# Class 13 Frequency domain analysis

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1 Class 13 Frequency domain analysis

2 The frequency response is the output of the system in steady state when the input of the system is sinusoidal Methods of system analysis by the frequency response, as the Bode Diagrams or the Nyquist Plot, are the most conventional techniques used in Engineering for analysis and project of Control Systems

3 The advantage of these methods of analysis of systems by the frequency response is that they allow us to find both the absolute and relative stability of linear systems in closed loop only with the knowledge of frequency response in open loop, which can be experimentally obtained with signal generators (sinusoidal) and precision measurements instruments (both easily available in laboratory)

4 Therefore, the analysis of complicated systems can be done through tests of frequency response without being necessary to determine the roots of the characteristic equation (i.e., the poles of the system) Linear time invariant system, zero initial conditions The output y(t) will have the same frequency of the input x(t), but, are, in general, different. the amplitude Y and the phase angle φ,

5 As a matter of fact, we have And doing s = 0 + jω, that is,

6 That is, doing s = 0 + jω in the transfer function G(s), one obtain G(jω) G(jω) = G(jω) e jϕ = G(jω) e j G(jω) where ϕ = G(jω) = arctg Im(G(jω) ( ) Re(G(jω)

7 As the input x(t) = sen(ωt) can be expressed as x(t) = sen(ωt) = then it can be shown that y ss, the output y(t) in steady state, is y ss and therefore, = G(jω) e jωt e jωt j (equação de Euler) e jωt e jωt y ss = G(jω) sen( ωt + G(jω) ) j Y φ

8 The most general case, the sinusoidal input with absolute value X and phase ϕ x(t) = X sen(ωt + φ) hence, y ss, the output y(t) in steady state, becomes y ss = X G(jω) sen( ωt + G(jω) + φ) Y Y(jω)

9 G(jω) = Y(jω) X(jω) = Y X ratio between the output and the input s amplitude ϕ = G(jω) = Y(jω) X(jω) difference between the phase angle of the output and the input Then, the characteristic of a system subject to a sinusoidal input can be obtained directly from G(jω) = Y(jω) X(jω) = G(jω) e jϕ ( F.T. ) sinusoidal

10 G(jω) = Y(jω) X(jω) = G(jω) e jϕ ( F.T. ) sinusoidal where ϕ = G(jω) = being -π < ϕ < π < 0 phase delay = 0 in phase > 0 phase in advance Se ϕ = π sen(ωt±π) = sen(ωt)

11 Análise no domínio da frequência In the next slides we shall see an EXAMPLE of the curves: Bode Diagram de G(jω) absolute value e phase (or angle) Nyquist Plot de G(jω) parametrized by frequency ω They are generic curves to EXEMPLIFY how the Bode Diagrams and Nyquist Plots are.

12 Bode Diagram (example) G(jω) db Obtained drawing a curve for the absolute value of G(jω) and another for the phase, or angle of G(jω). ω G(jω) G(jω) db is measured in dbs ω The ω axis are drawn in the logarithmic scale. G(jω) is measured in degrees

13 Nyquist Plot (example) Obtained drawing in the complex plane the values of G(jω) when ω varies from to +. Im {G(jω)} 0 Re {G(jω)} with the Nyquist Plot of G(jω) we can apply the Nyquist Criterion to determine the stability

14 Análise no domínio da frequência Next the first Example of Nyquist Plot

15 Example 1: Consider the closed loop system below K (s )(s + 4) thus, G(s) = (s K )(s + 4) and substituting s = jω, we obtain: G K ( ω + 8) K ω (jω) = + j ( ω + 4) ( ω + 16) ( ω + 4) ( ω + 16)

16 Example 1 (continued): K ( ω + 8) K ω G(jω) = + j ( ω + 4) ( ω + 16) ( ω + 4) ( ω + 16) Intersection with the real axis (imaginary part = 0) ω = 0 ω = 0 G(jω) = G(j0) = K/8 Intersection with the imaginary axis (real part = 0) ω real that makes null Re{G(jω)}, the real part of G(jω) this Nyquist Plot does not intercept the imaginary axis

17 Example 1 (continued): K ( ω + 8) K ω G(jω) = + j ( ω + 4) ( ω + 16) ( ω + 4) ( ω + 16) Limits at infinite (G(jω) for ω = ± ) G(j ) = 0 + j0 G( j ) = 0 + j0 + (3º quadrant) (º quadrant) and now we write down in the complex plane these points of intersection and the limits at infinite

18 Example 1 (continued): K ( ω + 8) K ω G(jω) = + j ( ω + 4) ( ω + 16) ( ω + 4) ( ω + 16) Im{G(jω)} K/8 0 Re{G(jω)} and then we draw a sketch of the Nyquist Plot of G(jω)

19 Example 1 (continued): K ( ω + 8) K ω G(jω) = + j ( ω + 4) ( ω + 16) ( ω + 4) ( ω + 16) Im{G(jω)} K/8 0 Re{G(jω)}

20 Example 1 (continued): K ( ω + 8) K ω G(jω) = + j ( ω + 4) ( ω + 16) ( ω + 4) ( ω + 16) But it still missing to give a direction for this Nyquist Plot to become complete Im{G(jω)} K/8 0 Re{G(jω)}

21 Example 1 (continued): K ( ω + 8) K ω G(jω) = + j ( ω + 4) ( ω + 16) ( ω + 4) ( ω + 16) This direction is done by following G(jω) when ω varies from + to Im{G(jω)} K/8 0 Re{G(jω)}

22 Análise no domínio da frequência In the next slides we shall see an EXAMPLE of the curves: direction and the number of encirclements of the Nyquist Plot de G(jω) It is a generic curve just for EXEMPLYING how to find the direction and how to calculate the number of encirclements of the Nyquist Plot.

23 The direction of the Nyquist Plot In order to give a direction to the Nyquist Plot, we follow G(jω) when ω varies from + to Im{G(jω)} 0 Re{G(jω)} To followg(jω) when ω varies from + to, it may be necessary to calculate additional points of ω as in this Nyquist Plot above

24 The direction of the Nyquist Plot Following G(jω) with only the points ω = 0, ± and ± it is not possible yet to determine the direction of the Nyquist Plot Im{G(jω)} 0 Re{G(jω)} However, by calculating some additional points, we can then follow G(jω) when ω varies from + to and then give a direction to this Nyquist Plot above

25 The direction of the Nyquist Plot By choosing a ω between+ and + (such as ω = 5) and another ω between + and 0 (such as ω = 1) it becomes possible to follow G(jω) and determine the direction of this Nyquist Plot ω = +5 Im{G(jω)} ω = +1 0 Re{G(jω)} Calculating G(jω) for ω = 5 and G(jω) for ω = 1 we can write down the points G(j5) and G(j1) in the Nyquist Plot above

26 The direction of the Nyquist Plot Although it is not necessary, let us also write down points in the plot G(jω) below: ω = 1e ω = 5 ω = +5 Im{G(jω)} ω = 1 ω = +1 ω = 5 0 Re{G(jω)} In that way we can better observe the property that the Nyquist Plot is always symmetric with respect to the real axis

27 The direction of the Nyquist Plot Now we can put arrows that indicate the direction of this Nyquist Plot G(jω) ω = +5 Im{G(jω)} ω = 1 0 Re{G(jω)} ω = +1 ω = 5

28 The direction of the Nyquist Plot Im{G(jω)} 0 Re{G(jω)} With the direction of G(jω) we can define the number of encirclements of a point at real axis that the Nyquist Plot does

29 The number of encirclements of the Nyquist Plot The number of encirclements of G(jω) around a point at the real axis has the signal positive if counter clockwise Im{G(jω)} negative if clockwise A B C 0 Re{G(jω)} For example: The number of encirclements of G(jω) around a point point A N A = 0 point B N B = + point C N C = +1

30 Análise no domínio da frequência Next more Example of Nyquist Plot

31 Example : Consider the closed loop system bellow (s 1)(s K + 4)(s + 5) thus, G(s) = (s 1)(s K + 4)(s + 5) and substituting s = jω, we obtain: (8ω + 0) K ( ω 11) ωk G(jω) = + j ( ω + 1) ( ω + 16)( ω + 5) ( ω + 1) ( ω + 16)( ω + 5)

32 Example (continued): G (8ω + 0) K ( ω 11) ωk (jω) = + j ( ω + 1) ( ω + 16)( ω + 5) ( ω + 1) ( ω + 16)( ω + 5) Intersection with real axis (imaginary part = 0) ω = 0 G(jω) = G(j0) = K/0 ω = 11 ω = ± 3,317 G(jω) = G(±j3,317) = K/108 Intersection with imaginary axis (real part = 0) ω real that makes null Re{G(jω)}, the real part of G(jω) this Nyquist Plot does not intercept the imaginary axis

33 Example (continued): G (8ω + 0) K ( ω 11) ωk (jω) = + j ( ω + 1) ( ω + 16)( ω + 5) ( ω + 1) ( ω + 16)( ω + 5) Limits at infinite (G(jω) for ω = ± ) G(j ) = 0 + j0 + G( j ) = 0 + j0 (º quadrant) (3º quadrant) and now we write down in the complex plane these points of intersection (already calculated) and the limits at infinite

34 Example (continued): G (8ω + 0) K ( ω 11) ωk (jω) = + j ( ω + 1) ( ω + 16)( ω + 5) ( ω + 1) ( ω + 16)( ω Im {G(jω)} + 5) ω = ± 3,317 K/0 K/108 0 Re {G(jω)} and then we draw a sketch of the Nyquist Plot of G(jω)

35 Example (continued): G (8ω + 0) K ( ω 11) ωk (jω) = + j ( ω + 1) ( ω + 16)( ω + 5) ( ω + 1) ( ω + 16)( ω Im {G(jω)} + 5) ω = ±j3,317 K/0 K/108 0 Re {G(jω)}

36 Example (continued): G (8ω + 0) K ( ω 11) ωk (jω) = + j ( ω + 1) ( ω + 16)( ω + 5) ( ω + 1) ( ω + 16)( ω Note that it has already put the arrows that indicate the direction of this Nyquist Plot Im {G(jω)} + 5) K/0 K/108 0 Re {G(jω)}

37 Example 3: Consider the closed loop system bellow: K (s ) (s + 1) thus, G(s) = K (s ) (s + 1) and substituting s = jω, we obtain: ( ω ) K 3ωK G(jω) = + j ( ω + 1) ( ω + 1)

38 Example 3 (continued): ( ω ) K 3ωK G(jω) = + j ( ω + 1) ( ω + 1) Intersection with real axis (imaginary part = 0) ω = 0 G(jω) = G(j0) = K = K / (½) Intersection with imaginary axis (real part = 0) ω = +1,41 G(jω) = +j1,41 K ω = ω = 1,41 G(jω) = j1,41 K

39 Example 3 (continued): ( ω ) K 3ωK G(jω) = + j ( ω + 1) ( ω + 1) Limits at infinite (G(jω) for ω = ± ) G(j ) = K + j0 + G( j ) = K + j0 (1º quadrant) (4º quadrant) and now we write down in the complex plane these points of intersection (already calculated) and the limits at infinite

40 Example 3 (continued): ( ω ) K 3ωK G(jω) = + j ( ω + 1) ( ω + 1) Im {G(jω)} ω = 1,41 K/(½) 0 K Re {G(jω)} and then we draw a sketch of the Nyquist Plot of G(jω) ω = 1,41

41 Example 3 (continued): ( ω ) K 3ωK G(jω) = + j ( ω + 1) ( ω + 1) Im {G(jω)} ω = 1,41 K/(½) 0 K Re {G(jω)} ω = 1,41

42 Example 3 (continued): ( ω ) K 3ωK G(jω) = + j ( ω + 1) ( ω + 1) Note that it has already put the arrows that indicate the direction of this Nyquist Plot Im {G(jω)} K/(½) 0 K Re {G(jω)}

43 Having the Nyquist Plot of G(jω) we can apply the Nyquist Criterion to determine the stability of the M.F. system

44 Nyquist Criterion (for stability) Closed loop system G(s) Clearly, the OLTF (open loop transfer function) is given by: G(s) = K G(s) H(s)

45 Nyquist Criterion (for stability) N 1 = P C.L. P O.L. Number of encirclements do Nyquist Plot around point 1 Number of open loop poles at RHP Number of closed loop poles at RHP N -1 = number of encirclements of G(jω) around point 1, which can be positive (if direction is counter clockwise), or negative (if direction is clockwise)

46 Nyquist Criterion (for stability) that is, P C.L. = N 1 + P O.L. P C.L. = Number of closed loop poles at RHP N 1 = Number of encirclements of the Nyquist Plot around point 1 P O.L. = Number of open loop poles at RHP And, of course, the closed loop system will be stable if P M.F. = 0

47 Application of Nyquist Criterion to system from Example 1

48 Example 1 (continued): Let us return to this problem G(s) which Nyquist Plot was K (s )(s + = then, 4) P O.L. = 1 By doing now an analysis to the number of encirclements of the Nyquist Plot around point 1, we get

49 Example 1 (continued): G(s) = K (s )(s + 4) P O.L. = 1 N -1 = 0 se K < 8 1 se K > 8 Now, by applying the Nyquist Criterion to determine the C.L. stability, we get

50 Example 1 (continued): G(s) = (s K )(s + 4) P M.A. = 1 P M.F. = = 1 se K < = 0 se K > 8 hence, the closed loop system is stable for K > 8

51 Application of Nyquist Criterion to system from Example

52 Let us now return to this problem Example (continued): which Nyquist Plot was G(s) = K (s 1)(s + 4)(s + 5) then, P M.A. = 1 By doing now an analysis to the number of encirclements of the Nyquist Plot around point 1, we get

53 Example (continued): G(s) = K (s 1)(s + 4)(s + 5) P M.A. = 1 N -1 = 0 se K < 0 1 se 0 < K < se K > 108 Now, by applying the Nyquist Criterion to determine the C.L. stability, we get

54 Example (continued): G(s) = K (s 1)(s + 4)(s + 5) P M.A. = = 1 se K < 0 P M.F. = = 0 se 0 < K < = se K > 108 hence, the closed loop system is stable for 0 < K < 108

55 Application of Nyquist Criterion to system from Example 3

56 Finally, let us now return to this problem Example 3 (continued): G(s) which Nyquist Plot was = K (s ) (s + 1) then, P M.A. = 0 By doing now an analysis to the number of encirclements of the Nyquist Plot around point 1, we get

57 Example 3 (continued): G(s) = K (s ) (s + 1) P M.A. = 0 N -1 = 0 se K < 0,5 1 se K > 0,5 Now, by applying the Nyquist Criterion to determine the C.L. stability, we get

58 Example 3 (continued): G(s) = K (s ) (s + 1) P M.A. = 0 P M.F. = = 0 se K < 0, = 1 se K > 0,5 hence, the closed loop system is stable for K < 0,5

59 Departamento de Engenharia Eletromecânica Thank you! Felippe de Souza

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