Feedback Control System

1. Feedback Control System THE ROOT-LOCUS DESIGN METHOD Chapter 5 Dr.-Ing. Erwin Sitompul http://zitompul.wordpress.com

2. Frequency Response • A linear system’s response to sinusoidal inputs is called the system’s frequency response. • Frequency response can be obtained from knowledge of the system’s pole and zero locations. Consider a system described by where With zero initial conditions, the output is given by

3. Frequency Response Assuming all the poles of G(s) are distinct, a partial fraction expansion of the previous equation will result in Transient response Steady-state response where

4. Frequency Response Examine Magnitude Phase But the input is Therefore • Meaning?

5. Frequency Response • A stable linear time-invariant system with transfer function G(s), excited by a sinusoid with unit amplitude and frequency ω0, will, after the response has reached steady-state, exhibit a sinusoidal output with a magnitude M(ω0) and a phase Φ(ω0) at the frequency ω0. • The magnitude M is given by |G(jω)| and the phase Φ is given by G(jω), which are the magnitude and the angle of the complex quantity G(s) evaluated with s taking the values along the imaginary axis (s = jω). • The frequency response of a system consists of the frequency functions |G(jω)| and G(jω), which describe how a system will respond to a sinusoidal input of any frequency.

6. Frequency Response • Frequency response analysis is interesting not only because it will help us to understand how a system responds to a sinusoidal input, but also because evaluating G(s) with s taking on values along jω axis will prove to be very useful in determining the stability of a closed-loop system. • As we know, jω axis is the boundary between stability and instability. Therefore, evaluating G(jω) along the frequency band will provide information that allows us to determine closed-loop stability from the open-loop G(s).

7. Frequency Response Given the transfer function of a lead compensation (a) Analytically determine its frequency response characteristics and discuss what you would expect from the result. Φ M = D • At low frequency, ω→0, M→K, Φ→0. • At high frequency, ω→∞, M→K/α, Φ→0. • At intermediate frequency, Φ> 0.

8. Frequency Response (b) Use MATLAB to plot D(jω) with K = 1, T = 1, and α = 0.1 for 0.1 ≤ ω ≤ 100 and verify the prediction from (a). Using bode(num,den), in this case bode([1 1],[0.1 1]), MATLAB produces the frequency response of the lead compensation

9. Frequency Response For second order system having the transfer function we already plotted the step response for various values of ζ. • The damping and rise time of a system can be determined from the transient-response curve.

10. Frequency Response The corresponding frequency response of the system can be found by replacing s = jω This G(jω) can be plotted along the frequency axis, for various values of ζ.

11. Frequency Response • The damping of a system can be determined from the peak in the magnitude of the frequency response curve. • The rise time can be estimated from the bandwidth, which is approximately equal to ωn. ? The transient-response curve and the frequency-response curve contain the same information.

12. Frequency Response • Bandwidth is defined as the maximum frequency at which the output of system will track an input sinusoid in a satisfactory manner. • By convention, the bandwidth is the frequency at which the output is attenuated to a factor of 0.707 times the input. • The maximum value of the frequency-response magnitude is referred to as the resonant peak Mr.

13. Bode Plot Techniques Advantages of working with frequency response in terms of Bode plots: Dynamic compensator design can be based entirely on Bode plots. Bode plots can be determined experimentally. Bode plots of systems in series can be simply added, which is quite convenient. The use of a logarithmic scale permits a much wider range of frequencies to be displayed on a single plot compared with the use of linear scales. Bode plot of a system is made of two curves, • The logarithm of magnitude vs. the logarithm of frequencylog M vs. log ω, or also Mdb vs. log ω. • The phase versus the logarithm of frequencyΦ vs. log ω.

14. Bode Plot Techniques For root locus design method, the open-loop transfer function is written in the form For frequency-response design method, s is replaced with jω to write the transfer function in the Bode form

15. Bode Plot Techniques To draw the Bode plot of a transfer function, it must be rewritten in magnitude equation and phase equation. For example Then Phase Equation Magnitude Equation Magnitude Equation (log) Magnitude Equation (db)

16. Bode Plot Techniques Examining all the transfer functions we have dealt with so far, all of them are the combinations of the following four terms: • Gain • Pole or zero at the origin • Simple pole or zero • Quadratic pole or zero Once we understand how to plot each term, it will be easy to draw the composite plot, since log M and Φ are the additive combination of the magnitude logarithms and the phases of all terms.

17. Bode Plot Techniques Gain Magnitude Phase

18. Bode Plot Techniques Pole or zero at the origin Magnitude Phase

19. Bode Plot Techniques As example, Bode plot of a zero (jω) at origin will be as follows:

20. Bode Plot Techniques Simple pole or zero • The point where ωτ = 1 or ω = 1/τ is called the break point.

21. Bode Plot Techniques As example, Bode magnitude plot of a simple zero (jωτ+1) is given below, with τ = 10. • The break point lies at ω = 1/τ = 0.1. • Correction of Asymptote

22. Bode Plot Techniques The corresponding Bode phase plot of a simple zero (jωτ+1) is given as: • Corrections of Asymptotes by 11°,at ω = 0.02 and ω = 0.5. • Corresponds to 1/5ωbreak and 5ωbreak

23. Bode Plot Techniques Quadratic pole or zero • Asymptotes can be used for rough sketch. • Afterwards, correction must be made according to the value of damping factor ζ.

24. Bode Plot Techniques

25. Bode Plot Techniques

26. Bode Plot: Example Plot the Bode magnitude and phase for the system with the transfer function • ωb1 = 0.5, ωb2 = 10, and ωb3 = 50 Convert the function to the Bode form, • One pole at the origin 5 terms will be drawn separately and finally composited

27. Bode Plot Techniques ωb1 = 0.5 ωb2 = 10 ωb3 = 50 60 –20 db/dec 40 0 db/dec 20 –20 db/dec db 0 –40 db/dec –20 –40 : Rough composite

28. Bode Plot Techniques ωb1 = 0.5 ωb2 = 10 ωb3 = 50 60 40 +3 db 20 –3 db db –3 db 0 –20 –40 Final Result : Rough composite : 3-db-corrected composite

29. Bode Plot Techniques ωb3 = 50 ωb2 = 10 ωb1 = 0.5 2 2.5 –11° –11° +11° +11° 10 0.1 –11° 50 +11° 250

30. Bode Plot Techniques ωb3 = 50 ωb2 = 10 ωb1 = 0.5 Final Result

31. Neutral Stability • Root-locus technique  determine the stability of a closed-loop system, given only the open-loop transfer function KG(s), by inspecting the denominator in factored form. • Frequency response technique  determine the stability of a closed-loop system, given only the open-loop transfer function KG(jω) by evaluating it and performing a test on it. The principle will be discussed now.

32. Neutral Stability Suppose we have a system defined by • System is unstable if K > 2. • The neutrally stable points lie on the imaginary axis, K = 2 and s = ±j. All points on the locus fulfill:

33. Neutral Stability • The frequency response of the system for various values of K is shown as follows. • At K = 2, the magnitude response passes through 1 at the frequency ω = 1 rad/sec (remember s = ±j), at which the phase passes through 180°. • After determining “the point of neutral stability”, we know that: • K < Kneutral stability stable, • K > Kneutral stability  unstable. • At (ω = 1 rad/sec, G(jω) = –180°), • |KG(jω)| < 1  stable K, • |KG(jω)| > 1  unstable K.

34. Neutral Stability • A trial stability condition can now be stated as follows If |KG(jω)| < 1 at G(jω) = –180°, then the system is stable. • This criterion holds for all system for which increasing gain leads to instability and |KG(jω)| crosses magnitude = 1 once. • For systems where increasing gain leads from instability to stability, the stability condition is If |KG(jω)| > 1 at G(jω) = 180°, then the system is stable. • For other more complicated cases, the Nyquist stability criterion can be used, which will be discussed next. • While Nyquist criterion is fairly complex, one should bear in mine, that for most systems a simple relationship exists between the closed-loop stability and the open-loop frequency response. “ ” “ ”

35. Homework 8 • No.1, FPE (5th Ed.), 6.3. • Hint: Draw the Bode plot in logarithmic and semi-logarithmic scale accordingly. • No.2. • Derive the transfer function of the electrical system given above. • If R1 = 10 kΩ , R2 = 5 kΩ and C = 0.1 μF, draw the Bode plot of the system.