MAE 40 inear ircuit Fall 2007 Final Intruction ) Thi exam i open book You may ue whatever written material you chooe, including your cla note and textbook You may ue a hand calculator with no communication capabilitie 2) You have 70 minute 3) On the quetion for which we have given the anwer, pleae provide detailed derivation Quetion Equivalent ircuit A B Figure : ircuit for Quetion (i) A + vc(0) B Figure 2: ircuit for Quetion (ii) Part (i) [5 mark] Auming zero initial condition, find the impedance equivalent to the circuit in Figure a een from terminal A and B The anwer hould be given a a ratio of two polynomial ompute [ ( Z() + // + )] + ( ) + 2 + ( + ) + 2 + (2 + ) + ( + ) 2 + 22 + ( + 2 ) + 2 + Part (ii) [5 mark] Auming that the initial condition of the capacitor i a indicated in the diagram, redraw the circuit hown in Figure 2 in -domain Then ue ource tranformation to find the -domain Thévenin equivalent to the circuit a een from terminal A and B
A A A / vc(0) Z() vc(0) Z() + V() B B B Figure 3: Sequence of Tranformation Quetion (ii) Firt tranform the circuit to the -domain to include the current ource I() v c (0) in parallel with /( ) Aociate the capacitor and reitor into the impedance Z() + + The Thévenin equivalent circuit i obtained after tranforming the current ource into a voltage ource V () Z()I() v c(0) + Thi equence of tranformation i hown in Figure 3 Quetion 2 aplace domain circuit analyi Figure 4: circuit for aplace analyi Part (i) [3 mark] onider the circuit depicted in Figure 4 The current ource i i a contant current upply, which i kept in place for a very long time until the witch i opened at time t 0 Show that the initial capacitor voltage i given by [Show your working] v (0 ) i 2 With a contant input and the witch cloed for a very long time, the circuit reache teady tate In teady tate, the circuit i decribed by all time derivative of ignal being zero In particular, dv (t) dt 0 Therefore, the teady tate current through the capacitor i zero, a i the current through the reitor So, all of the current i flow through reitor 2, yielding voltage acro that reitor of 2 i volt Since there i no current through, all of thi voltage drop appear acro capacitor Hence, v (0 ) i 2
Part (ii) [2 mark] Ue thi initial condition to tranform the circuit into the -domain for time t 0 [The ymbolic formulæ are hown in the text on page 449] [Show your working] With the initial voltage from Part (i), redraw the circuit for time t 0 a below Note that Figure 5: domain circuit the capacitor voltage V c () include both the equivalent impedance condition ource v (0 ) and the erie initial Part (iii) [5 mark] Ue -domain circuit analyi and invere aplace tranform to how that the capacitor voltage atifie, ( ) t v (t) i 2 exp u(t) ( + 2 ) [Show your working] The capacitor voltage V c () i the ame a the voltage acro the + 2 reitor pair Ue the voltage divider formula to yield V c () Hence, taking invere aplace tranform, ( v (t) i 2 exp + 2 + 2 + + 2 v (0 ), ( + 2 ) + i 2, i 2 + ( + 2) t ( + 2 ) ) u(t) Quetion 3 Active Filter Analyi and Deign Part (i) [3 mark] Show that the the tranfer function of the op-amp circuit in Figure 6 are given by T () 2 +, T () 3 4 + 3
Figure 6: (a) parallel op-amp circuit (b) parallel op-amp circuit The parallel impedance i given by Z () + + The op-amp circuit i in the inverting amplifier configuration and o it tranfer function i given by the ratio of impedance, with Z () in the numerator poition Hence, For the parallel impedance, T () 2 + Z () 3 3 + 3 + 3 We alo have an inverting op-amp configuration So, T () 3 4 + 3 Part (ii) [5 mark] Showing your reaoning, determine the nature of thee two filter frequency repone Further, determine the gain of the filter and their cut-off frequencie If 00nF, find value o that the filter ha cutoff frequency 5KHz and gain 5 If 0mH, find value o that the filter ha cutoff frequency KHz and gain 5 To generate the frequency repone, replace the aplace variable by jω, where ω i the radian frequency Thu, T (jω) 2 jω + Now compute the magnitude of thi frequency repone T (jω) jω + 2 2 ω 2 + () 2, learly the denominator trictly increae with frequency ω, while the numerator tay fixed Therefore the filter ha a frequency repone gain which decreae with frequency We have at low frequencie, lim ω 0 T (jω) 2
Thi i the reitive op-amp gain without the capacitor, becaue at low frequencie the capacitor i like an open circuit For very high frequencie, we have lim T (jω) 0 ω Therefore the filter i a low-pa filter The cut-off frequency i given by pole value, which i where the gain i 2 2 ω cutoff For the low-pa filter, the gain i given by T (0) 2 Similarly, for the filter T (jω) jω 3 4 jω + 3, 3 ω 4 ω2 + ( 3 /) 2 We have lim T (jω) 0, ω 0 and o i a high-pa filter The cutoff frequency i lim T (jω) 3, ω 4 ω cutoff 3 For the high-pa filter, the gain i given by lim ω T (jω) 3 4 To olve for the cut-off and gain value 5000 2π, or 38Ω, 3 000 2π, or 3 63Ω, 2 5, or 2 64Ω, 3 4 5, or 4 6Ω Part (iii) [2 mark] You are required to build a bandpa filter with; lower cut-off frequency of KHz, upper cut-off frequency of 5KHz, and paband gain of 25 Explain whether the two filter above can be ued to achieve thi deign pecification If o, then how how If not, then ugget how the above calculation might be changed to yield uch a deign To deign a bandpa filter a pecified, we need to reject high frequencie above 5KHz we will ue the low-pa filter to do thi and to reject low frequencie below KHz, for
which we will ue the high-pa filter Therefore, we have already deigned thee filter The bandpa filter i the cacade of the high-pa filter and the low-pa filter The order i unimportant The paband gain i then the product of the gain of the low-pa and high-pa filter Thi i achieved by thi deign G 2 3 4 Quetion 4 Op-Amp Analyi and Application The figure below how a circuit known a a gyrator or poitive impedance inverter It i primarily ued in active circuit deign to implement inductor, which are difficult to manufacture to pecification and within a mall volume Figure 7: (a) Gyrator circuit (b) -circuit Part (i) [5 mark] Uing the fundamental op-amp relationhip how that the voltage tranfer function in Figure 7(a) i given by T v () V o() V i () + Becaue the current into the p-terminal of the op-amp i zero, the, combination i a voltage divider on the input voltage v i Therefore, V p () + V i (), + ikewie, uing the op-amp relation v n v p and the property that the output terminal i directly connected to the n-terminal V o () V n () V p () + V i ()
Part (ii) [2 mark] Perform the ame calculation for the -circuit hown in Figure 7(b) to how that it tranfer function i T v () + Thu, the gyrator circuit effectively implement a circuit involving an inductor by uing a capacitor and an op-amp Determine the equivalent inductor value in the -circuit in term of the element value,,,, in the gyrator circuit For the -circuit we alo have a voltage divider Thu, directly we have Thi i the ame a for Part (i) with T v () + +, or Part (iii) [3 mark] Show that the equivalent impedance of the gyrator circuit een from the input, v i, i given by while, for the -circuit it i given by Z eq () +, + Z eq () + omputing the current i i flowing into the gyrator circuit in repone to the applied voltage v i, we have I i () (V i () V o ()) + cs(v i () V o ()), ( ) ( ) + V i () + V i (), ( ) ( ) + + V i (), ( + ) + V i () Inverting the relationhip to compute the impedance yield The -circuit reult i immediate Z eq () V i() I i () + + Part (iv) [3 mark, Bonu] Decribe the limitation in uing thi ubtitute circuit to realize an inductor in an application The main limitation to uing thi circuit and indeed it i ued very widely are a follow (a) The circuit only imulate an -circuit with the end of the inductor grounded In term of the poible filter application, it i only the differential amplifier (ubtractor) form which make ene Thi i a high-pa filter if the grounded element i an inductor
(b) Becaue the impedance of the gyrator circuit doe not match that of an -circuit, one cannot ue the gyrator a a drop-in replacement If one need an inductor a an impedance element, then thi circuit doe not achieve thi If the impedance i important to the operation of the input ide, then different approache need to be taken One might typically protect the input ide of the thi circuit with a voltage follower (c) Becaue the inductor i implemented a an op-amp circuit, all the limitation of opamp apply Notably the limit on the output voltage to the upply value to the op-amp i important Normally in operation, a phyical inductor i able to utain very high voltage due to rapidly changing current o-called flywheel effect The gyrator cannot do thi Quetion 5 Power Factor ompenation i(t) + v(t) Figure 8: Motor equivalent circuit Part (i) [3 mark] A large ingle phae alternating current (A) induction electric motor can be repreented by the circuit in Figure 8 Auming zero initial condition, find the equivalent impedance of the motor, Z(), and the equivalent admitance, Y () Z() ompute Y (jω), it real and imaginary part Thi i the admittance of the motor at the A upply frequency of ω radian per econd ompute the erie aociation of and, that i Z() +, Y () /( + ) Then Y (jω) + jω 2 + ω 2 2 + j ω 2 + ω 2 2 i(t) + v(t) Figure 9: Motor equivalent circuit with capacitor in parallel Part (ii) [3 mark] Now conider the circuit in Figure 9, in which a capacitor i added in parallel to the motor Auming zero initial condition, find the new equivalent admitance Y () ompute Y (jω), it real and imaginary part Ue thi to how that may be choen to make the imaginary part of the admittance zero at the upply frequency ω
ompute and then Y () + + Y (jω) + jω, + jω jω 2 + ω 2 + jω, 2 2 + ω 2 2 + jω ( ) 2 + ω 2 2 hooe 2 +ω 2 2 to have zero imaginary part Part (iii) [4 mark] If the upply voltage to the motor i v(t) V co(ωt), write the correponding inuoidal teady tate current i(t) drawn by the machine with admittance Y () Do thi in the time domain a a co function ompute the power p(t) v(t)i(t) and ue the relationhip co(a) co(b) 2 [co(a + B) + co(a B)], to derive an expreion for the power aborbed by the motor a a function of time Show that the average value of thi power contain a term in e(y (jω)) Y (jω) co[ (Y (jω))] If a capacitor i choen a in the previou item, doe the introduction of thi capacitor in parallel with the motor change the average power aborbed by the motor? Why? Becaue I() Y ()V (), by the frequency repone formula, i (t) V Y (jω) co(ωt + (Y (jω))) The power i given by the product above, whence p(t) V 2 Y (jω) co(ωt) co(ωt + (Y (jω))), V 2 2 Y (jω) [co(2ωt + (Y (jω))) + co( (Y (jω)))] The average of the firt term i zero, ince it i a co function The average of the econd term i < p > V 2 2 Y (jω) co( (Y (jω))) V 2 2 e(y (jω)) Becaue the introduction of the capacitor doe not change the real part of Y (jω), ie e(y (jω)) e(y (jω)), it will not affect the average power aborbed by the motor Part (iv) [3 mark, Bonu] Your power upplier (SDG&E and the like) will likely demand that you inert a capacitor a above They will enforce thi by meauring co[ (Y (jω))], aka power factor, and gently aking you to keep thi number a cloe to one a poible, like you did by chooing above Why do you think thee nice people would do that? A hown in the previou quetion, the capacitor will not affect the average power aborbed by the motor, that i but it will affect the magnitude of the current < p > V 2 2 V e(y (jω)) 2 2 e(y (jω)) i (t) V Y (jω) co(ωt + (Y (jω)))
Note that the power lot in the tranmiion wire p lo (t), ay with reitance lo, i which average to p lo (t) lo i 2 (t) 2 lov 2 Y (jω) 2 [co(2ωt + Y (jω)) + ] < p lo > 2 lov 2 Y (jω) 2 onequently, becaue 2 + ω Y (jω) 2 2 2 + ω 2 2 2 + ω 2 2 Y (jω), your power upplier will prefer to power a motor that behave like Y (jω) a compared to Y (jω), ince it can only charge for the power effectively delivered The power factor co( (Y (jω)) quantifie then a meaure of efficiency for the power company, relating the power it deliver and charge, proportional to Y (jω) co( (Y (jω)), with the power lot in the tranmiion, proportional to Y (jω) 2