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Schedule…

Schedule…. Ask. Alma 5:26   26 And now behold, I say unto you, my brethren, if ye have experienced a change of heart, and if ye have felt to sing the song of redeeming love, I would ask , can ye feel so now? . Lecture 20 – Exam 2 Review. Chapters 4 – 6, 8. Exam 2.

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Schedule…

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  1. Schedule… Discussion #20 – Exam 2 Review

  2. Ask Alma 5:26   26 And now behold, I say unto you, my brethren, if ye have experienced a change of heart, and if ye have felt to sing the song of redeeming love, I would ask, can ye feel so now? Discussion #20 – Exam 2 Review

  3. Lecture 20 – Exam 2 Review Chapters 4 – 6, 8 Discussion #20 – Exam 2 Review

  4. Exam 2 • 12 – 16 November (Monday – Friday) • Chapters 4 – 6 and 8 • 15 questions • 12 multiple choice (answer on bubble sheet!) • 1 point each • 3 long answer (show your work!) • 4 or 5 points each • Closed book! • One 3x5 card allowed • Calculators allowed • No time limit • Study lecture slides and homework Discussion #20 – Exam 2 Review

  5. Exam 2 Review…Overview • Capacitors and Inductors • Measuring Signal Strength • Phasors • Impedance • AC RLC Circuits • AC Equivalent Circuits • DC Transient Response • Frequency Response • Basic Filters • Op-Amps Discussion #20 – Exam 2 Review

  6. i + L – + C – i Capacitors & Inductors Discussion #20 – Exam 2 Review

  7. Capacitors & Inductors Discussion #20 – Exam 2 Review

  8. Capacitors & Inductors • What is the difference between the voltage and current behaviour of capacitors and inductors? Discussion #20 – Exam 2 Review

  9. Capacitors & Inductors • What is the difference between the voltage and current behaviour of capacitors and inductors? Capacitor current iC(t) Inductor voltage vL(t) NB: both can change instantaneously Capacitor voltage vC(t) Inductor current iL(t) NB: neither can change instantaneously Discussion #20 – Exam 2 Review

  10. Capacitors & Inductors • find the voltage v(t) for a capacitor C = 0.5F with the current as shown and v(0) = 0 Discussion #20 – Exam 2 Review

  11. Capacitors & Inductors • find the voltage v(t) for a capacitor C = 0.5F with the current as shown and v(0) = 0 Discussion #20 – Exam 2 Review

  12. Capacitors & Inductors • find the voltage v(t) for a capacitor C = 0.5F with the current as shown and v(0) = 0 Discussion #20 – Exam 2 Review

  13. Capacitors & Inductors • find the voltage v(t) for a capacitor C = 0.5F with the current as shown and v(0) = 0 Discussion #20 – Exam 2 Review

  14. Capacitors & Inductors • find the voltage v(t) for a capacitor C = 0.5F with the current as shown and v(0) = 0 • NB: The final value of the capacitor voltage after the current source has stopped charging the capacitor depends on two things: • The initial capacitor voltage • The history of the capacitor current Discussion #20 – Exam 2 Review

  15. Measuring Signal Strength • Compute the rms value of the sinusoidal current i(t) = I cos(ωt) Discussion #20 – Exam 2 Review

  16. Measuring Signal Strength • Compute the rms value of the sinusoidal current i(t) = I cos(ωt) Integrating a sinusoidal waveform over 2 periods equals zero Discussion #20 – Exam 2 Review

  17. Measuring Signal Strength • Compute the rms value of the sinusoidal current i(t) = I cos(ωt) The RMS value of any sinusoid signal is always equal to 0.707 times the peak value (regardless of phase or frequency) Discussion #20 – Exam 2 Review

  18. + – + – + – v1(t) v2(t) vs(t) ~ ~ ~ Phasors • compute the phasor voltage for the equivalent voltage vs(t) v1(t) = 15cos(377t+π/4) v2(t) = 15cos(377t+π/12) Discussion #20 – Exam 2 Review

  19. + – + – + – v1(t) v2(t) vs(t) ~ ~ ~ Phasors • compute the phasor voltage for the equivalent voltage vs(t) v1(t) = 15cos(377t+π/4) v2(t) = 15cos(377t+π/12) • Write voltages in phasor notation Discussion #20 – Exam 2 Review

  20. + – + – + – v1(t) v2(t) vs(t) ~ ~ ~ Phasors • compute the phasor voltage for the equivalent voltage vs(t) v1(t) = 15cos(377t+π/4) v2(t) = 15cos(377t+π/12) • Write voltages in phasor notation • Convert phasor voltages from polar to rectangular form (see Appendix A) Discussion #20 – Exam 2 Review

  21. + – + – + – v1(t) v2(t) vs(t) ~ ~ ~ Phasors • compute the phasor voltage for the equivalent voltage vs(t) v1(t) = 15cos(377t+π/4) v2(t) = 15cos(377t+π/12) • Write voltages in phasor notation • Convert phasor voltages from polar to rectangular form (see Appendix A) • Combine voltages Discussion #20 – Exam 2 Review

  22. + – + – + – v1(t) v2(t) vs(t) ~ ~ ~ Phasors • compute the phasor voltage for the equivalent voltage vs(t) v1(t) = 15cos(377t+π/4) v2(t) = 15cos(377t+π/12) • Write voltages in phasor notation • Convert phasor voltages from polar to rectangular form (see Appendix A) • Combine voltages • Convert rectangular back to polar Discussion #20 – Exam 2 Review

  23. + – + – + – v1(t) v2(t) vs(t) ~ ~ ~ Phasors • compute the phasor voltage for the equivalent voltage vs(t) v1(t) = 15cos(377t+π/4) v2(t) = 15cos(377t+π/12) • Write voltages in phasor notation • Convert phasor voltages from polar to rectangular form (see Appendix A) • Combine voltages • Convert rectangular back to polar • Convert from phasor to time domain NB: the answer is NOT simply the addition of the amplitudes of v1(t) and v2(t) (i.e. 15 + 15), and the addition of their phases (i.e. π/4 + π/12) Bring ωt back Discussion #20 – Exam 2 Review

  24. Im + – + – + – v1(t) v2(t) vs(t) ~ ~ ~ π/6 Vs(jω) 14.49 Re 25.10 Phasors • compute the phasor voltage for the equivalent voltage vs(t) v1(t) = 15cos(377t+π/4) v2(t) = 15cos(377t+π/12) Discussion #20 – Exam 2 Review

  25. i(t) i(t) i(t) + vR(t) – + vC(t) – + vL(t) – R C L + – + – + – + – Vs(jω) vs(t) vs(t) vs(t) ~ ~ ~ ~ I(jω) + VZ(jω) – Z Impedance Impedance: complex resistance (has no physical significance) • will allow us to use network analysis methods such as node voltage, mesh current, etc. • Capacitors and inductors act as frequency-dependent resistors Discussion #20 – Exam 2 Review

  26. Im ωL ZL π/2 Re R + – + R – Vs(jω) ~ ZR -π/2 + C – + L – ZC I(jω) -1/ωC + VZ(jω) – Z Impedance Impedance of resistors, inductors, and capacitors Phasor domain Discussion #20 – Exam 2 Review

  27. R1 L ZEQ R2 C Impedance • find the equivalent impedance (ZEQ) ω = 104 rads/s, C = 10uF, R1 = 100Ω, R2 = 50Ω, L = 10mH Discussion #20 – Exam 2 Review

  28. R1 L ZEQ R2 C Impedance • find the equivalent impedance (ZEQ) ω = 104 rads/s, C = 10uF, R1 = 100Ω, R2 = 50Ω, L = 10mH Discussion #20 – Exam 2 Review

  29. ZEQ1 ZEQ Impedance • find the equivalent impedance (ZEQ) ω = 104 rads/s, C = 10uF, R1 = 100Ω, R2 = 50Ω, L = 10mH R1 L NB: at this frequency (ω) the circuit has an inductive impedance (reactance or phase is positive) Discussion #20 – Exam 2 Review

  30. AC RLC Circuits AC Circuit Analysis • Identify the AC sources and note the excitation frequency (ω) • Convert all sources to the phasor domain • Represent each circuit element by its impedance • Solve the resulting phasor circuit using network analysis methods • Convert from the phasor domain back to the time domain Discussion #20 – Exam 2 Review

  31. R1 L vs(t) R2 C + – ~ ia(t) ib(t) AC RLC Circuits • find ia(t) and ib(t) vs(t) = 15cos(1500t)V, R1 = 100Ω, R2 = 75Ω, L = 0.5H, C = 1uF Discussion #20 – Exam 2 Review

  32. R1 L vs(t) R2 C + – ~ ia(t) ib(t) AC RLC Circuits • find ia(t) and ib(t) vs(t) = 15cos(1500t)V, R1 = 100Ω, R2 = 75Ω, L = 0.5H, C = 1uF • Note frequencies of AC sources Only one AC source - ω = 1500 rad/s Discussion #20 – Exam 2 Review

  33. R1 L ZR1 ZL vs(t) R2 C + – ZC ~ Vs(jω) ZR2 + – ia(t) ib(t) ~ Ia(jω) Ib(jω) AC RLC Circuits • find ia(t) and ib(t) vs(t) = 15cos(1500t)V, R1 = 100Ω, R2 = 75Ω, L = 0.5H, C = 1uF • Note frequencies of AC sources • Convert to phasor domain Discussion #20 – Exam 2 Review

  34. ZR1 ZL ZC Vs(jω) ZR2 + – ~ Ia(jω) Ib(jω) AC RLC Circuits • find ia(t) and ib(t) vs(t) = 15cos(1500t)V, R1 = 100Ω, R2 = 75Ω, L = 0.5H, C = 1uF • Note frequencies of AC sources • Convert to phasor domain • Represent each element by its impedance Discussion #20 – Exam 2 Review

  35. Vs(jω) + – ~ AC RLC Circuits • find ia(t) and ib(t) vs(t) = 15cos(1500t)V, R1 = 100Ω, R2 = 75Ω, L = 0.5H, C = 1uF +ZR1– +ZL– • Solve using network analysis • Mesh current + ZC – + ZR2 – Ia(jω) Ib(jω) Discussion #20 – Exam 2 Review

  36. Vs(jω) + – ~ AC RLC Circuits • find ia(t) and ib(t) vs(t) = 15cos(1500t)V, R1 = 100Ω, R2 = 75Ω, L = 0.5H, C = 1uF +ZR1– +ZL– • Solve using network analysis • Mesh current + ZC – + ZR2 – Ia(jω) Ib(jω) Discussion #20 – Exam 2 Review

  37. Vs(jω) + – ~ AC RLC Circuits • find ia(t) and ib(t) vs(t) = 15cos(1500t)V, R1 = 100Ω, R2 = 75Ω, L = 0.5H, C = 1uF +ZR1– +ZL– • Convert to Time domain + ZC – + ZR2 – Ia(jω) Ib(jω) Discussion #20 – Exam 2 Review

  38. I + V – Load Source Load Load I I VT(jω) IN(jω) + V – + V – ZT + – ZN AC Equivalent Circuits Thévenin and Norton equivalent circuits apply in AC analysis • Equivalent voltage/current will be complex and frequency dependent Thévenin Equivalent Norton Equivalent Discussion #20 – Exam 2 Review

  39. Z3 Z1 Z3 Z1 a a + – Z2 Z2 ZT Vs(jω) Z4 Z4 b b AC Equivalent Circuits Computation of Thévenin and Norton Impedances: • Remove the load (open circuit at load terminal) • Zero all independent sources • Voltage sources short circuit (v = 0) • Current sources open circuit (i = 0) • Compute equivalent impedance across load terminals (with load removed) ZL NB: same procedure as equivalent resistance Discussion #20 – Exam 2 Review

  40. Z3 Z1 a Z3 Z1 a + – Z2 + VT – Vs(jω) + – Z2 Vs(jω) Z4 Z4 b b AC Equivalent Circuits Computing Thévenin voltage: • Remove the load (open circuit at load terminals) • Define the open-circuit voltage (Voc) across the load terminals • Chose a network analysis method to find Voc • node, mesh, superposition, etc. • Thévenin voltage VT = Voc NB: same procedure as equivalent resistance Discussion #20 – Exam 2 Review

  41. Z3 Z1 a Z3 Z1 a + – Z2 IN Vs(jω) + – Z2 Vs(jω) Z4 Z4 b b AC Equivalent Circuits Computing Norton current: • Replace the load with a short circuit • Define the short-circuit current (Isc) across the load terminals • Chose a network analysis method to find Isc • node, mesh, superposition, etc. • Norton current IN = Isc NB: same procedure as equivalent resistance Discussion #20 – Exam 2 Review

  42. L Rs vs(t) + vL – + – C ~ RL AC Equivalent Circuits • find the Thévenin equivalent ω = 103 Hz, Rs = 50Ω, RL = 50Ω, L = 10mH, C = 0.1uF Discussion #20 – Exam 2 Review

  43. L Rs vs(t) + vL – + – C ~ RL AC Equivalent Circuits • find the Thévenin equivalent ω = 103 Hz, Rs = 50Ω, RL = 50Ω, L = 10mH, C = 0.1uF • Note frequencies of AC sources Only one AC source - ω = 103 rad/s Discussion #20 – Exam 2 Review

  44. L ZL Rs Zs vs(t) + vL – + – C ~ Vs(jω) RL ZC + – ~ ZLD AC Equivalent Circuits • find the Thévenin equivalent ω = 103 Hz, Rs = 50Ω, RL = 50Ω, L = 10mH, C = 0.1uF • Note frequencies of AC sources • Convert to phasor domain Discussion #20 – Exam 2 Review

  45. AC Equivalent Circuits • find the Thévenin equivalent ω = 103 Hz, Rs = 50Ω, RL = 50Ω, L = 10mH, C = 0.1uF • Note frequencies of AC sources • Convert to phasor domain • Find ZT • Remove load & zero sources ZL Zs ZC Discussion #20 – Exam 2 Review

  46. AC Equivalent Circuits • find the Thévenin equivalent ω = 103 Hz, Rs = 50Ω, RL = 50Ω, L = 10mH, C = 0.1uF ZL • Note frequencies of AC sources • Convert to phasor domain • Find ZT • Remove load & zero sources • Find VT(jω) • Remove load Zs Vs(jω) ZC + VT(jω) – + – ~ NB: Since no current flows in the circuit once the load is removed: Discussion #20 – Exam 2 Review

  47. ZT VT(jω) + – ~ ZLD AC Equivalent Circuits • find the Thévenin equivalent ω = 103 Hz, Rs = 50Ω, RL = 50Ω, L = 10mH, C = 0.1uF ZL Zs Vs(jω) ZC + – ~ ZLD Discussion #20 – Exam 2 Review

  48. vs R1 + – t = 0 R2 C DC Transient Response Transient response of a circuit consists of 3 parts: • Steady-state response prior to the switching on/off of a DC source • Transient response – the circuit adjusts to the DC source • Steady-state response following the transient response DC Source Energy element Switch Discussion #20 – Exam 2 Review

  49. R1 R1 t → ∞ t = 0 vs vs + – + – R2 R2 R3 R3 C C DC Transient Response – DC Steady-State Initial condition x(0): DC steady state before a switch is first activated • x(0–): right before the switch is closed • x(0+): right after the switch is closed Final condition x(∞): DC steady state a long time after a switch is activated Final condition Initial condition Discussion #20 – Exam 2 Review

  50. DC Transient Response – DC Steady-State Remember – capacitor voltages and inductor currents cannot change instantaneously • Capacitor voltages and inductor currents don’t change right before closing and right after closing a switch Discussion #20 – Exam 2 Review

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