SL semana 7 Clase 2 ZPK and Stability.pptx

Presentación representación de sistemas mediante polos, ceros y factor de escala

1. Zero-Pole-Gain Representation Poles and Waveforms Stability Linear Systems 1

2. 2 Let a linear time invariant system described by the th-order linear differential equation: Through Laplace Transform, the system’s transfer function becomes an alternative representation in the complex frequency domain:   ZPK Representation

3. ZPK Representation Let’s expand the summations: Now, factorize the coefficient of the largest derivative in the numerator and denominator: with: , and   3

4. ZPK Representation Note that both polynomials (numerator and denominator) can be written in a factorized form in terms of their roots (up to now, assuming all-different roots): is complex and make the output zero : . is also complex and make the output tend to plus/minus infinity, so it is known as pole : is a real gain . Above representation is known as the Zero - Pole - Gain (ZPK) representation.   4

5. ZPK Representation Since and are complex, can be plotted over the complex plane , usin g the symbol for zeros and for poles :   5 Important remark: Real systems hold transfer function with real coefficients. Consequently, if the complex is a root (at numerator or denominator), its conjugate must be also a root (at numerator or denominator).  

6. ZPK Representation E xercise 1 : Find the ZPK of a system modeled by the following: Solution: Step 1: Write the transfer function: Step 2: Normalize the polynomials to identify the gain:   6

7. ZPK Representation Step 3 : Find zeros and poles (TIP: You may use np.roots in Python or roots in Matlab for high order polynomials) :   7

8. ZPK Representation Step 4 : Draw poles and zeros over the complex plane:   8 https://www.geogebra.org/3d/za9q5dsg

9. ZPK Representation Exercise 1 : Sketch the zero-pole diagram of the following transfer function: Exercise 2 : Sketch the zero-pole diagram of the following transfer function: Exercise 2 : Determine the output of above systems for the input and .   9

10. Poles and Waveforms Recall the ZPK form of the transfer function: We can recover the impulse response by inverse-Laplace transforming the transfer function : Applying partial fractions to ease the inverse transform (assume all-different roots):   10

11. Poles and Waveforms Each exponential in the summation is known as a characteristic mode of the system ruled by the characteristic root or pole, :   11

12. Poles and Waveforms Each exponential in the summation is known as a characteristic mode of the system ruled by the characteristic root or pole, : A real negative pole A real positive pole   12

13. Poles and Waveforms A pole at the origin A double pole at origin   13

14. Poles and Waveforms Complex conjugate poles A negative real part A positive real part   14

15. 15 Poles and Waveforms Complex conjugate imaginary poles Complex conjugate imaginary double poles  

16. Internal (Asymptotic) Stability If, in the absence of an external input ( zero input ), a system remains in a particular state (or condition) indefinitely, then that state is said to be an equilibrium state of the system. For an LTI system, zero state , in which all initial conditions are zero, is an equilibrium state .   What happens to the ZIR of an LTI system with small nonzero initial conditions ( small disturbances)? 16 If the system is stable , when left to itself, every mode should approach to 0 as , so it should eventually return to zero state.   However, if even one of the modes GROWS INDEFINITELY , the system will NEVER RETURN to zero state, and the system would be identified as UNSTABLE . There is a borderline case, when some modes neither decay to zero nor grow indefinitely, while all the remaining modes decay to zero. Such a system is said to be marginally stable .

17. Internal (Asymptotic) Stability Note that this kind of stability is independent on the input. Hence, it is known as internal stability , asymptotic stability or stability in the sense of Lyapunov   Let     then If and do not have common factors, then the denominator of is identical to , the characteristic polynomial of the system. So, internal stability can be determined using the previous criteria.   17

18. Internal (Asymptotic) Stability If and have no common factors, the asymptotic stability criterion can be restated in terms of the poles of the transfer function of a system, as follows:   18 Compute poles Are ALL poles in the LHP? Asymptotically stable Are ANY poles in the RHP? Asymptotically unstable Are REPEATED poles in the IMAGINARY AXIS ?   Marginally stable YES NO NO YES YES NO

19. Internal (Asymptotic) Stability Exercise: Investigate the asymptotic stability of LTIC system described by the following equations, assuming that the equations are internal system descriptions:   19

20. Internal (Asymptotic) Stability Exercise : Find the transfer function of the circuit w.r.t. where is the output and is the input: Sketch the zero-pole diagram w.r.t. Inve stigate the system stability w.r.t.   20

21. Routh-Hurwitz Criteria Necessary conditions : Let the characteristic polynomial of an LTIC system. The following two conditions are necessary , but not sufficient for the system to be stable: All the coefficients of the polynomial must have the same sign. For instance: UNSTABLE: UNSTABLE: None of the coefficients vanishes, i.e. all the powers of s must be present in the characteristic equation . For instance: UNSTABLE: UNSTABLE: Necessary: If any of above conditions are not satisfied, the system is UNSTABLE. But, if those two conditions are satisfied, we need to further check it for stability through the Routh test .   21

22. Routh-Hurwitz Criteria ( cont ). Necessary and sufficient condition : The criteria were independently developed by Edward Routh and Adolf Hurwitz. R-H Criteria is a mathematical test that is a necessary and sufficient condition for the stability of an LTI system. For the test, we need to: Tabulate the coefficients of in a particular way. Compute the so-called Routh table (with two special cases). Interpret the Routh table to determine the number of poles in the RHP of the complex plane.   22

23. Routh-Hurwitz Criteria ( cont ). Tabulate the coefficients of in a particular way: Let the characteristic polynomial satisfying the two necessary conditions. Let’s tabulate the coefficients as follows:   23

24. Routh-Hurwitz Criteria ( cont ). Tabulate the coefficients of in a particular way: Examples : :   24

25. Routh-Hurwitz Criteria ( cont ). 2. Compute the so-called Routh table: Fill the table as follows:   25

26. Routh-Hurwitz Criteria ( cont ). 2. Compute the so-called Routh table: Fill the table as follows:   26

27. Routh-Hurwitz Criteria ( cont ). 2. Compute the so-called Routh table: Examples :   27

28. Routh-Hurwitz Criteria ( cont ). 2. Compute the so-called Routh table: Examples : Note that matches the last entry in the first column . That’s how you Know you’re doing well.   28

29. Routh-Hurwitz Criteria ( cont ). 3. Interpret the Routh table to determine the number of poles in the RHP of the complex plane: Check the necessary and sufficient conditions on the first column of the Routh table: If both conditions are satisfied, the system is STABLE. If any condition is not satisfied, the system is UNSTABLE, and the number of sign changes corresponds to the number of poles in the RHP. 29 All the coefficients in the first column must have the same sign. There should not be any sign change in the first column.

30. Routh-Hurwitz Criteria ( cont ). 3. Interpret the Routh table to determine the number of poles in the RHP of the complex plane: : :   30 All entries have with the same sign. There are no sign changes in the first column. The system is STABLE . There are two sign changes in the first column. The system is UNSTABLE . There are two pole in the RHP.

31. Routh-Hurwitz Criteria ( cont ). Exercise: Investigate the stability of LTIC system described by the following equations:   31

32. Routh-Hurwitz Criteria (cont.) Special case 1: The first element of any row in the Routh array is zero and the remaining row contains at least one non-zero element. Effect: The term in the next row becomes infinite an d the test fails: Exa mple: :   32

33. Routh-Hurwitz Criteria (cont.) Special case 1: The first element of any row in the Routh array is zero and the remaining row contains at least one non-zero element. Solution: Substitute the zero by a small positive and complete the array. Then, analyze by taking the limit Exa mple: :   33     There are two sign changes in the first column. The system is UNSTABLE .

34. Routh-Hurwitz Criteria (cont.) Special case 2: All elements of a row in the Routh array are zero. Effect: None terms in the next row can be computed an d the test fails: Exa mple: :   34

35. Routh-Hurwitz Criteria (cont.) Special case 2: All elements of a row in the Routh array are zero ( RoZ ). Solution: 1. Form a polynomial with the coefficients of the row just above the RoZ , known as Auxiliary Polynomial Exa mple: :   35  

36. Routh-Hurwitz Criteria (cont.) Special case 2: All elements of a row in the Routh array are zero ( RoZ ). Solution: 2. Replace the RoZ by the derivative of the Auxiliary Polynomial : Exa mple: :   36    

37. Routh-Hurwitz Criteria (cont.) Special case 2: All elements of a row in the Routh array are zero ( RoZ ). Solution: 3. Complete the Routh array: Exa mple: :   37

38. Routh-Hurwitz Criteria (cont.) Special case 2: All elements of a row in the Routh array are zero ( RoZ ). Solution: 4. Check stability with a little change in the criterion: Exa mple: :   38 If there are no sign changes in the first column, the system is MARGINALLY STABLE . If there are sign changes, the system is UNSTABLE .

39. Routh-Hurwitz Criteria (cont.) Special case 2: All elements of a row in the Routh array are zero ( RoZ ). Important remarks on special case 2: Auxiliary polynomial is ALWAYS built from an EVEN row. Auxiliary polynomial is always part of . That is, roots of are also roots of . The roots of are symmetric w.r.t the origin. Hence, a RoZ appears only in three cases of pole distribution: Case A: Real and symmetrical about the origin (from an UNSTABLE system). Case B: Imaginary and symmetrical about the origin (from a MARGINALLY STABLE system) . Case C: Quadran t and symmetrical about the origin (from an UNSTABLE system).   39