enee204 fall 2010 lectures 5&6

8/8/2019 ENEE204 Fall 2010 Lectures 5&6

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Lecture 5Sinusoids (Harmonic) Functions AC Quantities Sample Circuit Problem involving time harmonic function Complex Numbers and Phasors Solving Circuit Problem using Phasors

Reading: M&L 3.1-3.6: AC signals in steady state

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It is represented by sinusoidal function of the forms:

Used in all forms of electrical devices on a over a very broad range ofoperating frequencies

Electrical power distribution 50-Hz-60 Hz Submarine 3-3000 Hz Conventional Radio AM: 500-1500 kHz FM radio: 80-103 Mhz Cable (7 Mhz 800MHz) Cellphones: 300-1800 Mhz (e.g. GSM 850: 850 Mhz) Wireless Routers: 2.4 GHz

Frequency bands are allocated by governments. In the US, it is theFCC (Federal Communications Commission).

AC -alternating current- voltages and currents vary in a periodic* manner with TIME

What are typical AC circuits?


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Who owns those frequencies?

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www.ntia.doc.gov/osmhome/allochrt.pdf

Government auctions off these very variable real estate.

3 kHz

300 kHz

3 MHz

30 MHz

300 MHz

3 GHz

30 GHz


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More on phasor representation later 9/16/10

4

AC quantities are represented by sinusoidal functions of the forms:

Functions with sinusoidal functionof time

Or equivalently, by complex numbers called phasors .

f ( t ) = Re Ae j t e j f { }= Re Ae j f e j t { }


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Sinusoidal functions need only 3 parameters

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v(t) = V mcos ( t + v )1. Amplitude 2.Angular frequency* 3. Phase

*period = T

2 T

=

Angular frequency

T = 2

ex: =1

v(t)=cos(t)


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For example: v(t) = V mcos ( t + v )

amplitude, angular frequency and phase .

If you are asked to solve for an AC quantity, youneed to specify all 3 quantities.

Emphasis: Description ofAC Quantities involves 3 parameters

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small V m large V mVariation in Amplitude

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Variation in Frequency

, low frequency

v(t) = V mcos ( t + v )

DC (direct current) is a special case of AC with =0 .

Understanding amplitude and angularfrequency

, high frequency


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Varia%on of phase

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Range of phases: +/- /2

Common terms: quarter cycle : = /2 ; half cycle : = ; full cycle : =2

v = 0

in-phase

v > 0

phase is ahead

v < 0

phase is lagging

Understanding Phase v(t) = V

mcos ( t +

v )


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Useful concepts in dealing with %me

varying func%on

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1. RMS root mean square time average of asignal

2. Decomposition of a periodic function in sineand cosine components.


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v(t) = V o cos t ( )

v RMS = V o1

T cos t ( )[ ]

2

dt'0

T = V o 1T sin 2 t ( )

4 t +

20

T

= V o2

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Example:

Dene: time-averaged value of

a periodic wave: Root MeanSquare (RMS) of f(t)


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The Root Mean Square of f(t)The Average (time) of an AC Quantity

v RMS

= V

0

2=

170 V

2

= 120 V RMS


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Useful TheoremAny harmonic func%on of the frequency can be uniquely decomposed into the sine

and cosine components.

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= Pcos ( t)+ Q sin ( t)

t

K cos ( t + )

t

K

-K

P

-P

Q

-Q

K cos( t + )

= K cos( t )cos( ) K sin( t )sin( )

= P cos( t ) + Q sin( t )

P = K cos( ) and Q = K sin( )

K 2 = P 2 + Q 2

HOW: trigonometry

PROVIDED:

2cos( t + 3)

= 2cos( t )cos(

3)

2sin( t )sin(

3)

P = 212

= 1 Q = 23

2= 3

= cos( t ) 3sin( t )

Example: Decompose 2cos( t+ /3)


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Useful TheoremsInverse: Any sum of sine and cosine func%on of the frequency & phase=0 be

uniquely expressed as a cosine func%on of and .

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HOW:Example: what is cos( t)+sin( t)?

= Pcos ( t)+ Q sin ( t)

t

K cos ( t + )

t

K

-K

P

-P

Q

-Q

=


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cos( / 2)= 0 sin( /2) = 1

cos( ) = -1 sin( ) = 0

cos(3 / 2) = 0 sin(3 /2) = -1

cos(0) = cos(2 ) = 1 sin(0)= sin(2 ) = 0

0,2

/2

3 /2

sin ( - ) = sinA cosB cosA sinB sin ( + ) = sinA cosB + cosA sinB

cos ( + ) = cosA cosB sinA sinB cos ( - ) = cosA cosB + sinA sinB

Remember (memorize) trigonometric identities:

A DigressionReview of Elementary Trigonometry

Example:


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Voltage and Current are expressed as cosine function ofangular frequency and phase , (v , I )

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v(t) = V m cos ( t + V )

i(t) = I m cos ( t + I )

i(t)

+

_ v(t) Z could be a resistor, inductor, capacitor,or any combination of these elementsZ

What do these terminal relations imply?

Capacitor InductorResistor

AC Circuits


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+

_

v(t)

i(t)

Ci(t) = I o cos ( t + I )v(t) = V o cos ( t + v )

i( t ) = C dv ( t )

dt

I 0 cos ( t + I ) = - CV o sin( t + v )= CV

0cos ( t +

v+ /2 )

For a capacitor: V o = I o / C and V = I - /2

Current and voltage are NOT in-phase: Voltage lags the current by a

quarter cycle (/2 or 90o).

By using identity: cos( + /2 ) = - sin( )

Capacitors

v(t)i(t)

In this example, let V =0.

17

Z C

=

1

C Absolute value ofimpedance of capacitor:


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i(t) = I o cos ( t + I )v(t) = V o cos ( t + v )

V o cos ( t + V ) = - LI o sin( t + I )

= LI o cos ( t + I + /2 )

For an inductor:Vm = LIm, V = I + /2

Current and voltage are NOT in-phase:Voltage leads the current by a quarter

cycle

v ( t ) = Ldi ( t )

dt i(t)+

_

v(t) L

Inductors

Absolute value of impedance ofinductor:


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v s(t) = V 0 cos( t)

R

C

v R

vC

i R

iC i s

v s(t)

KCL: is = iR = iC = i

KVL: v R + v C - v s(t) = 0

i (t) = I o cos ( t + I )

v R(t) = V Ro cos ( t + Rv )

vC (t) = V Co cos ( t + Cv )

Put into AC quantities:

TR: v R = R i, i = C d vC /dt

GOAL: To nd i(t), v R(t), v C (t)

Need to solve for I m , I in terms of Vo, R,C, (known).

AC Circuit AnalysisAn RC Example


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v s(t) = V 0 cos( t)

R

C

v R

vC

i R

iC i s

v s(t)

KCL: is = iR = iC = i

KVL1+TR:

TR: v R = R i, i = C d vC /dt

First nd vC (t), others follow


v R + v C = v s ( t )

i R

R + v C = V o cos( t )

RC dv

C

dt + v C = V o cos( t )

Solve this differential equation


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R

C

v R

vC

i R

iC i s

v s(t)=V ocos( t)


RC dv

C

dt + v

C = V

o cos( t )

Let : v C = A cos( t + )dv

C

dt =

A

sin( t +

)

RC A sin( t + ) + A cos( t + ) = V 0 cos( t )

= Pcos ( t)+ Q sin ( t) K cos ( t + )But recall:

Or similarly: K cos ( t) = Pcos ( t + )+ Q sin ( t + )


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= tan 1 Q

P

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AC Circuit Analysis : need to nd A and in terms of known V 0 ,

A cos( t + ) RC A sin( t + ) = V 0 cos( t )

=V o cos ( t) Pcos ( t + )+ Q sin ( t + )

= P cos( t )cos( ) sin( t )sin( )( )+ Q sin( t )cos( ) sin( )cos( t )( ) V 0 = P cos( ) Q sin( ) and 0 = - P sin( ) + Q cos( )

V 0 = P2

+ Q2

But recall identity:

Therefore: V 0

= A 2 + A 2 R 2 C 2 2

A =V

0

1 + R 2 C 2 2

And:

= tan 1 RC ( )


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SEEMS LIKE A LOT OF WORK FOR A SIMPLE CIRCUIT. Easier way?9/16/10 23

R

C

v Ri R

iC i s

v s(t)=V ocos( t)

The solution ( i, v everywhere)Finally:

vC ( t ) =

V 0

1 + R 2C 2 2cos t + tan 1 RC ( )( )

iC ( t ) = C

dvC ( t )dt

= CV

0

1 + R 2C 2 2sin t + tan 1 RC ( )( )

v R ( t ) = iC R =

RCV 0

1 + R 2C 2 2sin t + tan 1 RC ( )( )


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A consequence of PHASOR is that we can denea generalized resistance for AC aka impedance Z:

Z resistor =R Z capacitor =1/j C

Z Total = Z R + Z C = R +1

j C =

1 + j RC j RC

R

CVs= V 0

I VR

VC

AC

The RC Example using Phasors

AC Circuit Analysis

vC

=V

0

2C 2 R 2 + 1

cos( t + tan 1 RC ( ))

vC = V 0 Z C Z T

= V 0

1 j C

R + 1 j C

=V o

j CR + 1=

V o (1 j CR )

2C 2 R 2 + 1


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Reviewing

Complex Numbersand Phasors

How?


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Lecture 6Introducing Phasors: How to reduce differential equations into systems of linear equations Solving Circuit Problem using Phasors

Reading: M&L 3.1-3.6: AC signals in steady state9/16/10 26


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Using Complex Numbers is the easiest method todescribe oscillatory motion.

A complex number is described by two numbers (a,b)which are themselves real.

1-=z = a +jb

Real part Imaginary part

Notation:

Re( z ) = a Im(z ) = b

Review: The Complex Number


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The Complex Number inGraphical Form

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x=Re( z )

y=Im( z )

a

b|z|

Complex numbers can be representedgraphically in vector form:

Length of vector

|z|= a2+b2

|z|2=z*z=(a-jb)(a+jb)

z*

complex conjugate

}

tan( ) =b

a=

Im( z)

Re( z)


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Addition of complex numbers

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Real and imaginaryParts do not mix.

Example: z 1=(3+j) z 2=(1-2j)-5 -4 -3 -2 -1 0 1 2 3 4 5 -5

-4 -3 -2 -1 0 1 2 3 4 5

z 1 z 2

Im (z)

Re(z) z 1+z 2

Multiplication of complex numbers

The Complex Number Additionand Multiplication


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The Complex Number

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Dividing Complex Numbers involves two steps:a. Multiply top and bottom with the complex conjugate* of denominator to

make denominator real.

b. Collect all terms in numerator into its real and imaginary parts.

conjugating* ComplexNumber simply involves

replacing j with j.

z1 z2

=ac + bd + j bc ad ( )

c 2 + d 2

z1 z2

=a + jb( )c + jd ( )

c jd ( )c + jd ( )*


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Absolute value of a complex number: |z|

a2

+b2

If a complex number and it s conjugate are known, it is easy to nd thereal and imaginary components.

a = Re {z} = (z + z*) / 2 b = Im{z}= (z - z*) / 2

Absolute Value of a Compex

Number, Its Real and ImaginaryParts

31


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Re(z)

Im(z)

a

br

(r, )

a = r cos( )b = r sin( )

r = a 2 + b 2

tan( ) =b

a=

Im( z)

Re( z)

This is because of the famous EULER S Theorem:

e j = cos ( )+ j sin ( )This is can be proven quite easily using series expansion.

z = a + jb

z = r cos( ) + jr sin( )

z = re j

Representing Complex Numbers in Polar Form makes them especially useful

in analyzing oscillating functions

The Cartesian and PolarRepresentation of Complex

Numbers


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e x = 1 + x + 1 / 2! x2 + 1 / 3! x3 + 1 / 4! x4 .

QED

A Digression: Euler s Thm ProofProving Eulers Theorem

Recall binomial expansion of e x.

Let x= j .


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We can convert from Cartesian to polar coordinates any time.

Remember Euler s Theorem!


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Converting from Cartesian formto polar representation

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Say: z = 2+j a=2 b=1

r= 2 2+1 2 = 5

=tan-1

(1/2) = 31.3o

= 31.3 o1

2

23

11-2 -1

-2

z = 2+j

z = 2+j = e j(31.3degs)5

a = 3cos( /3) = 1.5

b = 3 sin( /3) = 2.6

Converting from polar formto Cartesian representation.

z = 1.5+2.6j

= 60 o1

2

23

11-2 -1

-2

z = 1.5+2.6j

Conversion Examples


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-1

z 1 z 2

0

z3

z 4

y= Im (z):imaginary part

x= Re (z):real part

1

-1

1

Important CasesFor Imaginary Exponentials


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Multiplication of complex numbers in POLAR FORM:

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z1 z2 = r1ej 1 r2e

j 2 = r1r2ej ( 1 + 2 )

z1 = a + jb = r1ej 1

z * z=

re j

rej

=

r2

ej ( )

=

r2

Complex conjugate in POLAR FORM

z = a + jb = re j

z* = a jb = re j

Polar Form Multiplication is Easier


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Complex Notation for AC Circuits

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v(t) = V m cos ( t + v ) AC signal at a given frequencyis characterized by itsamplitude and phasev(t)

t

Recall Euler s Formula:

= t Let:

Re(e j t )=cos ( t ) and Im ( e j t )=sin ( t ) Therefore, know that

Re{e j( t+ v) }= cos ( t+ v) & Im( e j( t+ v) )= sin ( t+ v) Now it therefore follows that if we add v in the argument:


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Complex Notation for AC Circuits

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v(t) = V m cos ( t + v ) Typical AC signal

}e{ j Re t V = v(t)

Phasor is the complexsignal at t=0

:Phasor V

The phasor is a complex number that contains thevalue of the amplitude and the phase .

Using Phasors


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phasor real signal

V = 2 + 2 j v(t) = 2.83 cos ( t + /4)

I = 5 e -j/6 i (t) = 5cos ( t - /6)

Phasor

(convert rst to polar: get amplitude and phase)

Examples

Phasor


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Converting a Phasor to a Time-Domain Signal

It is very straightforward.

Phasor

Time-domain


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Z

i(t) +

_

v(t)

However,

: Dene Impedance

(in general)

and

= complex number

= generalized resistance for AC

Impedances and PhasorsFor Resistors, Capacitors and Inductors


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Vm = RI m v = I

A real number

i(t)

+

_ v(t)

v e V V j m

I e

I

I

m

Resistors and PhasorsA Resistors impedance is Real

Z R = R


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Complex Voltage: Ve j t

ComplexCurrent:

Ie j t

Resistors and Phasors

Resistor Voltage and Current are in Phase


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i(t)+

_ v(t)

Inductors and PhasorsAn Inductors Impedance is Positive Imaginary

(1)

(2)

(3)

Using Terminal Relation:

Comparing (1) and (3):

Therefore the impedance of the inductor:

Z L = j L


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ComplexCurrent:Ie j t

Inductors and PhasorsAn Inductors Voltage leads its Current by 90 degrees

Animation


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i(t)

Negative imaginary number

+

_ v(t)

Capacitors and PhasorsAn Capacitors Impedance is a negative imaginary number

(1)

(2)

Using Terminal Relation:

Comparing (2) and (3):

Therefore the impedance of the capacitor:

Z C = V C I C

=

V me j V

I me j I

=

1

C e

j 2 Z C =

1

j C = j

C


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ComplexCurrent:

Ie j t

Capacitors and PhasorsA Capacitors Voltage lags its Current by 90 degrees


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Resistor

Inductor

Capacitor

Z R

= R

Z L = j L

Z C

=

1

j C =

j C

To recap:Impedances of Passive Elements


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v s(t) = 170 cos (377t) volts

angular frequency = 2 60 s -1 =120 s-1

Im =V m/L = I m = 170/ (377 x 20x10 -3)

377 s-1

= 22.5 A !

frequency = 1/T = 60 Hz

power outlet 120 VAC:Vm = 1.41*120V = 170 VAC }

Fuse will blow!

ExampleAn Electric Motor, expressed as a 20mH inductor,

Is plugged into an AC socket. Will it blow a 10 Amp fuse?


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Inductor: Vm / Im = L

Capacitor: Vm / Im = 1/ C

DC HF

Short circuit(no resistance)

Short circuit(no resistance)

Open circuit(innite resistance)

Open circuit(innite resistance)

= 0

Resistor : Vm / Im = R

- > infinity

No effect No effect

(Absolute value)IMPEDANCE=V m/Im

Impedances in DC and High Frequencies


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Ci (t)vs(t)

I = 570 mA and = tan -10.57

0

=

2

i ( t ) = 570cos(6.28 x 10 8 t + 2

) mA

i ( t ) = 570sin 6.28 x 10 8 t ( )mA

ExampleA 30 pF Capacitor is connected across an RF transmitter (voltage source)

operating at 100 MHz carrying an amplitude of 30 Volts. Find i(t).

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i1(t)i2(t)

i3(t)

i4(t)

i5(t)

KCL

KVL I v s(t)

v1(t)

v2(t)

v6 (t)

Kirchhoff s Laws and Phasors


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R

Given: v s(t) = V 0 cos( t)

C

vR

vC

i R

i Ci s

vs(t) AC

Revisiting an ExampleAC Circuit Analysis


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R

v s(t) = V 0 cos( t) =Re (V 0 e j t )

CVs= V 0

Is

phasor

IR

VR IC

VC

Revisiting an ExampleConverting to Phasors


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KCL:

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Is IR IC= = I =

CVs= V 0

Is IR

IC

VC

R

VR + VC VR Vs =KVL:

VR

= R I

VC = (1/j C) I

Terminal Relations:

We only have to work with algebra! Life is good.

Revisiting an ExampleWorking with Phasors


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Revisiting an ExampleSimple Algebra


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9/16/10 59Same as we obtained in last lecture.

Revisiting the RC Circuit Example

I = tan1 b

a

= tan

1 1

RC

= 2

+ tan 1 RC ( )

i( t ) =V

o RC

1 + RC ( )2cos t + tan 1

1 RC

=

V o RC

1 + RC ( )2sin t + tan 1 RC ( )( )

I o

=V

O RC

1 + RC ( )2

enee204 fall 2010 lectures 5&6

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