Our objective is simple. We want to make sure that the controller settings will not lead to an unstable system. Consider the closed-loop system response that we derived in Section 5.2:
C = KmGcGaGp 1+ GmGcGaGp R+ GL 1+ GmGcGaGp L (7.1)
with the characteristic equation
1+ GmGcGaGp = 0. (7.2)
The closed-loop system is stable if all the roots of the characteristic polynomial have negative real parts. Or we can say that all the poles of the closed-loop transfer function lie in the left- hand plane (LHP). When we make this statement, the stability of the system is defined entirely on the inherent dynamics of the system and not on the input functions. In other words, the results apply to both servo and regulating problems.
We also see another common definition – bounded-input bounded-output (BIBO)
One illustration of this definition is to consider a hypothetical situation with a closed-loop pole atthe origin. In such a case, we know thatif we apply an impulse inputor a rectangular pulse input, the response remains bounded. However, if we apply a step input, which is bounded, the response is a ramp, which has no upper bound. For this reason, we cannot accept any control system that has closed-loop poles lying on the imaginary axis. They must be in the LHP.1
The addition of a feedback control loop can stabilize or destabilize a process. We will see plenty examples of the latter. For now, we use the classic example of trying to stabilize an open-loop unstable process.
Example 7.1: Consider the unstable process function Gp = [K/(s − a)], which may arise from a linearized model of an exothermic chemical reactor with an improper cooling design. The question is whether we can make a stable control system by using simply a proportional controller. For illustration, we consider a unity feedback loop with Gm = 1. We also take the actuator transfer function to be unity, Ga = 1.
With a proportional controller, Gc= Kc, the transfer function of this closed-loop servo system is Y R = GcGp 1+ GcGp = KcK s− a + KcK. The characteristic equation is
s− a + KcK = 0,
which means that if we want a stable system, the closed-loop poles must satisfy s= a − KcK< 0.
In other words, the closed-loop system is stable if Kc> a/K .
For a more complex problem, the characteristic polynomial will not be as simple, and we need tools to help us. The two techniques that we will learn are the Routh–Hurwitz criterion and root locus. Root locus is, by far, the more important and useful method, especially when we can use a computer. Where circumstances allow (i.e., the algebra is not too ferocious), we can also find the roots on the imaginary axis – the case of marginal stability. In the simple example above, this is where Kc= a/K . Of course, we have to be smart enough to pick Kc> a/K , and not Kc< a/K .
7.2. The Routh–Hurwitz Criterion
The time-honored (i.e., ancient!) Routh–Hurwitz criterion is introduced for stability testing. It is not proved here – hardly any text does anymore. Nonetheless, two general polynomials
1 Do not be confused by the integral control function; its pole at the origin is an open-loop pole. This
point should be cleared up when we get to the root-locus section. Furthermore, conjugate poles on the imaginary axis are BIBO stable – a step input leads to a sustained oscillation that is bounded in time. However, we do not consider this oscillatory steady state as stable, and hence we exclude the entire imaginary axis. In an advanced class, you should find more mathematical definitions of stability.
7.2. The Routh–Hurwitz Criterion
are used to illustrate some simple properties. First, consider a second-order polynomial with the leading coefficient a2= 1. If the polynomial has two real poles p1 and p2, itcan be
factored as
P(s)= s2+ a1s+ a0= (s − p1)(s− p2). (7.3)
We may observe that if a1 is zero, both roots,± j
√
a0, are on the imaginary axis. If a0 is
zero, one of the two roots is at the origin. We can expand the pole form to give
P(s)= s2− (p1+ p2)s+ p1p2 (7.4)
and compare the coefficients with the original polynomial in Eq. (7.3). If both p1and p2are
negative, the coefficients a1and a0must be positive definite, which is the mathematicians’
way of saying that a1> 0 and a0> 0.
Next, consider a third-order polynomial with leading coefficient a3= 1 and the form in
terms of the poles:
P(s)= s3+ a2s2+ a1s+ a0 = (s − p1)(s− p2)(s− p3). (7.5)
We expand the pole form to
P(s)= s3− (p1+ p2+ p3)s2+ (p1p2+ p1p3+ p2p3)s− p1p2p3. (7.6)
Once again, if all three poles are negative, the coefficients a2, a1, and a0mustbe positive
definite.
The idea is that the signs of the pole are related to the coefficients an, an−1, . . . , a0
of an nth-order characteristic polynomial. If we require all the poles to have negative-real parts, there must be some way that we can tell from the coefficients without actually having to solve for the roots. The idea is that all of the coefficients in the characteristic polynomial must be positive definite. We could develop a comprehensive theory, which Routh did. The attractiveness of the Routh criterion is that without solving for the closed-loop poles, we can derive inequalities that would provide a bound for stable controller design.
The complete Routh array analysis allows us to find, for example, the number of poles on the imaginary axis. Because we require that all poles lie in the LHP, we will not bother with these details (which are still in many control texts). Consider the fact that we can easily calculate the exact roots of a polynomial with MATLAB; we use the Routh criterion to the extent that it serves its purpose.2 That would be to derive inequality criteria for
proper selection of controller gains of relatively simple systems. The technique loses its attractiveness when the algebra becomes too messy. Now the simplified Routh–Hurwitz recipe without proof follows.
(1) Hurwitz test for the polynomial coefficients. For a given nth-order polynomial P(s)= ansn+ an−1sn−1+ · · · + a2s2+ a1s+ a0, (7.7)
all the roots are in the LHP if and only if all the coefficients a0, . . . , an are positive definite.
2 MATLAB does not even bother with a Routh function. Such an M-file is provided on the Web Support
If any one of the coefficients is negative, at least one root has a positive-real part [i.e., in the right hand plane (RHP)]. If any of the coefficients is zero, not all of the roots are in the LHP: it is likely that some of them are on the imaginary axis. Either way, stop. This test is a necessary condition for BIBO stability. There is no point in doing more other than to redesign the controller.
(2) Routh array construction. If the characteristic polynomial passes the coefficient test, we then construct the Routh array to find the necessary and sufficient conditions for stability. This is one of the few classical techniques that is not emphasized and the general formula is omitted. The array construction up to a fourth-order polynomial is used to illustrate the concept.
Generally, for an nth-order polynomial, we need (n+ 1) rows. The firsttwo rows are filled in with the coefficients of the polynomial in a column-wise order. The computation of the array entries is very much like the negative of a normalized determinant anchored by the first column. Even without the general formula, you may pick out the pattern as you read the following three examples.
The Routh criterion states that, in order to have a stable system, all the coefficients in the first column of the array must be positive definite. If any of the coefficients in the first column is negative, there is at least one root with a positive-real part. The number of sign changes is the number of positive poles.
Here is the array for a second-order polynomial, p(s)= a2s2+ a1s+ a0:
1: a2 a0
2: a1 0
3: b1=
a1a0− (0)a2
a1 = a0
In the case of a second-order system, the first column of the Routh array reduces to simply the coefficients of the polynomial. The coefficient test is sufficient in this case. Or we can say that both the coefficient test and the Routh array provide the same result.
The array for a third-order polynomial, p(s)= a3s3+ a2s2+ a1s+ a0, is
1: a3 a1 0 2: a2 a0 0 3: b1 = a2a1− a3a0 a2 b2= (a2)(0)− (a3)0 a2 = 0 4: c1= b1a0− b2a2 b1 = a 0 0
In this case, we have added one column of zeros; they are needed to show how b2is computed.
Because b2 = 0 and c1= a0, the Routh criterion adds one additional constraint in the case
of a third-order polynomial:
b1 =
a2a1− a3a0
a2 > 0.
7.2. The Routh–Hurwitz Criterion
We follow with the array for a fourth-order polynomial, p(s)= a4s4+ a3s3+ a2s2+ a1s+
a0: 1: a4 a2 a0 2: a3 a1 0 3: b1 = a3a2− a1a4 a3 b2 = a3a0− (0)a4 a3 = a 0 0 4: c1= b1a1− b2a3 b1 c2= b1(0)− (0)a3 b1 = 0 5: d1= c1b2− (0)b1 c1 = b 2= a0 0
The two additional constraints from the Routh array are hence
b1 = a3a2− a1a4 a3 > 0, (7.9) c1 = b1a1− b2a3 b1 = b1a1− a0a3 b1 > 0. (7.10)
Example 7.2: If we have only a proportional controller (i.e., one design parameter) and
negative-real open-loop poles, the Routh–Hurwitz criterion can be applied to a fairly high- order system with ease. For example, for the following closed-loop system characteristic equation,
1+ Kc
1
(s+ 3)(s + 2)(s + 1) = 0, find the stability criteria.
We expand and rearrange the equation to the polynomial form: s3+ 6s2+ 11s + (6 + Kc)= 0.
The Hurwitz test requires that Kc> −6 or simply Kc> 0 for positive controller gains. The Routh array is
1: 1 11
2: 6 6+ Kc
3: b1 0
4: 6+ Kc 0
The Routh criterion requires, as in Eq. (7.8), that b1 =
(6)(11)− (6 + Kc)
6 > 0 or 60 > Kc.
The range of proportional gain to maintain system stability is hence 0< Kc< 60.
Example 7.3: Consider a second-order process function Gp = [1/(s2+ 2s + 1)], which is critically damped. If we synthesize a control system with a PI controller, what are the stability constraints?
For illustration, we take Gm= Ga = 1 and the closed-loop transfer function for a servo problem is simply C R = GcGp 1+ GcGp.
In this problem, the closed-loop characteristic equation is
1+ GcGp= 1 + Kc 1+ 1 τIs 1 s2+ 2s + 1 = 0 or τIs3+ 2τIs2+ τI(1+ Kc)s+ Kc= 0.
With the Routh–Hurwitz criterion, we need immediatelyτI > 0 and Kc> 0. (The s term requires that Kc> −1, which is overridden by the last constant coefficient.) The Routh array for this third-order polynomial is
1: τI τI(1+ Kc) 2: 2τI Kc
3: b1 0
4: Kc
With the use of Eq. (7.8), we require that
b1 = 2τ2 I(1+ Kc)− τIKc 2τI = τ I(1+ Kc)− Kc 2 > 0. The inequality is rearranged to
τI > Kc 2(1+ Kc) or 2τI 1− 2τI > K c,
which can be interpreted in two ways. For a given Kc, there is a minimum integral time constant. If the proportional gain is sufficiently large such that Kc 1, the rough estimate for the integral time constant isτI > 1/2. Or if the value of τI is less than 0.5, there is an upper limiton how large Kccould be.
If the given value ofτIis larger than 0.5, the inequality simply infers that Kcmustbe larger than some negative number. To be more specific, if we pickτI = 1,3 the Routh criterion becomes
2> Kc (1+ Kc),
which of course can be satisfied for all Kc> 0. No new stability requirement is imposed in this case. Let us try another choice of τI = 0.1. In this case, the requirement for the proportional gain is
0.2(1 + Kc)> Kc or Kc< 0.25. 3 Note that, with this very specific case, ifτ
I = 1 is chosen, the open-loop zero introduced by the PI
controller cancels one of the open-loop poles of the process function at−1. If we do a root-locus plot, we will see how the root loci change to that of a purely second-order system. With respect to this example, the value is not important as long asτI > 1/2.
7.3. Direct-Substitution Analysis
The entire range of stability forτI = 0.1 is 0 < Kc< 0.25. We will revisitthis problem when we cover root-locus plots; we can make much better sense without doing any algebraic work!
7.3. Direct-Substitution Analysis
The closed-loop poles may lie on the imaginary axis at the moment a system becomes unstable. We can substitute s= jω into the closed-loop characteristic equation to find the proportional gain that corresponds to this stability limit (which may be called marginal unstable). The value of this specific proportional gain is called the critical or ultimate gain. The corresponding frequency is called the crossover or ultimate frequency.
Example 7.2A: Apply direct substitution to the characteristic equation in Example 7.2:
s3+ 6s2+ 11s + (6 + Kc)= 0. Substitution of s= jω leads to
− jω3− 6ω2+ 11ωj + (6 + K
c)= 0. The real- and imaginary-partequations are
Re: − 6ω2+ (6 + Kc)= 0 or Kc= 6(ω2− 1), Im: − ω3+ 11ω = 0 or ω(11 − ω2)= 0.
From the imaginary-part equation, the ultimate frequency isωu = √
11. Substituting this value into the real-part equation leads to the ultimate gain Kc,u= 60, which is consistent with the result of the Routh criterion.
If we have chosen the other possibility,ωu = 0, meaning that the closed-loop poles are on the real axis, the ultimate gain is Kc,u = −6, which is consistent with the other limit obtained with the Routh criterion.
Example 7.3A: Repeat Example 7.3 to find the condition for the ultimate gain.
If we make the s= jω substitution into τIs3+ 2τIs2+ τI(1+ Kc)s+ Kc= 0, itbecomes
−τIω3j− 2τIω2+ τI(1+ Kc)ωj + Kc= 0.
We have two equations after collecting all the real and the imaginary parts and requiring both to be zero:
Re: Kc− 2τIω2 = 0,
Im: τIω[−ω2+ (1 + Kc)]= 0.
Thus we have eitherω = 0 or −ω2+ (1 + K
c)= 0. Substitution of the real-part equation into the nontrivial imaginary-part equation leads to
−ω2+ 1 + 2τ
Iω2 = 0 or ω2u = 1 1− 2τI,
where in the second form, we have added a subscript to denote the ultimate frequency,ωu. Substitution of the ultimate frequency back into the real-part equation gives the relation for the ultimate proportional gain:
Kc,u= 2τI 1− 2τI.
Note that if we have chosen the other possibility ofωu = 0, meaning where the closed- loop poles are on the real axis, the ultimate gain is Kc,u= 0, which is consistent with the other limit obtained with the Routh criterion. The result of direct substitution confirms the inequality derived from the Routh criterion, which should not be a surprise.
We may question whether direct substitution is a better method. There is no clear-cut winner here. By and large, we are less prone to making algebraic errors when we apply the Routh–Hurwitz recipe, and the interpretation of the results is more straightforward. With direct substitution, we do not have to remember any formulas, and we can find the ultimate frequency, which, however, can be obtained with a root-locus plot or frequency-response analysis – techniques that are covered in Section 7.4 and Chap. 8.
When the system has dead-time, we must make an approximation, such as the Pad´e approximation, on the exponential dead-time function before we can apply the Routh– Hurwitz criterion. The result is hence only an estimate. Direct substitution allows us to solve for the ultimate gain and ultimate frequency exactly. The next example illustrates this point.
Example 7.4: Consider a system with a proportional controller and a first-order process but
with dead-time. The closed-loop characteristic equation is given as
1+ Kc 0.8e−2s
5s+ 1 = 0.
Find the stability criteria of this system.
Letus firstuse the first-order Pad´e approximation for the time-delay function and apply the Routh–Hurwitz criterion. The approximate equation becomes
1+ Kc 0.8 5s+ 1 (−s + 1) (s+ 1) = 0 or 5s2+ (6 − 0.8Kc)s+ (1 + 0.8Kc)= 0.
The Routh–Hurwitz criterion requires that 6− 0.8Kc> 0 or Kc< 7.5, and Kc> −1/0.8. When Kcis kept positive, the approximate range of the proportional gain for system stability is 0< Kc< 7.5.
We now repeat the problem with the s= jω substitution into the characteristic equation and rewrite the time-delay function with Euler’s identity:
(5ωj + 1) + 0.8Kc(cos 2ω − j sin 2ω) = 0.
Collecting terms of the real and the imaginary parts provides the two equations: Re: 1+ 0.8Kccos 2ω = 0 or Kc= −1/(0.8 cos 2ω),
7.4. Root-Locus Analysis
Substitution of the real-part equation into the imaginary-part equation gives 5ω + tan 2ω = 0.
The solution of this equation is the ultimate frequency ωu = 0.895, and, from the real- part equation, the corresponding ultimate proportional gain is Kc,u = 5.73. Thus the more accurate range of Kcthat provides system stability is 0< Kc< 5.73.
Note 1: This result is consistent with the use of frequency-response analysis in Chap. 8. Note 2: The iterative solution in solving the ultimate frequency is tricky. The equation has poor numerical properties – arising from the fact that tanθ “jumps” from infinity at θ = (π/2)−to negative infinity atθ = (π/2)+. To better see why, use MATLAB to make a
plot of the function (LHS of the equation) with 0< ω < 1. With MATLAB, we can solve the equation with thefzero()function. Create an M-file namedf.m, and enter these two statements in it:
function y=f(x) y = 5*x + tan(2*x);
After you have saved the file, enter, at the MATLAB prompt;
fzero('f',0.9)
where 0.9 is the initial guess. MATLAB should return 0.8953. If you shift the initial guess just a bit, say by using 0.8, you may get a “solution” of 0.7854. Note that (2)(0.7854) is π/2. If you blindly acceptthis incorrectvalue, Kc,uwill be infinity according to the real-part equation. MATLAB is handy, butitis notfoolproof!
7.4. Root-Locus Analysis
The idea of a root-locus plot is simple – if we have a computer. We pick one design parameter, say, the proportional gain Kc, and write a small program to calculate the roots of the characteristic polynomial for each chosen value of Kcas in 0, 1, 2, 3,. . . . , 100, . . . , etc. The results (the values of the roots) can be tabulated or, better yet, plotted on the complex plane. Even though the idea of plotting a root locus sounds so simple, it is one of the most powerful techniques in controller design and analysis when there is no time delay.
Root locus is a graphical representation of the roots of the closed-loop characteristic
polynomial (i.e., the closed-loop poles) as a chosen parameter is varied. Only the roots are plotted. The values of the parameter are not shown explicitly. The analysis most commonly uses the proportional gain as the parameter. The value of the proportional gain is varied from 0 to infinity, or in practice, just “large enough.” Now a simple example is needed to get this idea across.
Example 7.5: Construct the root-locus plots of a first- and a second-order system with a
proportional controller. See how the loci approach infinity.
(a) Consider the characteristic equation of a simple system with a first-order process and a proportional controller:
1+ Kc Kp τps+ 1 = 0.
(a) (b)
–1 τp –1 τ1 –1 τ2
Figure E7.5.
The solution, meaning the closed-loop poles of the system, is
s= −(1 + KcKp)
τp .
The root-locus plot (Fig. E7.5a) is simply a line on the real axis starting at s= −1/τp when Kc= 0 and extends to negative infinity as Kcapproaches infinity. As we increase the proportional gain, the system response becomes faster. Would there be an upper limit in reality? (Yes, saturation.)
(b) We repeat the exercise with a second-order overdamped process function. The closed- loop characteristic equation of the closed-loop system is
1+ Kc
Kp
(τ1s+ 1)(τ2s+ 1) = 0 or τ
1τ2s2+ (τ1+ τ2)s+ (1 + KcKp)= 0. The two closed-loop poles are
s= −(τ1+ τ2)±
(τ1+ τ2)2− 4τ1τ2(1+ KcKp) 2τ1τ2
.
In the mathematical limit of Kc= 0, we should find two negative real poles at s= −(τ1+ τ2)± (τ1− τ2)2 2τ1τ2 = −1 τ1 or −1 τ2 ,
which are the locations of the two open-loop poles. (This result is easy to spot if we use the first step without expanding the terms.) As Kcbecomes larger, we should come to a point where we have two repeated roots at
s= −(τ1+ τ2) 2τ1τ2 .
If we increase further the value of Kc, the closed-loop poles will branch off (or break away) from the real axis and become two complex conjugates (Fig. E7.5b). No matter how large Kc becomes, these two complex conjugates always have the same real part as given by the repeated root. Thus what we find are two vertical loci extending toward positive and negative infinity. In this analysis, we also see how, as we increase Kc, t he system changes from overdamped to become underdamped, butitis always stable.
This is the idea behind the plotting of the closed-loop poles – in other words, construction of root-locus plots. Of course, we need mathematical or computational tools when we have more complex systems. An important observation from Example 7.5 is that with simple first- and second-order systems with no open-loop zeros in the RHP, the closed-loop system is always stable.
7.4. Root-Locus Analysis
We can now state the problem in more general terms. Let us consider a closed-loop characteristic equation 1+ KcG0= 0, where KcG0is referred to as the “open-loop” transfer