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EE C128 / ME C134 Spring 2014 HW2 - Solutions UC Berkeley Homework 2 - Solutions 1. Transfer Functions of Electrical Networks with Operational Amplifiers Find the transfer function, G(s) = Vo (s)/Vi (s), for each operational amplifier circuit shown in the Figures below. (a) Solution: Calculating the feedback and feedforward impedances. Zf eedback (s) = Zf b (s) = 100 × 103 + 1 2 × 10−6 s Zf eedf orward (s) = Zf f (s) = 500 × 103 + 1 2 × 10−6 s Since the system is in the inverted amplifier configuration, we know that the gain is given by Zf b (s) Vo (s) 105 (2 × 10−6 s) + 1 =− =− Vi (s) Zf f (s) 5 × 105 (2 × 10−6 s) + 1 Vo (s) 0.2s + 1 =− Vi (s) s+1 (b) Solution: Rev. 1.0, 02/09/2014 1 of 5 EE C128 / ME C134 Spring 2014 HW2 - Solutions UC Berkeley Calculating the feedback and feedforward impedances. 5 (s + 10) 5 = 105 Zf eedback (s) = Zf b (s) = 10 1 + s+5 (s + 5) Zf eedf orward (s) = Zf f (s) = 105 5 +1 s = 105 (s + 5) s Since the system is in the inverted amplifier configuration, we know that the gain is given by Zf b (s) s(s + 10) Vo (s) =− =− Vi (s) Zf f (s) (s + 5)2 2. Transfer Functions of Translational Mechanical Systems For the system shown below, find the transfer function, G(s) = X1 (s)/F (s). Solution: Equation of motion (in s-domain) for mass M1 is given by (s2 + 6s + 9)X1 (s) − (3s + 5)X2 (s) = 0 (1) and equation of motion (in s-domain) for mass M2 is given by −(3s + 5)X1 (s) + (2s2 + 5s + 5)X2 (s) = F (s) ⇒ X2 (s) = F (s) + (3s + 5)X1 (s) 2s2 + 5s + 5 (2) Plugging Equation (2) in Equation (1) and rearranging yields X1 (s) = ⇒ Rev. 1.0, 02/09/2014 2s4 (3s + 5)F (s) + 17s3 + 44s2 + 45s + 20 X1 (s) 3s + 5 = 4 3 F (s) 2s + 17s + 44s2 + 45s + 20 2 of 5 EE C128 / ME C134 Spring 2014 HW2 - Solutions UC Berkeley 3. State Space Representation of Electrical Networks (a) Represent the electrical network shown below in state space, where iR (t) is the output. Solution: Write the differential equations for each energy storage element. i2 dv1 = dt 3 di3 vL = dt 2 " # v1 Therefore the state vector is x = . Now obtain vL and i2 in terms of the state variables, i3 vL = v1 − v2 = v1 − 3iR = v1 − 3(i3 + 4v1 ) = −11v1 − 3i3 1 1 1 i2 = i1 − i3 = (vi − v1 ) − i3 = − v1 − i3 + vi 3 3 3 Also, the output is y = iR = 4v1 + i3 Hence, − 19 ẋ = − 11 2 − 13 x + − 32 h i y= 4 1 x " # 1 9 0 vi (b) Find the state space representation of the network shown below if the output is vo (t). Rev. 1.0, 02/09/2014 3 of 5 EE C128 / ME C134 Spring 2014 HW2 - Solutions UC Berkeley Solution: Let C1 be the grounded capacitor and C2 be the other. Now, writing the equations for the energy storage components yields, diL = vi − vC1 dt dvC1 = i1 − i2 dt dvC2 = i2 − i3 dt (3) (4) (5) iL Thus the state vector is x = vC1 . Now, find the three loop currents in the terms of state variables v C2 and the input. Writing KVL around Loop 2 yields vC1 − vC2 = i2 Writing KVL around the outer loop yields i3 = vi − i2 = vi − vC1 + vC2 Also, i1 − i3 = iL . Hence, i1 = iL + i3 = iL + vi − vC1 + vC2 Substituting the loop currents in Equations (3), (4) and (5), yields the results in vector-matrix form, 0 −1 0 1 ẋ = 1 −2 2 x + 1 vi 0 2 −2 −1 Since v0 = i2 = vC1 − vC2 , the output equation is h i y = 0 1 −1 x 4. Transfer Function to Phase Variable Representation For the system shown below, write the state equations and the output equation for the phase-variable representation. Rev. 1.0, 02/09/2014 4 of 5 EE C128 / ME C134 Spring 2014 HW2 - Solutions Solution: Using the standard form 0 0 ẋ = 0 0 0 h c= 6 Rev. 1.0, 02/09/2014 UC Berkeley derived in the textbook, 1 0 0 0 0 0 0 1 0 0 0 0 1 0 x + 0 r(t) 0 0 0 1 0 0 −8 −13 −9 1 i 7 12 2 1 x 5 of 5