Chapter 5 Frequency-domain design

Chapter 5 Frequency-domain design Control Automático 3º Curso. Ing. Industrial Escuela Técnica Superior de Ingenieros Universidad de Sevilla

Outline of the presentation Introduction. Time response analysis in the frequency domain. Dynamic Systems as filters. Frequency domain Plots. Stability in frequency domain. Stability margins. Loop shaping. Frequency-domain specifications. Control design from a frequency domain point of view. Proportional Control (P) Proportional+Derivative control (PD) and Lead compensation Proportional+Integral control (PI) and Lag compensation Proportional+Integral+Derivative control (PID) and Lead-Lag compensation Conclussions and remarks. ESI.US.Introduction to Feedback Systems 2

Introduction Frequency domain methods are a classic tool in control engineering. The control loop is analyzed considering signals/systems described from a frequency description/response point of view. What will we learn in this chapter? Why frequency domain is a convenient tool to design/analyze controllers. How we can study stability with these new frequency tools. How specifications translate into the frequency domain. How we can design known control structures (P,PD,PID, etc) from a different point of view (some advantages & some drawbacks). ESI.US.Introduction to Feedback Systems 3

Time response analysis in the frequency domain Why frequency domain tools? Some basic notions Time and frequency domains are equivalent mathematical descriptions of the same reality. Both descriptions contain the same information ESI.US.Introduction to Feedback Systems 4

Time response analysis in the frequency domain Interpretation of frequency domain Example: Frequency domain (FD) representation of a pure sine. Information about amplitude and frequency of the time domain (TD) signal. What about more complex signals? A TD function can be described by a summation of waves with different amplitudes and phases (LINEARITY PROPERTY) ESI.US.Introduction to Feedback Systems 5

Time response analysis in the frequency domain How can we compute the frequency components of a time domain signal? Fourier transform provides the tool Remark: For causal systems, the continuous Fourier transform is equivalent to evaluating the Laplace transform with complex argument s = jω or s = 2πfj But, remember the Laplace transform? ESI.US.Introduction to Feedback Systems 6

Time response analysis in the frequency domain Laplace transform is our basic tool to describe LTI systems. There is a convenient and useful relationship between Time domain Laplace transform Frequency domain ESI.US.Introduction to Feedback Systems 7

Dynamic systems as filters A dynamic LTI system can be interpreted as a filter. Filters enhance some frequencies and attenuate others. 2 1.8 12 1.6 1.4 1.2 1 8 1 6.8.6 4.4 2.2-1 1 2 3 4 5 6 7 8 9 1-1 1 2 3 4 5 6 7 8 9 1 Low freq. amplified High freq. attenuated Frequency (Hz) ESI.US.Introduction to Feedback Systems 8

Dynamic systems as filters Types of filters from their frequency response Middle freq. amplified Low freq. attenuated Low freq. amplified Frequency (Hz) Low-pass filter High freq. attenuated Low & High freq. attenuated Frequency (Hz) Band-pass filter Frequency (Hz) High freq. amplified High-pass filter Cross-over frequency (w c ): Maximum frequency for which ESI.US.Introduction to Feedback Systems 9

Frequency domain plots How can we compute the frequency response of a filter (transfer functions)? Once again, the Laplace transform provides the tool Defines the amplification factor for frequency w Defines the phase shift for frequency w Important: These relations are only valid for stable systems G(s) ESI.US.Introduction to Feedback Systems 1

Frequency domain plots The information in G(jw) is usually displayed in different plots. Bode Diagram 25 2 15 1 Bode diagram Magnitude (db) 5-5 -1-15 -2 Phase (deg) -45-9 1-3 1-2 1-1 1 1 1 Frequency (rad/sec) 5 db Nyquist Diagram 4 3 2 2 db -2 db Nyquist Plot Imaginary Axis 4 db 1 6 db 1 db -1-2 -6 db -1 db -4 db -3-4 -5-2 2 4 6 8 1 Real Axis ESI.US.Introduction to Feedback Systems 11

Frequency domain plots Bode diagram Graphic plot of 2 semi-logarithmic scalar plots Bode Diagram Magnitude Magnitude (db) -2-4 -6-8 -1 Phase -12-45 Phase (deg) -9-135 -18 1-2 1-1 1 1 1 1 2 1 3 Frequenc y (rad/s ec ) ESI.US.Introduction to Feedback Systems 12

Graphic plots Nyquist diagram graphic plot of Polar plot Every point in the plot represents the value of G(jw) for a given frequency w Parameterized plot.1 -.1 -.2 -.3 -.4 -.5 -.6.2.4.6.8.1 ESI.US.Introduction to Feedback Systems 13

Quick review. How to plot a Bode diagram Express the system in Bode form ESI.US.Introduction to Feedback Systems 14

Bode of a pure gain ESI.US.Introduction to Feedback Systems 15

Bode of a minimum-phase zero ESI.US.Introduction to Feedback Systems 16

Bode of a non-minimum phase zero ESI.US.Introduction to Feedback Systems 17

Bode of an integrator ESI.US.Introduction to Feedback Systems 18

Bode of a stable pole ESI.US.Introduction to Feedback Systems 19

Bode of an unstable pole ESI.US.Introduction to Feedback Systems 2

Bode of a complex pole Depends on δ The magnitude is maximum for δ.77 Resonance peak ESI.US.Introduction to Feedback Systems 21

Bode of a complex pole ESI.US.Introduction to Feedback Systems 22

Bode plot example G( s) = ( s + 1) s( s + 1) 2-2 m1 m2 m4 m3 G(jω) 1-2 1-1 1 1 1 2 1 3 rad/s ESI.US.Introduction to Feedback Systems 23

Bode plot example 9º f4 º f3 f1 G(jω) -9º f2 1-2 1-1 1 1 1 2 1 3 rad/s ESI.US.Introduction to Feedback Systems 24

The meaning of a Bode plot Magnitude (db) 2 1-1 -2 Bode Diagram u F(s) y -3-4 -45 Phase (deg) -9-135 -18 1 1 1 1 2 Frequency (rad/sec) y u Close to resonance frequency: High output amplitudes Gain > db ESI.US.Introduction to Feedback Systems 25 y u High frequency: Small output amplitudes Gain < db

Nyquist stability criterion Is G bc (s) stable? Criterion based on G(s) Z= poles of G bc (s) inside C. P= poles of G(s) inside C. N= clockwise contour's encirclements of -1. N=Z-P G bc (s) is stable if and only if Z= (i.e., N = -P) ESI.US.Introduction to Feedback Systems 26

Example P=1 G(s) is stable if N=-1 ESI.US.Introduction to Feedback Systems 27

Reduced Nyquist criterion If the Nyquist diagram is closed by the right and P= (Open-loop stable minimum phase system) Phase margin Stability margins Gain margin G(s) is stable if M g and M f are positive ESI.US.Introduction to Feedback Systems 28

Reduced Nyquist criterion How can we measure Mg and Mf from the Bode diagram? ESI.US.Introduction to Feedback Systems 29

Loop shaping Control design in frequency domain can be posed as: Find a controller C(s) that modifies (shapes) the frequency response of the closed-loop systems such that it behaves according to specifications. R(s) - Loop shaping U(s) C(s) G(s) Y bc (s) Several questions arise from this definition How can we manipulate the closed-loop frequency response of the system through C(s)? How can we study time-domain specifications in frequency domain? What is an appropriate shape for the closed-loop frequency response? ESI.US.Introduction to Feedback Systems 31

Loop shaping Some facts: In general, it is difficult to find an exact relation on how the closed-loop frequency response changes with C(s). R(s) - U(s) C(s) G(s) Y bc (s) Open-loop and closed-loop frequency responses are related. The effect of C(s) in the open-loop chain is straighforward from it Bode Plot. Important information (stability, Mf, Mg, etc) can be extracted from the open-loop freq. response. C(s) G(s) Loop shaping is performed on the open-loop chain The controller C(s) modifies the frequency response of C(s)G(s), and indirectly the target closed loop-system ESI.US.Introduction to Feedback Systems 32

Loop-shaping Frequency response of the open-loop system determines the closed-loop time response to step change in the reference. Stability - Positive phase/gain margin ( if C(s)G(s) does not have poles in the right-half plane) SO(%) 6-Mf C(s) G(s) Raise time ts π / 2w c Steady state tracking error depends on the number of poles at the origin (integrators), and the open-loop Bode gain ESI.US.Introduction to Feedback Systems 33

Loop-shaping What is a good shape for C(s)G(s)? 1 Magnitude (db) 5-5 -1 Wc -15 Low frequency Steady state Crossover frequency Stability and transient High frequency Noise rejection ESI.US.Introduction to Feedback Systems 34

Frequency domain specifications Specifications for controlled systems are usually provided in time domain (SO, t s, t e, etc) It is necessary to transform specifications into frequency domain Remember the Root locus method: Time domain specifications were transformed into geometric restrictions in complex plane. Two different specifications Steady-state specifications Transient-regime specifications ESI.US.Introduction to Feedback Systems 35

Frequency domain specifications Specifications affect different frequencies 1 Magnitude (db) 5-5 -1 Wc -15 Low frequency Steady state Crossover frequency Stability and transient High frequency Noise rejection ESI.US.Introduction to Feedback Systems 36

Steady state specifications General error signal transfer function R(s) - E(s) U(s) C(s) G(s) Y bc (s) The error signal depends on the reference R(s). Different patterned references are used (step, ramp, etc) Example for step input Steady-state specifications usually translate into a minimum value for the gain (or minimum slope of the gain) at low frequencies. ESI.US.Introduction to Feedback Systems 37

Error Table (Review of chapter 3) Type of a system = Number of integrators Error Type 1 2 Step Ramp Parabolic ESI.US.Introduction to Feedback Systems 38

Transient regime specifications Conditions on the closed-loop system usually stablished for a reference step input. For the sake of simplifying controller design: Closed-loop response Shape of the open-loop system Simplifying hypothesis x The closed-loop system has a pair of complex poles (Dominant poles) x o Hypothesis must be verified x (Eventual influence of zeros and remaining poles) ESI.US.Introduction to Feedback Systems 39

Transient regime specifications Explicit expressions can be obtained for 2 nd order systems Closed-loop response 6 4 2-2 -4-6 -8-1 1-2 1-1 1 1 1 1 2-9 -135-18 1-2 1-1 1 1 1 1 2 Open-loop frequency response ESI.US.Introduction to Feedback Systems 4

Frequency domain specifications Overshoot (SO) vs. Phase margin (Mf) Qualitatively (Nyquist) 1 SO M f Quantitatively 8 7 6 SO(%) M f (grados) 9 M f (δ) - SO(δ) Graphic plot 5 4 3 ( M f (δ)~1 δ aprox.) 2 1.1.2.3.4.5.6.7.8.9 1 δ ESI.US.Introduction to Feedback Systems 41

Frequency domain specifications Bandwidth (wc) vs. Raise time (ts) Qualitatively (bandwidth): wc ts Quantitatively 3.4 3.2 3 2.8 2.6 2.4 2.2 w c ~ w n 2.1.2.3.4.5.6.7 ESI.US.Introduction to Feedback Systems 42

Frequency domain specifications Time-domain specifications can be translated into geometric restrictions in the open-loop frequency plot. 1 Magnitude (db) 5-5 -1 Wc -15 Steady state specs. Impose a minimum gain (or minimum gain slope) at low frequencies t s imposes a minimum Bandwidth (Wc) Decreasing gain at high frequencies advisable to limit noise effects ESI.US.Introduction to Feedback Systems 43

Time response analysis in the frequency domain Now we know how to translate time-domain specifications into open-loop frequency response. Thus, control design in frequency domain consists in modifying the open-loop frequency response of the system to fit specifications. Loop shaping Important: Loop-shaping in frequency domain methods can only be applied to stable systems or systems with integrators. This restriction does not hold for the Root Locus method in chapter 4. ESI.US.Introduction to Feedback Systems 44

Motivating Example Servomechanism Control system that converts a small mechanical motion into one requiring much greater power. The device is usually intended to track external references by means of a feedback control structure. ESI.US.Introduction to Feedback Systems 45

Motivating example Modeling the servomechanism Electric equations Mechanical equations Motor Part Gear Load Part ESI.US.Introduction to Feedback Systems 46

Motivating example Modelling the servomechanism (Transfer function) Overall system dynamics Transfer function with ESI.US.Introduction to Feedback Systems 47

Motivating example Block diagram Controller Plant R(s) U(s) Y bc (s) - Measured output Y m (s) Sensor filter Modified block diagram R(s) - Controller U(s) Plant Y bc (s) Y m (s) The measured variable is taken as output Closed-loop Defining ESI.US.Introduction to Feedback Systems 48

Motivating example The servomechanism Open-loop step response Angular position: Unstable behaviour θ (rad) 4.5 x 1-4 4 3.5 3 2.5 2 1.5 1.5 1 2 3 4 5 6 7 8 9 1 Time(s) 4.5 5 x 1-5 4 Angular speed: First order stable system d θ/dt (rad/s) 3.5 3 2.5 2 1.5 1.5 1 2 3 4 5 6 7 8 9 1 Time(s) ESI.US.Introduction to Feedback Systems 49

Motivating example Control of the servomechanism Let s consider the servomechanism system with parameters 2 Diagrama de Bode Módulo (db) 1-1 -2 1-4 1-2 1 1 2 ω (rad/s) Fase (grados) -5-1 -15-2 -25 1-4 1-2 1 1 2 ESI.US.Introduction to Feedback Systems 5 ω (rad/s)

Motivating example Control specifications (position tracking problem) Consider the following requirements for the closed-loop system y(t) S. O. = y( t p ) y( ) y( ) y( ) Time domain specifications t s t p t e Tiemp ESI.US.Introduction to Feedback Systems 51

Motivating example Control specifications Consider the following requirements for the closed-loop system Time domain specifications 1 9 8 7 6 SO(%) M f (grados) 5 4 3 2 1.1.2.3.4.5.6.7.8.9 1 δ Frequency domain specifications Static gain of controller > 33.3 = 3.5 db ESI.US.Introduction to Feedback Systems 52

Proportional control Control action is taken proportional to the output error to be corrected The equivalent transfer function: Magnitude (db) 21 2.8 2.6 2.4 2.2 2 19.8 19.6 19.4 19.2 19 1 Bode Diagram Modifies the static gain of G(s) Crossover frequency Stability Speed Phase (deg).5.5 1 1 1 1 Frequency (rad/sec) ESI.US.Introduction to Feedback Systems 54

Proportional control Effect on the system (Freq. domain) 2 Bode Diagram Magnitude shifts up or down depending on K Phase keeps unaffected Magnitude (db) 1-1 -2-3 -9 Phase (deg) -135-18 -225-27 1-3 1-2 1-1 1 1 1 1 2 1 3 Frequency (rad/sec) ESI.US.Introduction to Feedback Systems 55

Proportional control Effect on the system (Time domain) 1.8 1.6 1.4 1.2 1 y(t).8.6.4 High values of K may eventually lead to instability.2 1 2 3 4 5 6 7 8 9 1 Time (s) ESI.US.Introduction to Feedback Systems 56

Proportional control Loop-shaping analysis Magnitude (db) 1 5-5 - 1 Wc - 1 5 Low frequency: Steady state improves Steady-state error is reduced (Higher gains) Better tracking properties Crossover frequency - Stability worsens (Lower phase margin) - Transient speed increases Overshoot Increases High frequency Noise rejection worsens (Higher gain at high frequency) ESI.US.Introduction to Feedback Systems 57

Proportional control Design of a P controller for the servo example The gain K is adjusted to satisfy steady-state and raise time specifications. The most restrictive value (the highest) is taken. 1 Bode Diagram We take the maximum value Magnitude (db) 5-5 -1 Thus Unstable A Proportional controller can not Phase (deg) -15-9 -135-18 -225 achieve all control specifications -27 1-3 1-2 1-1 1 1 1 1 2 Frequency (rad/sec) ESI.US.Introduction to Feedback Systems 58

Proportional control Design of a P controller for the servo example As all specifications can not be met, it is also possible to design a proportional controller to satisfy only the overshoot (phase margin). 1 5 Bode Diagram Magnitude (db) -5-1 This gain is lower than the required value to satisfy steady-state and raise time specifications. Phase (deg) -15-2 -9-135 -18-225 -27 1-3 1-2 1-1 1 1 1 1 2 Frequency (rad/sec) ESI.US.Introduction to Feedback Systems 59

Proportional control Is it possible to design a controller such that both, raise time and overshoot, are simultaneously improved? 1.8 1.6 1.4 1.2 1 y(t).8.6.4.2 1 2 3 4 5 6 7 8 9 1 Tim e (s) ESI.US.Introduction to Feedback Systems 6

Proportional derivative control 4 Bode Diagram Low frequencies Gain is affected as K c (Proportional) Phase shift aprox. Magnitude (db) Phase (deg) 3 2 1 9 45 High frequencies Increasing gain with frequency (2 db/dec) Phase lead of 9º 1-2 1-1 1 1 1 1 2 Frequency (rad/sec) ESI.US.Introduction to Feedback Systems 62

Proportional derivative control Usual factorization R(s) U(s) Y m (s)=y(s) - PD Controller Plant Usually, K c can be obtained in first design steps from steady-state specs. Thus K c can be lumped into G(s) as: R(s) U(s) Y m (s)=y(s) - PD Controller Plant The rest of the design can be obtained from de Bode plot of K c G(s), so dependence of the controller in K c is dropped. ESI.US.Introduction to Feedback Systems 63

Proportional derivative control Effect on the system (Freq. domain) On low freq. magnitude shifts up or down depending on K c. Phase keeps unaffected On high freq. magnitude slope is increased with 2 db/dec. Phase lead of 9º is added. Magnitude (db) 15 1 5-5 Bode Diagram -1 9 For Phase (deg) 45-45 -9-135 -18-225 Faster and less oscilatory system -27 1-3 1-2 1-1 1 1 1 1 2 Frequency (rad/sec) ESI.US.Introduction to Feedback Systems 64

Proportional derivative control Effect on the system (Time domain) For 1.8 1.6 1.4 1.2 1.8.6 With proper design, both, faster response, and less overshoot can be achieved.4.2 1 2 3 4 5 6 7 8 9 1 ESI.US.Introduction to Feedback Systems 65

Proportional derivative control Loop-shaping analysis Magnitude (db) 1 5-5 - 1 Wc - 1 5 Low frequency: Low frequency response is modified as Kc. Same effect as proportional controller Crossover frequency - Stability improves - Increases phase margin - Transient: Speed increases Overshoot decreases High frequency Noise rejection worsens (Gain slope at highfrequency increases 2 db/dec) ESI.US.Introduction to Feedback Systems 66

Proportional derivative control A possible procedure to tune a PD controller Case a) (if Kc is fixed from the specifications) Step 1: K c is chosen to satisfy steady-state error and crossover frequency specifications. This value is lumped into the plant model, so that, the crossover frequency of, is computed. Remark: If the specifications do not impose any specific bound on Kc, it is possible to leave its adjustment to the final steps, where it is used to fix the crossover frequency (See case b) Step 2: T d is chosen to satisfy phase margin specifications once K c is fixed. The zero of the derivative action must be chosen to add the phase margin needed to satisfy the specifications. where represents the crossover frequency of the compensated system. ESI.US.Introduction to Feedback Systems 67

Proportional derivative control Important: In general, since the introduction of the PD action modifies the crossover frequency. Moreover is often unknown or difficult to compute, thus the phase margin correction is usually taken as Magnitude (db) 15 1 5-5 Bode Diagram -1 9 45 where accounts for the phase loss between and Phase (deg) -45-9 -135-18 -225 Typically -27 1-3 1-2 1-1 1 1 1 1 2 Frequency (rad/sec) ESI.US.Introduction to Feedback Systems 68

Proportional derivative control The design is slightly different depending on how much phase must be added If In this case 1/T d > ω c The crossover frequency is not modified, thus can be taken as Magnitude (db) 4 3 2 1 Bode Diagram 9 Choose the smallest T d that satisfies the specifications Phase (deg) 45 Minimize the amplification of high frequency noise 1-2 1-1 1 1 1 1 2 Frequency (rad/sec) ESI.US.Introduction to Feedback Systems 69

Proportional derivative control If In this case 1/T d < ω c 4 Bode Diagram The crossover frequency is Modified. Corrector factor must be taken into account. Magnitude (db) 3 2 1 9 Phase (deg) 45 1-2 1-1 1 1 1 1 2 Frequency (rad/sec) ESI.US.Introduction to Feedback Systems 7

PD Controller Design example: G( s) = ( s + 5) 2 ( s +.1)( s + 2s + 2) Specifications: erp for step input < 1% SO < 2% The uncontrolled system exhibits Mf=1 o and Mg= 4dB. 4 2-2 9-2 -1 1 2 1 1 1 1 1 A simple P controller is not enough to satisfy all specifications. -9-18 G ESI.US.Introduction to Feedback Systems 71

PD example G( s) = ( s + 5) 2 ( s +.1)( s + 2s + 2) ep rp < 1%, SO < 2% 4 2 From the steady-state specification on ep rp Kp>99. Since Kp=25 Kc Kc 4. A value Kc=4 is taken -2 Specification for SO M fd =5 º M f =-15 º, w c =2.6 rad/s Mf=5 º -(-15 º )+(1 º )=75 º Thus, it is a case b problem (1/Td<w c ) T d =tan(75 º )/2.6=1.43 1/T d =.7 rad/s -2-1 1 2 1 1 1 1 1 ESI.US.Introduction to Feedback Systems 72 9-9 -18 C L=CG KG

PD example The final result yields Mf= 66 o Mg= inf 4 2-2 w c = 7 rad/s -2-1 1 2 1 1 1 1 1 9 C -9-18 L=CG KG ESI.US.Introduction to Feedback Systems 73

PD example In time domain, the final controller provides: 1.2 1.8 r(t) y(t) SO= 13% t s =.28 s.6.4.2.5 1 1.5 2 2.5 3 ESI.US.Introduction to Feedback Systems 74

Proportional derivative control A possible procedure to tune a PD controller (Cont.) Case b) (if Kc is not fixed from the specifications) Step 1: Select the desired crossover frequency from the specifications (raise time or settling time for example). Remark: If the specifications do not impose any specific value on, it can be freely selected. A possible choice could be taking the fastest design possible, that is taking a such that the phase lead required approaches the PD theoretical limit (9º ) Step 2: Select T d to satisfy phase margin specifications at frequency. Step 3: Adjust Kc to guarantee that the crossover frequency is in. ESI.US.Introduction to Feedback Systems 75

PD Controller Design example: 5 Bode Diagram Specifications: erv for ramp input <.1 SO < 2% The uncontrolled system is unstable as its phase margin is negative. Notice that a P controller can not stabilize the system as no phase is added. Magnitude (db) Phase (deg) -5-1 -15-18 -225-27 1-1 1 1 1 1 2 Frequency (rad/sec) ESI.US.Introduction to Feedback Systems 76

PD example ev rp <.1, SO < 2% The system is type II, so steady state specs are automatically satisfied. No particular value for Kc is imposed. Thus, w c can be fixed to allow the PD to satisfy overshoot restrictions. For example, as M fd =5º, if we impose Mf=75º, then w c =2.3 rad/s. Magnitude (db) 15 1 5-5 -1-15 -9-135 Bode Diagram T d =tan(75 º )/2.3=1.62 1/T d =.62 rad/s Finally, Kc is adjusted to locate the crossover freq. in w c =2.3 rad/s. Kc G(jw c )(1+T d w c j) =db Kc=3.6dB = 1.51 Phase (deg) -18-225 -27 1-2 1-1 1 1 1 1 2 Frequency (rad/sec) ESI.US.Introduction to Feedback Systems 77

PD example In time domain, the final controller provides: SO= 26% t s =.9 s An overshoot slightly greater than desired is obtained, due to the additional zero from the PD controller. Redesign with more stringent overshoot specification is recommended. y(t) 1.4 1.2 1.8.6.4.2 5 1 15 Time (s) ESI.US.Introduction to Feedback Systems 78

Proportional derivative control Designing a proportional derivative controller for the servo system The gain is chosen to satisfy error specifications 1 Bode Diagram Unstable Magnitude (db) 5-5 -1 To stabilize we need to increase phase margin using a derivative term Phase (deg) -15-9 -135-18 -225-27 1-3 1-2 1-1 1 1 1 1 2 Frequency (rad/sec) ESI.US.Introduction to Feedback Systems 79

Proportional derivative control From Bode plot of 15 Bode Diagram Thus, the required phase lead Magnitude (db) 1 5-5 Taking -1 9 45 Thus Phase (deg) -45-9 -135 M f = 23.7º. System stable but does not satisfy specifications. -18-225 -27 1-3 1-2 1-1 1 1 1 1 2 Frequency (rad/sec) ESI.US.Introduction to Feedback Systems 8

Proportional derivative control The PD controller has several disadvantages in practice It is not realizable (It can no be physically implemented. Only approximations are possible) High frequencies (noise) are increasingly amplified. Can we improve the behaviour of a PD controller? The beneficial phase lead of a PD is only required around crossover frequency. There is no benefit in keeping high gains at high frequencies. Magnitude (db) 6 5 4 3 2 1 Bode Diagram PD C(s) A high frequency pole can be included to limit PD gain at high frequencies (Filtered derivative) Phase (deg) 9 45 1-2 1-1 1 1 1 1 2 1 3 Frequency (rad/sec) PD C(s) ESI.US.Introduction to Feedback Systems 81

Lead compensator Controller structure: Low frequencies.6 Módulo asintótico Gain is affected as K c (Proportional) Phase shifts aprox. Intermediate Freq. Gain slope of 2 db/dec. Phase leads a maximum db.4.2.1.2.3 ω ω (rad/s) m 9 75 K c 1/τ 2 db/dec Fase 1log(1/α) 1/(α τ) 2log(1/α) High frequencies Gain increases up to a constant value Phase shifts approx. grados 6 45 3 15 Φ m.1.2.3 ESI.US.Introduction to Feedback Systems 82

Lead compensator Some expressions: Maximum phase lead in.6 Módulo asintótico db.4.2 K c 2 db/dec 1log(1/α) 2log(1/α) Gain in.1.2.3 ω ω (rad/s) m 9 75 1/τ Fase 1/(α τ) 6 Maximum gain grados 45 3 15 Φ m.1.2.3 ESI.US.Introduction to Feedback Systems 83

Lead compensator Practical effects: The lead compensator adds phase on specific frequency range. Maximum gain is limited (Beneficial to prevent noise amplification) db.6.4.2 Módulo asintótico.1.2.3 ω ω (rad/s) m 9 K c 1/τ 2 db/dec Fase 1log(1/α) 1/(α τ) 2log(1/α) The effect is similar to PD controller with limited high frequency gain grados 75 6 45 3 15 Φ m.1.2.3 ESI.US.Introduction to Feedback Systems 84

Lead compensator Frequency domain effects: 4 Módulo asintótico 2 Faster and less oscilatory system db -1-2 -4-1log(1/α) -6 1/τ 1/(α τ) 1 ω 1 1 m =ω' =5.6 rad/s ω (rad/s) Time domain effect similar to PD controller with limited noise amplification (limited high-frequency gain ) grados -9-135 -18-225 Fase -27 1 1 1 ESI.US.Introduction to Feedback Systems 85

Lead compensator Loop-shaping analysis 1 Magnitude (db) 5-5 - 1 w c - 1 5 Low frequency: Magnitude is shifted as K c Same effect as Proportional control Crossover frequency - Stability improves (Higher Phase Margin) Overshoot decreases - Crossover freq. Increases (in general) Transient speed increases High frequency - Magnitude slope increases 2 db/dec. - Noise rejection worsens, but not as much as with PD control. (Limited highfrequency gain) ESI.US.Introduction to Feedback Systems 86

Lead compensator A procedure to tune a Lead controller Case a) (if Kc is fixed from the specifications) Step 1: K c is chosen to satisfy steady-state error and crossover frequency specifications. This value is lumped into the plant model, so that, (crossover frequency of ), and (phase margin) are computed. Remark: If the specifications do not impose any specific bound on Kc, it is possible to leave its adjustment to the final steps, where it is used to fix the crossover frequency (See case b) Step 2: Compute the required phase lead with Important: Maximum phase lead if is negative o greater than, the lead compensator can not be designed. ESI.US.Introduction to Feedback Systems 87

Lead compensator A procedure to tune a Lead controller Case a) (if Kc is fixed from the specifications) Step 3: Determine to guarantee that the peak phase lead,, happens at the crossover frequency. We know the peak phase lead happens at with So we look for frequency such that db ESI.US.Introduction to Feedback Systems 88

Lead compensator A procedure to tune a Lead controller Case a) (if Kc is fixed from the specifications) It is known that if, redesign is required. The easiest way is returning to step 1, and select such that the crossover freq of is. As is always greater than, the condition is satisfied (we are forcing ). Nonetheless, increasing, also reduces de phase margin of so higher values of are required. (Remember there is a limit in 78.5º) ESI.US.Introduction to Feedback Systems 89

Lead compensator Example: Specifications: G( s) Position error < 1% SO < 2% = ( s + 5) 2 ( s +.1)( s + 2s + 2) 1 9 8 7 6 SO(%) M f (grados) 5 4 3 2 1.1.2.3.4.5.6.7.8.9 1 δ Static gain of controller > 4 = 12.4 db ESI.US.Introduction to Feedback Systems 9

Lead compensator Example: G( s) = ( s + 5) 2 ( s +.1)( s + 2s + 2) Step 1 (Controller gain adjustment) 4 2 Bode Diagram Gm = -8.37 db (at 1.93 rad/sec), Pm = -15 deg (at 2.81 rad/sec) Step 2 (estimate required phase lead) Magnitude (db) -2-4 -6-8 required phase lead Phase (deg) -45-9 -135-18 -225 1-3 1-2 1-1 1 1 1 1 2 Frequency (rad/sec) ESI.US.Introduction to Feedback Systems 91

Lead compensator Step 3 (determination of ) Fix such that (desired).6 Módulo asintótico since we look for the frequency where db.4.2 K c 2 db/dec 1log(1/α) 2log(1/α).1.2.3 ω ω (rad/s) m 1/τ 1/(α τ) thus 9 75 6 Fase since grados 45 3 15 Φ m.1.2.3 ESI.US.Introduction to Feedback Systems 92

Lead compensator Final design 4 2 Bode Diagram Magnitude (db) -2-4 -6-8 -1 9 45 Phase (deg) -45-9 -135-18 -225 1-3 1-2 1-1 1 1 1 1 2 1 3 Frequency (rad/sec) ESI.US.Introduction to Feedback Systems 93

Lead compensator In time domain, the final controller provides: 1.4 1.2 1 SO= 23% t s =.28 s y(t).8.6.4.2.5 1 1.5 2 2.5 3 3.5 4 4.5 5 time (s) ESI.US.Introduction to Feedback Systems 94

Lead compensator A procedure to tune a Lead controller Case b) (if Kc is not fixed from the specifications) Step 1: Select the desired crossover frequency from the specifications (raise time or settling time for example). Remark: If the specifications do not impose any specific value on, it can be freely selected. A possible choice could be taking the fastest design possible, that is taking a such that the phase lead required approaches the Lead controller theoretical limit (78.5º ) Step 2: Compute the required phase lead. Important: Maximum phase lead if is negative o greater than, the lead compensator can not be designed ESI.US.Introduction to Feedback Systems 95

Lead compensator A procedure to tune a Lead controller (Cont.) Case b) (if Kc is not fixed from the specifications) Step 3: Determine to guarantee that the peak phase lead,, happens at the crossover frequency. Step 4: Adjust gain to fix the crossover frequency exactly in ESI.US.Introduction to Feedback Systems 96

Lead compensator Design example: 5 Bode Diagram Specifications: erv for ramp input <.1 SO < 2% The uncontrolled system is unstable as its phase margin is negative. Notice that a P controller can not stabilize the system as no phase is added. Magnitude (db) Phase (deg) -5-1 -15-18 -225-27 1-1 1 1 1 1 2 Frequency (rad/sec) ESI.US.Introduction to Feedback Systems 97

Lead compensator The system is type II, so steady state specs are automatically satisfied. No particular value for Kc is imposed. Thus, w c can be fixed to allow the Lead compensator to satisfy overshoot constraints. For example, as M fd =5º, if we impose Φ m =75º, then w c =2.3 rad/s. Magnitude (db) 15 1 5-5 -1-15 -9 Bode Diagram -135 Phase (deg) -18 Finally, Kc is adjusted to locate the crossover freq. in w c =2.3 rad/s. -225-27 1-2 1-1 1 1 1 1 2 1 3 Frequency (rad/sec) ESI.US.Introduction to Feedback Systems 98

Lead compensator In time domain, the final controller provides: SO= 23% t s =.8 s An overshoot slightly greater than desired is obtained. Redesign with more stringent overshoot specification is recommended. y(t) 1.4 1.2 1.8.6.4.2.5 1 1.5 2 2.5 3 3.5 4 4.5 5 time (s) ESI.US.Introduction to Feedback Systems 99

Lead compensator Example: Phase-lead controller for the servo system Step 1 (Controller gain adjustment) 1 Bode Diagram Gm = -25.9 db (at.5 rad/sec), Pm = -21.9 deg (at 2.14 rad/sec) 8 6 4 Step 2 (estimate required phase lead) Magnitude (db) 2-2 -4-6 -8-1 -9 required phase lead -135 Phase (deg) -18-225 DESIGN NOT POSSIBLE!! -27 1-3 1-2 1-1 1 1 1 1 2 Frequency (rad/sec) ESI.US.Introduction to Feedback Systems 1

Proportional integral control Módulo asintótico db - 2 db/dec K c 1/T i Fase -15 grados -3-45 -6 Effects: -75-9 Increases the number of poles at the origin Reduces the phase for frequencies lower than 1/T i ω (rad/s) ESI.US.Introduction to Feedback Systems 12

Proportional integral control Effect on the system On low frequencies the magnitude shifts up because of the pole at the origin. On high frequencies the magnitude is not modified. For frequencies lower than 1/T i the phase decreases. Magnitude (db) 2 15 1 5 5 1 15 Slope increases Bode Diagram Not modified 2 Improves disturbances rejection and reference tracking properties Phase (deg) 9 18 May induce instability (lower phase margin) Phase decreases 27 1 4 1 3 1 2 1 1 1 1 1 1 2 1 3 Frequency (rad/sec) ESI.US.Introduction to Feedback Systems 13

Proportional integral control Loop-shaping analysis -2 db/dec Magnitude (db) 1 5-5 Wc - 1-1 5 Low frequency: Low frequency response is improved. The slope is modified in -2dB/dec. Disturbance rejection increases Tracking improves Crossover frequency - Stability may worsen - Phase margin decreases for low frequencies. Overshoot may increase High frequency High frequency response is not modified ESI.US.Introduction to Feedback Systems 14

Proportional integral control Design procedure for a PI (1 + Ti s) C( s) = Kc T s i Note that the design problem can be addressed as a PD design problem Kc G( s) L( s) = C( s) G( s) = (1 + Ti s) = KG'( s)(1 + Ti s) T s i Tuning of K c and T i K c is not the bode gain of the controller! ESI.US.Introduction to Feedback Systems 15

PI design procedure Case a) K=K c /T i is not fixed from error specifications (1 + Ti s) C( s) = Kc T s Tuning Step 1: Choose to satisfy the transient response specifications Step 2 Choose to avoid decreasing too much the phase margin. Step 3: Determine that satisfies or approx. i Low 1/T i imply slow disturbance rejection and high settling time. Choose T i the smallest possible ESI.US.Introduction to Feedback Systems 16

PI example G( s) = 1 s( s + 1)( s + 1) Specifications: Velocity error.1 (steady state) SO 2% ESI.US.Introduction to Feedback Systems 17

PI example G( s) = 1 s( s + 1)( s + 1) Velocity error.1 This specification sets a lower limit on the bode gain of the controller C(s) 4 2-2 G(s) 1G(s) K v =.1K c 1 K B =1 Phase margin of 1G(s) = -4º SO 2% M fd 45º Proportional controller is not able to stabilize and satisfy velocity error specifications Increase phase margin using a derivative term? Mf = M fd M f + = 45-(-4)+1 =95º It is not possible with a PD controller (K c is too high) -2-1 1 2 1 1 1 1 1 ESI.US.Introduction to Feedback Systems 18-9 -18

PI example Solution: Use a PI controller PI controller increases the type of the open-loop transfer function so the velocity error specification is automatically satisfied. The parameters Kc and Ti can be chosen to satisfy the rest of the specifications. 4 2-2 w' c =.7rad/s In order to obtain a fast closed-loop system we choose the maximum crossover frequency that provides the desired phase margin. M fd =45 o, w c =.7rad/s, G(j w c ) =-18 db Kc is chosen to obtain the desired w' c Kc=18 db (Kc=8) -9-18 w -2-1 1 2 1 1 1 1 1 G Mf =45 o c ESI.US.Introduction to Feedback Systems 19

PI example Ti is chosen to guarantee that the integral term does not decreases the phase margin of the controlled system: 1/ Ti = w' c /1 =.5 Ti = 2 1 5-5 -1 C(s) = (K c /T i ) (1+T i s) /s = 8.5 (1+ 2s) / s =.4 (1+ 2s) / s Frequency response of C(s)G(s) M f = 48º M g = 22 db w c =.7 rad/s -9-18 -27 1-3 1-2 1-1 1 1 1 1 2 1 3 ESI.US.Introduction to Feedback Systems 11 C G L

PI example Step response of the closedloop system: SO= 22% ts= 2.7 s Note that the settling time is higher than rise time. The settling time depends on the low frequency response of L(s) = C(s)G(s) and in particular, on where the zero of the integral term is. Lower T i implies lower settling time (and more stability problems) 1.2 1.8.6.4.2 r(t) y(t) 1 2 3 4 ESI.US.Introduction to Feedback Systems 111

Proportional integral control Case b: K B = K c /T i is determined by the steady state specifications (position, velocity or acceleration errors) Tuning: Step 1: Choose K B to satisfy steady state error specifications Step 2: Design a controller C(s) for the system G'(s) Step 3: Once Ti is chosen, K c is obtained from K c = K B Ti Note: In general Ti is chosen such that 1/Ti is lower than the crossover frequency. Note that the bode plot of G'(s) is different from G(s)! ESI.US.Introduction to Feedback Systems 112

PI example G( s) = 1 s( s + 1)( s + 1) Specifications Acceleration error <.1 SO < 2% ESI.US.Introduction to Feedback Systems 113

PI example G( s) = 1 s( s + 1)( s + 1) Specifications Acceleration error <.1 SO < 2% 4 2-2 G(s) has M f =8 o, M g = 4dB G(s) is type 1 (one pole at the origin) In order to satisfy steady state specification, an integral term has to be used: PI controller -9-18 -2-1 1 2 1 1 1 1 1 G ESI.US.Introduction to Feedback Systems 114

PI example Acceleration error.1 fixes the minimum bode gain K B of the controller: Acceleration error (%) = 1/Ka * 1 Ka =.1Kc/Ti Kc/Ti 1 4 2-2 Frequency response of the modified system G = K B G/s Mf = -8 o, w c = 1 rad/s Mf=5-(-8)+ > 9 o -9-2 -1 1 2 1 1 1 1 1 Phase margin too small Impossible to satisfy the specifications -18 M f ESI.US.Introduction to Feedback Systems 115

PI example Specifications Velocity error <.1 SO < 6% ESI.US.Introduction to Feedback Systems 116

PI example Bode Diagram -1 Specifications Velocity error <.1 SO < 6% Magnitude (db) -2-3 -4-5 -6-7 G(s) has Mf=59 o, Mg= -8-9 -1 G(s) is type (No poles at the origin) In order to satisfy steady state specification, an integral term has to be used: PI controller Phase (deg) -45-9 -135-18 1-2 1-1 1 1 1 1 2 1 3 Frequency (rad/sec) ESI.US.Introduction to Feedback Systems 117

PI example Acceleration error.1 fixes the minimum bode gain K B of the controller: Acceleration error (%) = 1/Ka * 1 Ka = Kc/Ti Kc/Ti 1 1 5 Bode Diagram Gm = -19.2 db (at 3.16 rad/sec), Pm = -34.3 deg (at 8.66 rad/sec) Frequency response of the modified system G = K B G/s M f = -34.3 o, w c = 8.66 rad/s M f =2-(-34.3) +δ = 25-(-34.3) +1.7= 65º Magnitude (db) Phase (deg) -5-1 -15-9 -135-18 -225 This is the phase margin to be added with the zero of the PI controller -27 1-2 1-1 1 1 1 1 2 1 3 Frequency (rad/sec) ESI.US.Introduction to Feedback Systems 118

PI example SO 5% implies that the desired minimum phase margin is Mf = 2 o Mf= M fd Mf + δ =2-(-34.3)+ 1.7 => Mf =65 o Taking into account the Bode plot of a minimum phase zero at -1/T_i w c T i =tan( Mf) = tan(65 o ) =2.14 T i =2.14/8.66=.247 Finally, Kc = K B T i = 2.47 so the PI controller has the following transfer function Resulting phase margin Mf= 22.3º > 2º OK ESI.US.Introduction to Feedback Systems 119

PI example Frequency response of the open-loop transfer function M f = 22.3 o M g = db w c = 14.5 rad/s Magnitude (db) 1 5-5 G(s) C(s)G(s) Bode Diagram C(s) 1.6 Step Response -1 1.4-45 Amplitude 1.2 1.8.6 Closed-loop step response Phase (deg) -9-135.4.2 SO= 56% ts=.12 s.2.4.6.8 1 1.2 1.4 1.6 1.8 2 Time (sec) -18 1-2 1-1 1 1 1 1 2 1 3 Frequency (rad/sec) w c ESI.US.Introduction to Feedback Systems 12

Proportional integral control Designing a proportional integral controller for the servo system With a PI the system becomes type II. Thus e vrp = (Steady state specs satisfied without constraints in K c ) 15 1 Bode Diagram But, since the PI DOES NOT ADD PHASE the maximum achievable crossover frequency would be that where From the Bode Plot Magnitude (db) Phase (deg) 5-5 -1-15 -9-135 -18-225 -27 Pi design does not satisfy t s spec.!!! 1-3 1-2 1-1 1 1 1 1 2 Frequency (rad/sec) ESI.US.Introduction to Feedback Systems 121

Lag compensator Controller structure: Low frequencies Módulo asintótico K c Gain is affected as K c (Proportional) Phase shifts aprox. db -2 db/dec -2log(α) Intermediate Freq. Gain slope of -2 db/dec. Phase lags a maximum 1/(α τ) Fase 1/τ High frequencies -15-3 Gain decreases up to a constant value Phase shifts approx. grados -45-6 -75-9 ω (rad/s) ESI.US.Introduction to Feedback Systems 122

Lag compensator Some expressions: Maximum gain Módulo asintótico Minimum gain db K c -2 db/dec -2log(α) Phase lag is usually an undesirable effect. -15 1/(α τ) Fase 1/τ Phase lag compensators are designed to increase lowfrequency gains, while phase lag is pushed away from crossover frequency. grados -3-45 -6-75 -9 ESI.US.Introduction to Feedback Systems 123 ω (rad/s)

Lag compensator Practical effects: The lag compensator substracts phase on specific frequency range (not desirable). db K c Módulo asintótico -2 db/dec -2log(α) It can be used to increase low-frequency gain (Beneficial to improve steady-state behaviour) -15-3 1/(α τ) Fase 1/τ grados -45 The effect is similar to PI controller with limited low frequency gain -6-75 -9 ω (rad/s) ESI.US.Introduction to Feedback Systems 124

Lag compensator Frequency domain effects: Low-frequency gain is increased, preserving gain at intermediate-high frequencies => Steady state improvement Phase lag is usually located on frecuencies below crossover => Phase margin not affected Magnitude (db) 15 1 5-5 -1-15 Bode Diagram Time domain effect similar to PI controller with limited low-frequency gain (excesive controller outputs prevented) Phase (deg) -9-18 -27 1-3 1-2 1-1 1 1 1 1 2 1 3 Frequency (rad/sec) ESI.US.Introduction to Feedback Systems 125

Lag compensator Loop-shaping analysis Magnitude (db) 1 5-5 - 1 Wc - 1 5 Low frequency: Magnitude is increased Similar effect as Integral control Important: system type DOES NOT CHANGE Crossover frequency - Stability keeps approx. unaffected (slightly lower Phase Margin) - does not change (in general) Transient speed not affected High frequency - Not afected ESI.US.Introduction to Feedback Systems 126

Lag compensator design Assumption: From time domain specs, a minimum static gain for the controller, K cd, a minimum crossover freq., w cd, and a desired phase margin, M fd Design Procedure Step 1: Choose K c =K cd to satisfy steady-state specifications Step 2: Get the frequency w c were M fd= M f (KcG(s))+. That is, since the phase margin can not be increased with a lag compensator, we select the frequency were K c G(s) exhibits the desired phase margin (plus some additional that accounts for phase drop as the controller is introduced) If w c <w cd, the system can not be controlled with a lag compensator. ESI.US.Introduction to Feedback Systems 127

Lag compensator design Design Procedure (CONT.) Step 3: Compute α, to fix w c as crossover freq. C(jw c )/K c + K c G(jw c ) = 2log(α )= KcG(jw c ) db Step 4: Compute τ to induce small phase drop at crossover freq. 1/ τ [w c /3, w c /1] ESI.US.Introduction to Feedback Systems 128

Lag compensator Design Example: Specifications: G( s) = ( s + 5) 2 ( s +.1)( s + 2s + 2) Position error < 1% SO < 2% 1 9 8 7 6 SO(%) M f (grados) 5 4 3 2 1.1.2.3.4.5.6.7.8.9 1 δ Static gain of controller > 4 = 12.4 db ESI.US.Introduction to Feedback Systems 129

Lag compensator Example: G( s) = ( s + 5) 2 ( s +.1)( s + 2s + 2) Step 1 (Controller gain adjustment) Step 2 (Compute α) w c : Mf(w c )=M fd + = 48º+7º=55º thus w c =.6 rad/s db Módulo asintótico 14 12 1 8 6 4 2-2 -4-6 -8-1 -12 1-4 1-3 1-2 1-1 1 1 1 1 2 ω (rad/s) Fase 2log(α)= KcG(.6j) =45 db α=178 Step 3 (Compute τ) 1/ τ =.1 τ =1 grados -45-9 -135-18 -225 K c G(s) C(s)/K c G ba (s) -27 1-4 1-3 1-2 1-1 1 1 1 1 2 ESI.US.Introduction to Feedback Systems 13

Lag compensator Final design 1.2 RA RR 1 The response is slower compared to a lead controller (lower crossover freq). To: Y(1).8.6.4.2 1 2 3 4 5 6 7 8 9 1 ESI.US.Introduction to Feedback Systems 131

Lag compensator Design of a phase-lag controller for the servo system Step 1 (Controller gain adjustment) 15 1 Bode Diagram Step 2 (achievable crossover freq.) Find such that Magnitude (db) Phase (deg) 5-5 -1-15 -9-135 -18-225 thus -27 1-3 1-2 1-1 1 1 1 1 2 1 3 Frequency (rad/sec) ESI.US.Introduction to Feedback Systems 132

Lag compensator Design of a phase-lag controller Step 3 (determination of ) 15 1 Bode Diagram From the gain of the lag controller Magnitude (db) 5-5 -1-15 -9 thus Step 4 (determination of ) The zero is located such that Phase (deg) -135-18 -225-27 1-3 1-2 1-1 1 1 1 1 2 1 3 Frequency (rad/sec) ESI.US.Introduction to Feedback Systems 133

Lag compensator Design of a phase-lag controller Final design 3 2 Bode Diagram Magnitude (db) 1-1 -2-3 Not All specifications can be met Phase (deg) -9-18 -27 1-8 1-6 1-4 1-2 1 1 2 Frequency (rad/sec) ESI.US.Introduction to Feedback Systems 134

Review of Controllers PD Lead compensator 4 Bode Diagram.6 Módulo asintótico Magnitude (db) 3 2 1 9 db.4.2.3.1 1/τ ω ω (rad/s) m.2 1/(α τ) 9 K c 2 db/dec Fase 1log(1/α) 2log(1/α) 75 Phase (deg) 45 grados 6 45 3 Φ m 15 PI 1-2 1-1 1 1 1 1 2 Frequency (rad/sec) Módulo asintótico.1.2.3 Lag compensator Módulo asintótico K c db db -2 db/dec -2log(α) - 2 db/dec K c 1/T i 1/(α τ) 1/τ Fase Fase -15-15 -3-3 grados -45-6 grados -45-6 -75-75 -9-9 ω (rad/s) ESI.US.Introduction to Feedback Systems 135 ω (rad/s)

PID controller 2 1 TT s + T s + c d c Ti s Ti s i d i C( s) = K 1+ T s + = K 1 db - 2 db/dec Móduloasintótico 2 db/dec Real or complex zeros (if T i >4T d real zeros) T i and T d different from PD and PI PID has the same problem as PD Noise amplification Non realizable (Lead-Lag controller) grados 9 6 3-3 Fase -6-9 Effects: ω(rad/s) The type of the system is increased (steady state improved) Phase margin increases (lower overshoot) Crossover freq. increases (lower raise time) ESI.US.Introduction to Feedback Systems 137

PID controller Loop-shaping analysis Magnitude (db) 1 5-5 - 1 Wc - 1 5 Low frequency: Magnitude is increased Similar effect as PI controller Important: system type INCREASES Crossover frequency - Stability improves (slightly lower Phase Margin) - can be maximized (in general), by adding phase with the PD part of the controller Transient speed improved High frequency - Gain increased. (Noise is amplified) ESI.US.Introduction to Feedback Systems 138

PID controller design Case 1: Kc/Ti not fixed from steady-state specs. Express PID as PI cascaded with a PD Usual for stringent steady-state specifications 1+ T s C s K T s Choose w c and compute M f = M fd M f (w c ) 1 ( ) = (1 + 2 ) T1 s Design PD such that M fd is satisfied: w c T 2 =atan( M f ) Compute K to fix w c : K+2 log(w c T 2 )=- G(jw c ) Choose T 1 such that 1/ T 1 [w c /3, w c /1] Compute PID parameters T + K = K T, T = T + T, T = TT 1 2 1 2 c i 1 2 d T1 T1 + T2 ESI.US.Introduction to Feedback Systems 139

PID controller design Case 2: Kc/Ti is fixed from steady-state specs. 1 Express PID as 1 integrator and 2 PD C( s) = K (1 + T1 s)(1 + T2s) s Choose minimum K satisfying steady-state specs. Plot the Bode diagram of G (s)=k/s G(s) Compute the 2 PDs to verify transient specifications (SO and/or w c ) (Suggestion: Take T 1 =T 2 ) Compute the PID parameters K = K( T + T ), T = T + T, T = c 1 2 i 1 2 d TT 1 2 T + T 1 2 ESI.US.Introduction to Feedback Systems 14

PID example G( s) = 1 s( s + 1)( s + 1) Specifications: Parabolic steady state error.5 SO 4% ESI.US.Introduction to Feedback Systems 141

PID example G( s) = 1 s( s + 1)( s + 1) 6 4 módulo asintótico de G(w*j) Specifications: Parabolic steady state error.5 SO 4% 2-2 -4 KG(s)/s The steady state error specification suggests increasing the type of the system PI or PID with fixed gain -6-8 1-1 1 1 1 1 2-18 Fase de G(w*j) From steady state error, K=2 From the SO spec., Mfd=35º From the Bode plot of 2G(s)/s -225-27 -315 KG(s)/s Mf=-55º y w c =1.2 rad/s (Impossible design for a PI) -36 1-1 1 1 1 1 2 ESI.US.Introduction to Feedback Systems 142