) = Ki + K21e-pi'
+ K3e~p^
+ K22t
+...+
+...+
- ^ (r-l)\
i'-'e-'i'
Κη_,+ιϋ~ρ"-^.
(4.19)
We now consider that m > « in equation (4.14), which is the case (see section 2.4) when the system is improper; that is, it can happen when we consider idealised and physically non-realisable systems, such as resistanceless circuits. We then divide the numerator by the denominator until we obtain a proper fraction so that when applying a unit step input, we can write Y(s) as -m-ll + ^ + ^ - ' + . - Η J] Y(s) = K[Cl + c2s + ... + es"-™-1] + ——n „_, , · ' · , " s(s + a , s n _ 1 + . . . + a„)
(4.20)
where the cs, ds, K and K^ are all constants. The second term of the right-hand side is treated as in Case 1 or Case 2 above; the inverse Laplace transform of the first term involves the impulse function (see equation (2.7)) and various derivatives of it.
Example 4.2 Find the response of the system having a transfer function G(s) =
5(s2 + 5s + 6) s 3 + 6s 2 + 10s + 8
to a unit step input.
Solution In this case, 5(«2 + 5s + 6) Y(s) = s(s3 + 6s 2 + 10s + 8)
5(s + 2)(i + 3) s(s + 4)(s 2 + 2s + 2)
The term s2 + 2s + 2 = (s + 1 + i) (s + 1 — /'), hence K(s) can be written as the partial fraction expansion as
Y(S) = - + s
+ s+ 4
+ s + l + z'
s + 1 — /'
Sec. 4.5]
95
Steady state error
where K3 is the conjugate of K4. In fact, *i = J,K2
= -\,K3
= £(-7+i)
and
K4 = £ ( - 7 - / ) .
Hence 15 _ 1 y(t) = T4 ~ 74 e _ 4 ' 15 = -; 4
+
T e _ f I - 1 4 cos ί + 2 sin r]
1 _ 4f , V200 _ f . , , - „ . e Te + sinii + 352). 4 4
4.5 STEADY STATE ERROR It was stated in the previous section that one of the important considerations in system design is the steady state error e^ shown in Fig. 4.7. We would expect that e ss would depend on both the transfer function of the system and on the reference input. In this chapter we have used the notation u(t) to denote the input to a system - generally to denote a (unit) step input. We shall now change this notation to conform with the general usage and the fact that we consider different types of reference inputs. We shall use r(t) to denote the (reference) input and c(r) the output. A unity feedback system with an open loop transfer function A(s) will then be represented (as in Fig. 2.12) by Fig. 4.8. /?(s)
CIS)
7\ r
A[S)
Fig. 4.8 E(s),R(s) and C(s) are the Laplace transforms of e(t), r(t) and c(t) respectively.
We define the error runctionf e(t) = r(t) -
c{t).
(4.21)
Hence e ss = lim e(t). ■(This is assuming that rit) and c{t) are of the same dimensions, that is, are both voltages or speeds, etc., otherwise a non-unity feedback system must be used.
96
Transient and steady state response analysis
[Ch.4
Since E(s) = R(s) - A(s)E(s), it follows that E(s) = 1 + A(s) and by the final value theorem (equation (2.10)) e ss = lims^s) sR(s) = lim . **° 1 + A(s)
(4.22)
We now define three error coefficients which indicate the steady state error when the system is subjected to three different standard reference inputs r(t) (1) Step input, r(t) = ku(t) (k is a constant) (Fig. 4.9). e
sk/s _ ss = lim s+oi + A(s) l + \imA(s)
lim A(s) = Kp, called the position error constant, then Cec
1 + K„
(4.23)
Kp =
or
Hence if k and e ss are specified, the value of the transfer function at s = 0 is determined by equation (4.23).
Fig. 4.9
(2) Ramp input, r(t) = ktu{t) (Fig. 4.10). k k k In this case, R(s) = - j , so that e ss = — or Kv = — where Kv = lim sA(s), S
is called the velocity error constant.
Λν
é? ss
i*0
Sec. 4.5]
Steady state error
97
Fig. 4.10 (3) Parabolic input, r(t) = y kt2u(t) (Fig. 4.11). k k k R(s) = -3, so that e ss = — or Ka = — where Ka = lim s2A(s), called the acceleration error constant.
Fig. 4.11
Example 4.3 Find the (static) error coefficients for the system having the open loop transfer function A(s) =
s(4s + 2) '
98
Transient and steady state response analysis
[Ch. 4
Solution lim A(s) = oo Kv
lim sA(s) = 4 i+0
lim s2A(s) = 0. i+0
From the definitions of the error coefficients, it is seen that ess depends on the number of poles (or integrations) at s = 0 of the transfer function. This leads to the following classification. A transfer function is said to be of type N if it has N poles at the origin. Thus if A(s) =
K(s-zi)...(s-zm)
(4.24)
s'(s-Pl)...(s-p„)
where j = 0,A(s)isoitypeO, j = l,/l(s)is of type 1, / = 2, A(s) is of type 2. At s = 0,
A(s) = hm— *'0«'
where
K\ =
.
(4.25)
(ΓΡΰ-Λ-Ρη)
(Κι is called the gain of the transfer function.) Hence the steady state error eiS depends on / and r(t) as summarised in Table 4.1. Table 4.1 ^ss
/
System r(f)=ku(t)
r(t) = ktu(t)
r(f)= f'MO
0
Type 0
Finite
oo
oo
1
Type 1
0
Finite
oo
2
Type 2
0
0
Finite
It would seem that for a system to have a rapid response, capable of following various inputs with zero steady state error, the engineer must ensure that the open loop transfer function has a pole of high order at the origin. Unfortunately, such systems are prone to instability so that it is unusual to design a system having a pole at the origin of order greater than two (that is, / > 2).
Sec. 4.6]
99
Feedback control
4.6 FEEDBACK CONTROL In this section we shall discuss certain aspects of the effect of feedback on the system response. This is in fact the very basis of classical control theory, and much literature is devoted to it. We shall only consider certain general principles and not detailed system design considerations. We have already shown (see Fig. 2.12 and equation (2.16)) that the closed loop transfer function G(s) of the negative feedback system (Fig. 4.12) is related to the feed-forward transfer function A(s) and feed-back transfer function Bis) by A(s) G(s) = ——. (4.26) l+A(s)B(s)
Ris) - .
CIS)
EiS)
Als)
Ώ
Bis) Fig. 4.12
We consider a simple example of a first order system for which K as + 1 and
B(s) = c (a constant)
so that K/a
K U(S) —
as + Kc + l
—
Kc + l s+
We now take the inverse Laplace transform, that is, obtain the impulse response (see equation (2.14)) of the system, where (1 ) c = 0 (that is, response of the open loop system) K -t/a g(t) = - e-", a
and
100
Transient and steady state response analysis
[Ch.4
(2) οΦΟ K _«£±±t K _1 g(t) = - e a = - e <* a a where a =
a Kc + \
a and a are respectively the time-constants of the open loop and the closed loop systems. a is always positive (since we are considering a physical system), but since c can be made positive or negative, a may be either positive or negative. Fig. 4.13 shows how the time responses vary with different values of Kc. /Cc^-1
Fig. 4.13
If the impulse response does not decay to zero as t increases, the system is unstable. From Fig. 4.13 it is seen that the instability region is defined by Kc^-l In many applications the control system consists basically of a plant having the transfer function A(s) and a controller having a transfer function B(s), as in Fig. 4.14.
^
^
>
r
£,sl
Controller Bis)
Q{s)
l'ig. 4.14
Plant Als)
Cis)
Sec. 4.6]
Feedback control
101
The transfer function of the closed loop system can be found by application of equation (4.26) with feed-forward transfer function A(s)B(s) (see equation (2.15)) and a feed-back transfer function 1; then A(s)B(s) G(s) = —^-LL· . 1 + A(s)B(s)
(4.27)
The controllers can be of various types. (1) The on-offcontroller The action of such a controller is very simple. With reference to Fig. 4.14, the control depends on the actuating signal e(t). Thus if e(f) > 0
then
q(i) = ßi
and if e(r) < 0
then
q(t) = Q2
where q(t) is the output signal from the controller and ß x and β 2 are some constants. The on-off controller is obviously a nonlinear device and it cannot be described by a transfer function. An analysis of such systems can be accomplished by 'describing functions' which we do not consider in this book. (2) Proportional Controllers For this control action q(t) =
Kpe(t)
where Kp is a constant, called the gain of the controller. Thus the,transfer function for this controller is 2(i) Ä(s) = ~ = Kp. E(s)
(4.28)
(3) Integral Controller In this case q(t) = K \ e(t)dt,
hence
Jo B(s) = K/s.
(4.29)
102
Transient and steady state response analysis
[Ch. 4
(4) Derivative Controller In this case de q(t) = K—, dt
hence
B(s) = Ks. This type of controller is never used on its own, since its action is only effective while e(t) is changing, that is, during the transient state period, and has no effect in the steady state. (5) Proportional-plus-derivative (PD) Controller In this case de q{t) = K0e(t) + Ki—, dt
hence
B(s) = KM + - 1 s) F
*P
= Kp(l + Ks) where K =
(4.30) Kp
(6) Proportional-plus-integral (PI) Controller In this case ttf
e(t)dt,
1(0 = Kpe(t) + Kt
hence
Jo K B(s) = Kp(\ + ~) s
'where K =
(4.31) Kp
(7) Proportional-plus-derivative-plus-integral (PDI) Controller In this case
de q(t) = Kpe(t)
dt
B(s) = Kp(\ + kts + k2/s) Ki where ki = — Up
and
r
+ Ki— + K2
K2 k2 = — . Kp
e(t)dt,
Jo
hence
Sec. 4.6]
Feedback control
103
Controllers of the various types described above can be constructed for electrical, mechanical, and other systems. This was pointed out for electrical systems when discussing simulation in section 3.2. We shall not discuss the details of how such controllers are designed. For our purposes it is sufficient to appreciate that such controllers are at our disposal.
Example 4.4 Design a controller for a plant having the transfer function A(s) — l/(s + 2)) so that the resulting closed loop system has a zero steady state error to a reference ramp input. Solution From the Table 4.1 (on p. 98) we note that for zero steady state error to a ramp input, the system must De of type 2. Hence if we choose an integral controller with B(s) = K/s then the transfer function of the closed loop system including the plant and the controller (see Fig. 2.11) is A(s)B(s)
K 3
l + A(s)B(s) ~ s + 2s2 + K The response of this control system depends on the roots of the denominator polynomial s3 + 2s 2 + K = 0. It will be shown in the next chapter (example 5.3) that for all values of K at least one of the roots has a positive real part, which implies that the system is unstable. We cannot use an integral controller in this case. But if we use a PI (proportional-plus-integral) controller, say
B(s) = Kp(l + f ), the system is of type 2 and response of the system depends on the roots of s3 + 2s2 + Kps + KKP = 0. In this case, values of K and Kp can be found so that the system is stable (see example 5.3).
104
Transient and steady state response analysis
[Ch.
PROBLEMS (1) A unity feedback system has the open-loop transfer function C(s)
2 0
s + 5s2 + 6s
(i) Calculate the positional, velocity and acceleration error constants. (ii) Determine the steady-state error where the input for t > 0 is (a) r{t) = 2, (b) r(t) = 2t, (c) r(t) = \t2 and (d) r(t) = 3 - 2r. (2) Find the time constants of the system (i) whose behaviour is defined by the differential equation ay (a) 3 — + 2y = 2 dr , dy (b) i — + 3y = 3 di
given that at t = 0, y - 0. (ii) whose (open-loop) transfer function is (a) C(5) (b) G(s) =
l + 3s 1 l + 2s'
(iii) Unity feedback systems have open-loop transfer functions as in (ii) above. Determine the system time-constants. What is the effect on the response of the system on the addition of a closed loop? (3) Investigate the response to a unit step input of the system having the transfer function 4 GW = -■,
·
s J + 4s + 5 (4) Obtain the response to a unit step input of the system shown in Fig. 4.15 for K = l, 5.25, and 10.
— <
?\_ *
r
K
r,a G W
yy ( 5 )
1
-5.55*1
■
Fig. 4.15
4]
Problems
105
(5) The output of a system to a unit step input is a2 Y{s) = —1 · s(s + di s + a2) (a) Write this output (in its normalised form) as a function of the system natural frequency ωη and the damping ratio r. (b) The transfer function of the system is written as G(s) =
A:
. (s + pi)(s + p 2 ) Find the values of pt and p 2 in terms of the above two system constants. (6) A second order underdamped system has a response y(t) to a unit step input. Obtain the time of occurrence and the height of the maximum overshoot in terms of the system natural frequency ω„ and the damping ratio T.