Lecture Notes: EEEC4340318 Instrumentation and Control Systems - Stability and Routh-Hurwitz Condition
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This document discusses stability and the Routh-Hurwitz condition for determining stability of linear systems from their transfer functions. It begins by defining stability as a system having bounded outputs for bounded inputs. It then introduces the Routh-Hurwitz condition, which provides a necessary and sufficient test for stability by examining the signs of coefficients in a constructed table, without needing to find the system poles. The document proceeds to describe in detail how to construct the Routh-Hurwitz table and apply the condition to determine if a system is stable based on the number of sign changes in the first column of the table.
Lecture Notes: EEEC4340318 Instrumentation and Control Systems - Stability and Routh-Hurwitz Condition
- 1. EEEC4340318 INSTRUMENTATION AND CONTROL SYSTEMS Stability And Routh-Hurwitz Condition FACULTY OF ENGINEERING AND COMPUTER TECHNOLOGY DIPLOMA IN ELECTRICALAND ELECTRONIC ENGINEERING Ravandran Muttiah BEng (Hons) MSc MIET
- 2. 1 Stability A systems is said to be stable if all bounded inputs 𝑟 𝑡 give rise to bounded outputs 𝑦 𝑡 . Figure 1 𝑌 𝑠 𝑅 𝑠 𝐺 𝑠
- 3. 2 Conditions For Stability 𝑦 𝑡 = −∞ ∞ 𝑔 𝜏 𝑟 𝑡 − 𝜏 d𝜏 Let 𝑟 𝑡 be such that 𝑟 𝑡 ≤ 𝑟 𝑦 𝑡 = −∞ ∞ 𝑔 𝜏 𝑟 𝑡 − 𝜏 d𝜏 ≤ −∞ ∞ 𝑔 𝜏 𝑟 𝑡 − 𝜏 d𝜏 ≤ 𝑟 −∞ ∞ 𝑔 𝜏 d𝜏 Using this: system 𝐺 𝑠 is stable if −∞ ∞ 𝑔 𝜏 d𝜏 is finite Figure 2 𝑌 𝑠 𝑅 𝑠 𝐺 𝑠
- 4. 3 Condition In Terms Of Poles ? We want −∞ ∞ 𝑔 𝜏 d𝜏 to be finite Can we determine this from 𝐺 𝑠 We can write a general rational transfer function in the form 𝐺 𝑠 = 𝐾 𝑖 𝑠 + 𝑧𝑖 𝑠𝑁 𝑘 𝑠 + 𝜎𝑘 𝑚 𝑠2 + 2𝛼𝑚𝑠 + 𝛼𝑚 2 + 𝜔𝑚 2 Poles: 0, −𝜎𝑘, −𝛼𝑚 ± j𝜔𝑚 Assuming 𝑁 = 0 and no repeated roots, the impulse response is, 𝑔 𝑡 = 𝑘 𝐴𝑘𝑒−𝜎𝑘𝑡 + 𝑚 𝐵𝑚𝑒−𝛼𝑚𝑡 sin 𝜔𝑚𝑡 + 𝜃𝑚 Stability requires 𝑔 𝑡 d𝑡 to be bounded; that requires 𝜎𝑘 > 0, 𝜎𝑚 > 0 In fact, system is stable if poles have negative real parts.
- 5. 4 Marginal Stability Consider integrator: 𝐺 𝑠 = 1 s ; simple pole at origin, 𝑦 𝑡 = −∞ ∞ 𝑟 𝑡 d𝑡 If 𝑟 𝑡 = cos 𝑡 , which is bounded, then 𝑦 𝑡 = sin 𝑡 . Bounded. If 𝑟 𝑡 = 𝑢 𝑡 , which is bounded, then 𝑦 𝑡 = 𝑡. Not bounded. Consider 𝐺 𝑠 = 1 𝑠2+1 , simple poles at 𝑠 = ±j1 Unit step response: 𝑢 𝑡 − cos 𝑡 . Bounded What if 𝑟 𝑡 is a sinusoid of frequency 1 2π Hz ? Not bounded. If 𝐺 𝑠 has a pole with positive real part, or a repeated pole on j𝜔-axis, output is always unbounded.
- 6. 5 Routh-Hurwitz Condition We have seen how to determine stability from the poles. Much easier than having to find impulse response and then determining if 𝑔 𝜏 d𝜏 < ∞ Can we determine stability without having to determine the poles ? Yes! Routh-Hurwitz condition.
- 7. 6 Let 𝐺 𝑠 = 𝑝 𝑠 𝑞 𝑠 , where 𝑞 𝑠 = 𝑎𝑛𝑠𝑛 + 𝑎𝑛−1𝑠𝑛−1 + ⋯ 𝑎1𝑠 + 𝑎0 = 𝑎𝑛 𝑠 − 𝑟1 𝑠 − 𝑟2 ⋯ 𝑠 − 𝑟𝑛 where 𝑟𝑖 are the roots of 𝑞 𝑠 = 0. By multiplying out, 𝑞 𝑠 = 0 can be written as, 𝑞 𝑠 = 𝑎𝑛𝑠𝑛 − 𝑎𝑛 𝑟1 + 𝑟2 + ⋯ + 𝑟𝑛 𝑠𝑛−1 + 𝑎𝑛 𝑟1𝑟2 + 𝑟2𝑟3 + ⋯ 𝑠𝑛−2 − 𝑎𝑛 𝑟1𝑟2𝑟3 + 𝑟1𝑟2𝑟4 + ⋯ 𝑠𝑛−3 + ⋯ + −1 𝑛 𝑎𝑛 𝑟1𝑟2𝑟3 ⋯ 𝑟𝑛 = 0 If all 𝑟𝑖 are real and in left half plane, what is sign of coefficients of 𝑠𝑘 the same!
- 8. 7 That observation leads to a necessary condition. Hence, not that useful for design. A more sophisticated analysis leads to the Routh-Hurwitz condition, which is necessary and sufficient. Hence, can be quite useful for design.
- 9. 8 Routh-Hurwitz Condition: A First Look Consider 𝐺 𝑠 = 𝑝 𝑠 𝑞 𝑠 . Poles are solutions to 𝑞 𝑠 = 0; i.e., 𝑎𝑛𝑠𝑛 + 𝑎𝑛−1𝑠𝑛−1 + 𝑎𝑛−2𝑠𝑛−2 + ⋯ + 𝑎1𝑠 + 𝑎0 = 0 Construct a table of the form, 𝑠𝑛 𝑠𝑛−1 𝑠𝑛−2 𝑠𝑛−3 ⋮ 𝑠0 𝑎𝑛 𝑎𝑛−1 𝑏𝑛−1 𝑐𝑛−1 ⋮ ℎ𝑛−1 𝑎𝑛−2 𝑎𝑛−3 𝑏𝑛−3 𝑐𝑛−3 ⋮ 𝑎𝑛−4 ⋯ 𝑎𝑛−5 ⋯ 𝑏𝑛−5 ⋯ 𝑐𝑛−5 ⋯ ⋮ ⋯ where, 𝑏𝑛−1 = 𝑎𝑛−1𝑎𝑛−2 − 𝑎𝑛𝑎𝑛−3 𝑎𝑛−1 = −1 𝑎𝑛−1 𝑎𝑛 𝑎𝑛−2 𝑎𝑛−1 𝑎𝑛−3 𝑏𝑛−3 = −1 𝑎𝑛−1 𝑎𝑛 𝑎𝑛−4 𝑎𝑛−1 𝑎𝑛−5 𝑐𝑛−1 = −1 𝑏𝑛−1 𝑎𝑛−1 𝑎𝑛−3 𝑏𝑛−1 𝑏𝑛−3
- 10. 9 Now consider the table we have just constructed. 𝑠𝑛 𝑠𝑛−1 𝑠𝑛−2 𝑠𝑛−3 ⋮ 𝑠0 𝑎𝑛 𝑎𝑛−1 𝑏𝑛−1 𝑐𝑛−1 ⋮ ℎ𝑛−1 𝑎𝑛−2 𝑎𝑛−3 𝑏𝑛−3 𝑐𝑛−3 ⋮ 𝑎𝑛−4 ⋯ 𝑎𝑛−5 ⋯ 𝑏𝑛−5 ⋯ 𝑐𝑛−5 ⋯ ⋮ ⋯ Loosely speaking: Number of roots in the right half plane is equal to the number of sign changes in the first column of the table. Stability if no sign changes in the first column. Now let’s move towards a more sophisticated statement.
- 11. 10 Let 𝐺 𝑠 = 𝑝 𝑠 𝑞 𝑠 where 𝑞 𝑠 = 𝑎𝑛𝑠𝑛 + 𝑎𝑛−1𝑠𝑛−1 + ⋯ 𝑎1𝑠 + 𝑎0 System is stable if all poles of 𝐺 𝑠 have negative real parts. Recall, poles are solutions to 𝑞 𝑠 = 0. Can we find a necessary and sufficient condition that depends only on 𝑎𝑘 so that we don’t have to solve 𝑞 𝑠 = 0 ? Stability (Revision) Figure 3 𝑌 𝑠 𝑅 𝑠 𝐺 𝑠
- 12. 11 (1) Consider 𝑞 𝑠 with 𝑎𝑛 > 0 𝑎𝑛𝑠𝑛 + 𝑎𝑛−1𝑠𝑛−1 + 𝑎𝑛−2𝑠𝑛−2 + ⋯ 𝑎1𝑠 + 𝑎0 = 0 (2) Construct a table of the form, Row 𝑛 Row 𝑛 − 1 Row 𝑛 − 2 Row 𝑛 − 3 ⋮ Row 0 𝑎𝑛 𝑎𝑛−1 𝑏𝑛−1 𝑐𝑛−1 ⋮ ℎ𝑛−1 𝑎𝑛−2 𝑎𝑛−3 𝑏𝑛−3 𝑐𝑛−3 ⋮ 𝑎𝑛−4 ⋯ 𝑎𝑛−5 ⋯ 𝑏𝑛−5 ⋯ 𝑐𝑛−5 ⋯ ⋮ ⋯ Procedure provided on the following slides. (3) Count the sign changes in the first column. (4) That is the number of roots in the right half plane. Stability (poles in Left Half Plane if 𝑎𝑘 > 0 and all terms in first column > 0 Routh-Hurwitz Condition
- 13. 12 References (1) Tim Davidson, Introduction to Linear Control Systems, McMaster University, 2018.