Schedule. ECEN 301 Discussion #20 Exam 2 Review 1. Lab Due date. Title Chapters HW Due date. Date Day Class No. 10 Nov Mon 20 Exam Review.

Transcription

1 Schedule Date Day lass No. 0 Nov Mon 0 Exam Review Nov Tue Title hapters HW Due date Nov Wed Boolean Algebra ab Due date AB 7 Exam EXAM 3 Nov Thu 4 Nov Fri Recitation 5 Nov Sat 6 Nov Sun 7 Nov Mon ombinational ogic Nov Tue AB 0 EEN 30 Discussion #0 Exam Review

2 Ask Alma 5:6 6 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? EEN 30 Discussion #0 Exam Review

3 ecture 0 Exam Review hapters 4 6, 8 EEN 30 Discussion #0 Exam Review 3

4 Exam 6 November (Monday Friday hapters 4 6 and 8 5 questions multiple choice (answer on bubble sheet! point each 3 long answer (show your work! 4 or 5 points each losed book! One 3x5 card allowed alculators allowed No time limit Study lecture slides and homework EEN 30 Discussion #0 Exam Review 4

5 Exam Review Overview. apacitors and Inductors. Measuring Signal Strength 3. Phasors 4. Impedance 5. A R ircuits 6. A Equivalent ircuits 7. D Transient Response 8. Frequency Response 9. Basic Filters 0. Op-Amps EEN 30 Discussion #0 Exam Review 5

6 apacitors & Inductors Passive sign convention oltage urrent Power Inductors i di( t v( t dt t i( t v ( d i( t0 t0 di( t P ( t i( t dt apacitors i t v( t i ( d v( t0 t0 i( t dv( t dt P ( t dv( t v( t dt EEN 30 Discussion #0 Exam Review 6

7 apacitors & Inductors Energy An instantaneous change is not permitted in: Will permit an instantaneous change in: With D source element acts as a: Inductors apacitors W ( t i( t W ( t v( t urrent oltage oltage urrent Short ircuit Open ircuit EEN 30 Discussion #0 Exam Review 7

8 apacitors & Inductors. What is the difference between the voltage and current behaviour of capacitors and inductors? EEN 30 Discussion #0 Exam Review 8

9 apacitors & Inductors. What is the difference between the voltage and current behaviour of capacitors and inductors? apacitor voltage v (t Inductor current i (t NB: neither can change instantaneously apacitor current i (t Inductor voltage v (t NB: both can change instantaneously EEN 30 Discussion #0 Exam Review 9

10 current (A apacitors & Inductors. find the voltage v(t for a capacitor = 0.5F with the current as shown and v(0 = time (s EEN 30 Discussion #0 Exam Review 0

11 current (A apacitors & Inductors. find the voltage v(t for a capacitor = 0.5F with the current as shown and v(0 = current i(t i 0 t 0 in 4 intervals t 0 t : t time (s 0 t v( t 0 t id v(0 EEN 30 Discussion #0 Exam Review

12 current (A apacitors & Inductors. find the voltage v(t for a capacitor = 0.5F with the current as shown and v(0 = 0..0 voltage v(t in 4 intervals : v 0 t 0 d 0 0 t t time (s t ( d v( t 0 v( t v( t 0 t id v(0 EEN 30 Discussion #0 Exam Review

13 current (A apacitors & Inductors. find the voltage v(t for a capacitor = 0.5F with the current as shown and v(0 = voltage v 0 t v(t in 4 intervals t 0 0 t : time (s t 3 t t EEN 30 Discussion #0 Exam Review 3

14 current (A voltage ( apacitors & Inductors. find the voltage v(t for a capacitor = 0.5F with the current as shown and v(0 = time (s time (s 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 voltage v(t in 4 intervals v 0 t 0 t 0 t t t 3 t : EEN 30 Discussion #0 Exam Review 4

15 Measuring Signal Strength 3. ompute the rms value of the sinusoidal current i(t = I cos(ωt EEN 30 Discussion #0 Exam Review 5

16 EEN 30 Discussion #0 Exam Review 6 Measuring Signal Strength 0 cos( cos( ( cos ( / 0 / 0 / 0 0 I I d I I d I d I d i T i T rms Integrating a sinusoidal waveform over periods equals zero cos( ( cos t t 3. ompute the rms value of the sinusoidal current i(t = I cos(ωt

17 Measuring Signal Strength 3. ompute the rms value of the sinusoidal current i(t = I cos(ωt i rms T T 0 i ( d 0 / I cos ( d 0 / I cos( d The RMS value of any sinusoid signal is always equal to times the peak value (regardless of phase or frequency I I I 0 0 / I cos( d EEN 30 Discussion #0 Exam Review 7

18 Phasors 4. compute the phasor voltage for the equivalent voltage v s (t v (t = 5cos(377tπ/4 v (t = 5cos(377tπ/ v (t v (t ~ ~ v s (t ~ EEN 30 Discussion #0 Exam Review 8

19 Phasors 4. compute the phasor voltage for the equivalent voltage v s (t v (t = 5cos(377tπ/4 v (t = 5cos(377tπ/ v (t v (t ~ ~. Write voltages in phasor notation ( ( j j 5e 5 5e j j / 4 4 / v s (t ~ 5 EEN 30 Discussion #0 Exam Review 9

20 Phasors 4. compute the phasor voltage for the equivalent voltage v s (t v (t = 5cos(377tπ/4 v (t = 5cos(377tπ/ v (t v (t ~ ~ ( j onvert to 5 4 rectangula r :. Write voltages in phasor notation. onvert phasor voltages from polar to rectangular form (see Appendix A ( j onvert to 5 rectangula r : v s (t ~ ( j 5 cos 0.6 j5 sin 4 j0.6 4 ( j 5 cos 4.49 j5 sin j3.88 EEN 30 Discussion #0 Exam Review 0

21 Phasors 4. compute the phasor voltage for the equivalent voltage v s (t v (t = 5cos(377tπ/4 v (t = 5cos(377tπ/ v (t v (t ~ ~. Write voltages in phasor notation. onvert phasor voltages from polar to rectangular form (see Appendix A 3. ombine voltages v s (t ~ S ( j ( j ( j 5.0 j4.49 EEN 30 Discussion #0 Exam Review

22 Phasors 4. compute the phasor voltage for the equivalent voltage v s (t v (t v (t = 5cos(377tπ/4 v (t = 5cos(377tπ/ ~. Write voltages in phasor notation. onvert phasor voltages from polar to rectangular form (see Appendix A 3. ombine voltages 4. onvert rectangular back to polar v (t ~ S ( j onvert 5.0 to polar : j4.49 r (5.0 ( tan v s (t ~ S ( j EEN 30 Discussion #0 Exam Review

23 Phasors 4. compute the phasor voltage for the equivalent voltage v s (t v (t v (t v s (t v (t = 5cos(377tπ/4 v (t = 5cos(377tπ/ ~ ~ ~ S ( j v S ( t. Write voltages in phasor notation. onvert phasor voltages from polar to rectangular form (see Appendix A 3. ombine voltages 4. onvert rectangular back to polar 5. onvert from phasor to time domain cos 377 t 6 Bring ωt back NB: the answer is NOT simply the addition of the amplitudes of v (t and v (t (i.e. 5 5, and the addition of their phases (i.e. π/4 π/ EEN 30 Discussion #0 Exam Review 3

24 Phasors 4. compute the phasor voltage for the equivalent voltage v s (t v (t = 5cos(377tπ/4 v (t = 5cos(377tπ/ v (t ~ v (t v s (t ~ ~ S ( j v S ( t cos 377 t 6 Im 4.49 π/6 s(jω Re 5.0 EEN 30 Discussion #0 Exam Review 4

25 Impedance Impedance: complex resistance (has no physical significance will allow us to use network analysis methods such as node voltage, mesh current, etc. apacitors and inductors act as frequency-dependent resistors i(t i(t i(t v s (t ~ R v R (t v s (t ~ v (t v s (t ~ v (t I(jω s (jω ~ (jω EEN 30 Discussion #0 Exam Review 5

26 Impedance Impedance of resistors, inductors, and capacitors I(jω Im ω Phasor domain s (jω ~ (jω π/ R Re -π/ -/ω R R R ( j R ( j j ( j j EEN 30 Discussion #0 Exam Review 6

27 Impedance 5. find the equivalent impedance ( EQ ω = 0 4 rads/s, = 0uF, R = 00Ω, R = 50Ω, = 0mH R EQ R EEN 30 Discussion #0 Exam Review 7

28 Impedance 5. find the equivalent impedance ( EQ ω = 0 4 rads/s, = 0uF, R = 00Ω, R = 50Ω, = 0mH R R EQ EQ R R R (/ j (/ j R j R j(0 50 (0 50 j5.9 j ( (50 EEN 30 Discussion #0 Exam Review 8

29 Impedance 5. find the equivalent impedance ( EQ ω = 0 4 rads/s, = 0uF, R = 00Ω, R = 50Ω, = 0mH R EQ R EQ R 00 j 9.8 j(0 4 (0 (.37.9 j9.6 EQ 0.9 j90.38 EQ EQ 9.8 (.37 NB: at this frequency (ω the circuit has an inductive impedance (reactance or phase is positive EEN 30 Discussion #0 Exam Review 9

30 A R ircuits A ircuit Analysis. Identify the A sources and note the excitation frequency (ω. onvert 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. onvert from the phasor domain back to the time domain EEN 30 Discussion #0 Exam Review 30

31 A R ircuits 6. find i a (t and i b (t v s (t = 5cos(500t, R = 00Ω, R = 75Ω, = 0.5H, = uf R v s (t ~ i a (t i b (t R EEN 30 Discussion #0 Exam Review 3

32 A R ircuits 6. find i a (t and i b (t v s (t = 5cos(500t, R = 00Ω, R = 75Ω, = 0.5H, = uf R. Note frequencies of A sources v s (t ~ i a (t i b (t R Only one A source - ω = 500 rad/s EEN 30 Discussion #0 Exam Review 3

33 A R ircuits 6. find i a (t and i b (t v s (t = 5cos(500t, R = 00Ω, R = 75Ω, = 0.5H, = uf R. Note frequencies of A sources. onvert to phasor domain v s (t ~ i a (t i b (t R R s (jω ~ I a (jω I b (jω R EEN 30 Discussion #0 Exam Review 33

34 A R ircuits 6. find i a (t and i b (t v s (t = 5cos(500t, R = 00Ω, R = 75Ω, = 0.5H, = uf R. Note frequencies of A sources. onvert to phasor domain 3. Represent each element by its impedance s (jω ~ I a (jω I b (jω R s v s ( t ( j 5 cos( 5 0 t 5e j0 j / j R R 00 j(500(0.5 j750 R R 75 / j(500 (0 j667 6 EEN 30 Discussion #0 Exam Review 34

35 EEN 30 Discussion #0 Exam Review 35 A R ircuits 6. find i a (t and i b (t v s (t = 5cos(500t, R = 00Ω, R = 75Ω, = 0.5H, = uf 0 ( 0 ( 0 ( ( ( K at b : R b a R b b b a R I I I I I I j j j R s (jω ~ R I a (jω I b (jω j j R R 4. Solve using network analysis Mesh current s b R a s b a R a R s I I I I I j j j ( ( 0 ( ( ( K at a :

36 A R ircuits 6. find i a (t and i b (t v s (t = 5cos(500t, R = 00Ω, R = 75Ω, = 0.5H, = uf R 4. Solve using network analysis Mesh current s (jω ~ I a (jω I b (jω R I I a a ( 00 j667 I ( j667 ( j667 I (75 j83 b b 5 0 I I A A EEN 30 Discussion #0 Exam Review 36

37 A R ircuits 6. find i a (t and i b (t v s (t = 5cos(500t, R = 00Ω, R = 75Ω, = 0.5H, = uf R 5. onvert to Time domain s (jω ~ I A I R a (jω I (jω i ( t 3.cos(500 t 0.97 ma b I i ( t A 9 cos(500t.49 ma EEN 30 Discussion #0 Exam Review 37

38 A Equivalent ircuits Thévenin and Norton equivalent circuits apply in A analysis Equivalent voltage/current will be complex and frequency dependent I Thévenin Equivalent Source oad Norton Equivalent I I T (jω T oad I N (jω N oad EEN 30 Discussion #0 Exam Review 38

39 A Equivalent ircuits omputation of Thévenin and Norton Impedances:. Remove the load (open circuit at load terminal. ero all independent sources oltage sources short circuit (v = 0 urrent sources open circuit (i = 0 3. ompute equivalent impedance across load terminals (with load removed 3 a 3 a s (jω 4 4 T b b NB: same procedure as equivalent resistance EEN 30 Discussion #0 Exam Review 39

40 A Equivalent ircuits omputing Thévenin voltage:. Remove the load (open circuit at load terminals. Define the open-circuit voltage ( oc across the load terminals 3. hose a network analysis method to find oc node, mesh, superposition, etc. 4. Thévenin voltage T = oc 3 a 3 a s (jω 4 s (jω 4 T b NB: same procedure as equivalent resistance b EEN 30 Discussion #0 Exam Review 40

41 A Equivalent ircuits omputing Norton current:. Replace the load with a short circuit. Define the short-circuit current (I sc across the load terminals 3. hose a network analysis method to find I sc node, mesh, superposition, etc. 4. Norton current I N = I sc 3 a 3 a s (jω 4 s (jω 4 I N b NB: same procedure as equivalent resistance b EEN 30 Discussion #0 Exam Review 4

42 A Equivalent ircuits 7. find the Thévenin equivalent ω = 0 3 Hz, R s = 50Ω, R = 50Ω, = 0mH, = 0.uF R s v s (t ~ R v EEN 30 Discussion #0 Exam Review 4

43 A Equivalent ircuits 7. find the Thévenin equivalent ω = 0 3 Hz, R s = 50Ω, R = 50Ω, = 0mH, = 0.uF R s. Note frequencies of A sources Only one A source - ω = 0 3 rad/s v s (t ~ R v EEN 30 Discussion #0 Exam Review 43

44 A Equivalent ircuits 7. find the Thévenin equivalent ω = 0 3 Hz, R s = 50Ω, R = 50Ω, = 0mH, = 0.uF R s. Note frequencies of A sources. onvert to phasor domain v s (t ~ R v s (jω ~ s D EEN 30 Discussion #0 Exam Review 44

45 A Equivalent ircuits 7. find the Thévenin equivalent ω = 0 3 Hz, R s = 50Ω, R = 50Ω, = 0mH, = 0.uF s. Note frequencies of A sources. onvert to phasor domain 3. Find T Remove load & zero sources T R R S S S 50 ( j ( j (/ j (/ j j j EEN 30 Discussion #0 Exam Review 45

46 A Equivalent ircuits 7. find the Thévenin equivalent ω = 0 3 Hz, R s = 50Ω, R = 50Ω, = 0mH, = 0.uF s (jω ~ s T (jω. Note frequencies of A sources. onvert to phasor domain 3. Find T Remove load & zero sources 4. Find T (jω Remove load NB: Since no current flows in the circuit once the load is removed: T T S EEN 30 Discussion #0 Exam Review 46

47 A Equivalent ircuits 7. find the Thévenin equivalent ω = 0 3 Hz, R s = 50Ω, R = 50Ω, = 0mH, = 0.uF s T s (jω ~ D T (jω ~ D T S T EEN 30 Discussion #0 Exam Review 47

48 D Transient Response Transient response of a circuit consists of 3 parts:. Steady-state response prior to the switching on/off of a D source. Transient response the circuit adjusts to the D source 3. Steady-state response following the transient response R v s t = 0 R D Source Switch Energy element EEN 30 Discussion #0 Exam Review 48

49 D Transient Response D Steady-State Initial condition x(0: D 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( : D steady state a long time after a switch is activated R R t = 0 t v s R R 3 v s R R 3 Initial condition Final condition EEN 30 Discussion #0 Exam Review 49

50 D Transient Response D Steady-State Remember capacitor voltages and inductor currents cannot change instantaneously apacitor voltages and inductor currents don t change right before closing and right after closing a switch v (0 v (0 i (0 i (0 EEN 30 Discussion #0 Exam Review 50

51 D Transient Response D Steady-State 8. find the initial and final current conditions at the inductor i s = 0mA t = 0 i s i R EEN 30 Discussion #0 Exam Review 5

52 D Transient Response D Steady-State 8. find the initial and final current conditions at the inductor i s = 0mA. Initial conditions assume the current across the inductor is in steady-state. t = 0 i s i R i s i NB: in D steady state inductors act like short circuits, thus no current flows through R EEN 30 Discussion #0 Exam Review 5

53 D Transient Response D Steady-State 8. find the initial and final current conditions at the inductor i s = 0mA. Initial conditions assume the current across the inductor is in steady-state. i s i i (0 i s 0mA EEN 30 Discussion #0 Exam Review 53

54 D Transient Response D Steady-State 8. find the initial and final current conditions at the inductor i s = 0mA. Initial conditions assume the current across the inductor is in steady-state.. Throw the switch t = 0 i s i R i R NB: inductor current cannot change instantaneously EEN 30 Discussion #0 Exam Review 54

55 D Transient Response D Steady-State 8. find the initial and final current conditions at the inductor i s = 0mA. Initial conditions assume the current across the inductor is in steady-state.. Throw the switch t = 0 i s i R i R i (0 i (0 0mA NB: polarity of R NB: inductor current cannot change instantaneously EEN 30 Discussion #0 Exam Review 55

56 D Transient Response D Steady-State 8. find the initial and final current conditions at the inductor i s = 0mA t = 0. Initial conditions assume the current across the inductor is in steady-state.. Throw the switch 3. Final conditions i s i R i ( 0A NB: in D steady state inductors act like short circuits EEN 30 Discussion #0 Exam Review 56

57 D Transient Response Solving st order transient response:. Solve the D steady-state circuit: Initial condition x(0 : before switching (on/off Final condition x( : After any transients have died out (t. Identify x(0 : the circuit initial conditions apacitors: v (0 = v (0 Inductors: i (0 = i (0 3. Write a differential equation for the circuit at time t = 0 Reduce the circuit to its Thévenin or Norton equivalent The energy storage element (capacitor or inductor is the load The differential equation will be either in terms of v (t or i (t Reduce this equation to standard form 4. Solve for the time constant apacitive circuits: τ = R T Inductive circuits: τ = /R T 5. Write the complete response in the form: x(t = x( [x(0 - x( ]e -t/τ EEN 30 Discussion #0 Exam Review 57

58 D Transient Response 9. find v c (t for all t v s =, v (0 = 5, R = 000Ω, = 470uF R v s t = 0 i(t v (t EEN 30 Discussion #0 Exam Review 58

59 D Transient Response 9. find v c (t for all t v s =, v (0 = 5, R = 000Ω, = 470uF v s t = 0 i(t R v (t. D steady-state a Initial condition: v (0 b Final condition: v ( v ( t 0 v (0 5 NB: as t the capacitor acts like an open circuit thus v ( = v S v ( v S EEN 30 Discussion #0 Exam Review 59

60 D Transient Response 9. find v c (t for all t v s =, v (0 = 5, R = 000Ω, = 470uF R. ircuit initial conditions: v (0 v s t = 0 i(t v (t v (0 v 5 (0 EEN 30 Discussion #0 Exam Review 60

61 D Transient Response 9. find v c (t for all t v s =, v (0 = 5, R = 000Ω, = 470uF R 3. Write differential equation (already in Thévenin equivalent at t = 0 v s t = 0 i(t v (t v S R i v R ( t ( t R dv ( t dt K : v v v ( t ( t ( t 0 v v S S EEN 30 Discussion #0 Exam Review 6

62 D Transient Response 9. find v c (t for all t v s =, v (0 = 5, R = 000Ω, = 470uF R 4. Find the time constant τ v s t = 0 i(t v (t R (000( K S F v S EEN 30 Discussion #0 Exam Review 6

63 D Transient Response 9. find v c (t for all t v s =, v (0 = 5, R = 000Ω, = 470uF R 5. Write the complete response x(t = x( [x(0 - x( ]e -t/τ v s t = 0 i(t v (t v ( t v ( v (0 v ( e t / (5 e t / e t / 0.47 EEN 30 Discussion #0 Exam Review 63

64 EEN 30 Discussion #0 Exam Review 64 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 ( ( ( j j j H S ( ( ( j I j I j H S I ( ( ( j I j j H S

65 Frequency Response 0. compute the frequency response H (jω R = kω, R = 0kΩ, = 0uF R v s (t R EEN 30 Discussion #0 Exam Review 65

66 Frequency Response 0. compute the frequency response H (jω R = kω, R = 0kΩ, = 0uF R. Note frequencies of A sources v s (t R Only one A source so frequency response H (jω will be the function of a single frequency EEN 30 Discussion #0 Exam Review 66

67 Frequency Response 0. compute the frequency response H (jω R = kω, R = 0kΩ, = 0uF. Note frequencies of A sources. onvert to phasor domain R = R v s (t R s ~ =/jω D =R EEN 30 Discussion #0 Exam Review 67

68 Frequency Response 0. compute the frequency response H (jω R = kω, R = 0kΩ, = 0uF = R. Note frequencies of A sources. onvert to phasor domain 3. Solve using network analysis Thévenin equivalent =/jω T EEN 30 Discussion #0 Exam Review 68

69 Frequency Response 0. compute the frequency response H (jω R = kω, R = 0kΩ, = 0uF = R. Note frequencies of A sources. onvert to phasor domain 3. Solve using network analysis Thévenin equivalent s ~ =/jω T T ( j S ( j T EEN 30 Discussion #0 Exam Review 69

70 Frequency Response 0. compute the frequency response H (jω R = kω, R = 0kΩ, = 0uF T ~ T ( j S T T ( j D ( j. Note frequencies of A sources. onvert to phasor domain 3. Solve using network analysis Thévenin equivalent 4. Find an expression for the load voltage T ( S j ( j T D D D D EEN 30 Discussion #0 Exam Review 70

71 Frequency Response 0. compute the frequency response H (jω R = kω, R = 0kΩ, = 0uF T 5. Find an expression for the frequency response T ~ D H ( j S ( j ( j D D EEN 30 Discussion #0 Exam Review 7

72 Frequency Response 0. compute the frequency response H (jω R = kω, R = 0kΩ, = 0uF T 5. Find an expression for the frequency response T ~ D H ( j 0 D j D 0 /( j 4 D /( j 0 5 D /( j arctan 0 EEN 30 Discussion #0 Exam Review 7

73 Frequency Response 0. compute the frequency response H (jω R = kω, R = 0kΩ, = 0uF T 5. Find an expression for the frequency response ook at response for low frequencies (ω = 0 and high frequencies (ω = 0000 T ~ 00 D H ( j arctan 0 0 ω = 0 H ( j H ( j ω = EEN 30 Discussion #0 Exam Review 73

74 Basic Filters Electric circuit filter: attenuates (reduces or eliminates signals at unwanted frequencies ow-pass High-pass ω ω Band-pass Band-stop ω ω EEN 30 Discussion #0 Exam Review 74

75 Basic Filters Resonant Frequency Resonant Frequency (ω n : the frequency at which capacitive impedance and inductive impedance are equal and opposite (in nd order filters n Impedances in series Impedances in parallel v i (t v o (t v i (t v o (t EEN 30 Discussion #0 Exam Review 75

76 Basic Filters Resonant Frequency Resonant Frequency (ω n : the frequency at which capacitive impedance and inductive impedance are equal and opposite (in nd order filters Impedances in series i (jω EQ =0 o (jω j EQ 0 j o ( j 0 EEN 30 Discussion #0 Exam Review 76

77 Basic Filters Resonant Frequency Resonant Frequency (ω n : the frequency at which capacitive impedance and inductive impedance are equal and opposite (in nd order filters Impedances in parallel j j i (jω o EQ = ( j ( j i o (jω EQ / 0 EEN 30 Discussion #0 Exam Review 77

78 Basic Filters ow-pass Filters ow-pass Filters: only allow signals under the cutoff frequency (ω 0 to pass ow-pass R st Order v i (t v o (t ω. ω 0 ω ω 3 v i ( t cos( cos( cos( t 3 t t H (jω v o ( t cos( t v i (t R nd Order v o (t EEN 30 Discussion #0 Exam Review 78

79 Basic Filters High-Pass Filters High-pass Filters: only allow signals above the cutoff frequency (ω 0 to pass High-pass st Order v i (t R v o (t ω. ω 0 ω ω 3 v i ( t cos( cos( cos( t 3 t t H H (jω v o ( t cos( t 3 v i (t R nd Order v o (t EEN 30 Discussion #0 Exam Review 79

80 Basic Filters Band-Pass Filters Band-pass Filters: only allow signals between the passband (ω a to ω b to pass 0.0 Band-pass ω ω a ω. ω 3 ω b v i (t R nd Order v o (t v i ( t cos( cos( cos( t 3 t t H B (jω v o ( t cos( t EEN 30 Discussion #0 Exam Review 80

81 Basic Filters Band-Stop Filters Band-stop Filters: allow signals except those between the stopband (ω a to ω b to pass Band-stop nd Order 0.0 ω ω. a ω ω 3 ω b v i (t R v o (t v i ( t cos( cos( cos( t 3 t t H N (jω v o ( t cos( cos( t 3 t EEN 30 Discussion #0 Exam Review 8

82 Op-Amps Open-oop Mode Open-oop Model: an ideal op-amp acts like a difference amplifier (a device that amplifies the difference between two input voltages i v v v in i i i o v o v v v in R in R out A O v in v o NB: op-amps have near-infinite input resistance (R in and very small output resistance (R out v o A A O O v in ( v v A O open-loop voltage gain EEN 30 Discussion #0 Exam Review 8

83 Op-Amps losed-oop Mode ircuit Diagram A Inverting Amplifier R S v S R F v o v o A R R v F S S v S Summing Amplifier R S v S R S v S R Sn v Sn R F v o v o N A On n N RF n Rn v Sn v Sn EEN 30 Discussion #0 Exam Review 83

84 Op-Amps losed-oop Mode ircuit Diagram A Noninverting Amplifier R S R v S R F v o v o A v S R R F S v S oltage Follower v o v o A v s v s EEN 30 Discussion #0 Exam Review 84

85 Op-Amps losed-oop Mode ircuit Diagram A Differential Amplifier v R S R S v R F R F v o v o R R F S v v S S EEN 30 Discussion #0 Exam Review 85

86 Op-Amps losed-oop Mode ircuit Diagram A Ideal Integrator R S v S F v o (t v t o vs ( d R S F Ideal Differentiator v S S R F v o (t v o R F S dvs ( t dt EEN 30 Discussion #0 Exam Review 86

87 Op-Amps. find an expression for the gain F = /6 F, R = 3Ω, R = Ω, S = /6 F F v s (t R i F (t R v v o (t i (t i (t i S (t S i in v EEN 30 Discussion #0 Exam Review 87

88 Op-Amps. find an expression for the gain F = /6 F, R = 3Ω, R = Ω, S = /6 F s (jω I (jω Node a Node b I F (jω F =/jω F v I (jω I S (jω S I in v o (jω a. Transfer to frequency domain. Apply K at nodes a and b S F a o F o I a K at a : I F a F I o 0 0 S NB: v = v and I in = 0 a 3 j 6 o j 6 S 3 EEN 30 Discussion #0 Exam Review 88

89 Op-Amps. find an expression for the gain F = /6 F, R = 3Ω, R = Ω, S = /6 F s (jω I (jω Node a Node b I F (jω F =/jω F v I (jω I S (jω S I in v o (jω. Transfer to frequency domain. Apply K at nodes a and b a a o I o K at b : S I S o S I in a o j 6 0 EEN 30 Discussion #0 Exam Review 89

90 Op-Amps. find an expression for the gain F = /6 F, R = 3Ω, R = Ω, S = /6 F s (jω I (jω I F (jω F =/jω F v I (jω I S (jω S I in v o (jω. Transfer to frequency domain. Apply K at nodes a and b 3. Express o in terms of s 5 j 3 j a 3 3 j a o o 0 S o 6 j5 S 6 EEN 30 Discussion #0 Exam Review 90

91 Op-Amps. find an expression for the gain F = /6 F, R = 3Ω, R = Ω, S = /6 F s (jω I F (jω F =/jω F v o (jω. Transfer to frequency domain. Apply K at nodes a and b 3. Express o in terms of S 4. Find the gain ( o / S I (jω I (jω I S (jω S I in v o S 6 j5 6 EEN 30 Discussion #0 Exam Review 9