EECE 290 Exam 1 with Solution, Spring 2016

2 k 2 k + – 2 k 1 k VSRC VO + – – + 1 k 1 k 2 k + – VSRC VO + – – + 1 k 2 k 2 k 1 k + – vSRC vO + – – + 1  1 H vSRC(t), V t, s 0 1 3 5 2 4 A -A EECE 290 Quiz 1, February 20, 2016 4% (Primal Question) A. Determine VO assuming VSRC = 1 V. Solution: The two 1 k resistors in series with the op amp inputs are redundant since they do not carry any current. The 2 k resistors across VSRC and the op amp output are redundant since they are connected in parallel with ideal voltage sources. The circuit reduces to a noninverting configuration having Rf = 2 k and Rr = 1 k. It follows that VO = (1 + Rf/Rr)VSRC = 3VSRC V. Version 1: VSRC = 1 V, VO = 3 V Version 2: VSRC = 1.5 V, VO = 4.5 V Version 3: VSRC = 2 V, VO = 6 V Version 4: VSRC = 2.5 V, VO = 7.5 V Version 5: VSRC = 3 V, VO = 9 V 4% (Primal Question) B. Determine VO assuming VSRC = 1 V. Solution: The 1 k resistor in series with the noninverting input is redundant since it does not carry any current. The 2 k resistor connected between the inverting input and ground is redundant since this input is a virtual ground. The 2 k resistor at the op amp output is redundant since it is connected in parallel with an ideal voltage source. The circuit reduces to an inverting configuration having Rf = 2 k and Rr = 1 k. It follows that VO = -Rf/RrVSRC = -2VSRC V, Version 1: VSRC = 1 V, VO = -2 V Version 2: VSRC = 1.5 V, VO = -3 V Version 3: VSRC = 2 V, VO = -4 V Version 4: VSRC = 2.5 V, VO = -5 V Version 5: VSRC = 3 V, VO = -6 V 4% (Primal 1 vO(t), V t, s 0 1 3 5 2 4 A -A 1 mH ISRC(j) + –  VO(j) 1 mF R 2 k + – VSRC +15 V + – – + 1 k 1 k -15 V 3 k VO Question) C. Sketch vO(t) when vSRC(t) is the triangular waveform shown, assuming A = 1 V, Solution: ir(t) = vSRC(t)/1 = vSRC(t) A; vO(t) = -1dVSRC(t)/dt is the square waveform shown of amplitude A V/s. Version 1: A = 1 V Version 2: A = 2 V Version 3: A = 3 V Version 4: A = 4 V Version 5: A = 5 V 4% (Primal Question) D. Determine the type of response VO(j)/ISRC(j), its maximum magnitude, and the Q, assuming R = 10 . Solution: The response is bandpass. The maximum magnitude is at resonance, when ISRC(j) flows through R, so that the maximum response is R ; = 1000 rad/s; Qp = 0CR = 10310-3R = R. Version 1: R = 10 ; maximum response is 10 ; Q = 10 Version 2: R = 20 ; maximum response is 20 ; Q = 20 Version 3: R = 30 ; maximum response is 30 ; Q = 30 Version 4: R = 40 ; maximum response is 40 ; Q = 40 Version 5: R = 50 ; maximum response is 50 ; Q = 50 4% 1. Determine VO, assuming VSRC = 6 V. A. V B. V C. V D. 15 V E. V Solution: VP = (2/3)VSRC; VN = VP; the current flowing through the 1 k resistor to ground is (2/3)VSRC mA. VO = (2/3)VSRC + 3(2/3)VSRC = (8/3)VSRC = 16 V. If VSRC  6 V, the op amp is in positive saturation, so that VO = 15 V. Version 1: VSRC = 6 V, VO = 15 V Version 2: VSRC = 7 V, VO = 15 V 2 + – VSRC VO + – – + 1 k 2 k 1 k 1 mH 1 nF  + – VO + – – + 1 k 2 k 1 k 1 mH 1 nF VSRC VSRC /2 mA VSRC /2 mA VSRC /2 mA  Version 3: VSRC = 8 V, VO = 15 V Version 4: VSRC = 9 V, VO = 15 V Version 5: VSRC = 10 V, VO = 15 V 4% 2. Determine the maximum magnitude of VO and the phase angle at this magnitude, assuming VSRC = 10 V. A. V B. 1 V, 180 C. V D. V E. V Solution: Because of the virtual short circuit, maximum current occurs when the inductor or the capacitor acts as a short circuit, and is equal to VSRC/2 mA. This current flows through the 1 k resistors, giving VO = -2(VSRC/2) = -VSRC V. Version 1: VSRC = 1 V, VO = 1 V, 180 Version 2: VSRC = 2 V, VO = 2 V, 180 Version 3: VSRC = 3 V, VO = 3 V, 180 Version 4: VSRC = 4 V, VO = 4 V, 180 Version 5: VSRC = 5 V, VO = 5 V, 180 4% 3. A lowpass, first-order transfer function H1(j) has a maximum gain in the passband of -2 dB and a corner frequency of 50 kHz. Another lowpass, first-order transfer function H2(j) has a magnitude that is 8 times that of H1(j) and a corner frequency of 1 kHz. Determine |H1(j)H2(j)| in dB when it falls 3 dB below its maximum value at low frequencies. A. dB B. dB C. 11 dB D. dB E. dB Solution: The maximum value of 20log10|H1(j)| is -A dB. The maximum value of 20log10| H2(j)| is -A + 20log10(8) = -A + 18 dB. The maximum value of 20log10|H1(j)H2(j)| is -A + 18 – A = (18 – 2A). At a level -3 dB below this, the gain is (15 – 2A) dB. Version 1: A = 2 dB, gain = (15 – 2A) = 11 dB 3 – + VSRC(j) L 1 nF 90  10  VO(j) + – 1 H ISRC(j) + – + –  0.25 F 10  VO(j) (j0.3 + )VO(j) V Version 2: A = 3 dB, gain = (15 – 2A) = 9 dB Version 3: A = 4 dB, gain = (15 – 2A) = 7 dB Version 4: A = 6 dB, gain = (15 – 2A) = 3 dB Version 5: A = 7 dB, gain = (15 – 2A) = 1 dB. 4% 4. Determine the frequency at which the high-frequency asymptote of the response VO(j)/VSRC(j) crosses the 0 dB line, assuming L = 1 mH. A. 10 krad/s B. krad/s C. krad/s D. krad/s E. krad/s Solution: At high frequencies, the inductive reactance dominates. |I| = |VSRC(j)|/L; IVO(j)| = 10|I| = 10|VSRC(j)|/L; the high-frequency asymptote is |VO(j)/VSRC(j)| = 10/L, or, 20log10|VO(j)/VSRC(j)| = 20log10(10/L); this is zero at L = 10, or  = 10/L Version 1: L = 1 mH;  = 10/L = 10 krad/s Version 2: L = 2 mH;  = 10/L = 5 krad/s Version 3: L = 4 mH;  = 10/L = 2.5 krad/s Version 4: L = 5 mH;  = 10/L = 2 krad/s Version 5: L = 8 mH;  = 10/L = 1.25 krad/s. 4% 5. Determine the 3-dB bandwidth of the response VO(j)/ISRC(j), assuming  = 1. A. B. C. D. E. 5 rad/s Solution: BW = 0/Q = 0/(0Leff/Reff) = Reff/Leff . The current through the source in the direction of a voltage drop is IO(j) = VO(j)/10; the value of the dependent source is (j0.3 + )10IO(j) = (j3 + 10)IO(j). The dependent source is equivalent to (j3 + 10), that is, to an inductance of 3 H and a resistance of 10 . Hence, Leff = 4 H and Reff = 10( + 1) , so that BW = 2.5( + 1) rad/s. Version 1:  = 1, BW = 2.5( + 1) = 5 rad/s 4 + – VO + – 1 k 2 k 6 k 3 k 2 mA 1 mA 4 V 4 mA 3 mA x x x + – VO + – 1 k 2 k 6 k 3 k 2 mA 1 mA Version 2:  = 2, BW = 2.5( + 1) = 7.5 rad/s Version 3:  = 3, BW = 2.5( + 1) = 10 rad/s Version 4:  = 4, BW = 2.5( + 1) = 12.5 rad/s Version 5:  = 5, BW = 2.5( + 1) = 15 rad/s. 4% 6. Given the transfer function . Determine Q if the product of the two 3-dB cutoff frequencies is 106 (rad/s)2. A. B. C. 10 D. E. Solution: BW = 100 rad/s; ; Version 1: c1c2 = 106; = 103/100 = 10 Version 2: c1c2 = 4106; = 2103/100 = 20 Version 3: c1c2 = 9106; = 3103/100 = 30 Version 4: c1c2 = 16106; = 4103/100 = 40 Version 5: c1c2 = 25106; = 5103/100 = 50. 20% 7. Determine VO. Solution: VP = VN = 2(3|| 6) = 22 = 4 V. A current of 4 mA flows in the 1 k 5 2R Cf Rf – – + + 2R C – + VO(j) VSRC(j) 2R 1/sCf Rf – – + + 2R 1/sC – + VO(j) VSRC(j) Ir resistor towards ground. The current in The 2 k resistor is 3 mA. It Follows that VO = 23 + 4 = 10 V. 20% 8. (a) Show that the response VO(j)/VSRC(j) is the product of a first-order lowpass response and a first-order highpass response (9 grades). (b) Choose Rf and Cf so that the two first-order responses have the same 3-dB cutoff frequency, and the maximum magnitude of the overall response is 0 dB, assuming R = 2 k and C = 0.5 F (8 grades). (c) Specify the Q of the overall response (3 grades) Solution: (a) The two 2R resistors are effectively in parallel, the voltage across them being = Hence, . The parallel impedance of Rf and Cf is Rf/(1 + sCfRf). It follows that H(s) = , which is the product of a first-order lowpass response and a first-order highpass response. (b) If the two first-order responses have the same cutoff frequency c, H(s) = . Multiplying the numerator and denominator by , 6  ISRC(j) 2C 2C C C R 1 mH IO(j)  ISRC(j) 2C 2C C C IN IC IC 2IC 2IC  R 1 mH IO(j) 100 nF 3 ) (  j ISRC ; to have a unity magnitude of gain, Rf = 4R = 8 k. The time constant CR = 0.510-62103 = 10-3 s. Hence Cf = 10-3/(8103)  0.125 F. (c) The coefficient of s in the denominator in standard form is c/Q. It follows that Q = 0.5. 20% 9. (a) Determine R so that Q = 10, assuming C = 75 nF (7 grades). (b) Derive the transfer function IO(j)/ISRC(j) (8 grades). (c) Characterize the response in terms of the basic responses of second-order circuits (2 grades). (d) Determine the gain in dB as   0 and at  = 0 (1 grade and 2 grades). Solution: (a) Norton’s equivalent circuit of the capacitive bridge and current source is first determined. The current source divides into ISRC(j)/3 through C and 2ISRC(j)/3 through 2C. On short circuit, it is seen that IN = ISRC(j)/3. The capacitance looking into the capacitive bridge, with the current source replaced by an open circuit is twice C in series with 2C, which is 4C/3 = 100 nF. The circuit becomes the parallel circuit shown. = rad/s; k. (b) In terms of admittances, = . (c) The response is seen to be the sum of a bandpass and a lowpass response. (d) As s  0, H(j) = 1/3, 20log10(1/3) = -9.54 dB. At  = 0, s2 = -1010, and H(j0) = ; 20log10|H(j0)| = = 7 10.5 dB. 8