NPTEL Online Course: Control Engineering Ramkrishna Pasumarthy

Assignment-8 Solutions 1 1. Given an open loop transfer function G(s) = (s+2)(s+3) . Design a compensator Gc (s) = K s+a s+b . such that the closed-loop poles at s = −1 ± j. What is the value of K in the compensator? Hints: You may use a pole zero cancellation and also plot root locus on MATALB to verify.

Solution: Closed poles should be at s = −1 ± j. The characteristic equation should be s2 + 2s + 2 = 0. When this is written in unity feedback form, we get s2 + 2s + 2 = 0 2 1+ 2 =0 s + 2s 2 1+ =0 s(s + 2) Therefore the open loop transfer function with compensator is given by: 2 s(s + 2) K(s + a) 2 = (s + 2)(s + 3)(s + b) s(s + 2) G(s)Gc (s) =

As given in the hint, we can take a = 3 and cancel out the pole and zero as s = −3. By comparing the rest of the terms, we get K = 2 & b = 0. 2. In Problem 1, what is the value of a in the designed compensator ? Solution a=3 3. In Problem 1, what is the value of b in the designed compensator ?

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Solution b=0 1 4. Given an open loop transfer function G(s) = s(s+3)(s+6 , design a compensator which approximately meets the following specifications for unit step response:

(a) The settling time is ts ≤ 5 sec (b) The step response overshoot M p = 16.3% (c) The steady state error to a unit ramp is 10%. There is a proportional controller which will meet the transient specifications. Find that value of K (proportional controller). Solution Draw the root locus in MATLAB using the following commands: g = tf([1],[1 9 18 0]); rlocus(g) By moving the cursor on the root locus, we can find the value of K which meets the transient specifications of ts ≤ 5 sec and M p = 16.3% K = 28 (You will get marks for writing any value between 27 and 29) 5. What is the velocity error constant Kv of the gain adjusted system? Solution Kv = lim sKG(s) s→0

28 s→0 s(s + 3)(s + 6) 28 Kv = 18 Kv = 1.5 Kv = lim

(You will get marks for writing any value between 1.4 and 1.7) 6. For the gain compensated system in Problem No. 4, if we add a lag compensator, what should be the value of β ?

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Solution The value of β is given the ratio of desired Kv to the existing Kv of the system. Kv0 Kv 10 β= 1.5 ∴ β = 6.66 β=

(You will get marks for writing any value between 6 and 7) 7. In Problem No.6, if the compensator zero is added at −0.1, what should be the location of the compensator pole ? Solution zc pc zc ∴ pc = β −0.1 pc = 6.66 pc = −0.015 β=

(You will get marks for writing any value between -0.018 and -0.013) 8. Given the plant transfer function of a servomechanism to be G(s) =

10 s(s + 2)(s + 8)

Design a lead-lag compensator Gc (s) in unity feedback configuration to meet the following specification for step response: (a) M p = 16.3% (b) The rise time tr = 0.6046 sec (c) The steady state error to a unit ramp input must be equal 0.0125. What is the real part of the dominant poles of the compensated system? Solution √ The dominant poles are calculated to be sd = −2 ± 2 3 and the corresponding characteristic equation is s2 + 4s + 16. Therefore, the real part of the dominant pole is -2 9. In Problem No.8, we begin with design of lead compensator. What is the angle deficiency to be contributed by the lead compensator at the desired pole locations? 3

Solution The angle deficiency can be calculated using the angle criterion of the root locus. φ = −180 − (−∠sd − ∠(sd + 2) − ∠(sd + 8))  √   √   √  2 3 2 3 2 3 φ = −180 − (− tan−1 − tan−1 − tan−1 ) −2 −2 + 2 −2 + 8   √ 1 φ = −180 − (−(180 − tan−1 ( 3) − tan−1 (∞) − tan−1 √ ) 3 φ = −180 − (−120 − 90 − 30) φ = 60o 10. In Problem No.9, what is the value of γ Solution Taking ζ from the characteristic equation of dominant poles, we calculate θ = cos−1 (ζ ) = cos−1 (0.5) = 60o In Problem No.9,we got φ = 60o 1 γ = (180 − θ − φ ) 2 1 γ = (180 − 60 − 60) 2 γ = 30o 11. In Problem No.9, what is the location of the lead compensator zero ? Solution From the dominant pole characteristic equation, ωn = 4 The compensator zero is given by sin(γ) sin(θ + γ) sin(30) zc = 4 sin(60 + 30) zc = 2 zc = ωn

Since the zero is located in the LHP, its location is -2 It can be observed that there is a pole zero cancellation that happens in the compensated system. 12. In Problem No.9, what is the location of the lead compensator pole? 4

Solution The compensator pole is given by sin(γ + φ ) sin(θ + γ + φ ) sin(30 + 60) pc = 4 sin(60 + 30 + 60) pc = 8 pc = ωn

Since the pole is located in the LHP, its location is -8 13. In Problem No.9, what is the gain introduced by the lead compensator alone? Solution The open loop transfer function of the compensated system with gain K is given by: KG(s)Gc (s) = K

10 s(s + 8)2

The root locus for the compensated system is plotted in MATLAB using the following commands: g = tf([10],[1 16 64 0]); rlocus(g)

√ The gain of the system for the dominant closed loop poles to be sd = −2 ± 2 3 is determined to be 19.1 which is the gain to be provided by the compensator. (You will get marks for writing any value between 18 and 20) The open loop √ transfer function of the compensated system which has dominant closed loop poles sd = −2 ± 2 3 is given by: 191 KG(s)Gc (s) = s(s + 8)2 14. In Problem No.9, what is the velocity error constant of the lead compensated system ? Solution The open loop transfer function of the lead compensated system is: KG(s)Gc (s) =

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191 s(s + 8)2

The velocity error constant is given by: Kv = lim sKsG(s)Gc (s) s→0

Kv = lim s s→0

191 s(s + 8)2

191 64 Kv = 2.9

Kv =

(You will get marks for writing any value between 2.5 and 3.5) 15. For the lead compensated system in Problem No.9, we now add a lag compensator assuming αβ = 1. With this value of β , what is the new velocity error constant Kv ? Solution zc pc 2 α= 8 α = 0.25 α=

Given that αβ = 1, we get β=

1 α

1 0.25 β =4

β=

The new velocity error constant Kv is given by: Kv0 = β Kv Kv0 = 4 × 2.9 Kv0 = 11.6 (You will get marks for writing any value between 11 and 13) 16. n problem No.15, it is clear the design requirements are not met with the choice of β . We design a lag compensator independently. What should be the value of β such that steady state error specification is met ?

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Solution For the specification to be met, ess = 0.0125

Kv0 =

1 ess

1 0.0125 Kv0 = 80

Kv0 =

For Kv0 to be 80, the value of β should be: Kv0 Kv 80 β= 2.9 β = 27.59 β=

(You will get marks for writing any value between 25 and 28)

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