Electric ircuits Fall 05 Solution HW6 RULES: Please try to work on your own. Discussion is permissible, but identical submissions are unacceptable! Please show all intermediate steps: a correct solution without an explanation will get zero credit. Please submit on time. NO late submission will be accepted. Please prepare your submission in English only. No hinese submission will be accepted.. [6%] In the circuit of Fig., find: (a) v(0 + ) and i(0 + ) (b) (0 + )/ and di(0 + )/ (c) v( ) and i( ) u ut so that, (a) At t = 0 -, t and 0 i Fig. 0 4 / 3 5 0.5 A, an i Hence, i i 0 0 0.5 A 0 5 0.5 V. (b) i v v 0 0.5 V 0 / i 0 / or For t = 0 +, 4u(t) = 4 and 4u(-t) = 0. Since i and v cannot change abruptly, ir v 0 0 / 5.5 / 5 0.5A, For KL, i i ir 0 4 0 0 Hence, 0.5 4 i 0, which 0.5leads to i 0 4
Electric ircuits Fall 05 Solution Similarly, v L di L 0 / i 0 / 4 / 0. 40 V / s which leads to L L 3i 0 v 0 v 0 0 di 0 / v 0 / L or vl.5 v 0.5 0 0 4 L L di 0 / v 0 / L 4 / 0.5 6 A / s (c) i 5 4 / 3 5.5 A v 54 -.5 7.5 V. [8%] The circuit shown in Fig. has been in operation for a long time. At t = 0, the source voltage suddenly jumps to 50 V. Find vo(t) for t 0. Fig. Initial condition: i (0 ) i (0 ) i (0 ) 0A, v (0 ) v (0 ) 50V L L R 8000 As a series RL circuit, 5000rad/s 3 L 600 0 650 3 9 L 600 00 We find, so the circuit is critical damping Then, v () t V Ate A e 5000t 5000t 0 f When t =,we get Vf 6 50V When t = 0, apply Initial condition, Above all, 0 v (0) 50 A 50, so 00 0 (0) A 5000 A 0,so v t te e 6 5000 5000 0 ( ) 50 5 0 t t 00 V A 6 A 5 0 V
Electric ircuits Fall 05 Solution 3. [8%] The natural response for the circuit shown in Fig. 3 is known to be v(t) = -e -00t + 0e -400t V, t 0 If = μf and L =.5 H, find il(0 + ) in milliamperes. Fig. 3 400 00 500 From equation, = = = 50 6 0 Then, R 000 50 When t = 0 +, v 0 0 9V, 9 So, i R (0 ) 9mA 000 00t 400 Also we can get, 00e 8000e When t = 0 +, (0 ) i (0 ) 00 8000 6900V/s, 6 (0 ) 0 ( 6900) 3.8mA t For KL, we can get i (0 ) [ i (0 ) i (0 )] (9 3.8) 4.8mA L R 4. [4%] In the circuit in Fig. 4, the switch has been in position for all t < 0, no initial energy in L and. At t = 0, the switch is moved to position (and remains there for t > 0). a) Find v(t) for t > 0, determine which damped case it is. b) Let R = 0 Ω, find v(t) again and determine the damped case. c) Let R = 0 Ω, find v(t) again and determine the damped case. d) Sketch v(t) at these 3 cases on graph with MATLAB. (Your MATLAB script should be attached, containing XY axis and grid!) 3
Electric ircuits Fall 05 Solution Fig. 4 a) Initial condition: v (0 ) v (0 ) 0, (0 ) (0 ) (0 ) 0 For t>0, the circuit is a series RL, apply KVL dil VS RiL L v () As il, substitute this into () d v R v VS, il il R, gives, 0 L L L L L 40 For R=40 Ω, 0 0 0, overdamped 0.0 So, 00 3t 00 3t v () t V Ae A e, f V, f Hence, 00 3t 00 3t v ( t) Ae A e, 00 3t 00 3t ( t) 0 0 3 Ae 0 0 3 A e, Substitute initial condition: and 0 0 0 3 A 0 0 3 0 A A A 3 3 Get A, A,so 3 3 3 00 3t 3 00 3t v ( t) ( ) e ( ) e V 3 3 0 b) For R=0 Ω, 0 0 0, critically damped 0.0 Hence, v ( t) Ate A e, 0t 0t Similarly, A 0, A, So, v ( t) 0te e 0 t 0t 4
Electric ircuits Fall 05 Solution c) For R=0 Ω, 0 0, under damped. Hence, v ( t) Acos0 t Asin0 t, Similarly, A, A 0, So, v ( t) cos0t d) (4%) 4 3.5 overdamped criticaldamped underdamped 3.5 电压 /V.5 0.5 0 0 0.5.5.5 3 3.5 4 4.5 5 时间 t/s 5. [0%] The switch in the circuit of Fig. 5 has been in position a for a long time. At t = 0 the switch moves instantaneously to position b. Find a) vo(0 + ) b) o(0 + )/ c) vo(t) for t 0. a) t < 0: 0 40 60 io(0 ) io (0 ) 6mA 4000 6000 0000 v (0 ) v (0 ) 0 6000 0.006 6V 5
Electric ircuits Fall 05 Solution t = 0 + : v (0 ) i (0 ) 6 36 0.006 000 6 4V, o L o v (0 ) 0 v (0 ) 4V o b) For KVL, v ( t) 000i v o o di o ( t) 000 o So, 3 o dio vl(0 ) io (0 ) 4 60 (0 ) 000 (0 ) (0 ) 000 000 03680V/s 9 L 0.5 78.5 0 c) 6 0.560 600rad/s R, 000rad/s L L For 0, circuit is overdamped. s, 000 000 600 000 00 So, v () t V Ae A e o f 800t 300t V ( ) 0V f vo, Apply initial condition: 0 A A 4, 800 A 300 A 03680 So, A., A 35. v t e e, t >0 + 800t 300 t o( ) 0. 35. V 6
Electric ircuits Fall 05 Solution 6. [0%] Given the network in Fig. 6, find v(t) for t > 0. For t = 0-, v v 0 0 4 8V Fig. 6 For t > 0, the circuit becomes that shown in Figure below after source transformation. So, / L / / 5 5 o L R/ 6 / 3 s, 0 3 4 v t Vs Acos4 t Bsin4t e, Vs V 3t Thus, v 0 8 A or A 4 i, so, j t i / / 3 Acos4 t Bsin4 t e 4 Asin4 t Bcos4t e 3 3t i 0 3A 4B B 3 Finally, vt 3t 4cos4t 3sin4t e A, t > 0 7. [0%] The switch in the circuit shown in Fig. 7 has been closed for a long time before it is opened at t = 0. Assume that the circuit parameters are such that the response is underdamped. a) Derive the expression for vo(t) as a function of Vg,,d,, and R for t 0. b) Derive the expression for the value of t when the magnitude of vo is maximum. 7
Electric ircuits Fall 05 Solution Fig. 7 a) Let i be the current in the direction of the voltage drop vo(t). Then by hypothesis, circuit is underdamped i i Ae cos t A e sin t t t f d d Easy to find, if i( ) 0, i(0) A R t t Therefore, i Ae cos Ae sin di(0) For L 0, di [( d A A )cos ( A d A )sin ] e So, d A A 0, A A d Therefore, V g Vg R d t Vg t So, vo( t) e sinv, t>0 R d 8
Electric ircuits Fall 05 Solution b) o () t Vg t e ( d cos sin ), R When d o () t 0 So, arctan d t d, get cos t sin t 0 d d d 8. [0%] Obtain i and i for t > 0 in the circuit of Fig. 8. Fig. 8 At t = 0 -, i (0 ) i (0 ) 0, i(0 ) i(0 ) 0 () di Apply KL, 4 i i () di di Also, i ( ) 3 (3) d i di di Take the derivative of (), 0 (4) di di di From () and (3), 3i.5 3i Subsituting this into (4), get d i di, which gives 7 6i 4 s s s s 7 6 0 ( )( 6) Thus, i t I Ae Be I t 6t ( ) s [ ], s 4 / 6 4A i t Ae Be, and i (0) 4 A B (5) ( ) 4 [ t 6t ] di i 4 i 4 4 [ Ae Be ] 0.5[ Ae 6 Be ] 0.5Ae Be For i (0) 0.5 A B 0, combine with (5) t 6t t 6t t 6t 9
_ Electric ircuits Fall 05 Solution Get A 3., B 0.8, so i ( t) 4 3.e 0.8e t 6t, i ( t).6e t.6e 6t 9. [0%] The circuit in in Fig. 9 is the electrical analogue of a temperature control system. Fig. 9 Assuming A = F, B = 4 F, RA = Ω, RB = 4 Ω. is = K(V0 vb) where K = 5 A/V, V0 =. V. a) Write dynamical equations for this network in state form. Use va and vb as state variables. (As a check on your state equations, the stable steady-state value of vb is V. That is, you should have A/ = B/ = 0 for vb = V.) b) Now assume va = VA + va and vb = VB + vb, where VA and VB are the steady-state values and va and vb are small variations. Determine and draw a small-signal linear circuit model in which va and vb are the state variables. c) Is the zero-input response of the small-signal circuit under-damped, over-damped, or criticallydamped? a) Use KL for two node equations: v v A K V v A A B ( 0 B), RA v v v R R B B B A B B A b) Since i K V v, then the following small-signal approximation is valid: S 0 B is v b dis K( V0 vb) K( V0 VB), B See figure below for a small-signal model. i K( V V ) v. s 0 B b 0
Electric ircuits Fall 05 Solution c) First, write two new state equations using the small-signal model: v v v v v K( V v ) v, R R a a b A 0 B b RA We substitute in the numerical values given, and get the following: Get V V B from(a), then Eliminate va, getting the following: a va vb 5vb, b 4 0.5vb va vb. d vb This has the following characteristic equation: b 6 v b 0 6s s 0 Since 46 0, the system is underdamped. 0. [4%] Draw the dual of the circuit in Fig. 0. B b b a b B A Fig. 0
Electric ircuits Fall 05 Solution onstruct, Redraw:. [0%] a) Derive the differential equation that relates the output voltage to the input voltage for the circuit shown in Fig.. Fig. b) ompare the result with Eq. when R = R = R in Fig.. o vg () R R
Electric ircuits Fall 05 Solution Fig. c) What is the aantage of the circuit shown in Fig.? a) For KL on Va, V- and Vb va v a g va a v v a g 0 () R R R R d(0 vb ) 0 va b va 0 0 () R R b d( vb vo ) vb b vb o 0 (3) R R From (), v R b a and a d vb R Substitude into (), get d vb v b g d vb v b g R R R R (4) Differentiate (3), get d vb b d vo R (5) Substitute (4) into (5), We get o v g R 3
Electric ircuits Fall 05 Solution b) When R R R, o v g R The two equations are the same except for a reversal in algebraic sign. c) Two integrations of the input signal with one operational amplifier. 4