Routh-Hurwitz

Use the Routh-Hurwitz eriterion to determine the number of roots with positive real parts in the following eharaeteristie equations... 

In the same way that the Routh-Hurwitz criterion offers a simple method of determining the stability of continuous systems, the Jury (1958) stability test is employed in a similar manner to assess the stability of discrete systems. 

Stability analysis methods Routh-Hurwitz criterion Apply the Routh test on the closed-loop characteristic polynomial to find if there are closed-loop poles on the right-hand-plane. 

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 (/.< ., 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. 

We first introduce the time honored (/.< ., ancient ) Routh-Hurwitz criterion for stability testing. 

The complete Routh array analysis allows us to find, for example, the number of poles on the imaginary axis. Since BIBO stability requires that all poles lie in the left-hand plane, we will not bother with these details (which are still in many control texts). Consider the fact that we can calculate easily the exact roots of a polynomial with MATLAB, we use the Routh criterion to the extent that it serves its purpose.1 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. 

Example 7.2 If we have only a proportional controller (i.e., one design parameter) and real negative 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 ... 

With the Routh-Hurwitz criterion, we need immediately xr > 0 and Kc > 0. (The, v term requires Kc > -1, which is overridden by the last constant coefficient.) The Routh array for this third order polynomial is... 

One 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 ary formulas, and we can find the ultimate frequency, which however, can be obtained with a root locus plot or frequency response analysis—techniques that we will cover later. 

When the system has dead time, we must make an approximation, such as the Pade 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. 

Let us first use the first order Pade approximation for the time delay function and apply the Routh-Hurwitz criterion. The approximate equation becomes... 

The points at which the loci cross the imaginary axis can be found by the Routh-Hurwitz criterion or by substituting s = jco in the characteristic equation. (Of course, we can also use MATLAB to do that.)... 

For as instructive as root locus plots appear to be, this technique does have its limitations. The most important one is that it cannot handle dead time easily. When we have a system with dead time, we must make an approximation with the Pade formulas. This is the same limitation that applies to the Routh-Hurwitz criterion. 

With the Routh-Hurwitz analysis in Chapter 7, we should find that to have a stable system, we must keep Kc < 7.5. (You fill in the intermediate steps in the Review Problems. Other techniques such as root locus, direct substitution or frequency response in Chapter 8 should arrive at the same result.)... 

A Routh-Hurwitz analysis can confirm that. The key point is that with cascade control, the system becomes more stable and allows us to use a larger proportional gain in the primary controller. The main reason is the much faster response (smaller time constant) of the actuator in the inner loop.2... 

Show by means of the Routh-Hurwitz criterion that two conditions of controller parameters define upper and lower bounds on the stability of the feedback system incorporating this process. 

By using the Routh-Hurwitz criterion determine whether or not the system is stable. 

An inherently stable process can be destabilised by the addition of a feedback control loop—particularly where integral action is included. This is illustrated in the following example using the characteristic equation and the Routh-Hurwitz stability criterion. 

Hence with Kc= 1.8 and r, = 3.5, the polar plot of the open-loop transfer function passes through the point (-1, 0). This confirms the result obtained in Example 7.6 using the Routh-Hurwitz criterion, i.e. that with these controller parameters, the response of the controlled variable is conditionally stable. Figure 7.55 shows polar plots of the open-loop transfer function Gr(i) for different values of Kc and t>. 

In order to determine the number of roots of the z-transformed characteristic equation that lie outside the unit circle, a procedure analogous to the Routh-Hurwitz approach for continuous systems (Section 7.10.2) can be used. The Routh-Hurwitz criterion cannot be applied directly to the characteristic equation f(z) = 0. However, by mapping the interior of the unit circle in the z-piane on to the left half of a new complex variable -plane, the Routh-Hurwitz criterion can be applied as for continuous systems to the corresponding characteristic equation in terms of the new variable<4,). This mapping can be achieved using the bilinear transformation07 ... 

In order to apply the Routh-Hurwitz criterion, the transformation given by equation 7.230 must be applied. Hence equation 7.236 becomes ... 

Bilinear transformation to map the interior of the unit circle in the z-plane onto the left half of the complex variable -plane. (Application of the Routh-Hurwitz stability criterion). ... 

Cr Cr- l Capacitance of amplifier feedback circuit in hard-wired system generating PI action Capacitance of voltage input circuit in hard-wired system generating PD action i, etc. Elements in Routh-Hurwitz array F (farad) F (farad) M L-2T[Pg.731]

The stationary point of the system (5.2) is by a definition stable one, if all the roots of its characteristic equation (5.11) have the negative real parts. The Routh-Hurwitz criteria presented in Ref. [206] permits escaping the calculations of these roots to establish the simple relations between the coefficients ock, which allow to point out simple stability conditions. For instance, in the case of terpolymerization the positivity of both coefficients oq and oc2 is regarded to be a criteria of such stability, and as for four-component copolymerization the following non-equality a3 < oq < ot2 has also to be hold. At arbitrary number m of the components the positivity of all ak is regarded to be necessary (but not sufficient) stability condition. For the stability of the boundary SP of m-component system located inside the certain boundary 1-subsimplex of monomers Mk, M2,. .., M, the stability of the above SP in such subsimplex and negativity of all values of X, Xl+1,..., vm x (5.13) are needed. 

The proof of Theorem 5.3 consists in showing that the underlying semi-group is analytic (because of the degeneracy of the equation for a and t in (22)), and then in localizing the spectrum by the Routh-Hurwitz criterion. 

Since f p) < 0, the constant term is positive, so the Routh-Hurwitz criterion (Appendix A) says that Ec will be asymptotically stable if and only if... 

To show the instability of E, one must show that the inequality (4.8) is violated. This, however, is a very formidable task without further information about the function f p), which gives the effect of the inhibitor. In the next remark, a specific function is chosen, the function used originally by Lenski and Hattingh [LH]. The proof then reduces to the application of the Routh-Hurwitz condition, still a computationally formidable task. However, the point of interest is deriving any oscillation at all, given the tendency of the chemostat to approach a steady state. 

For stability at a rest point one wishes to show that the eigenvalues of the linearization lie in the left half of the complex plane. There is a totally general result, the Routh-Hurwitz criterion, that can determine this. It is an algorithm for determining the signs of the real parts of the zeros of a polynomial. Since the eigenvalues of a matrix A are the roots of a polynomial... 

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