Schedule Date Day Class Title Chapters HW Lab Exam No. Due - - PowerPoint PPT Presentation

ECEN 301 Discussion #20 – Exam 2 Review 1

Date Day Class No. Title Chapters HW Due date Lab Due date Exam 10 Nov Mon 20 Exam Review LAB 7 EXAM 2 11 Nov Tue 12 Nov Wed 21 Boolean Algebra 13.2 – 13.3 13 Nov Thu 14 Nov Fri Recitation 15 Nov Sat 16 Nov Sun 17 Nov Mon 22 Combinational Logic 13.3 – 13.5 LAB 10 18 Nov Tue

Schedule…

ECEN 301 Discussion #20 – Exam 2 Review 2

Ask

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

ECEN 301 Discussion #20 – Exam 2 Review 3

Lecture 20 – Exam 2 Review

Chapters 4 – 6, 8

ECEN 301 Discussion #20 – Exam 2 Review 4

Exam 2

12 – 16 November (Monday – Friday) Chapters 4 – 6 and 8 15 questions

 12 multiple choice (answer on bubble sheet!)

• 1 point each

 3 long answer (show your work!)

• 4 or 5 points each

Closed book!

 One 3x5 card allowed

Calculators allowed No time limit Study lecture slides and homework

ECEN 301 Discussion #20 – Exam 2 Review 5

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

ECEN 301 Discussion #20 – Exam 2 Review 6

Capacitors & Inductors

Inductors Capacitors

Passive sign convention Voltage Current Power

) ( ) ( 1 ) ( t i d v L t i

t t L

dt t di L t v ) ( ) (

+ L – i + C – i

) ( ) ( 1 ) ( t v d i C t v

t t C

dt t dv C t i ) ( ) (

dt t di t Li t P

L

) ( ) ( ) ( dt t dv t Cv t P

C

) ( ) ( ) (

ECEN 301 Discussion #20 – Exam 2 Review 7

Capacitors & Inductors

Inductors Capacitors

Energy An instantaneous change is not permitted in:

Current Voltage

Will permit an instantaneous change in:

Voltage Current

With DC source element acts as a:

Short Circuit Open Circuit

2

) ( 2 1 ) ( t Li t WL

2

) ( 2 1 ) ( t Cv t WC

ECEN 301 Discussion #20 – Exam 2 Review 8

Capacitors & Inductors

1. What is the difference between the voltage and current behaviour of capacitors and inductors?

ECEN 301 Discussion #20 – Exam 2 Review 9

Capacitors & Inductors

0.0 1.0 2.0 3.0 4.0 5.0 0.00 2.00 4.00 6.00 0.0 0.1 0.2 0.3 0.4 0.5 0.00 2.00 4.00 6.00

Capacitor voltage vC(t) Inductor current iL(t)

NB: neither can change instantaneously

Capacitor current iC(t) Inductor voltage vL(t)

NB: both can change instantaneously

1. What is the difference between the voltage and current behaviour of capacitors and inductors?

ECEN 301 Discussion #20 – Exam 2 Review 10

Capacitors & Inductors

2. find the voltage v(t) for a capacitor C = 0.5F with the current as shown and v(0) = 0

0.0 0.2 0.4 0.6 0.8 1.0 1.2 0.0 1.0 2.0 3.0 time (s) current (A)

ECEN 301 Discussion #20 – Exam 2 Review 11

Capacitors & Inductors

0.0 0.2 0.4 0.6 0.8 1.0 1.2 0.0 1.0 2.0 3.0 time (s) current (A)

t t t t t i 2 2 1 1 1 : intervals 4 in i(t) current

) ( 1 ) ( v id C t v

t

2. find the voltage v(t) for a capacitor C = 0.5F with the current as shown and v(0) = 0

ECEN 301 Discussion #20 – Exam 2 Review 12

Capacitors & Inductors

0.0 0.2 0.4 0.6 0.8 1.0 1.2 0.0 1.0 2.0 3.0 time (s) current (A)

t v t v d t d t v

t t

2 ) 2 ( 2 1 ) 1 ( ) 1 ( 2 1 2 : intervals 4 in v(t) voltage

1

) ( 1 ) ( v id C t v

t

2. find the voltage v(t) for a capacitor C = 0.5F with the current as shown and v(0) = 0

ECEN 301 Discussion #20 – Exam 2 Review 13

Capacitors & Inductors

0.0 0.2 0.4 0.6 0.8 1.0 1.2 0.0 1.0 2.0 3.0 time (s) current (A)

t t t t t t v 2 3 2 1 1 2 1 : intervals 4 in v(t) voltage

2

2. find the voltage v(t) for a capacitor C = 0.5F with the current as shown and v(0) = 0

ECEN 301 Discussion #20 – Exam 2 Review 14

Capacitors & Inductors

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 0.0 1.0 2.0 3.0 4.0 time (s) voltage (V) 0.0 0.2 0.4 0.6 0.8 1.0 1.2 0.0 1.0 2.0 3.0 time (s) current (A)

t t t t t t v 2 3 2 1 1 2 1 : intervals 4 in v(t) voltage

2

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

ECEN 301 Discussion #20 – Exam 2 Review 15

Measuring Signal Strength

• 3. Compute the rms value of the sinusoidal current

i(t) = I cos(ωt)

ECEN 301 Discussion #20 – Exam 2 Review 16

Measuring Signal Strength

2 2 1 ) 2 cos( 2 2 2 1 ) 2 cos( 2 1 2 1 2 ) ( cos 2 ) ( 1

2 / 2 2 2 / 2 2 / 2 2 2 2

I I d I I d I d I d i T i

T rms

Integrating a sinusoidal waveform

• ver 2 periods equals zero

2 1 ) 2 cos( ) ( cos2 t t

• 3. Compute the rms value of the sinusoidal current

i(t) = I cos(ωt)

ECEN 301 Discussion #20 – Exam 2 Review 17

Measuring Signal Strength

2 2 1 ) 2 cos( 2 2 2 1 ) 2 cos( 2 1 2 1 2 ) 2 ( cos 2 ) ( 1

2 / 2 2 2 / 2 2 / 2 2 2 2

I I d I I d I d I d i T i

T rms

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)

ECEN 301 Discussion #20 – Exam 2 Review 18

Phasors

4. compute the phasor voltage for the equivalent voltage vs(t)

v1(t) = 15cos(377t+π/4) v2(t) = 15cos(377t+π/12)

v1(t)

+ – ~

v2(t)

+ – ~

vs(t)

+ – ~

ECEN 301 Discussion #20 – Exam 2 Review 19

Phasors

v1(t)

+ – ~

v2(t)

+ – ~

vs(t)

+ – ~ 1. Write voltages in phasor notation

V e j V V e j V

j j

12 15 15 ) ( 4 15 15 ) (

12 / 2 4 / 1

4. compute the phasor voltage for the equivalent voltage vs(t)

v1(t) = 15cos(377t+π/4) v2(t) = 15cos(377t+π/12)

ECEN 301 Discussion #20 – Exam 2 Review 20

Phasors

V j j j V V j V 88 . 3 49 . 14 12 sin 15 12 cos 15 ) ( : r rectangula Convert to 12 15 ) (

2 2

v1(t)

+ – ~

v2(t)

+ – ~

vs(t)

+ – ~ 1. Write voltages in phasor notation 2. Convert phasor voltages from polar to rectangular form (see Appendix A) V j j j V V j V 61 . 10 61 . 10 4 sin 15 4 cos 15 ) ( : r rectangula Convert to 4 15 ) (

1 1

4. compute the phasor voltage for the equivalent voltage vs(t)

v1(t) = 15cos(377t+π/4) v2(t) = 15cos(377t+π/12)

ECEN 301 Discussion #20 – Exam 2 Review 21

Phasors

v1(t)

+ – ~

v2(t)

+ – ~

vs(t)

+ – ~ 1. Write voltages in phasor notation 2. Convert phasor voltages from polar to rectangular form (see Appendix A) 3. Combine voltages

49 . 14 10 . 25 ) ( ) ( ) (

2 1

j j V j V j VS

4. compute the phasor voltage for the equivalent voltage vs(t)

v1(t) = 15cos(377t+π/4) v2(t) = 15cos(377t+π/12)

ECEN 301 Discussion #20 – Exam 2 Review 22

Phasors

v1(t)

+ – ~

v2(t)

+ – ~

vs(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

6 98 . 28 ) ( 6 10 . 25 49 . 14 tan 98 . 28 (14.49) (25.10) r : polar to Convert 49 . 14 10 . 25 ) (

1 2 2

j V j j V

S S

4. compute the phasor voltage for the equivalent voltage vs(t)

v1(t) = 15cos(377t+π/4) v2(t) = 15cos(377t+π/12)

ECEN 301 Discussion #20 – Exam 2 Review 23

Phasors

v1(t)

+ – ~

v2(t)

+ – ~

vs(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

6 377 cos 98 . 28 ) ( 6 98 . 28 ) ( t t v j V

S S

Bring ωt back NB: the answer is NOT simply the addition of the amplitudes of v1(t) and v2(t) (i.e. 15 + 15), and the addition of their phases (i.e. π/4 + π/12)

4. compute the phasor voltage for the equivalent voltage vs(t)

v1(t) = 15cos(377t+π/4) v2(t) = 15cos(377t+π/12)

ECEN 301 Discussion #20 – Exam 2 Review 24

Phasors

v1(t)

+ – ~

v2(t)

+ – ~

vs(t)

+ – ~

6 377 cos 98 . 28 ) ( 6 98 . 28 ) ( t t v j V

S S

Re Im 14.49 π/6 25.10 Vs(jω)

4. compute the phasor voltage for the equivalent voltage vs(t)

v1(t) = 15cos(377t+π/4) v2(t) = 15cos(377t+π/12)

ECEN 301 Discussion #20 – Exam 2 Review 25

Impedance

Impedance: complex resistance (has no physical significance)

 will allow us to use network analysis methods such as node voltage, mesh current, etc.  Capacitors and inductors act as frequency-dependent resistors vs(t)

+ – ~ R + vR(t) – i(t)

vs(t)

+ – ~ C + vC(t) – i(t)

vs(t)

+ – ~ L + vL(t) – i(t)

Vs(jω)

+ – ~ + VZ(jω) – I(jω) Z

ECEN 301 Discussion #20 – Exam 2 Review 26

Impedance

L j j ZL ) (

R j ZR ) (

+ L – + C – Re Im

• π/2

π/2 R

• 1/ωC

ωL

ZR ZC ZL

Phasor domain + R –

C j j ZC 1 ) (

Vs(jω)

+ – ~ + VZ(jω) – I(jω) Z

Impedance of resistors, inductors, and capacitors

ECEN 301 Discussion #20 – Exam 2 Review 27

Impedance

• 5. find the equivalent impedance (ZEQ)

ω = 104 rads/s, C = 10uF, R1 = 100Ω, R2 = 50Ω, L = 10mH

ZEQ R2 C R1 L

ECEN 301 Discussion #20 – Exam 2 Review 28

Impedance

ZEQ R2 C R1 L

) 37 . 1 ( 81 . 9 62 . 9 92 . 1 5 1 50 ) 50 )( 10 10 )( 10 ( 1 50 1 ) / 1 ( ) / 1 ( ||

6 4 2 2 2 2 2 1

j j j CR j R C j R C j R Z Z Z

C R EQ

• 5. find the equivalent impedance (ZEQ)

ω = 104 rads/s, C = 10uF, R1 = 100Ω, R2 = 50Ω, L = 10mH

ECEN 301 Discussion #20 – Exam 2 Review 29

Impedance

723 . 2 . 136 38 . 90 92 . 101 62 . 9 92 . 1 ) 10 )( 10 ( 100 ) 37 . 1 ( 81 . 9

2 4 1 1 1

j j j L j R Z Z Z Z

EQ L R EQ

ZEQ R1 L ZEQ1

) 37 . 1 ( 81 . 9

1 EQ

Z

NB: at this frequency (ω) the circuit has an inductive impedance (reactance or phase is positive)

• 5. find the equivalent impedance (ZEQ)

ω = 104 rads/s, C = 10uF, R1 = 100Ω, R2 = 50Ω, L = 10mH

ECEN 301 Discussion #20 – Exam 2 Review 30

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

ECEN 301 Discussion #20 – Exam 2 Review 31

AC RLC Circuits

6. find ia(t) and ib(t)

vs(t) = 15cos(1500t)V, R1 = 100Ω, R2 = 75Ω, L = 0.5H, C = 1uF

R1

vs(t)

+ – ~ R2 L C ia(t) ib(t)

ECEN 301 Discussion #20 – Exam 2 Review 32

AC RLC Circuits

6. find ia(t) and ib(t)

vs(t) = 15cos(1500t)V, R1 = 100Ω, R2 = 75Ω, L = 0.5H, C = 1uF

R1

vs(t)

+ – ~ R2 L C ia(t) ib(t) 1. Note frequencies of AC sources Only one AC source - ω = 1500 rad/s

ECEN 301 Discussion #20 – Exam 2 Review 33

AC RLC Circuits

6. find ia(t) and ib(t)

vs(t) = 15cos(1500t)V, R1 = 100Ω, R2 = 75Ω, L = 0.5H, C = 1uF

R1

vs(t)

+ – ~ R2 L C ia(t) ib(t) 1. Note frequencies of AC sources 2. Convert to phasor domain ZR1

Vs(jω)

+ – ~ ZR2 Ia(jω) ZL ZC Ib(jω)

ECEN 301 Discussion #20 – Exam 2 Review 34

AC RLC Circuits

6. find ia(t) and ib(t)

vs(t) = 15cos(1500t)V, R1 = 100Ω, R2 = 75Ω, L = 0.5H, C = 1uF

ZR1

Vs(jω)

+ – ~ ZR2 Ia(jω) ZL ZC Ib(jω) 1. Note frequencies of AC sources 2. Convert to phasor domain 3. Represent each element by its impedance

100

1 1

R ZR

15 15 ) ( ) cos( 15 ) (

j s s

e j V t t v

75

2 2

R ZR

750 ) 5 . )( 1500 ( j j L j ZL 667 ) 10 )( 1500 ( / 1 / 1

6

j j C j ZC

ECEN 301 Discussion #20 – Exam 2 Review 35

AC RLC Circuits

6. find ia(t) and ib(t)

vs(t) = 15cos(1500t)V, R1 = 100Ω, R2 = 75Ω, L = 0.5H, C = 1uF ) ( ) ( ) ( ) ( ) ( : b at KVL

2 2 2 R L C b C a R b L b C b a R L C

Z Z Z I Z I Z I Z I Z I I j V j V j V

+ZR1–

Vs(jω)

+ – ~ + ZR2 – Ia(jω) +ZL– + ZC – Ib(jω)

667 750 75 100

2 1

j Z j Z Z Z

C L R R

4. Solve using network analysis

• Mesh current

s C b C R a s C b a R a C R s

V Z I Z Z I V Z I I Z I j V j V j V ) ( ) ( ) ( ) ( ) ( : a at KVL

1 1 1

ECEN 301 Discussion #20 – Exam 2 Review 36

AC RLC Circuits

6. find ia(t) and ib(t)

vs(t) = 15cos(1500t)V, R1 = 100Ω, R2 = 75Ω, L = 0.5H, C = 1uF

) 83 75 ( ) 667 ( j I j I

b a

+ZR1–

Vs(jω)

+ – ~ + ZR2 – Ia(jω) +ZL– + ZC – Ib(jω)

A I A I 49 . 1 019 . 917 . 0032 .

2 1

4. Solve using network analysis

• Mesh current

15 ) 667 ( ) 667 100 ( j I j I

b a

ECEN 301 Discussion #20 – Exam 2 Review 37

AC RLC Circuits

6. find ia(t) and ib(t)

vs(t) = 15cos(1500t)V, R1 = 100Ω, R2 = 75Ω, L = 0.5H, C = 1uF

+ZR1–

Vs(jω)

+ – ~ + ZR2 – Ia(jω) +ZL– + ZC – Ib(jω)

mA t t i A I ) 917 . 1500 cos( 2 . 3 ) ( 917 . 0032 .

1 1

5. Convert to Time domain

mA t t i A I ) 49 . 1 1500 cos( 19 ) ( 49 . 1 019 .

2 2

ECEN 301 Discussion #20 – Exam 2 Review 38

AC Equivalent Circuits

Thévenin and Norton equivalent circuits apply in AC analysis

 Equivalent voltage/current will be complex and frequency dependent

Load + V – I Source

VT(jω)

+ – ZT Load + V – I

IN(jω)

ZN Load + V – I Norton Equivalent Thévenin Equivalent

ECEN 301 Discussion #20 – Exam 2 Review 39

AC Equivalent Circuits

Computation of Thévenin and Norton Impedances:

1. Remove the load (open circuit at load terminal) 2. Zero all independent sources

 Voltage sources short circuit (v = 0)  Current sources

• pen circuit (i = 0)

3. Compute equivalent impedance across load terminals (with load removed) NB: same procedure as equivalent resistance ZL Z1

Vs(jω)

+ – Z3 Z2 Z4 a b Z1 Z3 Z2 Z4 a b ZT

ECEN 301 Discussion #20 – Exam 2 Review 40

AC Equivalent Circuits

Computing Thévenin voltage:

1. Remove the load (open circuit at load terminals) 2. Define the open-circuit voltage (Voc) across the load terminals 3. Chose a network analysis method to find Voc

 node, mesh, superposition, etc.

4. Thévenin voltage VT = Voc Z1

Vs(jω)

+ – Z3 Z2 Z4 a b Z1

Vs(jω)

+ – Z3 Z2 Z4 a b + VT – NB: same procedure as equivalent resistance

ECEN 301 Discussion #20 – Exam 2 Review 41

AC Equivalent Circuits

Computing Norton current:

1. Replace the load with a short circuit 2. Define the short-circuit current (Isc) across the load terminals 3. Chose a network analysis method to find Isc

 node, mesh, superposition, etc.

4. Norton current IN = Isc Z1

Vs(jω)

+ – Z3 Z2 Z4 a b Z1

Vs(jω)

+ – Z3 Z2 Z4 a b IN NB: same procedure as equivalent resistance

ECEN 301 Discussion #20 – Exam 2 Review 42

AC Equivalent Circuits

7. find the Thévenin equivalent

ω = 103 Hz, Rs = 50Ω, RL = 50Ω, L = 10mH, C = 0.1uF

Rs

vs(t)

+ – ~ RL L C + vL –

ECEN 301 Discussion #20 – Exam 2 Review 43

AC Equivalent Circuits

7. find the Thévenin equivalent

ω = 103 Hz, Rs = 50Ω, RL = 50Ω, L = 10mH, C = 0.1uF

Rs

vs(t)

+ – ~ RL L C + vL – 1. Note frequencies of AC sources Only one AC source - ω = 103 rad/s

ECEN 301 Discussion #20 – Exam 2 Review 44

AC Equivalent Circuits

7. find the Thévenin equivalent

ω = 103 Hz, Rs = 50Ω, RL = 50Ω, L = 10mH, C = 0.1uF

Rs

vs(t)

+ – ~ RL L C + vL – 1. Note frequencies of AC sources 2. Convert to phasor domain Zs ZLD + – ~ ZL ZC Vs(jω)

ECEN 301 Discussion #20 – Exam 2 Review 45

AC Equivalent Circuits

7. find the Thévenin equivalent

ω = 103 Hz, Rs = 50Ω, RL = 50Ω, L = 10mH, C = 0.1uF

1. Note frequencies of AC sources 2. Convert to phasor domain 3. Find ZT

• Remove load & zero sources

Zs ZL ZC

9182 . 33 . 82 414 . 65 50 1 ) / 1 ( ) ( ) / 1 )( ( ||

2

j LC L j R C j L j C j L j R Z Z Z Z

S S L C S T

ECEN 301 Discussion #20 – Exam 2 Review 46

AC Equivalent Circuits

7. find the Thévenin equivalent

ω = 103 Hz, Rs = 50Ω, RL = 50Ω, L = 10mH, C = 0.1uF

9182 . 33 . 82

T

Z

1. Note frequencies of AC sources 2. Convert to phasor domain 3. Find ZT

• Remove load & zero sources

4. Find VT(jω)

• Remove load

Zs + – ~ ZL ZC Vs(jω) + VT(jω) – NB: Since no current flows in the circuit once the load is removed:

S T

V V

ECEN 301 Discussion #20 – Exam 2 Review 47

AC Equivalent Circuits

7. find the Thévenin equivalent

ω = 103 Hz, Rs = 50Ω, RL = 50Ω, L = 10mH, C = 0.1uF

9182 . 33 . 82

T

Z

S T

V V

Zs ZLD + – ~ ZL ZC Vs(jω) ZT + – ~ VT(jω) ZLD

ECEN 301 Discussion #20 – Exam 2 Review 48

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

R1 R2 C

vs

+ –

t = 0

DC Source Switch Energy element

ECEN 301 Discussion #20 – Exam 2 Review 49

Initial condition x(0): DC steady state before a switch is first activated

 x(0–): right before the switch is closed  x(0+): right after the switch is closed

Final condition x(∞): DC steady state a long time after a switch is activated

R1 R2 C

vs

+ –

t = 0

R3 R1 R2 C

vs

+ –

t → ∞

R3 Initial condition Final condition

DC Transient Response – DC Steady-State

ECEN 301 Discussion #20 – Exam 2 Review 50

Remember – capacitor voltages and inductor currents cannot change instantaneously

Capacitor voltages and inductor currents don’t change right before closing and right after closing a switch

) ( ) ( ) ( ) (

L L C C

i i v v

DC Transient Response – DC Steady-State

ECEN 301 Discussion #20 – Exam 2 Review 51

DC Transient Response – DC Steady-State

8. find the initial and final current conditions at the inductor

is = 10mA

is

t = 0

R L iL

ECEN 301 Discussion #20 – Exam 2 Review 52

8. find the initial and final current conditions at the inductor

is = 10mA

is

t = 0

R L iL 1. Initial conditions – assume the current across the inductor is in steady-state.

is

iL NB: in DC steady state inductors act like short circuits, thus no current flows through R

DC Transient Response – DC Steady-State

ECEN 301 Discussion #20 – Exam 2 Review 53

8. find the initial and final current conditions at the inductor

is = 10mA

1. Initial conditions – assume the current across the inductor is in steady-state.

mA i i

s L

10 ) (

is

iL

DC Transient Response – DC Steady-State

ECEN 301 Discussion #20 – Exam 2 Review 54

8. find the initial and final current conditions at the inductor

is = 10mA

is

t = 0

R L iL 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 iL

DC Transient Response – DC Steady-State

ECEN 301 Discussion #20 – Exam 2 Review 55

• 8. find the initial and final current conditions at the

inductor

is = 10mA

is

t = 0

R L iL 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 iL NB: polarity of R

mA i i

L L

10 ) ( ) (

DC Transient Response – DC Steady-State

ECEN 301 Discussion #20 – Exam 2 Review 56

• 8. find the initial and final current conditions at the

inductor

is = 10mA

is

t = 0

R L iL 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

A iL ) (

DC Transient Response – DC Steady-State

ECEN 301 Discussion #20 – Exam 2 Review 57

DC Transient Response

Solving 1st order transient response:

1. Solve the DC steady-state circuit:

 Initial condition x(0–): before switching (on/off)  Final condition x(∞): After any transients have died out (t → ∞)

2. Identify x(0+): the circuit initial conditions

 Capacitors: vC(0+) = vC(0–)  Inductors: iL(0+) = iL(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 vC(t) or iL(t)  Reduce this equation to standard form

4. Solve for the time constant

 Capacitive circuits: τ = RTC  Inductive circuits: τ = L/RT

5. Write the complete response in the form:

 x(t) = x(∞) + [x(0) - x(∞)]e-t/τ

ECEN 301 Discussion #20 – Exam 2 Review 58

DC Transient Response

9. find vc(t) for all t

vs = 12V, vC(0–) = 5V, R = 1000Ω, C = 470uF

R C

vs

+ –

t = 0

i(t) + vC(t) –

ECEN 301 Discussion #20 – Exam 2 Review 59

9. find vc(t) for all t

vs = 12V, vC(0–) = 5V, R = 1000Ω, C = 470uF

V v v

S C

12 ) (

R C

vs

+ –

t = 0

i(t) + vC(t) – 1. DC steady-state a) Initial condition: vC(0) b) Final condition: vC(∞)

V v t v

C C

5 ) ( ) (

NB: as t → ∞ the capacitor acts like an open circuit thus vC(∞) = vS

DC Transient Response

ECEN 301 Discussion #20 – Exam 2 Review 60

9. find vc(t) for all t

vs = 12V, vC(0–) = 5V, R = 1000Ω, C = 470uF

R C

vs

+ –

t = 0

i(t) + vC(t) – 2. Circuit initial conditions: vC(0+)

V v v

C C

5 ) ( ) (

DC Transient Response

ECEN 301 Discussion #20 – Exam 2 Review 61

9. find vc(t) for all t

vs = 12V, vC(0–) = 5V, R = 1000Ω, C = 470uF

R C

vs

+ –

t = 0

i(t) + vC(t) – 3. Write differential equation (already in Thévenin equivalent) at t = 0

S C C S C C C R S

v t v dt t dv RC v t v R t i t v t v v ) ( ) ( ) ( ) ( ) ( ) ( : KVL

DC Transient Response

ECEN 301 Discussion #20 – Exam 2 Review 62

9. find vc(t) for all t

vs = 12V, vC(0–) = 5V, R = 1000Ω, C = 470uF

47 . ) 10 470 )( 1000 (

6

RC

R C

vs

+ –

t = 0

i(t) + vC(t) – 4. Find the time constant τ

1

S

K

12

S

v F

DC Transient Response

ECEN 301 Discussion #20 – Exam 2 Review 63

9. find vc(t) for all t

vs = 12V, vC(0–) = 5V, R = 1000Ω, C = 470uF

R C

vs

+ –

t = 0

i(t) + vC(t) – 5. Write the complete response x(t) = x(∞) + [x(0) - x(∞)]e-t/τ

47 . / 47 . / /

7 12 ) 12 5 ( 12 ) ( ) ( ) ( ) (

t t t C C C C

e e e v v v t v

DC Transient Response

ECEN 301 Discussion #20 – Exam 2 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 V j V j H

S L V

) ( ) ( ) ( j I j I j H

S L I

) ( ) ( ) ( j I j V j H

S L Z

ECEN 301 Discussion #20 – Exam 2 Review 65

Frequency Response

• 10. compute the frequency response HV(jω)

R1 = 1kΩ, RL = 10kΩ, C = 10uF

R1 C

vs(t)

+ – + RL –

ECEN 301 Discussion #20 – Exam 2 Review 66

Frequency Response

• 10. compute the frequency response HV(jω)

R1 = 1kΩ, RL = 10kΩ, C = 10uF

R1 C

vs(t)

+ – + RL – 1. Note frequencies of AC sources Only one AC source so frequency response HV(jω) will be the function of a single frequency

ECEN 301 Discussion #20 – Exam 2 Review 67

Frequency Response

• 10. compute the frequency response HV(jω)

R1 = 1kΩ, RL = 10kΩ, C = 10uF

R1 C

vs(t)

+ – + RL – 1. Note frequencies of AC sources 2. Convert to phasor domain Z1 = R1 ZLD=RL ZC=1/jωC

Vs

+ – ~

ECEN 301 Discussion #20 – Exam 2 Review 68

Frequency Response

• 10. compute the frequency response HV(jω)

R1 = 1kΩ, RL = 10kΩ, C = 10uF

1. Note frequencies of AC sources 2. Convert to phasor domain 3. Solve using network analysis

• Thévenin equivalent

Z1 = R1 ZC=1/jωC

C T

Z Z Z ||

1

ECEN 301 Discussion #20 – Exam 2 Review 69

Frequency Response

• 10. compute the frequency response HV(jω)

R1 = 1kΩ, RL = 10kΩ, C = 10uF

C C S T

Z Z Z j V j V

1

) ( ) (

1. Note frequencies of AC sources 2. Convert to phasor domain 3. Solve using network analysis

• Thévenin equivalent

Z1 = R1 + VT – ZC=1/jωC

Vs

+ – ~ C T

Z Z Z ||

1

ECEN 301 Discussion #20 – Exam 2 Review 70

Frequency Response

• 10. compute the frequency response HV(jω)

R1 = 1kΩ, RL = 10kΩ, C = 10uF

C C S T

Z Z Z j V j V

1

) ( ) (

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

C T

Z Z Z ||

1

ZT ZLD

VT

+ – ~ + VL –

LD C LD C C S LD T LD T L

Z Z Z Z Z Z Z j V Z Z Z j V j V || ) ( ) ( ) (

1 1

ECEN 301 Discussion #20 – Exam 2 Review 71

Frequency Response

• 10. compute the frequency response HV(jω)

R1 = 1kΩ, RL = 10kΩ, C = 10uF

5. Find an expression for the frequency response ZT ZLD

VT

+ – ~ + VL –

LD C LD C C S L V

Z Z Z Z Z Z Z j V j V j H || ) ( ) ( ) (

1 1

ECEN 301 Discussion #20 – Exam 2 Review 72

Frequency Response

• 10. compute the frequency response HV(jω)

R1 = 1kΩ, RL = 10kΩ, C = 10uF

5. Find an expression for the frequency response ZT ZLD

VT

+ – ~ + VL –

110 arctan 110 100 110 100 ) 10 /( 10 ) 10 /( 1 10 10 ) 10 /( 10 || ) (

2 2 5 3 5 3 4 5 4 1 1 1 1

j j j j Z Z Z Z Z Z Z Z Z Z Z Z Z Z j H

C C LD C LD LD C LD C C V

ECEN 301 Discussion #20 – Exam 2 Review 73

Frequency Response

• 10. compute the frequency response HV(jω)

R1 = 1kΩ, RL = 10kΩ, C = 10uF 2 010 . ) ( j HV

5. Find an expression for the frequency response

• Look at response for low frequencies (ω = 10)

and high frequencies (ω = 10000) ZT ZLD

VT

+ – ~ + VL –

110 arctan 110 100 ) (

2 2

j HV

0907 . 905 . ) ( j HV

ω = 10 ω = 10000

ECEN 301 Discussion #20 – Exam 2 Review 74

Basic Filters

Electric circuit filter: attenuates (reduces) or eliminates signals at unwanted frequencies

0.0

.

Low-pass

0.0

.

High-pass

0.0

.

Band-pass

0.0

.

Band-stop ω ω ω ω

ECEN 301 Discussion #20 – Exam 2 Review 75

Basic Filters – Resonant Frequency

Resonant Frequency (ωn): the frequency at which capacitive impedance and inductive impedance are equal and opposite (in 2nd order filters)

LC

n

1

C + vi(t) – + vo(t) – L C + vi(t) – + vo(t) – L Impedances in series Impedances in parallel

ECEN 301 Discussion #20 – Exam 2 Review 76

Basic Filters – Resonant Frequency

Resonant Frequency (ωn): the frequency at which capacitive impedance and inductive impedance are equal and opposite (in 2nd order filters)

) ( j Vo

C LC j ZL

+ Vi(jω) – + Vo(jω) – ZEQ=0

C LC j ZC

C L EQ

Z Z Z

Impedances in series

ECEN 301 Discussion #20 – Exam 2 Review 77

Basic Filters – Resonant Frequency

Resonant Frequency (ωn): the frequency at which capacitive impedance and inductive impedance are equal and opposite (in 2nd order filters)

) ( ) ( j V j V

i

• C

LC j ZL

+ Vi(jω) – + Vo(jω) – ZEQ=∞

C LC j ZC

/ || C L Z Z Z Z Z Z Z

C L C L C L EQ

Impedances in parallel

ECEN 301 Discussion #20 – Exam 2 Review 78

Basic Filters – Low-Pass Filters

Low-pass Filters: only allow signals under the cutoff frequency (ω0) to pass

) cos( ) (

1t

t vo

0.0

.

Low-pass ω1 ω0 ω2 ω3 ) cos( ) cos( ) cos( ) (

3 2 1

t t t t vi

HL(jω)

R C + vi(t) – + vo(t) – R C + vi(t) – + vo(t) – L 1st Order 2nd Order

ECEN 301 Discussion #20 – Exam 2 Review 79

Basic Filters – High-Pass Filters

High-pass Filters: only allow signals above the cutoff frequency (ω0) to pass

) cos( ) (

3t

t vo

) cos( ) cos( ) cos( ) (

3 2 1

t t t t vi

HH(jω)

0.0

.

High-pass ω1 ω0 ω2 ω3 + vi(t) – + vo(t) – 1st Order R C R + vi(t) – + vo(t) – 2nd Order C L

ECEN 301 Discussion #20 – Exam 2 Review 80

Basic Filters – Band-Pass Filters

Band-pass Filters: only allow signals between the passband (ωa to ωb) to pass

) cos( ) (

2t

t vo

) cos( ) cos( ) cos( ) (

3 2 1

t t t t vi

HB(jω)

0.0

.

Band-pass ω1 ωa ω2 ω3 ωb + vi(t) – + vo(t) – 2nd Order R C L

ECEN 301 Discussion #20 – Exam 2 Review 81

Basic Filters – Band-Stop Filters

Band-stop Filters: allow signals except those between the stopband (ωa to ωb) to pass

) cos( ) cos( ) (

3 1

t t t vo

) cos( ) cos( ) cos( ) (

3 2 1

t t t t vi

HN(jω)

+ vi(t) – + vo(t) – 2nd Order R C L

0.0

.

Band-stop ω1 ωa ω2 ω3 ωb

ECEN 301 Discussion #20 – Exam 2 Review 82

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– – + vo – io i2 i1 –

vin +

– +

+

–

Rout Rin i1 AOLvin + vo – – vin + v– v+ NB: op-amps have near-infinite input resistance (Rin) and very small output resistance (Rout)

) ( v v A v A v

OL in OL

• AOL – open-loop voltage gain

ECEN 301 Discussion #20 – Exam 2 Review 83

Op-Amps – Closed-Loop Mode

Circuit Diagram ACL

Inverting Amplifier Summing Amplifier

N n Sn n F N n Sn OLn

• v

R R v A v

1 1 – + + vo –

+ –

vS RS RF

S S F S CL

• v

R R v A v

– + + vo –

+ – + – + –

RSn RS2 RS1 vSn vS2 vS1 RF

ECEN 301 Discussion #20 – Exam 2 Review 84

Op-Amps – Closed-Loop Mode

Circuit Diagram ACL

Noninverting Amplifier Voltage Follower s s CL

• v

v A v

S S F S CL

• v

R R v A v 1

– + + vo –

+ –

R RS RF vS

– + + vo –

+ –

ECEN 301 Discussion #20 – Exam 2 Review 85

Op-Amps – Closed-Loop Mode

Circuit Diagram ACL

Differential Amplifier

1 2 S S S F

• v

v R R v

– + + vo –

+ –

+ –

RS RS RF RF v1 v2

ECEN 301 Discussion #20 – Exam 2 Review 86

Op-Amps – Closed-Loop Mode

Circuit Diagram ACL

Ideal Integrator Ideal Differentiator

dt t dv C R v

S S F

• )

(

t S F S

• d

v C R v ) ( 1

– + + vo(t) –

+ –

vS CS RF

– + + vo(t) –

+ –

vS RS CF

ECEN 301 Discussion #20 – Exam 2 Review 87

Op-Amps

• 11. find an expression for the gain

CF = 1/6 F, R1 = 3Ω, R2 = 2Ω, CS = 1/6 F

+

–

vo(t) iin

CF R2

v+ v– iF(t) i2(t)

R1

i1(t)

CS

iS(t) vs(t)

ECEN 301 Discussion #20 – Exam 2 Review 88

Op-Amps

• 11. find an expression for the gain

CF = 1/6 F, R1 = 3Ω, R2 = 2Ω, CS = 1/6 F

+

–

Vo(jω) Iin

ZF=1/jωCF Z2

v+ v– IF(jω) I2(jω)

Z1

I1(jω)

ZS

IS(jω) Vs(jω)

Node a Node b 3 1 6 2 1 2 1 6 3 1 1 1 1 1 1 1 : a at KCL

1 2 2 1 2 1 2 1 S

• a

S F

• F

a

• a

F a

• a

S F

V j V j V Z V Z Z V Z Z Z V Z V V Z V V Z V V I I I 1. Transfer to frequency domain 2. Apply KCL at nodes a and b

NB: v+ = v– and Iin = 0

ECEN 301 Discussion #20 – Exam 2 Review 89

Op-Amps

• 11. find an expression for the gain

CF = 1/6 F, R1 = 3Ω, R2 = 2Ω, CS = 1/6 F

+

–

Vo(jω) Iin

ZF=1/jωCF Z2

v+ v– IF(jω) I2(jω)

Z1

I1(jω)

ZS

IS(jω) Vs(jω)

Node a Node b

2 1 6 2 1 1 1 1 : b at KCL

2 2 2 2

j V V Z Z V Z V Z V Z V V I I I

• a

S

• a

S

• a

in S

1. Transfer to frequency domain 2. Apply KCL at nodes a and b

ECEN 301 Discussion #20 – Exam 2 Review 90

Op-Amps

• 11. find an expression for the gain

CF = 1/6 F, R1 = 3Ω, R2 = 2Ω, CS = 1/6 F

3 3 j V V

• a

+

–

Vo(jω) Iin

ZF=1/jωCF Z2

v+ v– IF(jω) I2(jω)

Z1

I1(jω)

ZS

IS(jω) Vs(jω)

1. Transfer to frequency domain 2. Apply KCL at nodes a and b 3. Express Vo in terms of Vs

S

• a

V j V j V 2 3 5

6 5 6

2

j V V

S

ECEN 301 Discussion #20 – Exam 2 Review 91

Op-Amps

• 11. find an expression for the gain

CF = 1/6 F, R1 = 3Ω, R2 = 2Ω, CS = 1/6 F

+

–

Vo(jω) Iin

ZF=1/jωCF Z2

v+ v– IF(jω) I2(jω)

Z1

I1(jω)

ZS

IS(jω) Vs(jω)

1. Transfer to frequency domain 2. Apply KCL at nodes a and b 3. Express Vo in terms of VS 4. Find the gain (Vo/VS)

6 5 6

2

j V V

S