Chapter 9: Controller design

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1 Chapter 9. Controller Design 9.1. Introduction 9.2. Effect of negative feedback on the network transfer functions Feedback reduces the transfer function from disturbances to the output Feedback causes the transfer function from the reference input to the output to be insensitive to variations in the gains in the forward path of the loop 9.3. Construction of the important quantities 1/(1T) and T/(1T) and the closed-loop transfer functions 1

2 Controller design 9.4. Stability The phase margin test The relation between phase margin and closed-loop damping factor Transient response vs. damping factor 9.5. Regulator design Lead (PD) compensator Lag (PI) compensator Combined (PID) compensator Design example 2

3 Controller design 9.6. Measurement of loop gains Voltage injection Current injection Measurement of unstable systems 9.7. Summary of key points 3

4 9.1. Introduction v g (t) Switching converter v(t) Load i load (t) Output voltage of a switching converter depends on duty cycle d, input voltage v g, and load current i load. δ(t) δ(t) Transistor gate driver Pulse-width modulator v c (t) v g (t) Switching converter v(t) = f(v g, i load, d) dt s T s t v(t) Disturbances i load (t) } d(t) } Control input 4

5 The dc regulator application Switching converter Objective: maintain constant output voltage v(t) = V, in spite of disturbances in v g (t) and i load (t). v g (t) i load (t) v(t) = f(v g, i load, d) } Disturbances v(t) Typical variation in v g (t): 100Hz or 120Hz ripple, produced by rectifier circuit. d(t) } Control input Load current variations: a significant step-change in load current, such as from 50% to 100% of rated value, may be applied. A typical output voltage regulation specification: 5V ± 0.1V. Circuit elements are constructed to some specified tolerance. In high volume manufacturing of converters, all output voltages must meet specifications. 5

6 The dc regulator application So we cannot expect to set the duty cycle to a single value, and obtain a given constant output voltage under all conditions. Negative feedback: build a circuit that automatically adjusts the duty cycle as necessary, to obtain the specified output voltage with high accuracy, regardless of disturbances or component tolerances. 6

7 Negative feedback: a switching regulator system Power input Switching converter Load i load v g v H Sensor gain Transistor gate driver δ Pulse-width modulator v c G c Compensator Error signal v e Hv Reference input v ref 7

8 Negative feedback Switching converter v g (t) v(t) = f(v g, i load, d) v ref Reference input Error signal v e (t) v c Compensator Pulse-width modulator i load (t) d(t) } Disturbances } Control input v(t) Sensor gain 8

9 9.2. Effect of negative feedback on the network transfer functions Small signal model: open-loop converter e d 1 : M(D) L e v g jd C v R i load Output voltage can be expressed as where v=g vd dg vg v g Z out i load G vd = v d v g =0 i load =0 G vg = v v g d =0 i load =0 Z out = v i load d =0 v g =0 9

10 Voltage regulator system small-signal model Use small-signal converter model Perturb and linearize remainder of feedback loop: v ref (t)=v ref v ref (t) v e (t)=v e v e (t) etc. v g v ref Reference input ed jd Error signal v e G c v c Compensator 1 : M(D) 1 V M Pulse-width modulator L e d C v R i load Hv H Sensor gain 10

11 Regulator system small-signal block diagram i load Load current variation ac line variation Pulse-width Compensator modulator 1 v ref v e v c d G c Reference input Error signal V M v g Duty cycle variation Z out G vg G vd Converter power stage v Output voltage variation H v H Sensor gain 11

12 Solution of block diagram Manipulate block diagram to solve for v. Result is v = v ref G c G vd / V M 1HG c G vd / V M v g G vg 1HG c G vd / V M i load Z out 1HG c G vd / V M which is of the form v = v ref 1 H T 1T v g G vg 1T i load Z out 1T with T=H G c G vd /V M ="loop gain" Loop gain T = products of the gains around the negative feedback loop. 12

13 Feedback reduces the transfer functions from disturbances to the output Original (open-loop) line-to-output transfer function: G vg = v v g d =0 i load =0 With addition of negative feedback, the line-to-output transfer function becomes: v v g v ref =0 i load =0 = G vg 1T Feedback reduces the line-to-output transfer function by a factor of 1 1T If T is large in magnitude, then the line-to-output transfer function becomes small. 13

14 Closed-loop output impedance Original (open-loop) output impedance: Z out = With addition of negative feedback, the output impedance becomes: v i load v ref =0 v g =0 Feedback reduces the output impedance by a factor of 1 1T v i load d =0 v g =0 = Z out 1T If T is large in magnitude, then the output impedance is greatly reduced in magnitude. 14

15 Feedback causes the transfer function from the reference input to the output to be insensitive to variations in the gains in the forward path of the loop Closed-loop transfer function from to v is: v ref v v ref v g =0 i load =0 = 1 H T 1T If the loop gain is large in magnitude, i.e., T >> 1, then (1T) T and T/(1T) T/T = 1. The transfer function then becomes v v ref 1 H which is independent of the gains in the forward path of the loop. This result applies equally well to dc values: V V ref = 1 H(0) T(0) 1T(0) 1 H(0) 15

16 9.3. Construction of the important quantities 1/(1T) and T/(1T) Example T 80 db 60 db T 0 db Q db T=T 0 1 s Qω p1 1 s ω z s ω p1 2 1 s ω p2 40 db f p1 40 db/decade 20 db 0 db f z 20 db/decade 20 db f c Crossover frequency f p2 40 db/decade 40 db 1 Hz 10 Hz 100 Hz 1 khz 10 khz 100 khz At the crossover frequency f c, T = 1 f 16

17 Approximating 1/(1T) and T/(1T) T 1T 1 for T >> 1 T for T << 1 1 1T 1 for T >> 1 T 1 for T << 1 17

18 Example: construction of T/(1T) 80 db 60 db T 1T 1 for T >> 1 T for T << 1 40 db f p1 T 20 db 0 db 20 db T 1T f z 20 db/decade Crossover frequency f c f p2 40 db/decade 40 db 1 Hz 10 Hz 100 Hz 1 khz 10 khz 100 khz f 18

19 Example: analytical expressions for approximate reference to output transfer function At frequencies sufficiently less that the crossover frequency, the loop gain T has large magnitude. The transfer function from the reference to the output becomes v v ref = 1 H v v ref = 1 H 19 T 1T 1 H This is the desired behavior: the output follows the reference according to the ideal gain 1/H. The feedback loop works well at frequencies where the loop gain T has large magnitude. At frequencies above the crossover frequency, T < 1. The quantity T/(1T) then has magnitude approximately equal to 1, and we obtain T 1T T H = G cg vd V M This coincides with the open-loop transfer function from the reference to the output. At frequencies where T < 1, the loop has essentially no effect on the transfer function from the reference to the output.

20 Same example: construction of 1/(1T) 80 db 60 db 40 db 20 db 0 db 20 db 40 db 60 db T 0 db f p1 40 db/decade 40 db/decade T 0 db f p1 Q db T f z 20 db/decade 20 db/decade f z Q db f c Crossover frequency 1 1T 1 1T f p2 1 for T >> 1 T 1 for T << 1 40 db/decade 80 db 1 Hz 10 Hz 100 Hz 1 khz 10 khz 100 khz f 20

21 Interpretation: how the loop rejects disturbances Below the crossover frequency: f < f c and T > 1 Then 1/(1T) 1/T, and disturbances are reduced in magnitude by 1/ T 1 1T 1 for T >> 1 T 1 for T << 1 Above the crossover frequency: f > f c and T < 1 Then 1/(1T) 1, and the feedback loop has essentially no effect on disturbances 21

22 Terminology: open-loop vs. closed-loop Original transfer functions, before introduction of feedback ( open-loop transfer functions ): G vd G vg Z out Upon introduction of feedback, these transfer functions become ( closed-loop transfer functions ): 1 H T 1T G vg 1T Z out 1T The loop gain: T 22

23 9.4. Stability Even though the original open-loop system is stable, the closed-loop transfer functions can be unstable and contain right half-plane poles. Even when the closed-loop system is stable, the transient response can exhibit undesirable ringing and overshoot, due to the high Q -factor of the closedloop poles in the vicinity of the crossover frequency. When feedback destabilizes the system, the denominator (1T) terms in the closed-loop transfer functions contain roots in the right half-plane (i.e., with positive real parts). If T is a rational fraction of the form N / D, where N and D are polynomials, then we can write T 1T = N D 1 N D 1 1T = 1 1 N D = = N ND D ND Could evaluate stability by evaluating N D, then factoring to evaluate roots. This is a lot of work, and is not very illuminating. 23

24 Determination of stability directly from T Nyquist stability theorem: general result. A special case of the Nyquist stability theorem: the phase margin test Allows determination of closed-loop stability (i.e., whether 1/(1T) contains RHP poles) directly from the magnitude and phase of T. A good design tool: yields insight into how T should be shaped, to obtain good performance in transfer functions containing 1/(1T) terms. 24

25 The phase margin test A test on T, to determine whether 1/(1T) contains RHP poles. The crossover frequency f c is defined as the frequency where T(j2πf c ) = 1 0dB The phase margin ϕ m is determined from the phase of T at f c, as follows: ϕ m = 180 T(j2πf c ) If there is exactly one crossover frequency, and if T contains no RHP poles, then the quantities T/(1T) and 1/(1T) contain no RHP poles whenever the phase margin ϕ m is positive. 25

26 Example: a loop gain leading to a stable closed-loop system T 60 db 40 db T T 20 db f p1 f z Crossover frequency 0 db T f c 0 20 db db ϕ m Hz 10 Hz 100 Hz 1 khz 10 khz 100 khz T(j2πf c ) = 112 ϕ m = = 68 f 26

27 Example: a loop gain leading to an unstable closed-loop system T 60 db 40 db T T 20 db f p1 Crossover frequency 0 db T f p2 f c 0 20 db db ϕ m (< 0) Hz 10 Hz 100 Hz 1 khz 10 khz 100 khz T(j2πf c ) = 230 ϕ m = = 50 f 27

28 The relation between phase margin and closed-loop damping factor How much phase margin is required? A small positive phase margin leads to a stable closed-loop system having complex poles near the crossover frequency with high Q. The transient response exhibits overshoot and ringing. Increasing the phase margin reduces the Q. Obtaining real poles, with no overshoot and ringing, requires a large phase margin. The relation between phase margin and closed-loop Q is quantified in this section. 28

29 A simple second-order system Consider the case where T can be wellapproximated in the vicinity of the crossover frequency as T = 1 s ω 0 1 s ω 2 40 db T f 0 T T f 20 db 0 db 20 db 40 db 20 db/decade T 90 f 0 f 2 /10 f f 2 ϕ m f 2 f 0 f 2 f 2 40 db/decade 10f

30 Closed-loop response If Then or, T = 1 s ω 0 1 s ω 2 T 1T = T T 1T = 1 1 s Qω c = 1 1 s ω 0 s2 ω 0 ω 2 s ω c 2 where ω c = ω 0 ω 2 =2π f c Q = ω 0 ω c = ω 0 ω 2 30

31 Low-Q case Q = ω 0 ω c = 40 db 20 db 0 db 20 db ω 0 ω 2 low-q approximation: Q ω c = ω ω c 0 Q = ω 2 T 20 db/decade T 1T f 0 f f 0 f c = f 0 f 2 Q = f0 / fc f 2 40 db f 0 f 2 f 2 40 db/decade f 31

32 High-Q case ω c = ω 0 ω 2 =2π f c Q = ω 0 ω c = ω 0 ω 2 60 db 40 db T 20 db/decade f 0 f 20 db f 2 0 db 20 db 40 db T 1T f c = f 0 f 2 f f 0 f 2 f 2 f 0 Q = f 0 /f c 40 db/decade 32

33 Q vs. ϕ m Solve for exact crossover frequency, evaluate phase margin, express as function of ϕ m. Result is: Q = cos ϕ m sin ϕ m ϕ m =tan Q 4 2Q 4 33

34 Q vs. ϕ m Q 20 db 15 db 10 db 5 db 0 db 5 db Q = 1 0 db ϕ m = 52 Q = db 10 db ϕ m = db 20 db ϕ m 34

35 Transient response vs. damping factor Unit-step response of second-order system T/(1T) v(t)=1 2Qe-ω c t/2q 4Q 2 1 sin 4Q 2 1 2Q ω c t tan -1 4Q 2 1 Q > 0.5 v(t)=1 ω 2 ω ω 2 ω e ω 1 t 1 1 ω 1 ω e ω 2 t 2 Q < 0.5 ω 1, ω 2 = ω c 2Q 1 ± 1 4Q2 For Q > 0.5, the peak value is peak v(t)=1e π / 4Q2 1 35

36 Transient response vs. damping factor v(t) Q = 50 Q = 10 Q = 4 Q = Q = 1 Q = 0.75 Q = 0.5 Q = 0.3 Q = 0.2 Q = 0.1 Q = 0.05 Q = ω c t, radians 36

37 9.5. Regulator design Typical specifications: Effect of load current variations on output voltage regulation This is a limit on the maximum allowable output impedance Effect of input voltage variations on the output voltage regulation This limits the maximum allowable line-to-output transfer function Transient response time This requires a sufficiently high crossover frequency Overshoot and ringing An adequate phase margin must be obtained The regulator design problem: add compensator network G c to modify T such that all specifications are met. 37

38 Lead (PD) compensator G c =G c0 1 s ω z 1 s ω p G c G c0 G c0 f p f z f p f ϕmax Improves phase margin 0 45 /decade f z /10 f z = f z f p f p /10 10f z 45 /decade G c f 38

39 Lead compensator: maximum phase lead Maximum phase lead f ϕmax = f z f p G c ( f ϕmax )=tan f p / f z f p f z 2 f p = 1sin θ f z 1 sin θ f z f p 39

40 Lead compensator design To optimally obtain a compensator phase lead of θ at frequency f c, the pole and zero frequencies should be chosen as follows: f z = f c f p = f c 1 sin θ 1sin θ 1sin θ 1 sin θ If it is desired that the magnitude of the compensator gain at f c be unity, then G c0 should be chosen as f G c0 = z f p G c 0 G c G c0 G c0 f z 45 /decade f z /10 f f p f z f p /10 f ϕmax = f z f p f p 10f z 45 /decade 40

41 Example: lead compensation 60 db 40 db T T 0 G c0 Original gain T T 20 db T 0 Compensated gain f 0 0 db f z f c 20 db 40 db T 0 Compensated phase asymptotes f p 0 Original phase asymptotes 90 ϕ m f 41

42 Lag (PI) compensation G c =G c 1 ω L s G c 20 db /decade G c Improves lowfrequency loop gain and regulation f L 10f L 0 G c 90 f L /10 45 /decade f 42

43 Example: lag compensation original (uncompensated) loop gain is T T u = u0 1 ω s 0 compensator: G c =G c 1 ω L s Design strategy: choose G c to obtain desired crossover frequency ω L sufficiently low to maintain adequate phase margin 40 db 20 db 0 db 20 db 40 db T u T u T T T u0 f L f 0 f 0 f G c T u0 10f L 10f 0 1 Hz 10 Hz 100 Hz 1 khz 10 khz 100 khz f c ϕ m

44 Example, continued Construction of 1/(1T), lag compensator example: 40 db T 20 db f L f 0 G c T u0 0 db f c 20 db 40 db 1 1T f L f 0 1 G c T u0 1 Hz 10 Hz 100 Hz 1 khz 10 khz 100 khz f 44

45 Combined (PID) compensator 40 db 20 db 0 db G c =G cm 1 ω L s 1 ω s z 1 ω s p1 1 ω s p2 G c G G c c G cm f L f z f c f p1 f p2 20 db 45 /decade 10f L 10f z f p2 / db G c 90 f L /10 f z /10 f p1 /10 90 /decade 90 /decade 10f p f 45

46 Design example L 50 µh i load v g (t) 28 V Transistor gate driver δ Pulse-width modulator C 500 µf f s = 100 khz V M = 4 V v c G c v(t) Compensator Error signal v e R 3 Ω v ref 5 V Hv H Sensor gain 46

47 Quiescent operating point Input voltage V g = 28V Output V = 15V, I load = 5A, R = 3Ω Quiescent duty cycle D = 15/28 = Reference voltage V ref = 5V Quiescent value of control voltage V c = DV M = 2.14V Gain H H = V ref /V = 5/15 = 1/3 47

48 Small-signal model V D 2 d 1 : D L v g V R d C v R i load v ref (= 0) Error signal v e G c v c Compensator 1 V M V M = 4 V d T H v H H =

49 Open-loop control-to-output transfer function G vd G vd = V D standard form: 1 1s L R s2 LC G vd =G 1 d0 1 s Q 0 ω s 0 ω 0 salient features: 2 60 dbv G vd G vd 40 dbv 20 dbv 0 dbv 20 dbv 40 dbv G vd G vd G d0 = 28 V 29 dbv f /2Q 0 f 0 = 900 Hz 10 1/2Q 0 f 0 = 1.1 khz Q 0 = db G d0 = D V =28V f 0 = ω 0 2π = 1 2π LC = 1kHz Q 0 = R C = dB L 1 Hz 10 Hz 100 Hz 1 khz 10 khz 100 khz f 49

50 Open-loop line-to-output transfer function and output impedance G vg =D 1 1s L R s2 LC same poles as control-to-output transfer function standard form: G vg =G 1 g0 1 s Q 0 ω s 0 ω 0 Output impedance: Z out =R 1 sc sl = 2 sl 1s L R s2 LC 50

51 System block diagram T=G c 1 VM G vd H T= G c H V 1 V M D 1 s Q 0 ω s 0 ω 0 2 v g ac line variation G vg i load Z out Load current variation v ref (=0) v e G c T v c V M = 4 V 1 V M H = 1 3 d Duty cycle variation G vd Converter power stage v H 51

52 Uncompensated loop gain (with G c = 1) With G c = 1, the loop gain is T u =T 1 u0 1 s Q 0 ω s 0 ω 0 T u0 = HV DV M = dB 2 40 db T u T u 20 db 0 db 20 db 40 db T u T u T u db f c = 1.8 khz, ϕ m = 5 f f 0 1 khz Q f 0 = 900 Hz Q f 0 = 1.1 khz Q 0 = db 40 db/decade 1 Hz 10 Hz 100 Hz 1 khz 10 khz 100 khz

53 Lead compensator design Obtain a crossover frequency of 5 khz, with phase margin of 52 T u has phase of approximately 180 at 5 khz, hence lead (PD) compensator is needed to increase phase margin. Lead compensator should have phase of 52 at 5 khz T u has magnitude of 20.6 db at 5 khz Lead compensator gain should have magnitude of 20.6 db at 5 khz Lead compensator pole and zero frequencies should be f z =(5kHz) f p =(5kHz) 1 sin (52 ) 1sin (52 ) = 1.7kHz 1sin (52 ) 1 sin (52 ) = 14.5kHz Compensator dc gain should be G c0 = f c f T u0 f z f p = dB 53

54 Lead compensator Bode plot 40 db G c G G c c0 f f G c z p 20 db 0 db G c0 fz f p f c = f z f p 20 db 40 db 0 f z /10 f p /10 10f z 90 0 G c Hz 10 Hz 100 Hz 1 khz 10 khz 100 khz f 54

55 Loop gain, with lead compensator 40 db 20 db T=T u0 G c0 1 s ω p 1 1 s ω z s Q 0 ω 0 s ω 0 T T T T 0 = db Q 0 = db 2 0 db 20 db 40 db T Hz f 0 f z 1 khz 1.7 khz f c 5 khz f p 900 Hz 14 khz khz 17 khz 1.1 khz ϕ m = Hz 10 Hz 100 Hz 1 khz 10 khz 100 khz f 55

56 1/(1T), with lead compensator 40 db 20 db T T 0 = db f 0 Q 0 = db need more low-frequency loop gain 0 db 20 db 1 1T 1/T 0 = db f z Q 0 f c f p hence, add inverted zero (PID controller) 40 db 1 Hz 10 Hz 100 Hz 1 khz 10 khz 100 khz f 56

57 Improved compensator (PID) 1 ω s 1 ω L z s G c =G cm 40 db 20 db 0 db 20 db 40 db f 1 s ω p G c G c G c G c 90 f z /10 f L /10 G f cm c f L f z 10f 90 L 10f z 45 /decade 45 /dec 0 f p /10 90 /decade Hz 10 Hz 100 Hz 1 khz 10 khz 100 khz f p add inverted zero to PD compensator, without changing dc gain or corner frequencies choose f L to be f c /10, so that phase margin is unchanged 57

58 T and 1/(1T), with PID compensator 60 db 40 db T 20 db Q 0 0 db f L f 0 fz f c 20 db 40 db 1 1T Q 0 f p 60 db 80 db 1 Hz 10 Hz 100 Hz 1 khz 10 khz 100 khz f 58

59 Line-to-output transfer function v v g 20 db 0 db G vg (0) = D Q 0 20 db 40 db 60 db Open-loop G vg 20 db/decade D T u0 G cm f L f 0 f z f c 80 db Closed-loop G vg 1T 40 db/decade 100 db 1 Hz 10 Hz 100 Hz 1 khz 10 khz 100 khz f 59

60 9.6. Measurement of loop gains Block 1 Block 2 A Z 1 v ref v e G 1 v e v x Z 2 G 2 v x = v T H Objective: experimentally determine loop gain T, by making measurements at point A Correct result is Z T=G 1 2 G 2 H Z 1 Z 2 60

61 Conventional approach: break loop, measure T as conventional transfer function V CC 0 Block 1 Block 2 dc bias Z 1 v ref v e G 1 v e v y v z v x Z 2 G 2 v x = v T m H measured gain is T m = v y v x v ref =0 v g =0 T m =G 1 G 2 H 61

62 Measured vs. actual loop gain Actual loop gain: T=G 1 Measured loop gain: Z 2 Z 1 Z 2 G 2 H T m =G 1 G 2 H Express T m as function of T: T m =T 1 Z 1 Z 2 T m T provided that Z 2 >> Z 1 62

63 Discussion Breaking the loop disrupts the loading of block 2 on block 1. A suitable injection point must be found, where loading is not significant. Breaking the loop disrupts the dc biasing and quiescent operating point. A potentiometer must be used, to correctly bias the input to block 2. In the common case where the dc loop gain is large, it is very difficult to correctly set the dc bias. It would be desirable to avoid breaking the loop, such that the biasing circuits of the system itself set the quiescent operating point. 63

64 Voltage injection 0 Block 1 Block 2 i Z 1 Z s v z v ref v e G 1 v e v y v x Z 2 G 2 v x = v T v H Ac injection source v z is connected between blocks 1 and 2 Dc bias is determined by biasing circuits of the system itself Injection source does modify loading of block 2 on block 1 64

65 Voltage injection: measured transfer function T v Network analyzer measures T v = v y v x v ref =0 v g =0 v ref 0 v e G 1 v e Block 1 v z Block 2 i Z 1 Z s v y T v v x Z 2 G 2 v x = v Solve block diagram: v e = H G 2 v x v y =G 1 v e i Z 1 Hence v y = v x G 2 H G 1 i Z 1 with i= v x Z 2 H Substitute: v y =v x G 1 G 2 H Z 1 Z 2 which leads to the measured gain T v =G 1 G 2 H Z 1 Z 2 65

66 Comparison of T v with T Actual loop gain is T=G 1 Z 2 Z 1 Z 2 G 2 H Gain measured via voltage injection: T v =G 1 G 2 H Z 1 Z 2 Express Tv in terms of T: T v =T 1 Z 1 Z 2 Z 1 Z 2 Condition for accurate measurement: T v T provided (i) Z 1 << Z 2, and (ii) T >> Z 1 Z 2 66

67 Example: voltage injection Block 1 Block 2 50 Ω v y v z v x 500 Ω Z 1 =50Ω Z 2 =500Ω Z 1 = dB Z 2 1 Z 1 Z 2 = dB suppose actual T= 1 s 2π 10Hz s 2π 100kHz 67

68 Example: measured T v and actual T 100 db 80 db T v =T 1 Z 1 Z 2 Z 1 Z 2 60 db 40 db T T v 20 db 0 db 20 db Z 1 Z 2 20dB T v 40 db 10 Hz 100 Hz 1 khz 10 khz 100 khz 1 MHz f T 68

69 Current injection T i = i y i x v ref =0 v g =0 0 Block 1 Block 2 i y i x Z 1 i z v ref v e G 1 v e Z s Z 2 G 2 v x = v T i H 69

70 Current injection It can be shown that T i =T 1 Z 2 Z 1 Z 2 Z 1 Injection source impedance Z s is irrelevant. We could inject using a Thevenin-equivalent voltage source: Conditions for obtaining accurate measurement: i y C b i z i x (i) Z 2 << Z 1, and R s (ii) T >> Z 2 Z 1 v z 70

71 Measurement of unstable systems Injection source impedance Z s does not affect measurement Increasing Z s reduces loop gain of circuit, tending to stabilize system Original (unstable) loop gain is measured (not including Z s ), while circuit operates stabily Block 1 Block 2 v z 0 Z 1 R ext Z s v ref v e G 1 v e v y L ext v x Z 2 G 2 v x = v T v H 71

72 9.7. Summary of key points 1. Negative feedback causes the system output to closely follow the reference input, according to the gain 1/H. The influence on the output of disturbances and variation of gains in the forward path is reduced. 2. The loop gain T is equal to the products of the gains in the forward and feedback paths. The loop gain is a measure of how well the feedback system works: a large loop gain leads to better regulation of the output. The crossover frequency f c is the frequency at which the loop gain T has unity magnitude, and is a measure of the bandwidth of the control system. 72

73 Summary of key points 3. The introduction of feedback causes the transfer functions from disturbances to the output to be multiplied by the factor 1/(1T). At frequencies where T is large in magnitude (i.e., below the crossover frequency), this factor is approximately equal to 1/T. Hence, the influence of low-frequency disturbances on the output is reduced by a factor of 1/T. At frequencies where T is small in magnitude (i.e., above the crossover frequency), the factor is approximately equal to 1. The feedback loop then has no effect. Closed-loop disturbance-tooutput transfer functions, such as the line-to-output transfer function or the output impedance, can easily be constructed using the algebra-onthe-graph method. 4. Stability can be assessed using the phase margin test. The phase of T is evaluated at the crossover frequency, and the stability of the important closed-loop quantities T/(1T) and 1/(1T) is then deduced. Inadequate phase margin leads to ringing and overshoot in the system transient response, and peaking in the closed-loop transfer functions. 73

74 Summary of key points 5. Compensators are added in the forward paths of feedback loops to shape the loop gain, such that desired performance is obtained. Lead compensators, or PD controllers, are added to improve the phase margin and extend the control system bandwidth. PI controllers are used to increase the low-frequency loop gain, to improve the rejection of low-frequency disturbances and reduce the steady-state error. 6. Loop gains can be experimentally measured by use of voltage or current injection. This approach avoids the problem of establishing the correct quiescent operating conditions in the system, a common difficulty in systems having a large dc loop gain. An injection point must be found where interstage loading is not significant. Unstable loop gains can also be measured. 74

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