Some special cases


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1 Lecture Notes on Control Systems/D. Ghose/ Some special cases Routh table is easy to form in most cases, but there could be some cases when we need to do some extra work. Case 1: The first element in any one row of the Routh array is zero, but the other elements are not. Why is this a problem? Because, subsequent rows have to be obtained by dividing by this element and if it is zero, we end up with. An Example: D(s) = s 3 3s +2 = s 3 +0s 2 3s +2 = a 0 s 3 + a 1 s 2 + a 2 s + a 3 Then the Routh array is given by, s s s 1 0 s 0 We can perhaps guess that there is at least one unstable pole. But how many are there? There are many ways to handle this situation. One is to multiply (s + a) to the original polynomial, where a could be any number. But remember that if you take a to be negative then you must account for it when you are counting the number of poles on the RHS of the splane. So, to avoid any confusion select a positive a. In the above example, let us select a = 3. Then, the new characteristic polynomial is, The Routh array is now, ˆD(s) = (s +3)D(s) =(s +3)(s 3 3s +2) = s 4 +3s 3 3s 2 7s +6 = a 0 s 4 + a 1 s 3 + a 2 s 2 + a 1 s + a 0
2 Lecture Notes on Control Systems/D. Ghose/ s s s s s So, there are 2 changes in sign and so there are two poles. Note that since a is positive, it does not affect the number of poles on the RHS of the splane and hence the number of sign changes. It is easy to verify that s 3 3s +2=(s 1) 2 (s +2) A word of caution here. The value of a has to be selected carefully or through trial and error. Not all values of a will serve our purpose. For example, consider a = 1. Then, (s 3 3s +2)(s +1)=s 4 + s 3 3s 2 s +2 The corresponding Rout table is s s s s s 0 So we end up with a row of zeros. There is another method to tackle the above situation. This is by putting a small number ɛ>0 in place of the zero in the Routh array, and then continue. If we do this then, the original Routh array becomes.
3 Lecture Notes on Control Systems/D. Ghose/ s s 2 ɛ 2 s 1 3ɛ 2 ɛ 0 s 0 2 Now, for ɛ>0, the term ( 3ɛ 2)/ɛ is a negative number that approaches 2/ɛ as ɛ 0. So, there are two sign changes. A word of caution: The ɛ method does not always work, especially when there are two pure imaginary roots. Case 2: When all the numbers in a row in the Routh array are zeros, then one of the following may be true. 1. There is a pair of real roots with opposite signs (See Figure 11.5(a)). 2. There is a pair of pure imaginary roots (See Figure 11.5(b)). 3. There are pairs of complex conjugate roots placed symmetrically about the origin of the splane (See Figure 11.5(c)). Figure 11.5: Pole locations for Case 2 To handle this case, follow the procedure given below. 1. Form an auxiliary equation by taking its coefficients from the row just above the rowofzeros 1. 1 If the auxiliary equation has order two, then there are two equal and opposite roots. If it has order four, then there must be two pairs of equal and opposite roots. If the auxiliary equation has odd order, then one root is on the origin and the rest are distributed according to the rules governing even order auxiliary equation. Moreover, all these roots can be obtained by simply solving for the roots of the auxiliary equation.
4 Lecture Notes on Control Systems/D. Ghose/ Take its derivative with respect to s. 3. Replace the row of zeros with the coefficients of the derivative of the auxiliary equation. 4. Continue with the Routh array. The number of sign changes indicates the number of poles on the RHS of the splane 2. An Example: The Routh array is, D(s) =s 4 + s 3 3s 2 s +2 s s s coefficients of auxiliary equation s new s 1 row s The s 1 row is all zeros, so the auxiliary equation is formed from the s 2 row as, which on differentiating, A(s) = 2s 2 +2 da(s) = 4s s This is used to obtain the new fourth row. So, there are two sign changes and the equation has two poles on the RHS. In fact, it can be verified that, D(s) =s 4 + s 3 3s 2 s +2=(s 1) 2 (s +2)(s +1) 2 However, note that there may not be any sign changes at all. In which case, it implies that there are no roots with positive real parts or no roots on the RHS of the splane. But the solution of the auxiliary equation will reveal that there are roots on the imaginary axis. So, just examining the sign of the first column is not enough but we need to also evaluate the roots of the auxiliary equation.
5 Lecture Notes on Control Systems/D. Ghose/ Another Example: The Routh array is, D(s) =s 6 + s 5 2s 4 3s 3 7s 2 4s 4 s s s auxiliary equation s new row s 2 6/ s 1 50/ s The auxiliary equation and its derivative are, A(s) = s 4 3s 2 4=0 da(s) s = 4s 3 6s From the Routh array, we can see that there is one sign change and so there is one root on the RHS of the splane. It can be verified that, D(s) = s 6 + s 5 2s 4 3s 3 7s 2 4s 4 = (s +2)(s 2)(s 2 +1)(s 2 + s +1) Note that there is a pair of roots on the imaginary axis. auxiliary equation. Examine the roots of the A(s) =s 4 3s 2 4=(s 2 4)(s 2 +1) Hence, some of the roots of the original equation (especially those that were causing the zero row!) can be identified by just solving the auxiliary equation.
6 Lecture Notes on Control Systems/D. Ghose/ A word of caution: RouthHurwitz criterion can be applied only when the coefficients are real (not complex). Another interesting result: If any element in any row of the Routh array is negative then the system is unstable Application of RouthHurwitz Criterion Consider the following feedback system. Let Figure 11.6: A feedback system with proportional control The closedloop transfer function is G(s) = The denominator polynomial is, The Routh array is, G 1 (s) = G 2 (s) = k 2 = = s +1 (s +2)(s 3) k 1 G 1 (s) 1+k 1 G 1 (s)g 2 (s) k 1 (s +1) (s +2)(s 3) + k 1 k 2 (s +1) k 1 (s +1) s 2 +(k 1 k 2 1)s +(k 1 k 2 6) D(s) =s 2 +(k 1 k 2 1)s +(k 1 k 2 6) s 2 1 k 1 k 2 6 s 1 k 1 k s 0 k 1 k 2 6 0
7 Lecture Notes on Control Systems/D. Ghose/ For the system to be stable k 1 k 2 > 6 and k 1 k 2 > 1 which implies that any value of k 1 and k 2 such that k 1 k 2 > 6 should make the system stable. Another Example: Let The overall transfer function is, Figure 11.7: A more complicated feedback system G(s) = The characteristic polynomial, G 1 (s) = G 2 (s) = s +1 s(s T ) 1 1+sQ (s + 1)(1 + sq) s(s T )(1 + sq)+(s +1) D(s) = s(s T )(1 + sq)+(s +1) = Qs 3 +(1 TQ)s 2 +(1 T )s +1 Just going by the necessary condition that says that all coefficients have to be positive in order for the system to be stable we can write Q>0, TQ < 1, and T<1 But this is not sufficient. So, let us form the Routh array, s 3 Q 1 T s 2 1 TQ 1 s 1 (1 T ) Q 1 TQ 0 s 0 1 0
8 Lecture Notes on Control Systems/D. Ghose/ So, suppose we assume Q>0 then, Q > 0 1 TQ > 0 TQ < 1 (1 T ) Q 1 TQ > 0 (1 T ) > Q 1 TQ Since TQ < 1, we can write the last inequality as, (1 T )(1 TQ) >Q (1 T ) (1 T )TQ > Q (1 T ) >Q[(1 T )T +1] 1 T Q< 1+T T 2 Can T>1? No, because then 1 T will be negative and the previous inequality will be violated as Q>0. Can T<0? Yes, in fact so long as 0 <T < 5 1 2, we have a feasible solution for Q. This can be easily shown. Look at the following table where we get the range of Q for some given values of T that will satisfy the inequality. T (1 T )(1 TQ) >Q Q 0 Q<1 0 <Q<1 0.1 Q< <Q< Q< <Q< Q< <Q< Q< <Q< Q< <Q<1.23
9 Lecture Notes on Control Systems/D. Ghose/ Figure 11.8: The feasible region in the (T,Q)space PROBLEM SET 5 1. Use the Hurwitz determinants to find the stability of systems having the following characteristic equations: (a) s 3 +20s 2 +9s = 0 (b) 3s 4 +10s 3 +5s 2 + s +2=0 (c) s 4 +2s 3 +6s 2 +8s +8=0 2. Use the RouthHurwitz criterion to find the stability of systems having the following characteristic equations: (a) s 3 +20s 2 +9s = 0 (c) s 6 +2s 5 +8s 4 +12s 3 +20s 2 +16s +16=0 3. Consider the following characteristic equations. Determine the values of K that corresponds to a stable system. Use the RouthHurwitz criterion. (a) s 4 +22s 3 +10s 2 +2s + K =0 (b) s 4 +20Ks 3 +5s 2 +(10+K)s +18=0 (c) s 3 +(K +0.5)s 2 +4Ks +50=0 4. Consider the following system with a unity feedback:
10 Lecture Notes on Control Systems/D. Ghose/ G(s) = K s(1 + Ts) (a) What would be the values of K and T to ensure stability? (b) Suppose we want the poles to be on the left of the line s = 3, then what would be the values of K and T to satisfy this requirement? (c) Plot the feasible values of K and T on the (K, T) plane. (d) Repeat (a), (b), and (c) when G(s) = K(s +1) s(1 + Ts)(1+2s)
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