# Professor Fearing EE C128 / ME C134 Problem Set 2 Solution Fall 2010 Jansen Sheng and Wenjie Chen, UC Berkeley

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1 Professor Fearing EE C128 / ME C134 Problem Set 2 Solution Fall 21 Jansen Sheng and Wenjie Chen, UC Berkeley 1. (15 pts) Partial fraction expansion (review) Find the inverse Laplace transform of the following function using partial fraction expansion: (s 2 +2s+2)(s+1) (1) The partial fraction expansion is: (s 2 +2s+2)(s+1) as+b s 2 +2s+2 + c s+1 (2) (as+b)(s+1)+c(s 2 +2s+2) (3) Solving for the unknowns, we get a 2, b 3, c 2. Now we can divide up our fractions into familiar inverse Laplace transforms: 2s+3 s 2 +2s s+1 2(s+1) (s+1) (s+1) s+1 (4) Finally, take the inverse Laplace transform: Note: You could also divide the fraction into three terms (2cost+sint 2)e t u(t) (5) (s 2 +2s+2)(s+1) a s+1 i + b s+1+i + c s+1, (6) and then continue to solve for the inverse Laplace transform from here and you would get the same result. 2. (15 pts) Equivalent models (review) For the mechanical circuit below, a. Write the differential equations describing the system s motion. b. Write the dynamic equations in state space form ẋ Ax+Bu, with input u F. c. Draw the equivalent electrical circuit, using F V and v i equivalances. M1 M2 F x1 K1 x2 K2 a. The system can be divided into two mass spring subsystems M1, K1 and M2, K2 with a force source attached. The dynamics of the first subsystem would be: The dynamics of the second subsystem would be: We can rearrange these equations into: F ẍ 1 +K 1 (x 1 x 2 ) (7) K 1 (x 1 x 2 ) M 2 ẍ 2 +K 2 x 2 (8) ẍ 1 1 ( K 1 x 1 +K 1 x 2 +F) ẍ 2 1 M 2 (K 1 x 1 (K 1 +K 2 )x 2 ) (9)

2 b. To write these equations into state space form, we need to use one variable for each derivative below the highest. In this case, we could use x [x 1 ẋ 1 x 2 ẋ 2 ] as our state (or a variety of other possibilities but this one is simple/standard). ẋ Ax+Bu 1 K1 K 1 M 2 K 1 1 K1+K2 M 2 x+ 1 F (1) c. To draw the equivalent circuit, remember that mechanical elements in parallel will have the same current in the circuit and mechanical elements connected in series will have the same voltage. This means that the inductor and capacitor corresponding to M2 and K2 will have the same velocity/current (in series). K1 separates M1 and the M2K2 subsystem so there will be a capacitor corresponding to K1 in between the two. The force source becomes a voltage source in parallel with everything. 3. (2 pts) 2nd order step response A memory system can be made using a mechanical head positioning system to read data stored on a surface.the head positioning system can be approximately modelled by the transfer function from applied force F(s) to output position X(s): X(s) F(s) 1 ms 2 +bs+k Assume that m 1 9 kg, b N sec m 1 and k 1 1 N m 1. a) Find the pole locations and sketch in the s-plane, and find ζ, ω n, θ sin 1 ζ, and ω d. b) For a 1 mn step, determine peak overshoot (µm), time to peak, and time for settling to within 1 µm of final value. c) Repeat b) for 1 µn step. a. The pole locations are the roots of the denominator, ms 2 + bs + k. These poles are given by the quadratic formula: X(s) F(s) 1 ms 2 +bs+k 1 k/m m(k/m) s 2 (11) +bs/m+k/m b± b 2 4mk 2m ( 2±j4 6) 1 3 (12) To solve for the values ζ, ω n, θ sin 1 ζ, and ω d, we can use the equations from FPE P111. s σ±jω d ( 2±j4 6) 1 3 σ (13) ω n 1 9 k/m 1 14 rad/s ζ σ ω n ω d ω n 1 ζ rad/s θ sin 1 ζ sin 1 (.2) (14)

3 b. The characteristics for a step response of a second order system are given in FPE P117. Since we are given a scaled step response (with magnitude 1 mn), the response will be scaled by both the input 1 mn and normalization factor 1/k. The peak overshoot is given by The time to peak is M p e πtanθ e πζ/ 1 ζ e π.2/ µm (15) And the time for settling to within 1µm is t p π π ω d 4.326ms (16) 6 13 e ζωnts e.2ts (1) 1 4 (17) t s 4.65ms (18) c. The response for a 1 µn step would have the same characteristics as in part b) but a different peak overshoot in µm (but same percentage) and different time for settling to 1 µm because the settling time is dependent on the percentage of the final value. M p e πtanθ e πζ/ 1 ζ e π.2/ µm (19) t p π π ω d 4.326ms (2) 6 13 And the time for settling to within 1µm (the response is scaled by both the input 1 µn and normalization factor 1/k) is 1 6 e ζωnts e.2ts (1) 1 2 (21) t s 2.33ms (22)

4 4. (25 pts) Stabilty - Routh Array The closed loop transfer function of a system is given by What is the range of K for stability? H(s) Ks+4s s 4 +4s 3 +Ks 2 +4s+1 (23) The characteristic equation for this closed-loop system is The Routh array for this polynomial is a(s) s 4 +4s 3 +Ks 2 +4s+1 (24) s 4 1 K 1 s s 2 4K K 4 K s 1 4() s (4 4 )1 () (25) For stability, the following conditions need to be satisfied K 1 > 4 4 K 1 > K > 2 (26) 5. (25 pts) Linearization A capacitive actuator has force given by F (d V 2 o x 1) where d o is the nominal capacitor plate gap, A is plate area, and V is applied voltage. The capacitive actuator has mass M and has a return spring with non-linear stiffness F k kx 3 1 and can be modelled as shown below. a. Write the dynamic equations in state space form ẋ f(x,u), with x 1 and ẋ 1 as the states. b. Write the dynamic equations in state space form ẋ Ax+Bu for the system linearized about a non-zero operating point V V o, x 1 x o, and x 2 ẋ 1. M F x 1 k a. The forces applied on the mass M are the actuator force F and the spring force F k. Application of Newton s Law yields Mẍ 1 F F k Mẍ 1 (d o x 1 ) V 2 kx 3 1 (27) Define the state as x [x 1 x 2 ] T [x 1 ẋ 1 ] T and the input as u V. Then ẋ ] [ [ẋ1 ẋ 2 k M x3 1 + x 2 M(d V 2 x 1) ] : [ ] f1 (x,u) f 2 (x,u) ẋ f(x,u) (28) b. The equilibrium point in this system is given as V V o, x 1 x o, and x 2 ẋ 1. Linearizing the system about this equilibrium point and input voltage yields ẋ Ax+Bu, where

5 A B [ f1 x 1 f 1 x 2 f 2 f 2 x 1 x 2 [ f1 ] u f 2 u x,ẋ,v ] [ ] [ ] 1 1 k M 3x2 1 + M(d x,ẋ x 1) V 2 k 2,V x,ẋ,v M 3x2 + M(d x ) V 2 2 [ ] [ ] (29) M(d 2V 2V x 1) M(d x ) x,ẋ,v

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