EECE 290 Exam 2 with Solution, Spring 2016

– + VSRC(j) -j10  j20  4 t f(t) 2 4 -2 -4 8 EECE 290 Quiz 2, Mar 19, 2016 4% (Primal Question) A. Given the transfer function for a normalized second-order Butterworth filter. Determine the transfer function of the filter when the frequency is scaled to 10 rad/s. Solution: Coefficient of s is 0/Q, where Q of the Butterworth filter is . The constant term is term is . Version 1: 0 = 10 rad/s, H(s) = Version 2: 0 = 20 rad/s, H(s) = Version 3: 0 = 30 rad/s, H(s) = Version 4: 0 = 40 rad/s, H(s) = Version 5: 0 = 50 rad/s, H(s) = . 4% (Primal Question) B. Enter between the parenthesis (T) or (F) according to whether the statement is true or false, respectively: ( T )The average value of f(t) is 4 ( F )The ac component of the FSE of f(t) consists of even harmonic only ( T )The trigonometric terms of the FSE are cosine functions only ( F )The harmonics of the FSE of f(t) attenuate at a rate of approximately 1/n (Note: 1 grade for each correct answer. No penalty for wrong answers) 4% (Primal Question) C. Determine the reactive power absorbed by the inductor and the capacitor, assuming VSRC(j) = 10-45 V rms. Solution: QL = -|VSRC(j)|2BL = -|VSRC(j)|2×[-(1/20)]; QC = -|VSRC(j)|2BC = -|VSRC(j)|2×[(1/10)]. 1 Version 1: |VSRC(j)| = 10; QL = 100/20 = 5 VAR, QC = -100/10 = -10 VAR Version 2: |VSRC(j)| = 20; QL = 400/20 = 20 VAR, QC = -400/10 = -40 VAR Version 3: |VSRC(j)| = 30; QL = 900/20 = 45 VAR, QC = -900/10 = -90 VAR Version 4: |VSRC(j)| = 40; QL = 1600/20 = 80 VAR, QC = -1600/10 = -160 VAR Version 5: |VSRC(j)| = 50; QL = 2500/20 = 125 VAR, QC = -2500/10 = -250 VAR 4% (Primal Question) D. A source has an open circuit voltage of 10-140 V rms, a source inductance of 10 mH and a source resistance of 10  , the frequency being 400 Hz. Determine the load that will absorb maximum power from this source and the value of this power. Solution: The source reactance is 2×400×10-2 = 8 . For maximum power transfer, the load resistance should be 10  and the reactance should be -8 , which gives a capacitance of -1/(8×800)  15.8 F. The power delivered is PL = |VSRC|2/(4*10) = Version 1: |VSRC| = 10; PL = 100/40 = 2.5 W Version 2: |VSRC| = 8; PL = 64/40 = 1.6 W Version 3: |VSRC| = 6; PL = 36/40 = 0.9 W Version 4: |VSRC| = 5; PL = 25/40 = 0.625 W Version 5: |VSRC| = 4; PL = 16/40 = 0.4 W. 4% 1. A highpass 4th order Butterworth filter has a passband gain of 10 and a 3-dB cutoff frequency of 10 kHz. Determine the frequency at which the gain is 10-6. A. Hz B. Hz C. Hz D. Hz E. 177.8 Hz Solution: For a 4th order highpass Butterworth filter of a passband gain of 10, . The frequency fx at which the gain is A×10-6 is small compared to fch if A is relatively small. Hence, (fch/fx) >> 1, so that , or = , or, = 177.8 Hz if A = 1. Alternatively, fx will lie on the LF asymptote, whose slope is 80 dB/dec. The gain of 10-6 is 10-7 below the passband gain, which corresponds to -140 dB. This is 140/80 = 7/4 decades less than 10 2 t, s 1 2 -1 -2 A vO + – – + – + – + 1  1 F 1  1 H vSRC vSRC kHz, which gives fX = 104×10-7/4 = 177.8 Hz Version 1: A = 1, = 177.8 Hz Version 2: A = 2, = 211.5 Hz Version 3: A = 3, = 234.0 Hz Version 4: A = 4, = 251.5 Hz Version 5: A = 5, = 265.9 Hz 4% 2. Determine the type of response VO(j)/VSRC(j). A. V B. Lowpass C. V D. V E. V Solution: Let the RC combination be denoted by Zf and the RL combination be denoted by Zr. Since the op amp is ideal, the voltage across the RL combination is VN = VSRC, because of the virtual short circuit. The current through Zr is VSRC/Zr. It follows that VO = VN – VSRC + VSRCZf/Zr = VSRCZf/Zr, where Zf = R/(1 + jCR) = 1/(s + 1), and Zr = (R + jL) = (s + 1), so that . The response is second-order lowpass. 4% 3. f1(t) and f2(t) are, respectively, the square and triangular waveforms shown, each having an amplitude A and zero average value. Determine the component of (f1(t) + f2(t)) having a frequency of 3 /2 rad/s, assuming A = 1. A. B. C. D. -0.42cos1.5t E. 3 t s 0 1 3 2 4 5 + A -5 + A Solution: The period of f1(t) is 4 s, so 01 = 2 /4 =  /2. The period of f2(t) is 2 s, so 02 = 2/2 = . Both waveforms are quarter-wave symmetric and have odd harmonics only, so that the harmonics of f1(t) are 3 /2, 5 /2, etc. The harmonics of f2(t) are 3 , 5 , etc. The FSE of the sum of the two FSEs will have a component at a frequency of 3 /2 due to the third harmonic of f1(t) only. Hence, . It follows that the component of frequency 3 /2 is . Version 1: A = 1, a3 = -4/3 = -0.42cos1.5t Version 2: A = 2, a3 = -8/3 = -0.85cos1.5t Version 3: A = 3, a3 = -12/3 = -1.27cos1.5t Version 4: A = 4, a3 = -16/3 = -1.70cos1.5t Version 5: A = 5, a3 = -20/3 = -2.12cos1.5t. 4% 4. f(t) is a periodic waveform of period 4 s, its ac component is quarter-wave symmetric, with f(t) = (5 + A) + kt2, 0 < t  1, where A = 1 and k is such that the rms value of f(t) is 8. Determine the rms value of the ac component of |f(t)| when delayed by 1 s. A. 5.92 B. C. D. E. Solution: The required rms value is the same as that of the ac component of f(t). The average value of f(t) is A. Hence, , so that Version 1: A = 1, f(t)rms = 6; Version 2: A = 2, f(t)rms = 6; Version 3: A = 3, f(t)rms = 6; Version 4: A = 4, f(t)rms = 7; Version 5: A = 5, f(t)rms = 8, 6.24. 4 – + vSRC j20  20  XL 10  ISRC XL  Load 1 Load 2 4% 5. Load 1 absorbs a complex power of VA, whereas load 2 absorbs a complex power of VA. Determine XL, assuming ISRC = 0.25-30 A. Note that the other circuit parameters need not be specified. A.  B.  C. 400  D.  E.  Solution: The total complex power delivered by the source = (10 + j10) + (5 – j5) = 15 + j5 VA. = 250/|ISRC|2; reactive power absorbed by XL is 10 = 250/(XL|ISRC|2); hence, XL = 25/|ISRC|2. Version 1: |ISRC| = 0.25, XL = 25/|ISRC|2 = 25/0.0625 = 400  Version 2: |ISRC| = 0.5, XL = 25/|ISRC|2 = 25/0.25 = 100  Version 3: |ISRC| = 1.25, XL = 25/|ISRC|2 = 25/1.5625 = 16  Version 4: |ISRC| = 2, XL = 25/|ISRC|2 = 25/4 = 6.25  Version 5: |ISRC| = 2.5, XL = 25/|ISRC|2 = 25/6.25 = 4  4% 6. If XL can be varied between -30 and -50 , determine the maximum power that can be delivered to the 10  resistor, assuming VSRC = 1030 V rms. A. B. C. D. E. 0.8 W Solution: The voltage source and the series impedance are transformed to a current source ISRC in parallel with an admittance Ysrc, where ISRC = A, |ISRC| = A rms, , so that Gsrc = 1/40 S and Bsrc = -1/40 S. For maximum power transfer, BL = 1/40 S, or XL = -40 , which is within the given range of XL. 5 50 k vO vI + – + – + – + – + – 50 k 0.2 nF 50 k 50 k 10 k 10 k 100 k 0.16 F 4 5 -4 -3 t  f(t) - The circuit reduces to ISRC in parallel with 40  and 10 . The current in the 10  resistor is ISRC×40/50 = 0.8ISRC. The power absorbed by the 10  resistor is PL = 0.64|ISRC|2×10 = Version 1: |VSRC| = 10; PL = |VSRC|2/125 = 100/125 = 0.8 W Version 2: |VSRC| = 15; PL = |VSRC|2/125 = 225/125 = 1.8 W Version 3: |VSRC| = 20; PL = |VSRC|2/125 = 400/125 = 3.2 W Version 4: |VSRC| = 25; PL = |VSRC|2/125 = 625/125 = 5 W Version 5: |VSRC| = 30; PL = |VSRC|2/125 = 900/125 = 7.2 W. 20% 7. (a) Derive the transfer function VO(j)/VI(j) (12 grades). (b) Characterize the response (3 grades). (c) Determine the maximum gain (2 grades), the minimum gain (3 grades), both in dB, and the frequencies at which they occur. Solution: (a) The lower 3-dB cutoff frequency Is = 125 rad/s; the upper 3-dB cutoff frequency is krad/s. The transfer function is . (b) The response is a combination of second-order bandpass and second-order bandstop responses. (c) The maximum gain is 10, or 20 dB, and occurs as   0, or,   . The minimum gain is 10×250/100125  0.025  20log10(0.025) = -32 dB 20% 8. (a) Derive the first three terms of the FSE of f(t), where f(t) = 6 -j40  20  50 A rms  j18  4  a b -j5  20  80 – j40 V rms j10  10  a b – + 4 + sin 2t, -/2 < t < /2 and f(t) = -4 + sin 2t, -  t < -/2 and /2 < t   (16 grades); (b) determine these terms when f(t) is advanced by a quarter period (4 grades). Solution: (a) f(t) is neither odd nor even, is not half-wave symmetric, and is of zero average. It is the sum of sin2t and an even square wave of amplitude 4, zero average, and period 2, or 0 = 1 rad/s. The square wave is quarter-wave symmetric so it consists of odd cosine terms only. . The fundamental is , and the third harmonic is . The sinusoidal function is its own FSE, so that the first three terms of the FSE of f(t) are: + sin2t . (b) If the function is advanced by a quarter period, the phase angle is increased by n0ta, where 0 = 1 rad/s, and ta = /2 s. The FSE becomes + sin(2t + ) = – sin2t . The function becomes odd. 20% 9. A load RL + jXL is connected between terminals ‘ab’, where RL can be varies between 0 and 10  and XL can be varied between 5 . Determine: (a) the values of RL and XL that result in maximum power dissipated in RL (15 grades); (b) the complex power absorbed by the load under these conditions (5 grades). Solution: The parallel impedance of the resistor and capacitor is -j8(1 + j2) = 16 – j8 ; VTh = 5(16 – j8) = 80 – j40 V rms. The total source impedance is 16 – j8 + 4 + j18 = 20 + j10 . (a) XL that maximizes power transfer is that which minimizes the total reactance. Hence, XL = -5 . RL for maximum power transfer = 20.62 . Hence, RL = 10  should be used. The circuit becomes as shown. |I|2 = 7 A2. The real power dissipated in RL is |I|2RL = 86.5 W and the reactive power absorbed by XL is -5|I|2 = -43.2 VAR. Hence, SL = 86.5 – j43.2 VA. 8