EECE 290 Exam 2 with solution, Spring 2015

EECE 290 Quiz 2, Mach 21, 2015 4% (Primal Question) A. Draw the circuit diagram of a first-order, lowpass active filter, using a 0.1 F capacitor, and having a 3-dB cutoff frequency of 1 krad/s and a gain of 0 dB. Indicate the values of all the components. Solution: For a 3-dB cutoff frequency of  rad/s, k, where  is in krad/s. For 0 dB gain, Rr = Rf. Version 1:  = 1 krad/s, Rf = Rr = 10 k Version 2:  = 2 krad/s, Rf = Rr = 5 k Version 3:  = 4 krad/s, Rf = Rr = 2.5 k Version 4:  = 5 krad/s, Rf = Rr = 2 k Version 5:  = 10 krad/s, Rf = Rr = 1 k. 4% (Primal Question) B. Given the RL filter shown. Determine the frequency and magnitude scaling factors to have a 3-dB cutoff frequency of 10 krad/s and an inductance of 1 mH. Solution: The 3-dB cutoff frequency of the given filter is 1 rad/s. To have a 3-dB cutoff frequency of c rad/s, kf = c/1 = c. To have a scaled inductance L, requires km = kfL = cL. Version 1: c = 10 krad/s, L = 1 mH, kf = 104, km = 10410-3 = 10 Version 2: c = 20 krad/s, L = 2 mH, kf = 2104, km = 2104210-3 = 40 Version 3: c = 30 krad/s, L = 3 mH, kf = 3104, km = 3104310-3 = 90 Version 4: c = 40 krad/s, L = 4 mH, kf = 4104, km = 4104410-3 = 160 Version 5: c = 50 krad/s, L = 5 mH, kf = 5104, km = 5104510-3 = 250. 4% (Primal Question) C. (a) Determine the average value of the periodic function f(t), assuming A = 2 units; (b) specify all the symmetry properties of the ac component of f(t). Solution: (a) The area of a 2-s half period is 2(A + 1)/2 = (A + 1), and the average value is (A + 1)/2. (b) ac component is even, half-wave symmetric, and quarter-wave symmetric. 1 Rr 0.1 F Rf vI – + – – + vO + 1 A 2 6 4 f(t) 8 t, s vO + – vI + – 1 H 1  – + – + 500  1 mH 1 k – + VO(j) VSRC(j) 2 nF Version 1: A = 2, a0 = 1.5 Version 2: A = 3, a0 = 2 Version 3: A = 4, a0 = 2.5 Version 4: A = 5, a0 = 3 Version 5: A = 6, a0 = 3.5. 4% (Primal Question) D. Determine RL and C for maximum power transfer to RL, assuming RL and C can be varied independently over an arbitrary range, Rsrc = 100  and  = 100 krad/s. Solution: For maximum power transfer, RL.= Rsrc, and L = 1/C, or . Version 1: Rsrc = 100 ,  = 100 krad/s, RL = 100 , C = 10/2 = 10/1010 = 10-9  1 nF Version 2: Rsrc = 200 ,  = 50 krad/s, RL = 200 , C = 10/2 = 10/(0.251010)  4 nF Version 3: Rsrc = 300 ,  = 20 krad/s, RL = 300 , C = 10/2 = 10/(0.041010)  25 nF Version 4: Rsrc = 400 ,  = 10 krad/s, RL = 400 , C = 10/2 = 10/(108)  0.1 F Version 5: Rsrc = 500 ,  = 5 krad/s, RL = 500 , C = 10/2 = 10/(0.25108)  0.4 F. 4% 1. Specify the type of response VO(j)/VSRC(j), assuming an ideal op amp. A. Highpass B. Perfect differentiator C. Allpass D. Lowpass E. Perfect integrator Solution: ; . . But . Hence, , which is the response of a perfect integrator. 4% 2. The low-frequency asymptote of a highpass Butterworth response of maximum gain 9 dB has a slope of 80 dB/decade and intersects the 9 dB line at a frequency of 1 kHz. 2 – + vSRC 0.1 H C RL Rsrc 10 k vSRC vO + _ + _ – +  10 k 1 mH 2 mH L Determine the frequency at which the gain is 6 dB. A. 5 kHz B. 1 kHz C. 4 kHz D. 2 kHz E. 3 kHz Solution: The required frequency is the 3-dB cutoff frequency, which is the same as the corner frequency, that is, 1 kHz. Version 1: Corner frequency = 1 kHz, fch = 1 kHz Version 2: Corner frequency = 2 kHz, fch = 2 kHz Version 3: Corner frequency = 3 kHz, fch = 3 kHz Version 4: Corner frequency = 4 kHz, fch = 4 kHz Version 5: Corner frequency = 5 kHz, fch = 5 kHz. 4% 3. Determine the type of filter shown and the passband gain, assuming L = 2 mH,  = 1 krad/s, and an ideal op amp. A. Bandstop B. Bandpass C. Highpass D. Lowpass E. Allpass Solution: As   , the op amp output is zero. As   0 the circuit reduces to an inverting amplifier, with gain -L/2. The response is lowpass. Version 1: L = 2 mH, gain = -1 Version 2: L = 3 mH, gain = -1.5 Version 3: L = 4 mH, gain = -2 Version 4: L = 5 mH, gain = -2.5 Version 5: L = 6 mH, gain = -3. 4% 4. Determine the rms value of the periodic function shown, assuming A = 1 unit. A. 1.29 3 A + 1 1 2 f(t) t A -A – + vSRC j15  -jX 7  5  Load B. 3.27 C. 2.27 D. 5.26 E. 4.26 Solution: For 0  t  1, f(t) = t + A. The area under the square is + A21 = = . The mean square is , and Frms = . Version 1: A = 1, Frms = = 1.29 Version 2: A = 2, Frms = = 2.27 Version 3: A = 3, Frms = = 3.27 Version 4: A = 4, Frms = = 4.26 Version 5: A = 5, Frms = = 5.26. 4% 5. If X can be varied between 5  and 10 , determine the value of X that results in maximum power transfer to the load and calculate this power, assuming vSRC = 5sin(200t + 45) V. A. X = 10 , PL = 0.75 W B. X = 10 , PL = 1.33 W C. X = 10 , PL = 0.52 W D. X = 10 , PL = 1.33 W E. X = 10 , PL = 1.68 W. Solution: X for maximum power transfer is that which results in the smallest load reactance, which is X = 10 . The load reactance is j15 – j10 = j5 . The total impedance becomes 12 + j5 . The current amplitude is = Vm/13 A. The real power transferred to 4 – + 10 V rms -j15  -j10  j10  2  1:a RL j15  + – + _ VI(s) + _ VO(s) Rf 1/sC R the load is PL = . Version 1: Vm = 5 V, PL = 25/48.286 = 0.52 W Version 2: Vm = 6 V, PL = 36/48.286 = 0.75 W Version 3: Vm = 7 V, PL = 49/48.286 = 1.01 W Version 4: Vm = 8 V, PL = 64/48.286 = 1.33 W Version 5: Vm = 9 V, PL = 81/48.286 = 1.68 W. 4% 6. Determine RL for maximum power transfer to it and calculate this power, assuming a = 2. A. 18 , 12.5 W B. 72 , 12.5 W C. 8 , 12.5 W D. 32 , 12.5 W E. 8 , 12.5 W Solution: For maximum power transfer to RL, RL = 2a2. The power transferred is PL = 100/(42) = 12.5 W.. Version 1: a = 2, RL = 2a2 = 8 , PL = 12.5 W Version 2: a = 3, RL = 2a2 = 18 , PL = 12.5 W Version 3: a = 4, RL = 2a2 = 32 , PL = 12.5 W Version 4: a = 5, RL = 2a2 = 50 , PL = 12.5 W Version 5: a = 6, RL = 2a2 = 72 , PL = 12.5 W. 20% 7. (a) Derive the transfer function VO(j)/VI(j), assuming R = 1 k and an ideal op amp. Represent the transfer function so as include an expression of the form (1 + j/c) (10 grades). (b) Determine the low-frequency and high-frequency asymptotes of the Bode plot, the corner frequency, and the gain at this frequency in dB (2 grades each). (c) Sketch the Bode magnitude and phase plots (2 grades each). Solution: (a) ; from virtual short 5 + – + _ VI(j) + _ VO(j) R 1 k 1 mF Frequency, Hz 0.01 0.03 0.1 0.3 1.0 3.0 10 0 10 20 30 40 dB 0d 50d 100d Frequency, Hz 0.01 0.03 0.1 0.3 1.0 3.0 10 2A 1 3 2 f(t) 4 t, s A -A -2A circuit, ; hence, H(j) = K + j10-3103 = K(1 + j/K), where K = (1 + 1/R), R is in k. (b) ; when  = 0, |H(j)| = K; the low-frequency asymptote is 20log10| H(j)| = 20log10K. As   , |H(j)| = ; the high-frequency asymptote is 20log10|H(j)| = 20log10, slope = +20 dB/decade. The corner frequency is when 20log10 = 20log10K, or c = K rad/s. At this frequency, 20log10|H(j)| = 20log10 = 20log10K + 3 dB. (c) The Bode magnitude and phase plots are as shown. Version 1: R = 1 k, K = (1 + 1/R) = 2, LF asymptote: 20log102 = 6 dB; c = 2 rad/s; gain = 6 + 3 = 9 dB Version 2: R = 1/2 k, K = (1 + 2) = 3, LF asymptote: 20log103 = 9.5 dB; c = 3 rad/s; gain = 9.5 + 3 = 12.5 dB Version 3: R = 1/3 k, K = (1 + 3) = 4, LF asymptote: 20log104 = 12 dB; c = 4 rad/s; gain = 12 + 3 = 15 dB Version 4: R = 1/4 k, K = (1 + 4) = 5, LF asymptote: 20log105 = 14 dB; c = 5 rad/s; gain = 14 + 3 = 17 dB Version 5: R = 1/5 k, K = (1 + 5) = 6, LF asymptote: 20log106 = 15.6 dB; c = 6 rad/s; gain = 15.6 + 3 = 18.6 dB. 20% 8. The first period of the periodic function f(t) is shown, where A = 1 unit. Determine: (a) C0, C1, C2, and C3 (4 grades each); (b) Derive the amplitude and phase angle of the third harmonic component of f(t) (2 grades each). Solution: (a) f(t) is half-wave symmetric, 0 = 2/T = /2 rad/s. C0 = 0; 6 = = = for n odd. If n = 1, ; C2 = 0, since f(t) is half-wave symmetric; if n = 3, . (b) a3 = , b3 = ; amplitude is 0.671A; the third harmonic is: cos30t + sin30t = 0.67Asin(30t + ), where tan() = -1/3, so  = tan-1(-1/3) = -18.4. Version 1: A = 1, c3 = 0.67, c3 = -18.4 Version 2: A = 2, c3 = 1.34, c3 = -18.4 Version 3: A = 3, c3 = 2.01, c3 = -18.4 Version 4: A = 4, c3 = 2.68, c3 = -18.4 Version 5: A = 5, c3 = 3.36, c3 = -18.4. 20% 9. (a) A load L1 consisting of a 12 k resistor in series with a 40 H inductor is connected in parallel with a load L2 consisting of a 75 nF capacitor in parallel with a resistor having a conductance of 40 S, the frequency being 400 rad/s. If the 40 S resistor dissipates a real power 4 kW, (i) determine the complex power absorbed by L1 and L2 (6 grades), (ii) draw the power phasor diagrams for the two loads (2 grades each), and (iii) calculate the p.f. of the combined load of L1 and L2 (2 grades). (b) The 7 + _ L1 L2 jX vSRC P1 Q1 P2 Q2 voltage across L1 and L2 is kept the same as in (a), but a reactance X is connected between L1 and L2 and the source, as shown. Determine: (i) the value of X that results in a p.f. of unity at the source (4 grades), and (ii) the rms value of the source voltage (4 grades). Solution: (a) 4000 = , VLrms = 104 V; QC = -1084007510-9 = -3000 VAR; the magnitude of current through L1 is A rms; the real power dissipated in the 12 k resistor is 12103(0.5)2 = 3000 W; the reactive power absorbed by the inductor is (40040)(0.5)2 = 4000 VAR. The power phasor diagram is as shown. The complex power of the load is 7000 + j1000, and the p.f. is . (b) X should absorb -j1000 VAR; 7000 + j1000 = 10000 , where IS is the source current; |ISrms| = A. Power absorbed by X is: -1000 = , so that X = -2000 . Power delivered by the source is 7000 W = VSRC1/ , which gives VSRC =  9900 V rms. 8