# ME 375 Final Examination Thursday, May 7, 2015 SOLUTION

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1 ME 375 Final Examination Thursday, May 7, 2015 SOLUTION

2 POBLEM 1 (25%) negligible mass wheels negligible mass wheels v motor no slip ω r r F D O no slip e in Motor% Cart%with%motor%a,ached% The coupled electro-mechanical equations for an electric motor (ignoring armature inductance and armature damping) are given by: J!ω K T i a τ L i a + K T ω e i t where ω is the rotation rate of the motor, J is the mass moment of inertia of the armature, K T is the motor constant, i a is the armature current, τ L is the applied motor torque, e i ( t) is the input voltage applied to the armature and is the armature resistance. An electric motor governed by the above equations is used to propel a cart along a horizontal roadway. In this application, a disk attached to the output shaft of the motor drives one of the rear wheels of the cart through a no-slip frictional contact. The cart moves with no slipping of the wheels on the roadway and experiences a drag force of f D cv to the right as it moves to the left, where v is the speed of the cart to the left. Let M represent the total mass of the motor and cart. Ignore the mass of all of the wheels. For this problem: a) Draw free body diagrams of the driven rear wheel and of the total system. From these develop the differential equation of motion of the cart in terms of the speed v. b) Develop the transfer function relating the input voltage to the cart speed, G( s) V ( s) / E i ( s). c) The motor is given a step input voltage e i (t) e 0 h(t). Determine the steady-state speed attained by the cart as it moves along the roadway.

3 SOLUTION r F D O F M Oy O O x Let W be the radius of the rear wheels. From FBD of the wheel: f f M O F M W f W 0 f F M From the FBD of the cart/wheel: F f F D M!v F M τ L M!v + cv r (1) Taking LT of (1) gives: T L (s) r( Ms + c)v (s) (2) Taking LT of the motor equations gives: JsΩ K T I a T L (3) E i I a + K T Ω (4) For kinematics, we have: ω W r ω Ω W r Ω W W v W ω W V W Ω W rω Ω V / r (5) where ω W is the angular speed of the driven wheel. Combining above gives: (4) I a E i K T Ω (3),(5) & (6) Js V r K T ( E i r K T V ) r 2 ( Ms + c)v E i K T V / r r ( Ms + c )V JsV K T ( J + r 2 M )s + r 2 2 c + K T V K T re i J + Mr 2 And: G(s) V (s) E i (s) K T r / ( J + Mr 2 )s + r 2 c + K 2 T / From FVT: v ss lim sv (s) s 0 lim s e 0 G(s) K G(0) T re 0 s 0 s r 2 2 c + K T!v + r2 c + K T 2 / v K T re i (t) / (6)

4 POBLEM 2 (25%) plant ( s) K ( s) Y ( s) Consider the proportional feedback control system shown above. The plant transfer function s has a static gain of 10, has no zeros and has the three poles shown below. a) Determine the plant transfer function s. b) For the closed-loop root locus with K 0 : i. Determine the K asymptotes and their real-axis intersections. ii. Determine the breakaway/breakin (re-entry) points on the real axis. iii. Determine the departure angles from the open-loop poles. iv. Determine the values of K, if any, at which the root locus crosses into the right-half plane (unstable). v. Sketch the root locus on the plot provided below. vi. For what range of values for the gain K, if any, would the closed-loop response have a 2% settling time t s < 0.5AND a percent overshoot of %OS < 4% for a step input? K θ K 0 p2 p 2 11/ 3 p 3 K 0 θ p2 60 K 0 60 p K

5 SOLUTION A (s) ( s p 1 ) s p 2 A ( s +1 ) s + 5 Since G(0) 10, we have: ( s p 3 ) A ( s +1) ( s j) ( s j) (0) 10 A 29 A 290 A s 3 +11s s + 29 N(s) D(s) ule #2: The L has starting points at the three open-loop poles and end points at infinity. ule #4: The L exists on the real axis to the left of the real pole p 1 1. ule #5: The asymptotes have real axis intercepts and angles of: p i z i 1+ ( j) + ( 5 2 j) 0 σ 0 11 N P N z θ k ( 2k 1) ( 2k 1) 180 ( 2k 1)60 60,180,300 N P N z 3 0 ule #6: BA/E points on the real axis are found from: N dd ds D dn A( 3s s + 39) 0 ds s 1,2 22 ± 222 (4)(3)(39) (2)(3) 22 ± 4 6 3, 13 3 ule #7: From the general equation: ( s z 1 ) ( s p 1 ) + ( s p 2 ) 180 we can write the following for OL pole p 1 : j +1 +θ p2 + ( p 2 p 3 ) p 2 p 1 +θ p2 + ( j j) +θ p2 + ( 4 j) j 180 tan θ p2 + tan θ p θ p2 63.4

6 ule #8: The characteristic equation for the closed-loop system is: 0 D(s) + KN(s) s 3 +11s s K Substituting s jω into the CLCE gives: 3 +11( jω ) ( jω ) K 0 jω jω 3 11ω 2 + j(39ω ) K K 11ω 2 + j( 39 ω 2 )ω Imag : ( 39 ω 2 )ω 0 ω 2 0,39 eal : K 11ω 2 0 K ω ,1.38 For t s ζω n ζω n 8. It is not possible to have all of the roots to the left of 8. Therefore, it is not possible to meet the settling time criteria.

7 POBLEM 3 (25%) A plant with a transfer function s 1 s s + 1 is to be controlled with unity PD feedback control where the controller has a transfer function G C ( s) K P + K D s. a) Derive the closed-loop transfer function and closed-loop characteristic equation for this control system. b) Determine the gain values K P and K D corresponding to a 2% settling time of t s 4 and a %OS 4 for a step input. c) If K P 4, what value of the gain K D produces the smallest possible 2% settling time for a step input? SOLUTION The CL transfer function and characteristic equation are given by: G CL (s) G C 1+ G C D CL (s) 1+ G C 1+ K P + K D s s s +1 0 s 2 + ( 1+ K D )s + K P 0 respectively. The general form of the characteristic equation for a second-order system is: s 2 + 2ζω n s + ω n 2 0 For %OS 4 andt s 4, we have: %OS 100e ζπ / 1 ζ 2 ζ + π 2 ln %OS / 100 ln 2 %OS / 100 t s ζω ζω n 1 ω n n Balancing coefficients in the CLCE s: K P ω n K D 2ζω n K D 2ζω n 1 2 ( 1) π 2 ln 0.04 ln The smallest possible settling time occurs when ζ 1. Since ω n K P 4 2, we can write: K D 2ζω n

8 POBLEM 4 PAT A (10%) Consider the electrical circuit shown below. Develop an expression for the impedance V i ( s) I ( s) Z s. Express your final answer as a single fraction. a b i v i (t) + - L d e C SOLUTION Inductor be and capacitor de are connected in series. That connection is in parallel with resistor bd. In turn, that is in series with resistor ab. Therefore: Z be Ls and: Z de 1 Cs Z bd Z bde Z bd Z be + Z de + 1 Ls +1/ Cs 1 + Cs LCs 2 +1 LCs Cs LCs 2 + LCs 2 Z bde + LCs Cs which leads to: LCs 2 + Z(s) Z a + Z bde + LCs 2 + Cs +1 2LCs Cs + 2 LCs 2 + Cs +1

9 POBLEM 4 (continued) PAT B (10%) s plant K ( s) Y ( s) The closed-loop transfer function for the above system has the following characteristic equation: D CL (s) s 4 + 5Ks 3 + (3+ K)s 2 + 6s K 0 a) Determine the plant transfer function (s). b) Determine the magnitude of the steady-state response of the closed-loop system to a unit harmonic input r(t) sin5t for K 1. SOLUTION 0 s 4 + 5Ks 3 + (3+ K)s 2 + 6s K s 4 + 3s 2 + 6s K 5s 3 + s s 3 + s K s 4 + 3s 2 + 6s K (s) The CL transfer function is written as: For K 1: G CL (s) K (s) K 5s 3 + s K (s) s 4 + 3s 2 + 6s K 5s 3 + s s 3 + s 2 +1 G CL (s) s 4 + 5s 3 + 4s 2 + 6s + 5 Magnitude of steady-state response for harmonic input: G CL (5 j) 5(5 j) 3 + (5 j) 2 +1 (5 j) 4 + 5(5 j) 3 + 4(5 j) 2 + 6(5 j) j j

10 POBLEM 4 (continued) PAT C (5%) G 1 (s) s Y ( s) G 2 (s) Consider the feedback control system above, with: s 2 +1 G 1 (s) s 3 + 3s +1 G 2 (s) s s + 4 Determine the single input-output differential equation governing the output y(t) for an input r(t). What is the order of the closed-loop system? SOLUTION From block diagram: Y G 1 G 2 Y Y G 1 (s 2 +1) / (s 3 + 3s +1) 1+ G 1 G 2 1+ s(s 2 +1) / (s 3 + 3s +1)(s + 4) (s 2 +1)(s + 4) (s 3 + 3s +1)(s + 4) + s(s 2 +1) s 3 + 4s 2 + s + 4 s 4 + 5s 3 + 3s 2 +14s + 4 ( s 4 + 5s 3 + 3s 2 +14s + 4)Y ( s 3 + 4s 2 + s + 4) Performing an inverse LT on the above:!!!! y + 5!!! y + 3!!y +14!y + 4y!!! r (t) + 4!! r(t) +!r(t) + 4r(t) Fourth-order system.

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