EECE 290 Final, Spring 2015

t, s 10 2 4 vSRC1, V t, s -A  2 4 vSRC2, V + – vO 1 k – + 2 k – + 0.1 F vSRC1 vSRC2 – + + – 10  – + 1 F vSRC – + vO t, s f(t) 1 1 2 -1 -2 3 4 2 1 3 2 4 t, s 1 -1 -2 5 6 0 EECE 290 Final Exam, May 13, 2015 3% A. Determine vO(t), assuming vSRC = 0.1u(t) V. Solution: vO(t) = -CRdvSRC/dt = -100.1(t) = -(t) V 3% B. Determine the rms value of the periodic function whose first period is shown. Solution: The area under the square of the first half cycle Is 1 + 4 + 1 = 6. The mean is 2 and the rms value is . 3% C. Convolve tu(t) with tu(t). Solution: . 3% D. Derive the Fourier transform of f(t). Solution: f(t) = u(-t) + u(t – 2); F(j) = + . 3% 1. vSRC1(t) and vSRC2(t) are the periodic functions shown. Determine  so that vO(t) is periodic, assuming A = 16 V. A. 1 s B. 0.2 s C. 0.5 s D. 1.25 s E. 2.5 s Solution: For vO(t) to be periodic, the average current through the capacitor must be zero. p1/8 C1 C2 C3 C4 t = 0 – + 2 V – + 4 V + – v3 + – v4 2 F 4 F 1 F 2 F ffw(t) t, s 1 1 2 -1 -2 1 2 -1 -2 1 0.6 (a) (b) t, s 1 mH C – + VSRC This requires that , where VSRC1 and VSRC2 are the respective dc components, that is, VSRC1 = 5 V and VSRC2 = -A/2. This gives: , or  = 20/A s. Version 1: A = 16;  = 20/16 = 1.25 s Version 2: A = 20;  = 20/20 = 1 s Version 3: A = 40;  = 20/40 = 0.5 s Version 4: A = 80;  = 20/80 = 0.25 s Version 5: A = 100;  = 20/100 = 0.2 s. 3% 2. The full-wave rectified waveform in (a) has , n = 1, 2, 3,… Determine C1b for the modified full- wave rectified waveform in (b). A. 0.042 B. 0.085 C. -0.085 D. -0.042 E. 0.064 Solution: C1a = ; when shifted to the right or left by half a period, C1a is multiplied by ; when negated, the sign of C1b remains positive. The magnitude of C1b is 0.4 times that of C1a. It follows that C1b = = 0.085. 3% 3. The source delivers a complex power of j8 kVA, with VSRC = 10030 V rms. Determine C, assuming  = 1 krad/s. A. 0.6 mF B. 0.4 mF C. 0.2 mF D. 0.5 mF E. 0.3 mF Solution: |S| = |VSRC|2; C = 1 – = 1 – 0.1S, where S is the magnitude of the complex power in kVA, or C = (1 – 0.1S) mF. Version 1: S = 8; C = (1 – 0.8) = 0.2 mF Version 2: S = 7; C = (1 – 0.7) = 0.3 mF Version 3: S = 6; C = (1 – 0.6) = 0.4 mF Version 4: S = 5; C = (1 – 0.5) = 0.5 mF Version 5: S = 4; C = (1 – 0.4) = 0.6 mF. 3% 4. The switch is closed at t = 0, with initial voltages on C1 p2/8 – + 2 H 2 H 1 H 1 H L1 L2 L3 L4 1A 3A 3A i3 VSRC and C2, and with q4(0-) = 4 C. Initial voltage on C3 is not shown. Determine the charge through the switch when it closes. A. 12 C B. 6 C C. 9 C D. 15 C E. 10 C Solution: Ceq = 2/3 + 4/3 = 2 F. Veq = 6 V. Hence, qeq = 62 = 12 C, which is the charge that passes through the switch. 3% 5. In the preceding problem, determine the final value of v4. A. -0.5 V B. 1 V C. 0.5 V D. -1.5 V E. -1V Solution: Ceq3,4 = 4/3 F, qeq3,4 = 64/3 = 8 C. This charge flows through C3 and C4 CCW. The final charge on C4 is 4 – 8 = -4 C, so the final voltage on C4 is -1 V. 3% 6. The initial currents in L1 and L4 are shown, those in L2 and L3 are not shown. Determine i3(0+), assuming VSRC = 3(t) V. A. 4 A B. 6 A C. 3 A D. 7 A E. 5 A Solution: The circuit is equivalent to a 1 H inductor in series with a 0.5 H inductor, with the impulse applied across the series combination. Considering the inductors to be uncharged, an impulse 2K(t)/3 appears across each of the 2 H inductors, resulting in a current K/3 A through each inductor, that adds to the initial current in the inductor. From KCL, i3(0-) = 2 A. It follows that i3(0+) = (K/3 + 2) A. Version 1: K = 3; i3(0+) = (K/3 + 2) = 3 A Version 2: K = 6; i3(0+) = (K/3 + 2) = 4 A Version 3: K = 9; i3(0+) = (K/3 + 2) = 5 A Version 4: K = 12; i3(0+) = (K/3 + 2) = 6 A Version 5: K = 15; i3(0+) = (K/3 + 2) = 7 A. p3/8 2 F 3 k – + vO 4 F – + 2 V 4(t) A + – V20 C1 C2 5  – + vO 2 H 4 H 6 H iSRC i2 5  – + vO 10 H 6 H 4 H iSRC 3% 7. Determine vO(t), assuming an initial voltage of 2 V on C1 and an initial voltage V20 = 2 V on C2. A. V B. V C. V D. V E. V Solution: The impulse passes through C1, depositing a charge of 4 C, which reduces the voltage across C1 to zero. Hence, vO(0+) = V20; vO() = 0; and  = 38/6 = 4 ms. It follows that V, t is in ms. Version 1: V20 = 2 V; V, t is in ms Version 2: V20 = 3 V; V, t is in ms Version 3: V20 = 4 V; V, t is in ms Version 4: V20 = 5 V; V, t is in ms Version 5: V20 = 6 V; V, t is in ms. 3% 8. Determine vO(t), assuming iSRC = 2u(t) A and no initial energy storage. A. V, t is in s B. V, t is in s C. V, t is in s D. V, t is in s E. V, t is in s Solution: From the T-equivalent circuit, the step current ISRC produces a voltage impulse K(t) between the middle and lower nodes that instantaneously changes the currents in the 4 H and 6 H inductors. Since the flux linkage is the same, the currents in these inductors are inversely proportional to the inductances, so that i2(0+) = ISRC4/(6 + 4) = 0.4ISRC; hence, vO(0+) = 2ISRC; vO() = 0; and  = 10/5 = 2 s. It follows that V, t is in s. Alternatively, vO = MdiSRC/dt = MISRC(t). hence, i2(0+) = (M/L2)ISRC = 0.4ISRC; vO(0+) = 2ISRC, as before. Version 1: ISRC = 2 A; V, t is in s Version 2: ISRC = 3 A; V, t is in s Version 3: ISRC = 4 A; V, t is in s p4/8 Version 4: ISRC = 6 A; V, t is in s Version 5: ISRC = 8 A; V, t is in s. 3% 9. Given . Determine f(0+). A. 0 B. infinite C. 1 D. 2 E. Initial-value theorem does not apply Solution: Multiplying by s and letting s  , gives f(0+) = 0. Check: the PFE is: ; ; . 3% 10. The transfer function of a circuit is . If the steady-state output of the circuit is: vO(t) = 4 + 10cos(100t + 45) V, determine the input vI(t). A. V B. V C. V D. V E. V Solution: . Under dc conditions,  = 0, so that VIDC = . For  = 100 rad/s, ; it follows that vI(t) = 0.8 + . 3% 11. Given v(t) = 2sinc(2t) V. Determine the energy dissipated in a 2  resistor over all time. A. 7.07 J B. 3.14 J C. 1.07 J D. 3.53 J E. 6.28 J Solution: FT of v(t) is or, FT of is . Hence, FT of is . The energy dissipated in a 1  resistor is 4 = = J. The energy dissipated in a 2  resistor is W = J. Version 1: Vm = 2 V; W = = 3.14 J Version 2: Vm = 3 V; W = = 7.07 J Version 3: Vm = 4 V; W = = 12.57 J Version 4: Vm = 5 V; W = = 19.63 J Version 5: Vm = 6 V; W = = 28.27 J. p5/8 I1 I2 Two-port circuit – + VSRC + – V2 + – V1  H1(s) + – + – VO(s) H2(s) VI(s) x1 3% 12. The two-port circuit is described by its g parameters as: g11 = (1 – j) S, g12 = (-1 – j), g21 = (1 + j), g22 = (1 + j) . Determine the complex power delivered by the source, assuming that VSRC = 245 V rms. The g-parameter equations are: I1 = g11V1 + g12I2; V2 = g21V1 + g22I2. A. 8 W B. 845 VA C. 16 W D. 1645 VA E. 8-45 VA Solution: With V2 = 0, I2 = -g21V1/g22 = -V1; substituting in the first equation, I1 = V1(g11 – g12) = 2V1; S = . Version 1: Vm = 2 V; S = 8 W Version 2: Vm = 3 V; S = 18 W Version 3: Vm = 4 V; S = 32 W Version 4: Vm = 5 V; S = 50 W Version 5: Vm = 6 V; S = 72 W. 13% 13. Two circuits are cascaded, with isolation. The first circuit, of open-circuit transfer function H1(s), is a first-order, normalized, lowpass Butterworth filter of passband gain of 2. The overall open-circuit transfer function H(s) = VO(s)/VI(s) is that of a second-order, normalized, lowpass Butterworth response of passband gain of 1/2. (a) Derive the open-circuit transfer function H2(s) and specify its zeros and poles (3 grades); (b) sketch the magnitude Bode plots of H1(j) and H(j) (3 grades); (c) determine |H2(j)| and evaluate |H2(j1)| in dB (3 grades); (d) derive the equations of the low-frequency and high-frequency asymptotes of |H2(j)| and evaluate the 3-dB cutoff frequency (4 grades). Solution: (a) ; ; it has a zero at s = -1, in addition to the zero at infinity, and a pair of complex conjugate poles at s = . (b) , low-pass gain = 20log10(2) = 6 dB, 3-dB cutoff frequency is 1 rad/s, slope of high-frequency asymptote is -20 dB/decade; ; low-pass gain = 20log10(1/2) = -6 dB, 3-dB cutoff frequency is 1 rad/s, slope of high-frequency asymptote is -40 dB/decade. (c) , |H2(j1)| = 20log10(1/4) = -12 dB. (d) the low-frequency asymptote as   0 is 20log10(1/4) = -12 dB; the high-frequency asymptote as    is -12 dB – 20log10, the 3-dB cutoff frequency is given by: , or 1 + = 2 p6/8 t, s T  T 3T 5T 5T  3T  f(t) 1 5 F 2 mH + – 32 V 32  t = 0 + – vC 1/sC sL 32/s 32  I(s) – + – + L + – VC(s) 500 turns 1000 turns - j 4  4  – + 100 V rms ZL  9  - j 9  – + 150 V rms ZL 9  9  + 2 , or – 2 – 1 = 0, which gives , or c = 1.55 rad/s. 13% 14. The switch is opened at t = 0 after being closed for a long time. (a) Represent the circuit in the s domain for t  0+ (5 grades) ; (b) derive VC(s) (4 grades); (c) determine vC(t), t  0+ (4 grades). Solution: (a) The initial voltage across C is 32 V and the initial current through L is 1 A. The circuit in the s domain is as shown. (b) ; = == = . (c) To obtain vC(t), VC(s) is expressed as: VC(s) = = . Hence, vC(t) = u(t) V, t is in ms, or, vC(t) = u(t) V. 13% 15. (a) Reflect the circuit on the primary (source) side to the secondary (load) side and determine the z parameters of the two-port circuit between the source and the load (6 grades); (b) determine ZL for maximum power transfer to ZL (4 grades) and calculate this power (3 grades). The z-parameter equations are: V1 = z11I1 + z12I2; V2 = z21I1 + z22I2. Solution: (a) the effective turns ratio from the primary side to the secondary side is 1.5. The source becomes 15 V and impedances are multiplied by (1.5)2 = 2.25 to become 9  and – j9 . The circuit on the secondary side becomes as shown. The circuit is symmetric, so that z12 = z21 = -j9 , and z11 = z22 = 9 – j9 = 9(1 – j) . (b) VTh at the load terminals is ; ZTh = 9 + 9 + ; ZLm = 13.5 + j4.5 ; the power transferred to the load is W. 13% 16. f(t) consists of three pulses of unit amplitude centered p7/8 at the origin, each pulse being of duration T/2 and separated from the adjacent pulse by T/2. It is required to derive the convolution function y(t) = f(t)cos using the convolution-in-time property of the Fourier transform. Solution: The FT of the middle pulse is . Adding a delayed pulse and an advanced pulse gives: = . Multiplying by the FT of cos: . This product is non-zero only at . The term is even in  and evaluates at to . The product of the two FTs becomes: = . This is the FT of y(t). It follows that y(t) = . Check: direct application of the convolution integral gives the same result: + + . p8/8