1. ECEN 301Discussion #20 – Exam 2 Review1 DateDayClass No. TitleChaptersHW Due date Lab Due date Exam 10 NovMon20Exam Review LAB 7 EXAM 2 11 NovTue 12 NovWed21Boolean Algebra13.2 – 13.3 13 NovThu 14 NovFriRecitation 15 NovSat 16 NovSun 17 NovMon22Combinational Logic13.3 – 13.5 LAB 10 18 NovTue Schedule…

2. ECEN 301Discussion #20 – Exam 2 Review2 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?

3. ECEN 301Discussion #20 – Exam 2 Review3 Lecture 20 – Exam 2 Review Chapters 4 – 6, 8

4. ECEN 301Discussion #20 – Exam 2 Review4 Exam 2 u12 – 16 November (Monday – Friday) uChapters 4 – 6 and 8 u15 questions Ù12 multiple choice (answer on bubble sheet!) 1 point each Ù3 long answer (show your work!) 4 or 5 points each uClosed book! ÙOne 3x5 card allowed uCalculators allowed uNo time limit uStudy lecture slides and homework

5. ECEN 301Discussion #20 – Exam 2 Review5 Exam 2 Review…Overview 1.Capacitors and Inductors 2.Measuring Signal Strength 3.Phasors 4.Impedance 5.AC RLC Circuits 6.AC Equivalent Circuits 7.DC Transient Response 8.Frequency Response 9.Basic Filters 10.Op-Amps

6. ECEN 301Discussion #20 – Exam 2 Review6 Capacitors & Inductors InductorsCapacitors Passive sign convention Voltage Current Power + L – i + C – i

7. ECEN 301Discussion #20 – Exam 2 Review7 Capacitors & Inductors InductorsCapacitors Energy An instantaneous change is not permitted in: CurrentVoltage Will permit an instantaneous change in: VoltageCurrent With DC source element acts as a: Short CircuitOpen Circuit

8. ECEN 301Discussion #20 – Exam 2 Review8 Capacitors & Inductors 1.What is the difference between the voltage and current behaviour of capacitors and inductors?

9. ECEN 301Discussion #20 – Exam 2 Review9 Capacitors & Inductors Capacitor voltage v C (t) Inductor current i L (t) NB: neither can change instantaneously Capacitor current i C (t) Inductor voltage v L (t) NB: both can change instantaneously 1.What is the difference between the voltage and current behaviour of capacitors and inductors?

10. ECEN 301Discussion #20 – Exam 2 Review10 Capacitors & Inductors 2. find the voltage v(t) for a capacitor C = 0.5F with the current as shown and v(0) = 0

11. ECEN 301Discussion #20 – Exam 2 Review11 Capacitors & Inductors 2. find the voltage v(t) for a capacitor C = 0.5F with the current as shown and v(0) = 0

12. ECEN 301Discussion #20 – Exam 2 Review12 Capacitors & Inductors 2. find the voltage v(t) for a capacitor C = 0.5F with the current as shown and v(0) = 0

13. ECEN 301Discussion #20 – Exam 2 Review13 Capacitors & Inductors 2. find the voltage v(t) for a capacitor C = 0.5F with the current as shown and v(0) = 0

14. ECEN 301Discussion #20 – Exam 2 Review14 Capacitors & Inductors NB: The final value of the capacitor voltage after the current source has stopped charging the capacitor depends on two things: 1.The initial capacitor voltage 2.The history of the capacitor current 2. find the voltage v(t) for a capacitor C = 0.5F with the current as shown and v(0) = 0

15. ECEN 301Discussion #20 – Exam 2 Review15 Measuring Signal Strength 3.Compute the rms value of the sinusoidal current i(t) = I cos(ωt)

16. ECEN 301Discussion #20 – Exam 2 Review16 Measuring Signal Strength Integrating a sinusoidal waveform over 2 periods equals zero 3.Compute the rms value of the sinusoidal current i(t) = I cos(ωt)

17. ECEN 301Discussion #20 – Exam 2 Review17 Measuring Signal Strength The RMS value of any sinusoid signal is always equal to 0.707 times the peak value (regardless of phase or frequency) 3.Compute the rms value of the sinusoidal current i(t) = I cos(ωt)

18. ECEN 301Discussion #20 – Exam 2 Review18 Phasors 4. compute the phasor voltage for the equivalent voltage v s (t) v 1 (t) = 15cos(377t+π/4) v 2 (t) = 15cos(377t+π/12) v 1 (t) +–+– ~ v 2 (t) +–+– ~ v s (t) +–+– ~

19. ECEN 301Discussion #20 – Exam 2 Review19 Phasors v 1 (t) +–+– ~ v 2 (t) +–+– ~ v s (t) +–+– ~ 1.Write voltages in phasor notation 4. compute the phasor voltage for the equivalent voltage v s (t) v 1 (t) = 15cos(377t+π/4) v 2 (t) = 15cos(377t+π/12)

20. ECEN 301Discussion #20 – Exam 2 Review20 Phasors v 1 (t) +–+– ~ v 2 (t) +–+– ~ v s (t) +–+– ~ 1.Write voltages in phasor notation 2.Convert phasor voltages from polar to rectangular form (see Appendix A) 4. compute the phasor voltage for the equivalent voltage v s (t) v 1 (t) = 15cos(377t+π/4) v 2 (t) = 15cos(377t+π/12)

21. ECEN 301Discussion #20 – Exam 2 Review21 Phasors v 1 (t) +–+– ~ v 2 (t) +–+– ~ v s (t) +–+– ~ 1.Write voltages in phasor notation 2.Convert phasor voltages from polar to rectangular form (see Appendix A) 3.Combine voltages 4. compute the phasor voltage for the equivalent voltage v s (t) v 1 (t) = 15cos(377t+π/4) v 2 (t) = 15cos(377t+π/12)

22. ECEN 301Discussion #20 – Exam 2 Review22 Phasors v 1 (t) +–+– ~ v 2 (t) +–+– ~ v s (t) +–+– ~ 1.Write voltages in phasor notation 2.Convert phasor voltages from polar to rectangular form (see Appendix A) 3.Combine voltages 4.Convert rectangular back to polar 4. compute the phasor voltage for the equivalent voltage v s (t) v 1 (t) = 15cos(377t+π/4) v 2 (t) = 15cos(377t+π/12)

23. ECEN 301Discussion #20 – Exam 2 Review23 Phasors v 1 (t) +–+– ~ v 2 (t) +–+– ~ v s (t) +–+– ~ 1.Write voltages in phasor notation 2.Convert phasor voltages from polar to rectangular form (see Appendix A) 3.Combine voltages 4.Convert rectangular back to polar 5.Convert from phasor to time domain Bring ωt back NB: the answer is NOT simply the addition of the amplitudes of v 1 (t) and v 2 (t) (i.e. 15 + 15), and the addition of their phases (i.e. π/4 + π/12) 4. compute the phasor voltage for the equivalent voltage v s (t) v 1 (t) = 15cos(377t+π/4) v 2 (t) = 15cos(377t+π/12)

24. ECEN 301Discussion #20 – Exam 2 Review24 Phasors v 1 (t) +–+– ~ v 2 (t) +–+– ~ v s (t) +–+– ~ Re Im 14.49 π/6 25.10 Vs(jω) 4. compute the phasor voltage for the equivalent voltage v s (t) v 1 (t) = 15cos(377t+π/4) v 2 (t) = 15cos(377t+π/12)

25. ECEN 301Discussion #20 – Exam 2 Review25 Impedance Impedance: complex resistance (has no physical significance) Ùwill allow us to use network analysis methods such as node voltage, mesh current, etc. ÙCapacitors and inductors act as frequency-dependent resistors v s (t) +–+– ~ R + v R (t) – i(t) v s (t) +–+– ~ C + v C (t) – i(t) v s (t) +–+– ~ L + v L (t) – i(t) V s (jω) +–+– ~ + V Z (jω) – I(jω) Z

26. ECEN 301Discussion #20 – Exam 2 Review26 Impedance +L–+L– +C–+C– Re Im -π/2 π/2 R -1/ωC ωLωL ZRZR ZCZC ZLZL Phasor domain +R–+R– V s (jω) +–+– ~ + V Z (jω) – I(jω) Z Impedance of resistors, inductors, and capacitors

27. ECEN 301Discussion #20 – Exam 2 Review27 Impedance 5.find the equivalent impedance (Z EQ ) ω = 10 4 rads/s, C = 10uF, R 1 = 100Ω, R 2 = 50Ω, L = 10mH Z EQ R2R2 C R1R1 L

28. ECEN 301Discussion #20 – Exam 2 Review28 Impedance Z EQ R2R2 C R1R1 L 5.find the equivalent impedance (Z EQ ) ω = 10 4 rads/s, C = 10uF, R 1 = 100Ω, R 2 = 50Ω, L = 10mH

29. ECEN 301Discussion #20 – Exam 2 Review29 Impedance Z EQ R1R1 L Z EQ1 NB: at this frequency (ω) the circuit has an inductive impedance (reactance or phase is positive) 5.find the equivalent impedance (Z EQ ) ω = 10 4 rads/s, C = 10uF, R 1 = 100Ω, R 2 = 50Ω, L = 10mH

30. ECEN 301Discussion #20 – Exam 2 Review30 AC RLC Circuits AC Circuit Analysis 1.Identify the AC sources and note the excitation frequency (ω) 2.Convert all sources to the phasor domain 3.Represent each circuit element by its impedance 4.Solve the resulting phasor circuit using network analysis methods 5.Convert from the phasor domain back to the time domain

31. ECEN 301Discussion #20 – Exam 2 Review31 AC RLC Circuits 6.find i a (t) and i b (t) v s (t) = 15cos(1500t)V, R 1 = 100Ω, R 2 = 75Ω, L = 0.5H, C = 1uF R 1 v s (t) +–+– ~ R2R2 L C i a (t) i b (t)

32. ECEN 301Discussion #20 – Exam 2 Review32 AC RLC Circuits 6. find i a (t) and i b (t) v s (t) = 15cos(1500t)V, R 1 = 100Ω, R 2 = 75Ω, L = 0.5H, C = 1uF R 1 v s (t) +–+– ~ R2R2 L C i a (t) i b (t) 1.Note frequencies of AC sources Only one AC source - ω = 1500 rad/s

33. ECEN 301Discussion #20 – Exam 2 Review33 AC RLC Circuits 6.find i a (t) and i b (t) v s (t) = 15cos(1500t)V, R 1 = 100Ω, R 2 = 75Ω, L = 0.5H, C = 1uF R 1 v s (t) +–+– ~ R2R2 L C i a (t) i b (t) 1.Note frequencies of AC sources 2.Convert to phasor domain Z R1 V s (jω) +–+– ~ Z R2 I a (jω) Z L Z C I b (jω)

34. ECEN 301Discussion #20 – Exam 2 Review34 AC RLC Circuits 6.find i a (t) and i b (t) v s (t) = 15cos(1500t)V, R 1 = 100Ω, R 2 = 75Ω, L = 0.5H, C = 1uF Z R1 V s (jω) +–+– ~ Z R2 I a (jω) Z L Z C I b (jω) 1.Note frequencies of AC sources 2.Convert to phasor domain 3.Represent each element by its impedance

35. ECEN 301Discussion #20 – Exam 2 Review35 AC RLC Circuits 6.find i a (t) and i b (t) v s (t) = 15cos(1500t)V, R 1 = 100Ω, R 2 = 75Ω, L = 0.5H, C = 1uF +Z R1 – V s (jω) +–+– ~ + Z R2 – I a (jω) +Z L – + Z C – I b (jω) 4.Solve using network analysis Mesh current

36. ECEN 301Discussion #20 – Exam 2 Review36 AC RLC Circuits 6.find i a (t) and i b (t) v s (t) = 15cos(1500t)V, R 1 = 100Ω, R 2 = 75Ω, L = 0.5H, C = 1uF +Z R1 – V s (jω) +–+– ~ + Z R2 – I a (jω) +Z L – + Z C – I b (jω) 4.Solve using network analysis Mesh current

37. ECEN 301Discussion #20 – Exam 2 Review37 AC RLC Circuits 6.find i a (t) and i b (t) v s (t) = 15cos(1500t)V, R 1 = 100Ω, R 2 = 75Ω, L = 0.5H, C = 1uF +Z R1 – V s (jω) +–+– ~ + Z R2 – I a (jω) +Z L – + Z C – I b (jω) 5.Convert to Time domain

38. ECEN 301Discussion #20 – Exam 2 Review38 AC Equivalent Circuits Thévenin and Norton equivalent circuits apply in AC analysis ÙEquivalent voltage/current will be complex and frequency dependent Load +V–+V– I Source V T (jω) +–+– ZTZT Load +V–+V– I I N (jω) ZNZN Load +V–+V– I Norton Equivalent Thévenin Equivalent

39. ECEN 301Discussion #20 – Exam 2 Review39 AC Equivalent Circuits Computation of Thévenin and Norton Impedances: 1.Remove the load (open circuit at load terminal) 2.Zero all independent sources ÙVoltage sources short circuit (v = 0) ÙCurrent sources open circuit (i = 0) 3.Compute equivalent impedance across load terminals (with load removed) NB: same procedure as equivalent resistance Z L Z 1 V s (j ω) +–+– Z 3 Z2Z2 Z 4 a b Z 1 Z 3 Z2Z2 Z 4 a b ZTZT

40. ECEN 301Discussion #20 – Exam 2 Review40 AC Equivalent Circuits Computing Thévenin voltage: 1.Remove the load (open circuit at load terminals) 2.Define the open-circuit voltage (V oc ) across the load terminals 3.Chose a network analysis method to find V oc Ùnode, mesh, superposition, etc. 4.Thévenin voltage V T = V oc Z 1 V s (j ω) +–+– Z 3 Z2Z2 Z 4 a b Z 1 V s (j ω) +–+– Z 3 Z2Z2 Z 4 a b +VT–+VT– NB: same procedure as equivalent resistance

41. ECEN 301Discussion #20 – Exam 2 Review41 AC Equivalent Circuits Computing Norton current: 1.Replace the load with a short circuit 2.Define the short-circuit current (I sc ) across the load terminals 3.Chose a network analysis method to find I sc Ùnode, mesh, superposition, etc. 4.Norton current I N = I sc Z 1 V s (j ω) +–+– Z 3 Z2Z2 Z 4 a b Z 1 V s (j ω) +–+– Z 3 Z2Z2 Z 4 a b ININ NB: same procedure as equivalent resistance

42. ECEN 301Discussion #20 – Exam 2 Review42 AC Equivalent Circuits 7.find the Thévenin equivalent ω = 10 3 Hz, R s = 50Ω, R L = 50Ω, L = 10mH, C = 0.1uF R s v s (t) +–+– ~ RLRL L C +vL–+vL–

43. ECEN 301Discussion #20 – Exam 2 Review43 AC Equivalent Circuits 7.find the Thévenin equivalent ω = 10 3 Hz, R s = 50Ω, R L = 50Ω, L = 10mH, C = 0.1uF R s v s (t) +–+– ~ RLRL L C +vL–+vL– 1.Note frequencies of AC sources Only one AC source - ω = 10 3 rad/s

44. ECEN 301Discussion #20 – Exam 2 Review44 AC Equivalent Circuits 7.find the Thévenin equivalent ω = 10 3 Hz, R s = 50Ω, R L = 50Ω, L = 10mH, C = 0.1uF R s v s (t) +–+– ~ RLRL L C +vL–+vL– 1.Note frequencies of AC sources 2.Convert to phasor domain Z s Z LD +–+– ~ ZLZL ZCZC V s (jω)

45. ECEN 301Discussion #20 – Exam 2 Review45 AC Equivalent Circuits 7.find the Thévenin equivalent ω = 10 3 Hz, R s = 50Ω, R L = 50Ω, L = 10mH, C = 0.1uF 1.Note frequencies of AC sources 2.Convert to phasor domain 3.Find Z T Remove load & zero sources Z s ZLZL ZCZC

46. ECEN 301Discussion #20 – Exam 2 Review46 AC Equivalent Circuits 7.find the Thévenin equivalent ω = 10 3 Hz, R s = 50Ω, R L = 50Ω, L = 10mH, C = 0.1uF 1.Note frequencies of AC sources 2.Convert to phasor domain 3.Find Z T Remove load & zero sources 4.Find V T (jω) Remove load Z s +–+– ~ ZLZL ZCZC V s (jω) + V T (jω) – NB: Since no current flows in the circuit once the load is removed:

47. ECEN 301Discussion #20 – Exam 2 Review47 AC Equivalent Circuits 7.find the Thévenin equivalent ω = 10 3 Hz, R s = 50Ω, R L = 50Ω, L = 10mH, C = 0.1uF Z s Z LD +–+– ~ ZLZL ZCZC V s (jω) Z T +–+– ~ V T (jω) Z LD

48. ECEN 301Discussion #20 – Exam 2 Review48 DC Transient Response Transient response of a circuit consists of 3 parts: 1.Steady-state response prior to the switching on/off of a DC source 2.Transient response – the circuit adjusts to the DC source 3.Steady-state response following the transient response R 1 R2R2 C vsvs +–+– t = 0 DC Source Switch Energy element

49. ECEN 301Discussion #20 – Exam 2 Review49 Initial condition x(0): DC steady state before a switch is first activated Ùx(0 – ): right before the switch is closed Ùx(0 + ): right after the switch is closed Final condition x(∞): DC steady state a long time after a switch is activated R 1 R2R2 C vsvs +–+– t = 0 R3R3 R 1 R2R2 C vsvs +–+– t → ∞ R3R3 Initial condition Final condition DC Transient Response – DC Steady-State

50. ECEN 301Discussion #20 – Exam 2 Review50 Remember – capacitor voltages and inductor currents cannot change instantaneously ÙCapacitor voltages and inductor currents don’t change right before closing and right after closing a switch DC Transient Response – DC Steady-State

51. ECEN 301Discussion #20 – Exam 2 Review51 DC Transient Response – DC Steady-State 8.find the initial and final current conditions at the inductor i s = 10mA isis t = 0 R L iLiL

52. ECEN 301Discussion #20 – Exam 2 Review52 8.find the initial and final current conditions at the inductor i s = 10mA isis t = 0 R L iLiL 1.Initial conditions – assume the current across the inductor is in steady-state. isis iLiL NB: in DC steady state inductors act like short circuits, thus no current flows through R DC Transient Response – DC Steady-State

53. ECEN 301Discussion #20 – Exam 2 Review53 8.find the initial and final current conditions at the inductor i s = 10mA 1.Initial conditions – assume the current across the inductor is in steady-state. isis iLiL DC Transient Response – DC Steady-State

54. ECEN 301Discussion #20 – Exam 2 Review54 8.find the initial and final current conditions at the inductor i s = 10mA isis t = 0 R L iLiL 1.Initial conditions – assume the current across the inductor is in steady-state. 2.Throw the switch NB: inductor current cannot change instantaneously R L iLiL DC Transient Response – DC Steady-State

55. ECEN 301Discussion #20 – Exam 2 Review55 8.find the initial and final current conditions at the inductor i s = 10mA isis t = 0 R L iLiL 1.Initial conditions – assume the current across the inductor is in steady-state. 2.Throw the switch NB: inductor current cannot change instantaneously –R+–R+ L iLiL NB: polarity of R DC Transient Response – DC Steady-State

56. ECEN 301Discussion #20 – Exam 2 Review56 8.find the initial and final current conditions at the inductor i s = 10mA isis t = 0 R L iLiL 1.Initial conditions – assume the current across the inductor is in steady-state. 2.Throw the switch 3.Final conditions NB: in DC steady state inductors act like short circuits DC Transient Response – DC Steady-State

57. ECEN 301Discussion #20 – Exam 2 Review57 DC Transient Response Solving 1 st order transient response: 1.Solve the DC steady-state circuit: ÙInitial condition x(0 – ): before switching (on/off) ÙFinal condition x(∞): After any transients have died out (t → ∞) 2.Identify x(0 + ): the circuit initial conditions ÙCapacitors: v C (0 + ) = v C (0 – ) ÙInductors: i L (0 + ) = i L (0 – ) 3.Write a differential equation for the circuit at time t = 0 + ÙReduce the circuit to its Thévenin or Norton equivalent ÙThe energy storage element (capacitor or inductor) is the load ÙThe differential equation will be either in terms of v C (t) or i L (t) ÙReduce this equation to standard form 4.Solve for the time constant ÙCapacitive circuits: τ = R T C ÙInductive circuits: τ = L/R T 5.Write the complete response in the form: Ùx(t) = x(∞) + [x(0) - x(∞)]e -t/τ

58. ECEN 301Discussion #20 – Exam 2 Review58 DC Transient Response 9.find v c (t) for all t v s = 12V, v C (0 – ) = 5V, R = 1000Ω, C = 470uF R C vsvs +–+– t = 0 i(t) + v C (t) –

59. ECEN 301Discussion #20 – Exam 2 Review59 9.find v c (t) for all t v s = 12V, v C (0 – ) = 5V, R = 1000Ω, C = 470uF R C vsvs +–+– t = 0 i(t) + v C (t) – 1.DC steady-state a)Initial condition: v C (0) b)Final condition: v C (∞) NB: as t → ∞ the capacitor acts like an open circuit thus v C (∞) = v S DC Transient Response

60. ECEN 301Discussion #20 – Exam 2 Review60 9.find v c (t) for all t v s = 12V, v C (0 – ) = 5V, R = 1000Ω, C = 470uF R C vsvs +–+– t = 0 i(t) + v C (t) – 2.Circuit initial conditions: v C (0 + ) DC Transient Response

61. ECEN 301Discussion #20 – Exam 2 Review61 9.find v c (t) for all t v s = 12V, v C (0 – ) = 5V, R = 1000Ω, C = 470uF R C vsvs +–+– t = 0 i(t) + v C (t) – 3.Write differential equation (already in Thévenin equivalent) at t = 0 DC Transient Response

62. ECEN 301Discussion #20 – Exam 2 Review62 9.find v c (t) for all t v s = 12V, v C (0 – ) = 5V, R = 1000Ω, C = 470uF R C vsvs +–+– t = 0 i(t) + v C (t) – 4.Find the time constant τ DC Transient Response

63. ECEN 301Discussion #20 – Exam 2 Review63 9.find v c (t) for all t v s = 12V, v C (0 – ) = 5V, R = 1000Ω, C = 470uF R C vsvs +–+– t = 0 i(t) + v C (t) – 5.Write the complete response x(t) = x(∞) + [x(0) - x(∞)]e -t/τ DC Transient Response

64. ECEN 301Discussion #20 – Exam 2 Review64 Frequency Response Frequency Response H(jω): a measure of how the voltage/current/impedance of a load responds to the voltage/current of a source

65. ECEN 301Discussion #20 – Exam 2 Review65 Frequency Response 10.compute the frequency response H V (jω) R 1 = 1kΩ, R L = 10kΩ, C = 10uF R 1 C v s (t) +–+– +RL–+RL–

66. ECEN 301Discussion #20 – Exam 2 Review66 Frequency Response 10.compute the frequency response H V (jω) R 1 = 1kΩ, R L = 10kΩ, C = 10uF R 1 C v s (t) +–+– +RL–+RL– 1.Note frequencies of AC sources Only one AC source so frequency response H V (jω) will be the function of a single frequency

67. ECEN 301Discussion #20 – Exam 2 Review67 Frequency Response 10.compute the frequency response H V (jω) R 1 = 1kΩ, R L = 10kΩ, C = 10uF R 1 C v s (t) +–+– +RL–+RL– 1.Note frequencies of AC sources 2.Convert to phasor domain Z 1 = R 1 Z LD =R L Z C =1/jωC VsVs +–+– ~

68. ECEN 301Discussion #20 – Exam 2 Review68 Frequency Response 10.compute the frequency response H V (jω) R 1 = 1kΩ, R L = 10kΩ, C = 10uF 1.Note frequencies of AC sources 2.Convert to phasor domain 3.Solve using network analysis Thévenin equivalent Z 1 = R 1 Z C =1/jωC

69. ECEN 301Discussion #20 – Exam 2 Review69 Frequency Response 10.compute the frequency response H V (jω) R 1 = 1kΩ, R L = 10kΩ, C = 10uF 1.Note frequencies of AC sources 2.Convert to phasor domain 3.Solve using network analysis Thévenin equivalent Z 1 = R 1 +VT–+VT– Z C =1/jωC VsVs +–+– ~

70. ECEN 301Discussion #20 – Exam 2 Review70 Frequency Response 10.compute the frequency response H V (jω) R 1 = 1kΩ, R L = 10kΩ, C = 10uF 1.Note frequencies of AC sources 2.Convert to phasor domain 3.Solve using network analysis Thévenin equivalent 4.Find an expression for the load voltage Z T Z LD VTVT +–+– ~ +VL–+VL–

71. ECEN 301Discussion #20 – Exam 2 Review71 Frequency Response 10.compute the frequency response H V (jω) R 1 = 1kΩ, R L = 10kΩ, C = 10uF 5.Find an expression for the frequency response Z T Z LD VTVT +–+– ~ +VL–+VL–

72. ECEN 301Discussion #20 – Exam 2 Review72 Frequency Response 10.compute the frequency response H V (jω) R 1 = 1kΩ, R L = 10kΩ, C = 10uF 5.Find an expression for the frequency response Z T Z LD VTVT +–+– ~ +VL–+VL–

73. ECEN 301Discussion #20 – Exam 2 Review73 Frequency Response 10.compute the frequency response H V (jω) R 1 = 1kΩ, R L = 10kΩ, C = 10uF 5.Find an expression for the frequency response Look at response for low frequencies (ω = 10) and high frequencies (ω = 10000) Z T Z LD VTVT +–+– ~ +VL–+VL– ω = 10 ω = 10000

74. ECEN 301Discussion #20 – Exam 2 Review74 Basic Filters Electric circuit filter: attenuates (reduces) or eliminates signals at unwanted frequencies Low-pass High-pass Band-pass Band-stop ωω ω ω

75. ECEN 301Discussion #20 – Exam 2 Review75 Basic Filters – Resonant Frequency Resonant Frequency (ω n ): the frequency at which capacitive impedance and inductive impedance are equal and opposite (in 2 nd order filters) C + v i (t) – + v o (t) – L C + v i (t) – + v o (t) – L Impedances in seriesImpedances in parallel

76. ECEN 301Discussion #20 – Exam 2 Review76 Basic Filters – Resonant Frequency Resonant Frequency (ω n ): the frequency at which capacitive impedance and inductive impedance are equal and opposite (in 2 nd order filters) + V i (jω) – + V o (jω) – Z EQ =0 Impedances in series

77. ECEN 301Discussion #20 – Exam 2 Review77 Basic Filters – Resonant Frequency Resonant Frequency (ω n ): the frequency at which capacitive impedance and inductive impedance are equal and opposite (in 2 nd order filters) + V i (jω) – + V o (jω) – Z EQ =∞ Impedances in parallel

78. ECEN 301Discussion #20 – Exam 2 Review78 Basic Filters – Low-Pass Filters Low-pass Filters: only allow signals under the cutoff frequency (ω 0 ) to pass Low-pass ω1ω1 ω0ω0 ω2ω2 ω3ω3 H L (jω) R C + v i (t) – + v o (t) – R C + v i (t) – + v o (t) – L 1 st Order 2 nd Order

79. ECEN 301Discussion #20 – Exam 2 Review79 Basic Filters – High-Pass Filters High-pass Filters: only allow signals above the cutoff frequency (ω 0 ) to pass H H (jω) High-pass ω1ω1 ω0ω0 ω2ω2 ω3ω3 + v i (t) – + v o (t) – 1 st Order R C R + v i (t) – + v o (t) – 2 nd Order C L

80. ECEN 301Discussion #20 – Exam 2 Review80 Basic Filters – Band-Pass Filters Band-pass Filters: only allow signals between the passband (ω a to ω b ) to pass H B (jω) Band-pass ω1ω1 ωaωa ω2ω2 ω3ω3 ωbωb + v i (t) – + v o (t) – 2 nd Order R C L

81. ECEN 301Discussion #20 – Exam 2 Review81 Basic Filters – Band-Stop Filters Band-stop Filters: allow signals except those between the stopband (ω a to ω b ) to pass H N (jω) + v i (t) – + v o (t) – 2 nd Order R C L Band-stop ω1ω1 ωaωa ω2ω2 ω3ω3 ωbωb

82. ECEN 301Discussion #20 – Exam 2 Review82 Op-Amps – Open-Loop Mode Open-Loop Model: an ideal op-amp acts like a difference amplifier (a device that amplifies the difference between two input voltages) – + +v+–+v+– +v––+v–– +vo–+vo– ioio i2i2 i1i1 – v in + – + +–+– R out R in i1i1 A OL v in +vo–+vo– – v in + v–v– v+v+ NB: op-amps have near-infinite input resistance (R in ) and very small output resistance (R out ) A OL – open-loop voltage gain

83. ECEN 301Discussion #20 – Exam 2 Review83 Op-Amps – Closed-Loop Mode Circuit DiagramA CL Inverting Amplifier Summing Amplifier – + +vo–+vo– +–+– vSvS RSRS RFRF – + +vo–+vo– +–+– +–+– +–+– R Sn R S2 R S1 v Sn v S2 v S1 RFRF

84. ECEN 301Discussion #20 – Exam 2 Review84 Op-Amps – Closed-Loop Mode Circuit DiagramA CL Noninverting Amplifier Voltage Follower – + +vo–+vo– +–+– R RSRS RFRF vSvS – + +vo–+vo– +–+–

85. ECEN 301Discussion #20 – Exam 2 Review85 Op-Amps – Closed-Loop Mode Circuit DiagramA CL Differential Amplifier – + +vo–+vo– +–+– +–+– RSRS RSRS RFRF RFRF v1v1 v2v2

86. ECEN 301Discussion #20 – Exam 2 Review86 Op-Amps – Closed-Loop Mode Circuit DiagramA CL Ideal Integrator Ideal Differentiator – + + v o (t) – +–+– vSvS CSCS RFRF – + + v o (t) – +–+– vSvS RSRS CFCF

87. ECEN 301Discussion #20 – Exam 2 Review87 Op-Amps 11.find an expression for the gain C F = 1/6 F, R 1 = 3Ω, R 2 = 2Ω, C S = 1/6 F + – v o (t) i in CFCF R2R2 v+v+ v–v– i F (t) i 2 (t) R1R1 i 1 (t) CSCS i S (t) v s (t)

88. ECEN 301Discussion #20 – Exam 2 Review88 Op-Amps 11.find an expression for the gain C F = 1/6 F, R 1 = 3Ω, R 2 = 2Ω, C S = 1/6 F + – Vo(jω)Vo(jω) I in Z F =1/jωC F Z2Z2 v+v+ v–v– IF(jω)IF(jω) I2(jω)I2(jω) Z1Z1 I 1 (jω) ZSZS IS(jω)IS(jω) Vs(jω)Vs(jω) Node a Node b 1.Transfer to frequency domain 2.Apply KCL at nodes a and b NB: v + = v – and I in = 0

89. ECEN 301Discussion #20 – Exam 2 Review89 Op-Amps 11.find an expression for the gain C F = 1/6 F, R 1 = 3Ω, R 2 = 2Ω, C S = 1/6 F + – Vo(jω)Vo(jω) I in Z F =1/jωC F Z2Z2 v+v+ v–v– IF(jω)IF(jω) I2(jω)I2(jω) Z1Z1 I 1 (jω) ZSZS IS(jω)IS(jω) Vs(jω)Vs(jω) Node a Node b 1.Transfer to frequency domain 2.Apply KCL at nodes a and b

90. ECEN 301Discussion #20 – Exam 2 Review90 Op-Amps 11.find an expression for the gain C F = 1/6 F, R 1 = 3Ω, R 2 = 2Ω, C S = 1/6 F + – Vo(jω)Vo(jω) I in Z F =1/jωC F Z2Z2 v+v+ v–v– IF(jω)IF(jω) I2(jω)I2(jω) Z1Z1 I 1 (jω) ZSZS IS(jω)IS(jω) Vs(jω)Vs(jω) 1.Transfer to frequency domain 2.Apply KCL at nodes a and b 3.Express V o in terms of V s

91. ECEN 301Discussion #20 – Exam 2 Review91 Op-Amps 11.find an expression for the gain C F = 1/6 F, R 1 = 3Ω, R 2 = 2Ω, C S = 1/6 F + – Vo(jω)Vo(jω) I in Z F =1/jωC F Z2Z2 v+v+ v–v– IF(jω)IF(jω) I2(jω)I2(jω) Z1Z1 I 1 (jω) ZSZS IS(jω)IS(jω) Vs(jω)Vs(jω) 1.Transfer to frequency domain 2.Apply KCL at nodes a and b 3.Express V o in terms of V S 4.Find the gain (V o /V S )