MAE40 Linear Circuit Fall 202 Final, December 3th 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. Do not forget to write your name and tudent number. 4. Thi exam ha 5 quetion, for a total of 50 point and 6 bonu point.. Circuit Equivalence In thi quetion you will find condition for the circuit in Figure (a) and (b) to be equivalent. Thi i known a the Y equivalence. In the quetion below, AB i the equivalent reitance a een from terminal A and B, and o on. Anwer the following quetion: (a) (3 point) For the circuit in Figure (a), how that AB = + 2, BC = 2 + 3, AC = + 3. Calculating AB, the reitance 3 ee no current and therefore can be taken out of the circuit. Conequently and 2 are in erie and AB = + 2. Calculating BC, the reitance ee no current and therefore can be taken out of the circuit. Conequently 2 and 3 are in erie and BC = 2 + 3. Calculating AC, the reitance 2 ee no current and therefore can be taken out of the circuit. Conequently and 3 are in erie and AC = + 3. (b) (3 point) For the circuit in Figure (b), how that AB = B( A + C ) A + B + C, BC = C( A + B ) A + B + C, AC = A( B + C ) A + B + C. A 2 B A B B 3 D 3 A C 2 3 3 C C E (a) Y -circuit (b) -circuit (c) Bridge circuit Figure : Circuit for Quetion.
Calculating AB, the reitance B i in parallel with the erie aociation of A and C, hence AB = B( A + C ) B + ( A + C ) = B( A + C ) A + B + C Calculating AB, the reitance C i in parallel with the erie aociation of A and B, hence AB = C( A + B ) C + ( A + B ) = C( A + B ) A + B + C Calculating AC, the reitance A i in parallel with the erie aociation of B and C, hence AB = A( B + C ) A + ( B + C ) = A( B + C ) A + B + C NOTE TO GADE: Aign full point if a tudent work out in detail one cae and correctly indicate the other. (c) (2 point (bonu)) Show that if the to Y formula: = A B A + B + C, 2 = B C A + B + C, 3 = A C A + B + C hold then the two circuit in Figure (a) and (b) are equivalent. Hint: Note that in the Y circuit: 2 = AB + AC BC. From the Y circuit: and therefore AB = + 2, BC = 2 + 3, AC = + 3. which i the hint. From the circuit: Hence 2 = AB + AC BC 2 = AB + AC BC = B( A + C ) + A( B + C ) C( A + B ) A + B + C A + B + C A + B + C 2 A B =. A + B + C = A B A + B + C. A imilar proof work for the other two reitance becaue 2 2 = AB + BC AC, 2 3 = BC + AC AB. NOTE TO GADE: Aign full point if a tudent work out in detail one cae and correctly indicate the other. Page 2
(d) (4 point) Ue the to Y formula from part (c) to find the equivalent reitance a een from terminal D and E in the bridge circuit of Figure (c). Identify a circuit in the original circuit, for example: B =3 D 2 A =3 C=3 in which A = B = C = 3. The equivalent Y circuit i one where E = 2 = 3 = 92 9 =. that i = 2 = D 2 3= The equivalent reitance i therefore E eq = + 3 2 = + = + 62 5 (3) (2) 3 + 2 =. 5 Page 3
v A i c A 0V i a v B L i i b C v C Figure 2: Circuit for Quetion 2. 2. Nodal and Meh Analyi In your anwer to thi problem, clearly indicate the voltage, current and equation that need to be olved in each part. There i no need to actually olve any equation for thi exam. All ource are contant. The initial current through the inductor i zero. The initial voltage acro the capacitor i v C (0) = 0V. (a) (4 point) Formulate node-voltage equation in the -domain for the circuit in Figure 2. Ue the reference node and other label a hown in the figure. Ue the initial condition provided. Tranform initial condition into current ource. Tranforming the circuit in the -domain we obtain V A () 0V V B () L I() A V C () 0 C By inpection at node V B and V C : 2 + L 0 L V A () ( ) L 0 V B () = 0. L + V C () + 0 C along with the equation determining voltage V A (): (+2 point) V A () = 0V. Page 4
(b) (4 point) Formulate meh-current equation in the -domain for the circuit in Figure 2. Ue the meh current hown in the figure. Ue the initial condition provided. Tranform initial condition into voltage ource. Tranforming the circuit in the -domain we obtain 0V I a () I c () L I() I b () A 0 By inpection at mehe I a and I b : [ ] 0 2 I a () + L + L I b () = I c () 0 (+2 point) along with the equation determining the meh current I c (): I c () = 0.A. (c) (2 point) Expre the -domain current I(), indicated in Figure 2 in the time domain a i, in term of the -domain node voltage and the meh current. You do not need to olve any equation. I() = V B() V C (), L I() = I b () I c (). 3. Frequency epone and OpAmp Circuit A tudent built a tranitor pre-amplifier to connect hi guitar to a Digital Signal Proceor (DSP). The input voltage to the DSP hould be between 0V and 0V. The output of the tudent pre-amplifier i (t) = 0V + v x (t) where v x (t) i the pre-amplified ound coming from hi guitar. The ignal v x (t) varie from V to V and only have component in frequencie higher than 20Hz. In order to maximize Page 5
C v o v o 9 9 (a) Solution #0 (b) Solution # C v o C 0 9 v o (c) Solution #2 (d) Solution #3 Figure 3: Circuit for Quetion 3. the reolution of the DSP, the tudent want to build another amplifier uing OpAmp that can amplify only the ignal v x (t) by multiplying it by a factor of 0. That i, the output of the OpAmp circuit hould be v o (t) = 0 v x (t). The tudent tried Solution #0 in Figure 3(b) but it did not work. A friend uggeted that he added a capacitor to block the DC component from (t). That i Solution #, which alo did not work. Other friend uggeted variation of thi idea, which are Solution #2 and Solution #3. Anwer the following quetion for each of the circuit in Figure 3: (a) (2 point) Aume zero initial condition and convert the circuit into the -domain. The circuit in the -domain are: Solution #0: V () 9 V o () Solution #: V () V o () (+0.5 point) 9 Solution #2: (+0.5 point) Page 6
V () 9 V o () Solution #3: V () 0 (+0.5 point) V o () (+0.5 point) (b) (2 point) Compute the output voltage V o () in term of the input voltage V () and the correponding tranfer function G() = V o ()/V (). Solution #0. Thi i a tandard non-inverting amplifier where V o () = 9 + V () = 0V (), G() = V o() V () = 0. (+0.5 point) Solution #. Becaue I + () = 0 the voltage acro the capacitor i zero and V + () = V (). From thi point on thi i a tandard non-inverting amplifier where V o () = 9 + V +() = 0V (), G() = V o() V () = 0. (+0.5 point) Solution #2. Becaue I + () = 0 the voltage V + () can be computed uing voltage diviion V + () = + V () = + V () C From thi point on thi i a tandard non-inverting amplifier where V o () = 9 + V +() = 0 + V (), C G() = V o() V () = 0 +. (+0.5 point) C Solution #3. Thi i a tandard inverting amplifier where V o () = 0 + V () = 0 + V (), C G() = V o() V () = 0 +. (+0.5 point) C Page 7
(c) (3 point) Sketch the magnitude plot of the frequency repone, that i G(jω), a a function of ω. Solution #0 and # have the ame tranfer function G() = 0 which i contant for all frequencie, i.e. G(jω) = 0. A ketch of the repone i: G(jω) 0 ω Solution #2 and #3 have tranfer function of the form G() = ±0 + ω c, ω c = C, which have the ame magnitude repone G(jω) = 0 ω ω 2 + ωc 2 Thee are high-pa filter with cut-off frequency ω c = /(C) and gain equal to 0. A ketch of the repone i: G(jω) 0 7 ω c ω Then anwer thee quetion for ome bonu point: (d) ( point) Explain why Solution #0 doe not work? Becaue the DC part of alo get amplified. The output would be omething like: v o (t) = 00V + 0 v x (t), which i not in the range ±0V and will likely aturate the OpAmp. NOTE TO GADES: Any reaonable anwer earn a point. (e) ( point) Compute value for and C uch that the magnitude of the frequency repone for Solution #2 and #3 be greater than 0 2 / 2 7 for any frequency component higher than 20Hz preent in the ignal v x. All that i needed to atify thi requirement i that ω c be low enough but not Page 8
zero: which implie For example 0 < ω c = C 2π20 C 0.008 = 8 0 3 2π20 (+.5 point) C = nf = 0 9, = 8MΩ = 8 0 6. (+.5 point) NOTE TO GADES: Aign a full point if only mitake i forgetting to multiply by 2π. (f) ( point) Why Solution #2 and #3 work? Solution #2 and #3 work becaue they have a zero gain at ω = 0. In the -domain, the repone to the input ignal i equal to V o () = ±0 jω jω + ω c V x (). If ω c i low enough V o () ±0V x () = v o (t) ±0v x (t) NOTE TO GADES: Any reaonable anwer that mention the highpa feature earn a point. (g) ( point (bonu)) Explain why Solution #2 work but Solution # doe not work? Solution #2 and #3 are very imilar. However, becaue of the very high input impedance of the OpAmp, there i not enough current through the capacitor to make it work a a highpa filter in Solution #2. Thi i olved in Solution #3 by explicitely adding a path to ground through the reitor. NOTE TO GADES: Any reaonable anwer earn a point. Page 9
4. OpAmp Deign In thi quetion you will deign an OpAmp circuit that olve a problem imilar to the one in Quetion 3. ecall that the output of the pre-amplifier i of the form (t) = 0V + v x (t) where the ignal v x (t) varie from V to V. In thi quetion you will deign an OpAmp circuit uing only reitor that ha a input (t) and produce a output v o (t) = 5V ± 5 v x (t). Aume that voltage upplie +5V and 5V are available. Do not add voltage or current ource to the circuit other than (t), +5V and 5V. Ue voltage diviion or a gain if you need to produce a contant voltage from the ±5V upplie. The ymbol ± mean it can be either poitive or negative. Anwer the following quetion: (a) (3 point) Deign an OpAmp circuit that ha (t) a input and produce a it output. v (t) = ±[V ± v x (t)] One olution i to ue the ubtractor circuit: 3 4 +5V v 2 The output of thi circuit i v = + 2 4 5V 2 3 + 4 ( + 2 4 = 5V ) 2 0V 3 + 4 2 v x. In order to obtain v = V v x we need to have = 2 = 2 = + 2 = 2, and = + 2 4 3 + 4 5 2 0 = 4 3 + 4 30 0 Page 0
or For example + 3 / 4 30 = 9 = 3 4 = 0 3 = 7 3. = 2 =, 3 = 7, 4 = 3, for ome, ay MΩ. NOTE TO GADES: Any correct anwer earn the 3 point. (b) (3 point) Deign an OpAmp circuit that ha v (t) a input and produce a it output. v 2 (t) = ±5 v (t) Many eay olution here. One i baed on the inverting OpAmp 5 6 v o The output of thi circuit i v 2 = 5 6 v. In order to obtain v o = 5v we need to have 6 = 5 5 = 6 5 = 5. For example 5 =, 6 = 5, for ome, ay MΩ. NOTE TO GADES: Any correct anwer earn the 3 point. (c) (4 point) Show how the circuit in part (a)-(b) can be connected to produce having a input (t). v o (t) = 5V ± 5 v x (t). Page
One poible olution i to imply connect which in thi cae lead to v = V v x, v o = v 2 = 5v v o = 5( V v x ) = 5V + 5 v x The combined circuit look like: 3 4 +5V 5 6 v v o = v 2 2 where, 2, 3, 4, 5, and 6 are choen a before, ay: = 2 =, 3 = 7, 4 = 3, 5 =, 6 = 5, for ome, ay MΩ. NOTE TO GADES: Any correct anwer earn the 4 point. (d) (2 point (bonu)) Show how the anwer to thi quetion can be implemented uing a ingle OpAmp ubtractor circuit (differential amplifier circuit) with one OpAmp. Let u reviit the ubtractor circuit olution: 3 4 +5V v o 2 The output of thi circuit i v o = + 2 4 5V 2 3 + 4 ( + 2 4 = 5V ) 2 0V 3 + 4 2 v x. In order to obtain v = 5V 5v x we need to have 2 = 5, = 2 = 5, + 2 = 6, Page 2
and 5 = + 2 4 3 + 4 5 2 0 = 4 3 + 4 6 5 50 or + 3 / 4 6 5 = 55 = 3 = 8 4 = 7. For example = 2 =, 3 = 7, 4 =, for ome, ay MΩ. NOTE TO GADES: Any correct anwer earn the 2 point. 5. Circuit with Switche 0V C + vc Figure 4: Circuit for Quetion 5. In the circuit of Figure 4, the witch ha been help open for a long time. At t = 0 the witch i cloed and remain cloed until t = 0 when it i opened again. Aume C = 0 and anwer the following quetion: (a) ( point) Convert the circuit to the -domain for t 0. Clearly indicate how you handle the initial condition on the capacitor. After the witch i cloed the circuit, already in the -domain, become: 0 V + VC() (+.5 point) The initial voltage at the capacitor, v C (0), i zero becaue the witch ha been open for a long time. (+.5 point) (b) (4 point) Show that v C (t) = ( e 0.t )0V, 0 t 0. Page 3
Calculate v C (0). Hint: /e 0.37. The voltage at the capacitor, V C (), can be computed uing voltage diviion: V C () = 0 + V = ω c 0 + ω c V, ω c = C = 0.. Expanding in partial fraction: V C () = α + α 2 + 0. where α = lim 0 + 0. = 0V, α 2 = lim 0. = 0V. we obtain V C () = 0 0 + 0. Back to the t-domain: v C (t) = ( e 0.t )0V, t 0. At t = 0 v C (0) = ( e )0V ( 3.7)0V = 6.3V. (c) ( point) Convert the circuit to the -domain for t 0. Clearly indicate how you handle the initial condition on the capacitor. Hint: Shift the origin of time to t = 0 and apply the Laplace tranform a uual. We hift the origin to time to t = 0 and with the witch opened derive the -domain equivalent to the circuit: 6.3 V (+.5 point) The initial voltage at the capacitor, v C (0) = 6.3V, i obtained from v C (0) computed in part (b). (+.5 point) Page 4
(d) (4 point) Show that v C (t) e 0.05(t 0) 6.3V, t 0. Hint: If you ued my hint in part (c) do not forget to hift your anwer back in time. We aociate the two reitor 6.3 V VC() + 2 and compute V C () through voltage diviion: V C () = 2 6.3 2 + V = 0 V, ω c = + ω c 2C = 0.05. Back to the t-domain: v C (t) e 0.05t 6.3V, t 0. Shifting back in time 0: (e) ( point (bonu)) Sketch v C (t) for t 0. A ketch of the olution v C (t) i hown below: v C (t) e 0.05(t 0) 6.3V, t 0. Page 5
7 6 5 Vc(t) (V) 4 3 2 0 0 5 0 5 20 25 30 t () Note the tranition at t = 0 from one circuit to another. Page 6