EE3CL4: Introduction to Linear Control Systems

Transcription

1 1 / 30 EE3CL4: Introduction to Linear Control Systems Section 9: of and using Techniques McMaster University Winter 2017

2 2 / 30 Outline 1 2 3

3 4 / 30 domain analysis Analyze closed loop using open loop transfer function L(s) = G c (s)g(s)h(s). Nyquist s stability criterion Gain margin: 1 L(jω x ), where ω x is the frequency at which L(jω) reaches 180 Phase margin, φ pm : L(jω c ), where ω c is the frequency at which L(jω) equals 1 Damping ratio: φ pm = f (ζ) Roughly speaking, settling time decreases with increasing bandwidth of the closed loop

4 5 / 30 Bode diagram L(jω) = Gain margin 15 db Phase margin 43 1 jω(1 + jω)(1 + jω/5)

5 6 / 30 and Bode diagram We have seen the importance of phase margin If G(s) does not have the desired margin, how should we choose G c (s) so that L(s) = G c (s)g(s) does? To begin, how does G c (s) affect the Bode diagram Magnitude: ( 20 log 10 Gc (jω)g(jω) ) ( = 20 log 10 ( Gc (jω) ) ( ) + 20 log 10 G(jω) Phase: G c (jω)g(jω) = G c (jω) + G(jω)

6 8 / 30 G c (s) = Kc(s+z) s+p, with z < p, alternatively, G c (s) = Kc 1+sατ α 1+sτ, where p = 1/τ and α = p/z > 1 Bode diagram (in the figure, K 1 = K c /α):

7 9 / 30 Compensation What will lead compensation, do? Phase is positive: might be able to increase phase margin φ pm Slope is positive: might be able to increase the cross-over frequency, ω c, (and the bandwidth)

8 10 / 30 Compensation G c (s) = Kc 1+sατ α 1+sτ By making the denom. real, can show that G c (jω) = atan ( ) ωτ(α 1) 1+α(ωτ) 2 Max. occurs when ω = ω m = 1 τ α = zp Max. phase angle satisfies tan(φ m ) = α 1 2 α Equivalently, sin(φ m ) = α 1 α+1 At ω = ω m, we have G c (jω m ) = K c / α

9 11 / 30 Bode Principles (lead) Set the loop gain so that desired steady-state error constants are obtained Insert the compensator to modify the transient properties: Damping: through phase margin Response time: through bandwidth Compensate for the attenuation of the lead network, if appropriate To maximize impact of phase lead, want peak of phase near ω c of the compensated open loop

10 12 / 30 Guidelines 1 For uncompensated (i.e., proportionally controlled) closed loop, set gain K p so that steady-state error constants of the closed loop meet specifications 2 Evaluate the phase margin, and the amount of phase lead required. 3 Add a little safety margin to the amount of phase lead 4 From this, determine α using sin(φ m ) = α 1 α+1 5 To maintain steady-state error const s, set K c = K p α 6 Determine (or approximate) the frequency at which K p G(jω) has magnitude 10 log 10 (α). 7 If we set ω m of the compensator to be this frequency, then G c (jω m )G(jω m ) = 1 (or 1). Hence, the compensator will provide its maximum phase contribution at the appropriate frequency 8 Choose τ = 1/(ω m α). Hence, p = ωm α. 9 Set z = p/α. 10 : G c (s) = Kc(s+z) s+p.

11 13 / 30 Example Type 1 plant of order 2: G(s) = 5 s(s+2) goals: Steady-state error due to a ramp input less than 5% of velocity of ramp Phase margin at least 45 (implies a damping ratio) Steady state error requirement implies K v = 20. For prop. controlled Type 1 plant: K v = lim s 0 sk p G(s). Hence K p = 8. To find phase margin of prop. controlled loop we need to find ω c, where K p G(jω c ) = 40 = 1 ω c 6.2rad/s jω c(jω c+2) Evaluate K p G(jω) = 90 atan(ω/2) at ω = ω c Hence φ pm, prop = 18

12 14 / 30 Example φ pm, prop = 18. Hence, need 27 of phase lead Let s go for a little more, say 30 So, want peak phase of lead comp. to be 30 Solving α 1 α+1 = sin(30 ) yields α = 3. Set K c = 3 8 Since 10 log 10 (3) ( = 4.8 db we should choose ω m to be where 20 log 40 ) 10 = 4.8 db jω m(jω m+2) Solving this equation yields ω m = 8.4rad/s Therefore z = ω m / α = 4.8, p = αz = 14.4 G c (s) = 24(s+4.8) s+14.4 G c (s)g(s) = 120(s+4.8) s(s+2)(s+14.4), actual φ pm = 43.6 Goal can be achieved by using a larger target for additional phase, e.g., α = 3.5

13 15 / 30 Bode Diagram

14 16 / 30 Step Response

15 17 / 30 Ramp Response

16 18 / 30 Ramp Response, detail

17 20 / 30 G c (s) = Kc(s+z) s+p, with p < z, alternatively, G c (s) = Kcα(1+sτ) 1+sατ, where z = 1/τ and α = z/p > 1 Bode diagrams of lag compensators for two different αs, in the case where K c = 1/α

18 21 / 30 What will lag compensation do? Since zero and pole are typically close to the origin, phase lag aspect is not really used. What is useful is the attenuation above ω = 1/τ: gain is 20 log 10 (α), with little phase lag Can reduce cross-over frequency, ω c, without adding much phase lag Tends to reduce bandwidth

19 22 / 30 Qualitative example Uncompensated system has small phase margin Phase lag of compensator does not play a large role Attenuation of compensator does: ω c reduced by about a factor of a bit more than 3 Increased phase margin is due to the natural phase characteristic of the plant

20 23 / 30 Bode Principles (lag) For lag compensators: Set the loop gain so that desired steady-state error constants are obtained Insert the compensator to modify the phase margin: Do this by reducing the cross-over frequency Observe the impact on response time Basic principle: Set attenuation to reduce ω c far enough so that uncompensated open loop has desired phase margin

21 24 / 30 Guidelines 1 For uncompensated (i.e., proportionally controlled) closed loop, set gain K p so that steady-state error constants of the closed loop meet specifications 2 Evaluate the phase margin, analytically, or using a Bode diagram. If that is insufficient... 3 Determine ω c, the frequency at which the uncompensated open loop, K p G(jω), has a phase margin equal to the desired phase margin plus 5. 4 a lag comp. so that the gain of the compensated open loop, G c (jω)g(jω), at ω = ω c is 0 db Choose K c = K p /α so that steady-state error const s are maintained Place zero of the comp. around ω c/10 so that at ω c we get almost all the attenuation available from the comp. Choose α so that 20 log 10 (α) = 20 log 10 ( K p G(jω c) ). With that choice and K c = K p /α, G c (jω c)g(jω c) 1 Place the pole at p = z/α : G c (s) = Kc(s+z) s+p

22 25 / 30 Example, same set up as lead design Type 1 plant of order 2: G(s) = 5 s(s+2) goals: Steady-state error due to a ramp input less than 5% of velocity of ramp Phase margin at least 45 (implies a damping ratio) Steady state error requirement implies K v = 20. For prop. controlled Type 1 plant: K v = lim s 0 sk p G(s). Hence K p = 8. To find phase margin of prop. controlled loop we need to find ω c, where K p G(jω c ) = 40 = 1 ω c 6.2rad/s jω c(jω c+2) Evaluate K p G(jω) = 90 atan(ω/2) at ω = ω c Hence φ pm, prop = 18

23 26 / 30 Example Since want phase margin to be 45, we set ω c such that G(jω c) = = 130. = ω c 1.5 To make the open loop gain at this frequency equal to 0 db, the required attenuation is 20 db. Actual curves are around 2 db lower than the straight line approximation shown Hence α = 10. Set K c = K p /α = 0.8 Zero set to be one decade below ω c; z = 0.15 Pole is z/α = Hence G c (s) = 0.8(s+0.15) s+0.015

24 27 / 30 Example: Comp d open loop Compensated open loop: G c (s)g(s) = Numerical evaluation: new ω c = 1.58 new phase margin = 46.8 By design, K v remains 20 4(s+0.15) s(s+2)(s+0.015)

25 28 / 30 Step Response

26 29 / 30 Ramp Response

27 30 / 30 Ramp Response, detail