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.


 Marcus Arnold
 3 years ago
 Views:
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. OpAmps 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 frequencydependent 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 opencircuit 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 shortcircuit 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:. Steadystate response prior to the switching on/off of a D source. Transient response the circuit adjusts to the D source 3. Steadystate 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 SteadyState 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 SteadyState 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 SteadyState 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 SteadyState 8. find the initial and final current conditions at the inductor i s = 0mA. Initial conditions assume the current across the inductor is in steadystate. 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 SteadyState 8. find the initial and final current conditions at the inductor i s = 0mA. Initial conditions assume the current across the inductor is in steadystate. i s i i (0 i s 0mA EEN 30 Discussion #0 Exam Review 53
54 D Transient Response D SteadyState 8. find the initial and final current conditions at the inductor i s = 0mA. Initial conditions assume the current across the inductor is in steadystate.. 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 SteadyState 8. find the initial and final current conditions at the inductor i s = 0mA. Initial conditions assume the current across the inductor is in steadystate.. 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 SteadyState 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 steadystate.. 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 steadystate 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 steadystate 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 owpass Highpass ω ω Bandpass Bandstop 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 owpass Filters owpass Filters: only allow signals under the cutoff frequency (ω 0 to pass owpass 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 HighPass Filters Highpass Filters: only allow signals above the cutoff frequency (ω 0 to pass Highpass 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 BandPass Filters Bandpass Filters: only allow signals between the passband (ω a to ω b to pass 0.0 Bandpass 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 BandStop Filters Bandstop Filters: allow signals except those between the stopband (ω a to ω b to pass Bandstop 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 OpAmps Openoop Mode Openoop Model: an ideal opamp 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: opamps have nearinfinite input resistance (R in and very small output resistance (R out v o A A O O v in ( v v A O openloop voltage gain EEN 30 Discussion #0 Exam Review 8
83 OpAmps losedoop 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 OpAmps losedoop 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 OpAmps losedoop 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 OpAmps losedoop 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 OpAmps. 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 OpAmps. 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 OpAmps. 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 OpAmps. 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 OpAmps. 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
4/27 Friday. I have all the old homework if you need to collect them.
4/27 Friday Last HW: do not need to turn it. Solution will be posted on the web. I have all the old homework if you need to collect them. Final exam: 79pm, Monday, 4/30 at Lambert Fieldhouse F101 Calculator
More informationCourse Updates. Reminders: 1) Assignment #10 due Today. 2) Quiz # 5 Friday (Chap 29, 30) 3) Start AC Circuits
ourse Updates http://www.phys.hawaii.edu/~varner/phys272spr10/physics272.html eminders: 1) Assignment #10 due Today 2) Quiz # 5 Friday (hap 29, 30) 3) Start A ircuits Alternating urrents (hap 31) In this
More informationNotes on Electric Circuits (Dr. Ramakant Srivastava)
Notes on Electric ircuits (Dr. Ramakant Srivastava) Passive Sign onvention (PS) Passive sign convention deals with the designation of the polarity of the voltage and the direction of the current arrow
More informationChapter 10 AC Analysis Using Phasors
Chapter 10 AC Analysis Using Phasors 10.1 Introduction We would like to use our linear circuit theorems (Nodal analysis, Mesh analysis, Thevenin and Norton equivalent circuits, Superposition, etc.) to
More informationSinusoidal Steady State Analysis (AC Analysis) Part I
Sinusoidal Steady State Analysis (AC Analysis) Part I Amin Electronics and Electrical Communications Engineering Department (EECE) Cairo University elc.n102.eng@gmail.com http://scholar.cu.edu.eg/refky/
More informationChapter 10: Sinusoidal SteadyState Analysis
Chapter 10: Sinusoidal SteadyState Analysis 1 Objectives : sinusoidal functions Impedance use phasors to determine the forced response of a circuit subjected to sinusoidal excitation Apply techniques
More informationChapter 9 Objectives
Chapter 9 Engr8 Circuit Analysis Dr Curtis Nelson Chapter 9 Objectives Understand the concept of a phasor; Be able to transform a circuit with a sinusoidal source into the frequency domain using phasor
More informationECE 201 Fall 2009 Final Exam
ECE 01 Fall 009 Final Exam December 16, 009 Division 0101: Tan (11:30am) Division 001: Clark (7:30 am) Division 0301: Elliott (1:30 pm) Instructions 1. DO NOT START UNTIL TOLD TO DO SO.. Write your Name,
More informationLectures 16 & 17 Sinusoidal Signals, Complex Numbers, Phasors, Impedance & AC Circuits. Nov. 7 & 9, 2011
Lectures 16 & 17 Sinusoidal Signals, Complex Numbers, Phasors, Impedance & AC Circuits Nov. 7 & 9, 2011 Material from Textbook by Alexander & Sadiku and Electrical Engineering: Principles & Applications,
More informationSinusoids and Phasors
CHAPTER 9 Sinusoids and Phasors We now begins the analysis of circuits in which the voltage or current sources are timevarying. In this chapter, we are particularly interested in sinusoidally timevarying
More informationModule 4. Singlephase AC Circuits. Version 2 EE IIT, Kharagpur 1
Module 4 Singlephase A ircuits ersion EE IIT, Kharagpur esson 4 Solution of urrent in  Series ircuits ersion EE IIT, Kharagpur In the last lesson, two points were described:. How to represent a sinusoidal
More informationEE 205 Dr. A. Zidouri. Electric Circuits II. Frequency Selective Circuits (Filters) Low Pass Filter. Lecture #36
EE 05 Dr. A. Zidouri Electric ircuits II Frequency Selective ircuits (Filters) ow Pass Filter ecture #36   EE 05 Dr. A. Zidouri The material to be covered in this lecture is as follows: o Introduction
More information12 Chapter Driven RLC Circuits
hapter Driven ircuits. A Sources... . A ircuits with a Source and One ircuit Element... 3.. Purely esistive oad... 3.. Purely Inductive oad... 6..3 Purely apacitive oad... 8.3 The Series ircuit...
More informationIntroduction to AC Circuits (Capacitors and Inductors)
Introduction to AC Circuits (Capacitors and Inductors) Amin Electronics and Electrical Communications Engineering Department (EECE) Cairo University elc.n102.eng@gmail.com http://scholar.cu.edu.eg/refky/
More information1 Phasors and Alternating Currents
Physics 4 Chapter : Alternating Current 0/5 Phasors and Alternating Currents alternating current: current that varies sinusoidally with time ac source: any device that supplies a sinusoidally varying potential
More informationDesigning Information Devices and Systems II Spring 2016 Anant Sahai and Michel Maharbiz Midterm 2
EECS 16B Designing Information Devices and Systems II Spring 2016 Anant Sahai and Michel Maharbiz Midterm 2 Exam location: 145 Dwinelle (SIDs ending in 1 and 5) PRINT your student ID: PRINT AND SIGN your
More informationEE313 Fall 2013 Exam #1 (100 pts) Thursday, September 26, 2013 Name. 1) [6 pts] Convert the following timedomain circuit to the RMS Phasor Domain.
Name If you have any questions ask them. Remember to include all units on your answers (V, A, etc). Clearly indicate your answers. All angles must be in the range 0 to +180 or 0 to 180 degrees. 1) [6 pts]
More informationSinusoidal Steady State Analysis (AC Analysis) Part II
Sinusoidal Steady State Analysis (AC Analysis) Part II Amin Electronics and Electrical Communications Engineering Department (EECE) Cairo University elc.n102.eng@gmail.com http://scholar.cu.edu.eg/refky/
More informationPhysics 142 AC Circuits Page 1. AC Circuits. I ve had a perfectly lovely evening but this wasn t it. Groucho Marx
Physics 142 A ircuits Page 1 A ircuits I ve had a perfectly lovely evening but this wasn t it. Groucho Marx Alternating current: generators and values It is relatively easy to devise a source (a generator
More informationBME/ISE 3511 Bioelectronics  Test Six Course Notes Fall 2016
BME/ISE 35 Bioelectronics  Test Six ourse Notes Fall 06 Alternating urrent apacitive & Inductive Reactance and omplex Impedance R & R ircuit Analyses (D Transients, Time onstants, Steady State) Electrical
More informationChapter 33. Alternating Current Circuits
Chapter 33 Alternating Current Circuits 1 Capacitor Resistor + Q = C V = I R R I + + Inductance d I Vab = L dt AC power source The AC power source provides an alternative voltage, Notation  Lower case
More informationEE 3120 Electric Energy Systems Study Guide for Prerequisite Test Wednesday, Jan 18, pm, Room TBA
EE 3120 Electric Energy Systems Study Guide for Prerequisite Test Wednesday, Jan 18, 2006 67 pm, Room TBA First retrieve your EE2110 final and other course papers and notes! The test will be closed book
More informationAC Circuits Homework Set
Problem 1. In an oscillating LC circuit in which C=4.0 μf, the maximum potential difference across the capacitor during the oscillations is 1.50 V and the maximum current through the inductor is 50.0 ma.
More informationEIT Review. Electrical Circuits DC Circuits. Lecturer: Russ Tatro. Presented by Tau Beta Pi The Engineering Honor Society 10/3/2006 1
EIT Review Electrical Circuits DC Circuits Lecturer: Russ Tatro Presented by Tau Beta Pi The Engineering Honor Society 10/3/2006 1 Session Outline Basic Concepts Basic Laws Methods of Analysis Circuit
More informationPhasors: Impedance and Circuit Anlysis. Phasors
Phasors: Impedance and Circuit Anlysis Lecture 6, 0/07/05 OUTLINE Phasor ReCap Capacitor/Inductor Example Arithmetic with Complex Numbers Complex Impedance Circuit Analysis with Complex Impedance Phasor
More informationChapter 10: Sinusoids and Phasors
Chapter 10: Sinusoids and Phasors 1. Motivation 2. Sinusoid Features 3. Phasors 4. Phasor Relationships for Circuit Elements 5. Impedance and Admittance 6. Kirchhoff s Laws in the Frequency Domain 7. Impedance
More informationSinusoidal SteadyState Analysis
Sinusoidal SteadyState Analysis Almost all electrical systems, whether signal or power, operate with alternating currents and voltages. We have seen that when any circuit is disturbed (switched on or
More informationVer 3537 E1.1 Analysis of Circuits (2014) E1.1 Circuit Analysis. Problem Sheet 1 (Lectures 1 & 2)
Ver 3537 E. Analysis of Circuits () Key: [A]= easy... [E]=hard E. Circuit Analysis Problem Sheet (Lectures & ). [A] One of the following circuits is a series circuit and the other is a parallel circuit.
More informationSinusoidal SteadyState Analysis
Chapter 4 Sinusoidal SteadyState Analysis In this unit, we consider circuits in which the sources are sinusoidal in nature. The review section of this unit covers most of section 9.1 9.9 of the text.
More informationREACTANCE. By: Enzo Paterno Date: 03/2013
REACTANCE REACTANCE By: Enzo Paterno Date: 03/2013 5/2007 Enzo Paterno 1 RESISTANCE  R i R (t R A resistor for all practical purposes is unaffected by the frequency of the applied sinusoidal voltage or
More informationALTERNATING CURRENT
ATENATING UENT Important oints:. The alternating current (A) is generally expressed as ( ) I I sin ω t + φ Where i peak value of alternating current.. emf of an alternating current source is generally
More informationCIRCUIT ANALYSIS II. (AC Circuits)
Will Moore MT & MT CIRCUIT ANALYSIS II (AC Circuits) Syllabus Complex impedance, power factor, frequency response of AC networks including Bode diagrams, secondorder and resonant circuits, damping and
More informationELEC 2501 AB. Come to the PASS workshop with your mock exam complete. During the workshop you can work with other students to review your work.
It is most beneficial to you to write this mock midterm UNDER EXAM CONDITIONS. This means: Complete the midterm in 3 hour(s). Work on your own. Keep your notes and textbook closed. Attempt every question.
More informationBME/ISE 3511 Bioelectronics  Test Five Review Notes Fall 2015
BME/ISE 35 Bioelectronics  Test Five Review Notes Fall 205 Test Five Topics: RMS Resistive Power oss (I 2 R) A Reactance, Impedance, Power Factor R ircuit Analysis alculate Series R Impedance alculate
More informationLecture 11  AC Power
 AC Power 11/17/2015 Reading: Chapter 11 1 Outline Instantaneous power Complex power Average (real) power Reactive power Apparent power Maximum power transfer Power factor correction 2 Power in AC Circuits
More informationEE292: Fundamentals of ECE
EE292: Fundamentals of ECE Fall 2012 TTh 10:0011:15 SEB 1242 Lecture 20 121101 http://www.ee.unlv.edu/~b1morris/ee292/ 2 Outline Chapters 13 Circuit Analysis Techniques Chapter 10 Diodes Ideal Model
More informationLecture 24. Impedance of AC Circuits.
Lecture 4. Impedance of AC Circuits. Don t forget to complete course evaluations: https://sakai.rutgers.edu/portal/site/sirs Posttest. You are required to attend one of the lectures on Thursday, Dec.
More informationFirst Order RC and RL Transient Circuits
First Order R and RL Transient ircuits Objectives To introduce the transients phenomena. To analyze step and natural responses of first order R circuits. To analyze step and natural responses of first
More informationElectric Circuit Theory
Electric Circuit Theory Nam Ki Min nkmin@korea.ac.kr 01094192320 Chapter 11 Sinusoidal SteadyState Analysis Nam Ki Min nkmin@korea.ac.kr 01094192320 Contents and Objectives 3 Chapter Contents 11.1
More informationPhysics272 Lecture 20. AC Power Resonant Circuits Phasors (2dim vectors, amplitude and phase)
Physics7 ecture 0 A Power esonant ircuits Phasors (dim vectors, amplitude and phase) What is reactance? You can think of it as a frequencydependent resistance. 1 ω For high ω, χ ~0  apacitor looks
More informationmywbut.com Lesson 16 Solution of Current in AC Parallel and Seriesparallel
esson 6 Solution of urrent in Parallel and Seriesparallel ircuits n the last lesson, the following points were described:. How to compute the total impedance/admittance in series/parallel circuits?. How
More informationDriven RLC Circuits Challenge Problem Solutions
Driven LC Circuits Challenge Problem Solutions Problem : Using the same circuit as in problem 6, only this time leaving the function generator on and driving below resonance, which in the following pairs
More informationModule 4. Singlephase AC Circuits
Module 4 Singlephase AC Circuits Lesson 14 Solution of Current in RLC Series Circuits In the last lesson, two points were described: 1. How to represent a sinusoidal (ac) quantity, i.e. voltage/current
More informationECE Spring 2015 Final Exam
ECE 20100 Spring 2015 Final Exam May 7, 2015 Section (circle below) Jung (1:30) 0001 Qi (12:30) 0002 Peleato (9:30) 0004 Allen (10:30) 0005 Zhu (4:30) 0006 Name PUID Instructions 1. DO NOT START UNTIL
More informationEECE 2150 Circuits and Signals Final Exam Fall 2016 Dec 16
EECE 2150 Circuits and Signals Final Exam Fall 2016 Dec 16 Instructions: Write your name and section number on all pages Closed book, closed notes; Computers and cell phones are not allowed You can use
More informationFigure Circuit for Question 1. Figure Circuit for Question 2
Exercises 10.7 Exercises Multiple Choice 1. For the circuit of Figure 10.44 the time constant is A. 0.5 ms 71.43 µs 2, 000 s D. 0.2 ms 4 Ω 2 Ω 12 Ω 1 mh 12u 0 () t V Figure 10.44. Circuit for Question
More informationINTC 1307 Instrumentation Test Equipment Teaching Unit 6 AC Bridges
IHLAN OLLEGE chool of Engineering & Technology ev. 0 W. lonecker ev. (8/6/0) J. Bradbury INT 307 Instrumentation Test Equipment Teaching Unit 6 A Bridges Unit 6: A Bridges OBJETIVE:. To explain the operation
More informationSinusoidal Steady State Analysis
Sinusoidal Steady State Analysis 9 Assessment Problems AP 9. [a] V = 70/ 40 V [b] 0 sin(000t +20 ) = 0 cos(000t 70 ).. I = 0/ 70 A [c] I =5/36.87 + 0/ 53.3 =4+j3+6 j8 =0 j5 =.8/ 26.57 A [d] sin(20,000πt
More informationChapter 3: Capacitors, Inductors, and Complex Impedance
hapter 3: apacitors, Inductors, and omplex Impedance In this chapter we introduce the concept of complex resistance, or impedance, by studying two reactive circuit elements, the capacitor and the inductor.
More informationSinusoidal Response of RLC Circuits
Sinusoidal Response of RLC Circuits Series RL circuit Series RC circuit Series RLC circuit Parallel RL circuit Parallel RC circuit RL Series Circuit RL Series Circuit RL Series Circuit Instantaneous
More informationUC DAVIS. Circuits I Course Outline
UC DAVIS Circuits I Course Outline ENG 17 Professor Spencer Fall 2010 2041 Kemper Hall Lecture: MWF 4:105:00, 1003 Giedt Hall 7526885 Discussion Section 1: W 1:102:00, 55 Roessler CRN: 61417 Discussion
More informationChapter 31: RLC Circuits. PHY2049: Chapter 31 1
hapter 31: RL ircuits PHY049: hapter 31 1 L Oscillations onservation of energy Topics Damped oscillations in RL circuits Energy loss A current RMS quantities Forced oscillations Resistance, reactance,
More informationProf. Anyes Taffard. Physics 120/220. Voltage Divider Capacitor RC circuits
Prof. Anyes Taffard Physics 120/220 Voltage Divider Capacitor RC circuits Voltage Divider The figure is called a voltage divider. It s one of the most useful and important circuit elements we will encounter.
More informationC R. Consider from point of view of energy! Consider the RC and LC series circuits shown:
ircuits onsider the R and series circuits shown: ++++  R ++++  Suppose that the circuits are formed at t with the capacitor charged to value. There is a qualitative difference in the time development
More informationChapter 6 Objectives
hapter 6 Engr8 ircuit Analysis Dr urtis Nelson hapter 6 Objectives Understand relationships between voltage, current, power, and energy in inductors and capacitors; Know that current must be continuous
More informationECE 205: Intro Elec & Electr Circuits
ECE 205: Intro Elec & Electr Circuits Final Exam Study Guide Version 1.00 Created by Charles Feng http://www.fenguin.net ECE 205: Intro Elec & Electr Circuits Final Exam Study Guide 1 Contents 1 Introductory
More informationPhysics 116A Notes Fall 2004
Physics 116A Notes Fall 2004 David E. Pellett Draft v.0.9 Notes Copyright 2004 David E. Pellett unless stated otherwise. References: Text for course: Fundamentals of Electrical Engineering, second edition,
More informationHomework Assignment 11
Homework Assignment Question State and then explain in 2 3 sentences, the advantage of switched capacitor filters compared to continuoustime active filters. (3 points) Continuous time filters use resistors
More informationOPERATIONAL AMPLIFIER APPLICATIONS
OPERATIONAL AMPLIFIER APPLICATIONS 2.1 The Ideal Op Amp (Chapter 2.1) Amplifier Applications 2.2 The Inverting Configuration (Chapter 2.2) 2.3 The Noninverting Configuration (Chapter 2.3) 2.4 Difference
More informationALTERNATING CURRENT. with X C = 0.34 A. SET UP: The specified value is the rootmeansquare current; I. EXECUTE: (a) V = (0.34 A) = 0.12 A.
ATENATING UENT 3 3 IDENTIFY: i Icosωt and I I/ SET UP: The specified value is the rootmeansquare current; I 34 A EXEUTE: (a) I 34 A (b) I I (34 A) 48 A (c) Since the current is positive half of the time
More informationTo find the step response of an RC circuit
To find the step response of an RC circuit v( t) v( ) [ v( t) v( )] e tt The time constant = RC The final capacitor voltage v() The initial capacitor voltage v(t ) To find the step response of an RL circuit
More informationChapter 10 Sinusoidal Steady State Analysis Chapter Objectives:
Chapter 10 Sinusoidal Steady State Analysis Chapter Objectives: Apply previously learn circuit techniques to sinusoidal steadystate analysis. Learn how to apply nodal and mesh analysis in the frequency
More informationEECE 2510 Circuits and Signals, Biomedical Applications Final Exam Section 3. Name:
EECE 2510 Circuits and Signals, Biomedical Applications Final Exam Section 3 Instructions: Closed book, closed notes; Computers and cell phones are not allowed Scientific calculators are allowed Complete
More informationEIT Review 1. FE/EIT Review. Circuits. John A. Camara, Electrical Engineering Reference Manual, 6 th edition, Professional Publications, Inc, 2002.
FE/EIT eview Circuits Instructor: uss Tatro eferences John A. Camara, Electrical Engineering eference Manual, 6 th edition, Professional Publications, Inc, 00. John A. Camara, Practice Problems for the
More informationElectric Circuits II Sinusoidal Steady State Analysis. Dr. Firas Obeidat
Electric Circuits II Sinusoidal Steady State Analysis Dr. Firas Obeidat 1 Table of Contents 1 2 3 4 5 Nodal Analysis Mesh Analysis Superposition Theorem Source Transformation Thevenin and Norton Equivalent
More informationBasics of Network Theory (PartI)
Basics of Network Theory (PartI). A square waveform as shown in figure is applied across mh ideal inductor. The current through the inductor is a. wave of peak amplitude. V 0 0.5 t (m sec) [Gate 987: Marks]
More informationEE40 Lecture 11 Josh Hug 7/19/2010
EE40 Lecture Josh 7/9/200 Logistical Things Lab 4 tomorrow Lab 5 (active filter lab) on Wednesday Prototype for future lab for EE40 Prelab is very short, sorry. Please give us our feedback Google docs
More informationEE221 Circuits II. Chapter 14 Frequency Response
EE22 Circuits II Chapter 4 Frequency Response Frequency Response Chapter 4 4. Introduction 4.2 Transfer Function 4.3 Bode Plots 4.4 Series Resonance 4.5 Parallel Resonance 4.6 Passive Filters 4.7 Active
More informationECE2262 Electric Circuits. Chapter 6: Capacitance and Inductance
ECE2262 Electric Circuits Chapter 6: Capacitance and Inductance Capacitors Inductors Capacitor and Inductor Combinations OpAmp Integrator and OpAmp Differentiator 1 CAPACITANCE AND INDUCTANCE Introduces
More informationSCHOOL OF MATHEMATICS MATHEMATICS FOR PART I ENGINEERING. Selfpaced Course
SCHOOL OF MATHEMATICS MATHEMATICS FOR PART I ENGINEERING Selfpaced Course MODULE 26 APPLICATIONS TO ELECTRICAL CIRCUITS Module Topics 1. Complex numbers and alternating currents 2. Complex impedance 3.
More informationEE292: Fundamentals of ECE
EE292: Fundamentals of ECE Fall 2012 TTh 10:0011:15 SEB 1242 Lecture 18 121025 http://www.ee.unlv.edu/~b1morris/ee292/ 2 Outline Review RMS Values Complex Numbers Phasors Complex Impedance Circuit Analysis
More informationECE Circuit Theory. Final Examination. December 5, 2008
ECE 212 H1F Pg 1 of 12 ECE 212  Circuit Theory Final Examination December 5, 2008 1. Policy: closed book, calculators allowed. Show all work. 2. Work in the provided space. 3. The exam has 3 problems
More informationELEC273 Lecture Notes Set 11 AC Circuit Theorems
ELEC273 Lecture Notes Set C Circuit Theorems The course web site is: http://users.encs.concordia.ca/~trueman/web_page_273.htm Final Exam (confirmed): Friday December 5, 207 from 9:00 to 2:00 (confirmed)
More informationECE 202 Fall 2013 Final Exam
ECE 202 Fall 2013 Final Exam December 12, 2013 Circle your division: Division 0101: Furgason (8:30 am) Division 0201: Bermel (9:30 am) Name (Last, First) Purdue ID # There are 18 multiple choice problems
More informationBasic RL and RC Circuits RL TRANSIENTS: STORAGE CYCLE. Engineering Collage Electrical Engineering Dep. Dr. Ibrahim Aljubouri
st Class Basic RL and RC Circuits The RL circuit with D.C (steady state) The inductor is short time at Calculate the inductor current for circuits shown below. I L E R A I L E R R 3 R R 3 I L I L R 3 R
More informationChapter 10: Sinusoidal SteadyState Analysis
Chapter 10: Sinusoidal SteadyState Analysis 10.1 10.2 10.3 10.4 10.5 10.6 10.9 Basic Approach Nodal Analysis Mesh Analysis Superposition Theorem Source Transformation Thevenin & Norton Equivalent Circuits
More informationChapter 5 SteadyState Sinusoidal Analysis
Chapter 5 SteadyState Sinusoidal Analysis Chapter 5 SteadyState Sinusoidal Analysis 1. Identify the frequency, angular frequency, peak value, rms value, and phase of a sinusoidal signal. 2. Solve steadystate
More informationHomework 2 SJTU233. Part A. Part B. Problem 2. Part A. Problem 1. Find the impedance Zab in the circuit seen in the figure. Suppose that R = 5 Ω.
Homework 2 SJTU233 Problem 1 Find the impedance Zab in the circuit seen in the figure. Suppose that R = 5 Ω. Express Zab in polar form. Enter your answer using polar notation. Express argument in degrees.
More informationGeneral Physics (PHY 2140)
General Physics (PHY 2140) Lecture 10 6/12/2007 Electricity and Magnetism Induced voltages and induction SelfInductance RL Circuits Energy in magnetic fields AC circuits and EM waves Resistors, capacitors
More informationBasic Electronics. Introductory Lecture Course for. Technology and Instrumentation in Particle Physics Chicago, Illinois June 914, 2011
Basic Electronics Introductory Lecture Course for Technology and Instrumentation in Particle Physics 2011 Chicago, Illinois June 914, 2011 Presented By Gary Drake Argonne National Laboratory Session 2
More informationEIT QuickReview Electrical Prof. Frank Merat
CIRCUITS 4 The power supplied by the 0 volt source is (a) 2 watts (b) 0 watts (c) 2 watts (d) 6 watts (e) 6 watts 4Ω 2Ω 0V i i 2 2Ω 20V Call the clockwise loop currents i and i 2 as shown in the drawing
More informationEE 40: Introduction to Microelectronic Circuits Spring 2008: Midterm 2
EE 4: Introduction to Microelectronic Circuits Spring 8: Midterm Venkat Anantharam 3/9/8 Total Time Allotted : min Total Points:. This is a closed book exam. However, you are allowed to bring two pages
More informationElectrical Circuit & Network
Electrical Circuit & Network January 1 2017 Website: www.electricaledu.com Electrical Engg.(MCQ) Question and Answer for the students of SSC(JE), PSC(JE), BSNL(JE), WBSEDCL, WBSETCL, WBPDCL, CPWD and State
More informationBasic. Theory. ircuit. Charles A. Desoer. Ernest S. Kuh. and. McGrawHill Book Company
Basic C m ш ircuit Theory Charles A. Desoer and Ernest S. Kuh Department of Electrical Engineering and Computer Sciences University of California, Berkeley McGrawHill Book Company New York St. Louis San
More informationModule 4. Singlephase AC circuits. Version 2 EE IIT, Kharagpur
Module 4 Singlephase circuits ersion EE T, Kharagpur esson 6 Solution of urrent in Parallel and Seriesparallel ircuits ersion EE T, Kharagpur n the last lesson, the following points were described:. How
More informationPhysics 2112 Unit 20. Outline: Driven AC Circuits Phase of V and I Conceputally Mathematically With phasors
Physics 2112 Unit 20 Outline: Driven A ircuits Phase of V and I onceputally Mathematically With phasors Electricity & Magnetism ecture 20, Slide 1 Your omments it just got real this stuff is confusing
More information11. AC Circuit Power Analysis
. AC Circuit Power Analysis Often an integral part of circuit analysis is the determination of either power delivered or power absorbed (or both). In this chapter First, we begin by considering instantaneous
More informationCircuit Theorems Overview Linearity Superposition Source Transformation Thévenin and Norton Equivalents Maximum Power Transfer
Circuit Theorems Overview Linearity Superposition Source Transformation Thévenin and Norton Equivalents Maximum Power Transfer J. McNames Portland State University ECE 221 Circuit Theorems Ver. 1.36 1
More informationPhysics 212. Lecture 11. RC Circuits. Change in schedule Exam 2 will be on Thursday, July 12 from 8 9:30 AM. Physics 212 Lecture 11, Slide 1
Physics 212 Lecture 11 ircuits hange in schedule Exam 2 will be on Thursday, July 12 from 8 9:30 AM. Physics 212 Lecture 11, Slide 1 ircuit harging apacitor uncharged, switch is moved to position a Kirchoff
More informationEE221 Circuits II. Chapter 14 Frequency Response
EE22 Circuits II Chapter 4 Frequency Response Frequency Response Chapter 4 4. Introduction 4.2 Transfer Function 4.3 Bode Plots 4.4 Series Resonance 4.5 Parallel Resonance 4.6 Passive Filters 4.7 Active
More informationAn op amp consisting of a complex arrangement of resistors, transistors, capacitors, and diodes. Here, we ignore the details.
CHAPTER 5 Operational Amplifiers In this chapter, we learn how to use a new circuit element called op amp to build circuits that can perform various kinds of mathematical operations. Op amp is a building
More informationENGG Fundamentals of Electrical Circuits and Machines Final Examination
Name: Lecture Section: ID#: ENGG 225  Fundamentals of Electrical Circuits and Machines Final Examination Thursday, April 17, 2014 Time: 8:0011:00 AM Red Gymnasium (L01L03) Auxiliary Gymnasium (L04)
More information09/29/2009 Reading: Hambley Chapter 5 and Appendix A
EE40 Lec 10 Complex Numbers and Phasors Prof. Nathan Cheung 09/29/2009 Reading: Hambley Chapter 5 and Appendix A Slide 1 OUTLINE Phasors as notation for Sinusoids Arithmetic with Complex Numbers Complex
More informationSolved Problems. Electric Circuits & Components. 11 Write the KVL equation for the circuit shown.
Solved Problems Electric Circuits & Components 11 Write the KVL equation for the circuit shown. 12 Write the KCL equation for the principal node shown. 12A In the DC circuit given in Fig. 1, find (i)
More informationLecture 15. LC Circuit. LC Oscillation  Qualitative. LC Oscillator
Lecture 5 Phys. 07: Waves and Light Physics Department Yarmouk University 63 Irbid Jordan &KDSWHUElectromagnetic Oscillations and Alternating urrent L ircuit In this chapter you will see how the electric
More informationFUNDAMENTALS. Introduction. 1.1 Electric Signals. Chapter 1
hapter FUNDAMENTALS Introduction This chapter analyses the fundamental notions that everyone needs when exploring the field of electronics. We will start the chapter with the beginning, talking about electric
More informationChapter 5. Department of Mechanical Engineering
Source Transformation By KVL: V s =ir s + v By KCL: i s =i + v/r p is=v s /R s R s =R p V s /R s =i + v/r s i s =i + v/r p Two circuits have the same terminal voltage and current Source Transformation
More informationEECE 2150 Circuits and Signals, Biomedical Applications Final Exam Section 3
EECE 2150 Circuits and Signals, Biomedical Applications Final Exam Section 3 Instructions: Closed book, closed notes; Computers and cell phones are not allowed You may use the equation sheet provided but
More informationECE Spring 2017 Final Exam
ECE 20100 Spring 2017 Final Exam May 2, 2017 Section (circle below) Qi (12:30) 0001 Tan (10:30) 0004 Hosseini (7:30) 0005 Cui (1:30) 0006 Jung (11:30) 0007 Lin (9:30) 0008 PeleatoInarrea (2:30) 0009 Name
More informationSingle Phase Parallel AC Circuits
Single Phase Parallel AC Circuits 1 Single Phase Parallel A.C. Circuits (Much of this material has come from Electrical & Electronic Principles & Technology by John Bird) n parallel a.c. circuits similar
More information