EE C28 / ME C34 Problem Set Solution (Fall 200) Wenjie Chen and Janen Sheng, UC Berkeley. (0 pt) BIBO tability The ytem h(t) = co(t)u(t) i not BIBO table. What i the region of convergence for H()? A bounded input x(t) = in(t)u(t) give an unbounded output for thi ytem. Determine y(t) = x(t) h(t) uing Laplace tranform propertie. The Laplace tranform of co t i H = co t u(t)e t dt. Since co t u t i bounded by a contant, the region of convergence for H() i for all poitive, or Re > 0. To calculate the convolution y(t) = x(t) h(t), we can calculate the tranform of each function and ue the propertie of Laplace tranform (FPE6e P758). H =, X = 2 + 2 + Now, convolution in time become a product in the Laplace domain o Y = ( 2 +) = 2 + 2 + 2. To convert Y() back into the time domain, notice we can ue the property that multiplication by time become the negative derivative in the -domain (tf(t) become d d F()). Y = 2 d d d = X() o y t = t in t u(t), which i unbounded a t. 2 + 2 d 2 2. (25 pt) Laplace tranform review For each tranfer function below (all are caual and at leat marginally table), determine h(t) and ketch the impule repone. (Pay attention to cale, dynamic, etc.) i) H = +0 (+) = 9 + +0 9 + h t = 9 e 0t u(t) + 9 e t u(t) ii) H 2 = +0 (+) = + + + 9 + = + 9 + h 2 t = δ t + 9e t u(t)
iii) H 3 = 0 = 20 +0 (+) 9 +0 9 h 3 t = 20 9 e 0t u(t) 9 e t u(t) + iv) H 4 = = 2 +2+0 (+) 2 +0 2 h 4 t = 0 e t in (0t)u(t) v) H 5 = 2 +2+0 = (+) 2 +0 2 = + (+) 2 +0 2 (+) 2 +0 2 h 5 t = e t co 0t u(t) 0 e t in (0t)u(t) 2
3. (20 pt) Equivalent model The figure below how a mechanical model for a imple reonant wing drive ytem, where a bending actuator drive a wing through a pulley. i) Write the tranfer function relating input force F SL to output velocity. ii) Draw the equivalent electrical circuit and determine tranfer function from voltage input to current output for the circuit. i) Define going toward the right of the page a the poitive x-direction. By obervation, we can ee that a poitive hift from equilibrium would create two negative force from the two pring, a poitive velocity on the ma would alo create two negative force from the two damper, and a poitive force from the actuator would create a poitive force on the ma. Now we can write the differential equation defining the dynamic of the ytem a: F SL B SL x B L x K SL x K L x = M L x. To calculate the tranfer function, ue the Laplace tranform and olve for Y() X () =, R() F SL () F SL () B SL X() B L X() K SL X() K L X() = 2 M L X() X() F SL () = 2 M L + (B SL + B L ) + K SL + K L ii) Uing voltage a our input and current a our output, we can convert the mechanical model into an electrical model. Force become voltage, velocity become current, mae become inductor, pring become capacitor, and damper become reitor. Component in parallel become thing in erie and vice vera. Following thee rule, we obtain the equivalent electrical circuit: 3
To olve for the tranfer function, we can ue Ohm law with impedance and implify V = I B SL + I B L + K SLI + K SI + M L I I() V = B SL + B L + K SL + K S + M L I V = 2 M L + B SL + B L + K SL + K S Which we can ee i the ame reult a above, verifying that our model are equivalent. 4. (20 pt) Electromechanical ytem example A DC motor ha electrical contant K e, torque contant k t, reitance R m, inertia J m and vicou damping b m. The motor i connected to an inertial load J L through a haft with pring contant k L and the inertial load ha vicou damping B L. Write the equation of motion for the ytem. (The DC motor dynamic can be found in FPE6e P47-48.) Here, the free-body diagram for thi electromechanical ytem with the load inertia i hown in Fig.4. Figure 4.: Electromechanical Sytem (Standard DC Motor & Inertia Load) where θ m and θ l are the rotation angle of the motor and load inertia, repectively. V i //the input voltage of DC motor, L i the motor inductance. All other variable are defined a in the problem tatement. 4
The dynamic of the ytem can be derived individually for each part of the ytem. Firt we aume the following: L R m, o we diregard the motor inductance (treat a wire, L = 0) Perfect efficiency of the motor and haft tranmiion The motor torque output to the rotor i T = k t I m, and the back emf voltage i V e = K e θ m. Analyi of the electric circuit and the application of Kirchhoff voltage law yield V = R m I m + L di m + V dt e = R m I m + K e θ m (4.) The motor output torque T i ued to drive the rotor directly, accounting for the rotor rotation, vicou damping and the haft torion. The torque reulted from the haft torion i k L θ l θ m, which i to drive the load inertia. Application of Newton law yield J m θ m + b m θ m = T k L θ l θ m = k t I m k L θ l θ m (4.2) J L θ l + B L θ l = k L (θ l θ m ) (4.3) Solve I m in (4.) and then ubtitute into (4.2). It follow J m θ m + b m + k tk e R m θ m = k tv R m k L θ l θ m (4.4) J L θ l + B L θ l = k L (θ l θ m ) (4.5) 5. (5 pt) Electrical circuit example For the circuit below, uing ideal op-amp aumption, determine Z in = V in () I in (), where V in and I in are voltage acro and current into Z in node. Applying the ideal op-amp aumption, we have (FPE6e P43) i + = i = 0, V + = V (5.) The current through the reitance R L i I L = V in V R L, the current through the capacitor C i I C = C d V in V +, and the current through the reitance R i I dt R = V +. Since i R + = 0, application of Kirchhoff current law yield I C = I R C d V in V + dt Take Laplace tranform of both ide of (5.2), reulting Now, the current through Z in node i = V + R C V in V + () = V + ()/R V + = (5.2) RC RC+ V in () (5.3) 5
I in = I L + I C = V in V + V + R L R Subtituting V () = V + () = RC RC+ in () into (5.4), we obtain R L C + V in = RR L C + R L I in () (5.5) Z in = V in () LC+R L I in () R L C+ (5.6) 6. (0 pt) Block Diagram Manipulation Uing block diagram manipulation, determine Y ()/R(). Hint- tart reduction at pt. A. (5.4) Uing block diagram manipulation: Starting from Point A, replace the firt inner negative feedback loop with it equivalent tranfer function. Again, at Point A, replace inner negative feedback loop with it equivalent tranfer function. Now replace the erial connection of the haded tranfer function and G 2 with it equivalent tranfer function. 6
Then replace the negative feedback loop with it equivalent tranfer function, by moving the feedback point to the inner loop and adding another feedforward path. Now replace the negative feedback loop with it equivalent tranfer function. Now reduce the parallel combination part (H 3 G 3 ) with it equivalent tranfer function. Then replace the erial connection of the haded tranfer function with it equivalent tranfer function. Now replace the parallel combination part with it equivalent tranfer function. Finally, the haded tranfer function correpond to Y() R(). 7
We can alo determine Y() by olving the equation from the block diagram. R() Aign the variable T, T 2, T 3 a in the following diagram: We have the relation from the diagram: T = R H 3 Y H 2 T 3 T 2 = T H T 3 T 3 = G T 2 Y = G 2 T 3 + G 3 R Solving the above equation, we obtain Y() R() = G G 2 + G 3 + G H G 3 + G H 2 G 3 + G H + G H 2 + G G 2 H 3 8