MAE140 - Linear Circuits - Winter 16 Final, March 16, 2016

Transcription

1 ME140 - Lnear rcuts - Wnter 16 Fnal, March 16, 2016 Instructons () The exam s open book. You may use your class notes and textbook. You may use a hand calculator wth no communcaton capabltes. () You have 180 mnutes () Do not forget to wrte your name and student number (v) On the questons for whch the answers are gven, please provde detaled dervatons (v) The exam has 5 questons for a total of 50 ponts and 2 bonus ponts Good luck! a v x (t) L v x Fgure 1: rcut for Queston Equvalent rcuts Part I: [2 ponts] ssumng zero ntal condtons, transform the crcut n Fgure 1 nto the s-doman. oluton: nce all ntal condtons are zero, t s easy to transform the crcut to the s-doman. a V x (2 ponts) I(s) V x

2 Part II: [4 ponts] Fnd the open-crcut voltage n the crcut obtaned n Part I (as seen from termnals and ). e mndful of the presence of the dependent source. oluton: a V x oth current sources are n parallel wth mpedances, so we use source transformaton twce to get the crcut on the rght I(s) The open-crcut voltage V (s) s exactly the voltage drop seen by the vertcal mpedance. y voltage dvson, we deduce that V x V (s) = 3 + (I (s) av x (s)) We need to determne what the value of V x (s) s (.e., we need to handle the dependent source). Ths s n fact easy to do by notng that, by voltage dvson, V x (s) = 3 + (I (s) av x (s)) (In fact, V x (s) = V (s)). olvng for V x, we obtan Therefore, the open-crcut voltage s V x (s) = 3 + a 2 + I (s) V (s) = 3 + a 2 + I (s) Part III: [4 ponts] Fnd the s-doman Thévenn equvalent of the crcut obtaned n Part I as seen from termnals and. e mndful of the presence of the dependent source. oluton: We need to fnd V T and Z T. It turns out that we computed the open-crcut voltage n Part II, so n fact we know that V T (s) = 3 + a 2 + I (s) To fnd the Thévenn mpedance, we need to compute the short-crcut current (turnng off sources s not an opton because of the presence of the dependent source) Page 2

3 The crcut then looks lke the plot on the rght a V x I(s) I(s) sc V x a V x The same source transformatons as we dd n Part II get us the equvalent crcut on the rght I(s) I(s) sc V x nce the vertcal mpedance s n parallel wth a short crcut, t does not play any role (no current flows through t). Therefore, we are actually dealng wth I(s) a V x I(s) sc Therefore, the short-crcut current can be expressed as V x I sc (s) = I (s) av x (s) 2 + s before, we need to determne V x (.e., handle the presence of the dependent source). From the crcut, and usng voltage dvson, we see that V x (s) = 2 + (I (s) av x (s)) olvng for V x, we obtan V x (s) = Therefore, the short-crcut current s 2 + a 2 + I (s) I sc (s) = I (s) av x (s) 2 + = 2 + a 2 + I (s) Page 3

4 Fnally, the Thévenn mpedance s Z T (s) = V T (s) I sc (s) = (2 + a2 + ) 3 + a 2 + V(s) T Z(s) T Page 4

5 1 1 (t) v(0)=v v (t) L D (0)=0 L Fgure 2: Nodal and Mesh nalyss rcut for Queston Nodal and Mesh nalyss Part I: [5 ponts] Formulate node-voltage equatons n the s-doman for the crcut n Fgure 2. Use the reference node and other labels as shown n the fgure. Use the ntal condtons ndcated n the fgure and transform them nto current sources. Make sure your fnal answer has the same number of ndependent equatons as unknown varables. No need to solve any equatons! oluton: 1 v 1/s I (s) V (s) 2 D Page 5

6 In the above fgure, we have transformed the crcut nto the s-doman, takng good care of respectng the polarty of the capactor. nce the ntal condton of the nductor s zero, no need to worry about addng a source for t. (1 pont for correct crcut; 1 pont for correct ntal condton) For nodal analyss, the presence of the voltage source poses a problem. However, the choce of ground provdes a soluton for t. In fact, we have V D = 0 and V (s) = V (s) (ths s method #2). We next wrte KL node equatons for nodes and. For convenence, we use the shorthand notaton G 1 = 1/ 1 and G 2 = 1/ 2. For node, we have s(v (s) V (s)) + G 2 V (s) = v I (s) For node, we have G 1 (V (s) V (s)) + 1 V (s) = I (s) Ths gves a total of 3 ndependent equatons n 3 unknowns (V (s), V (s), V (s)). lternatvely, one can take the expresson for V (s) and substtute t n the other equatons to arrve at 2 ndependent equatons n 2 unknowns (V (s), V (s)). Part II: [5 ponts] Formulate mesh-current equatons n the s-doman for the crcut n Fgure 2. Use the currents shown n the fgure. Use the ntal condtons ndcated n the fgure and transform them nto voltage sources. Make sure your fnal answer has the same number of ndependent equatons as unknown varables. No need to solve any equatons! oluton: 1 I 1 1/s v /s I (s) V (s) I 2 I 3 2 In the above fgure, we have transformed the crcut nto the s-doman, takng good care of respectng the polarty of the capactor. gan, no need to worry about the ntal condton of Page 6

7 the nductor because t s zero. (1 pont for correct crcut; 1 pont for correct ntal condton) For mesh-current analyss, the presence of the current source s a problem that must be dealt wth. In ths case, we need to use a supermesh (because the current source s not n parallel wth an mpedance and because t belongs to more than one mesh). Therefore, we set I 3 (s) I 1 (s) = I (s) (ths s method #3). KVL for the supermesh looks lke 1 I 1 (s) + I 3 (s) + 2 (I 3 (s) I 2 (s)) + 1 s (I 1(s) I 2 (s)) v s = 0 For mesh 2, KVL takes the form 2 (I 2 (s) I 3 (s)) V (s) + 1 s (I 2(s) I 1 (s)) + v s = 0 Ths gves a total of 3 ndependent equatons n 3 unknowns (I 1 (s), I 2 (s), I 3 (s)). Part III: [1 bonus pont] Express the transform of the capactor voltage n terms of your unknown varables of Part I and also n terms of your unknown varables of Part II. oluton: We just need to be careful to not lose track of the transform of the capactor voltage. In the case of Part I, because we use a current source to account for the ntal condton, we have V capactor (s) = V (s) V (s) (.5 bonus pont) In the case of Part II, because we use a voltage source to account for the ntal condton, we actually have V capactor (s) = 1 s (I 2(s) I 1 (s)) + v s (.5 bonus pont) Page 7

8 /2 /2 v v(t) o L v(t) v(0)=0 - + Fgure 3: L crcut for Laplace nalyss for Queston Laplace Doman rcut nalyss Part I: [2 ponts] onsder the crcut depcted n Fgure 3. The value v of the voltage source at the top s constant. The swtch s kept n poston for a very long tme. t t = 0 t s moved to poston. how that the ntal condton for the nductor s gven by [how your work] L (0 ) = v. oluton: To fnd the ntal condton, we substtute the nductor by a short crcut. [1 pont for correct crcut; 1 pont for substtutng nductor by short crcut] /2 /2 L v Therefore, we deduce that L (0 ) = v Part II: [4 ponts] Use ths ntal condton to transform the crcut nto the s-doman for t 0. Use an equvalent model for the nductor n whch the ntal condton appears as a voltage source. Page 8

9 Do you recognze the resultng crcut as one of the basc op-amp buldng blocks? Express the output response transform V o (s) as a functon of V (s) and v. oluton: We add one voltage source n seres for the nductor to take care of ts ntal condton, payng specal attenton to the drecton of the current. No need to worry about the ntal condton of the capactor because t s zero. /2 V(s) o Lv / - + V(s) 1/s (1 pont for correct crcut; 1 pont for correct polarty) We recognze ths crcut as one of the basc op-amp buldng blocks. In fact, ths s a dfferental amplfer. Gven the above, the output response transform s V o (s) = 1/s 3/2 + 1/s + /2 + V L (s) /2 + v 3 + 2Ls = (1 + s)( + 2Ls) V 2L (s) + 2Ls v Part III: [4 ponts] Use partal fractons and nverse Laplace transforms to show that the output voltage v o (t) when v = 1 V, v (t) = e 1500t u(t) V, = 1 mf, L = 1 mh, and = 1 Ω s v o (t) = (e 500t 2e 1000t )u(t). oluton: From our answer to Part II, and substtutng the values for the mpedances and the sources, the Laplace transform of the output voltage s s 1 V o (s) = ( s)( s) s s 1000 = ( s)(500 + s) s Usng partal fractons, we get V o (s) = s s s Page 9

10 Usng the resdue method to compute and, we obtan 2 V o (s) = s s s = s s The output voltage s then v o (t) = (e 500t 2e 1000t )u(t) Page 10

11 v(t) v(t) o Fgure 4: Frequency esponse nalyss for Queston Frequency esponse nalyss Part I: [1 pont] ssumng zero ntal condtons, transform the crcut n Fgure 4 nto the s-doman. oluton: nce all ntal condtons are zero, there s no need to add an ndependent source for the capactors. Therefore, the crcut n the s-doman looks lke V(s) 1/s 1 1/s 2 V(s) o - + Part II: [2 ponts] how that the transfer functon from V (s) to V o (s) s gven by T (s) = V o(s) V (s) = 1s s. [how your work] Page 11

12 oluton: Ths s clearly an nvertng op-amp. Therefore the output response transform can be expressed as from whch the answer follows. V o (s) = 1/s 2 1/s 1 V (s) = 1s s V (s). Part III [5.5 ponts] Let = 1 kω, 1 = 20 µf and 2 = 1 µf. ompute the gan and phase functons of T (s). What are the D gan and the -freq gan? What are the correspondng values of the phase functon? What s the cut-off frequency ω c and ts phase? ketch plots for the gan and phase functons. What type of flter s ths one? [Explan your answer] oluton: For the gven values of, 1 and 2, the transfer functon takes the form T (s) = s s = 20s s The frequency response s then the complex functon Its magntude s the gan functon, T (jω) = 20jω jω, ω 0 T (jω) = 20jω jω = 20ω ω 2 nd ts phase s T (jω) = ( 20jω) ( jω) = 3π ( ω ) 2 arctan 1000 t ω = 0, we obtan T (j0) = 0, T (j0) = 3π 2 rad (correct D-gan gets.5 pont, correct phase gets.5 pont) t ω =, we obtan The cut-off frequency s defned by T (j ) = 20, T (j ) = π rad (correct -freq gan gets.5 pont, correct phase gets.5 pont) T (jω c ) = T max 2 = olvng for t, we fnd ω c = 1000 rad/s. The phase at ω c s T (j1000) = 5π 4 rad. Wth the values obtaned above, you can sketch the magntude of the frequency response as Page 12

13 Magntude (d) ode Dagram Phase (deg) Frequency (rad/s) Ths crcut s a hgh-pass flter. (.5 pont for gan plot,.5 pont for phase plot) Part IV [1.5 ponts] Usng what you know about frequency response, compute the steady state response vo (t) of ths crcut when v (t) = 1 4 cos(500t + π 4 ) usng the same values of, 1, and 2 as n Part III. oluton: To compute the steady-state response to the nput v (t) = 1 4 cos(500t + π 4 ), we use the frequency response values for ω = 500, Therefore, T (j500) = 4 5 T (j500) = 3π 2 arctan ( 1) rad. 2 v o (t) = 1 4 T (j500) cos ( 500t + π 4 + T (j500) ) (1 pont for correct expresson) = 5 cos (500t ) (.5 pont for correct values) Page 13

14 L v(t) v(t) o Fgure 5: rcut for Queston Loadng and the han ule onsder the crcut n Fgure 5. Part I: [2 ponts] ssumng zero ntal condtons, plot the crcut n the s-doman. Fnd the transfer functon T (s) and determne ts poles and zeros assumng > 2 L/. oluton: nce there are no ntal condtons to take care of, the crcut n the s-doman looks lke V(s) 1/s V(s) o We can easly determne the transfer functon by usng voltage dvson. T (s) = V 1 o(s) V (s) = s 1 s + + = s + Ls 2 There are no fnte zeros. The transfer functon has two zeros at s =. The poles are of T are p 1 = + () 2 4L 2L nce > 2 L/, both poles are real and negatve., p 2 = () 2 4L. 2L Part II: [3 ponts] student wth a rusty recollecton of ME140 connected two dentcal copes of ths crcut n seres and was surprsed to observe that the transfer functon of the resultng crcut s not T (s) T (s) = T (s) 2. ould you explan to hm why ths s so and suggest an easy way to solve the problem so that he obtans T (s) 2 as the transfer functon of the resultng crcut? Properly justfy your answer. oluton: The problem s loadng. If we connect two exact copes of the crcut n Fgure 5 n seres, stage 2 wll draw current from stage 1, and load t, nvaldatng the chan rule. Page 14

15 n easy way to solve the problem s to add a voltage follower n between the two copes of the crcut n Fgure 5, as depcted on v(t) the rght. The nfnte nput mpedance of the op-amp avods loadng stage 1. The zero output mpedance of the op-amp avods stage 3 loadng stage 2. Therefore, the chan rule apples. nce the transfer functon of the voltage follower s 1, the overall transfer functon s T (s) 1 T (s) = T (s) 2. L + - L v(t) o Part III: [2 ponts] We could help the student of Part II even further by desgnng a dfferent crcut wth the same transfer functon that does not have the same problem. To make the computatons concrete, let = 100 Ω, = 10 µf and L = 5 mh, and compute the correspondng numercal values for the poles of the transfer functon T (s). Next, wrte T (s) as a product of the form T (s) = s + α s + β and determne approprate values for > 0, > 0, α > 0, and β > 0. oluton: Wth the values of,, and L, the transfer functon can be wrtten as The poles are then T (s) = s s 2 = s + s 2 p 1 = 10 4 ( 1 +.8) , p 2 = 10 4 ( 1.8) Therefore, s + s 2 = (s p 1 )(s p 2 ). We use ths nformaton to factorze the denomnator of the transfer functon as T (s) = s p 1 s p 2 o we choose α = p 1 and β = p 2. ny and whose product s wll do, for nstance, = and = Part IV: [3 ponts] Desgn an nvertng op-amp crcut whose transfer functon s s+α and another nvertng op-amp crcut whose transfer functon s s+β, wth the numercal values for,, α, β you chose n Part III. What s the transfer functon of the seres connecton of these two nvertng op-amps? oluton: There are several ways to do ths. For nstance, to get s+α, we use an nductor and two resstors together wth the op-amp, lke ths Page 15

16 L wth 1 = α Ohms, 2 = Ohms, and L = 1 H. To get 1+β/s, we change our desgn and use two capactors and one resstor together wth the op-amp, lke ths s+β = /s wth = 1 Ohms, 1 = 1/β F, and 2 = 1/ F. The transfer functon of the seres connecton of the two nvertng op-amps s the product of both (there s no loadng snce stage 2 s connected to the output of an op-amp n stage 1), whch s precsely T (s)! Part V: [1 bonus pont] Let us call crcut the seres connecton of the two nvertng op-amps that you obtaned n Part IV. Would the student wth the rusty recollecton of ME140 have the same problem he had n Part II f he were to connect two dentcal copes of crcut n seres? Why? oluton: No, he would not, because the zero output mpedance of the op-amp avods loadng. o f the student connected two dentcal copes of crcut n seres, he would obtan a crcut whose transfer functon s n fact T (s) T (s) = T (s) 2, as he orgnally ntended. Page 16