Fall 2000 EE201 AC Steady-State Analysis 1

Chapter 8 AC Steady-State Analysis

We want to analyze the behavior of circuits to x(t) = XMsin(t); why?

(1) This is the format of signals generated by power companies such as AEP.

(2) Using a tool called Fourier Series, we can represent other signals (i.e. digital signals) as a sum of sinusoidal signals.

Fall 2000 EE201 AC Steady-State Analysis 2

x(t) = XMsin(t);

t

XM

-XM

/2

3/2 2

The function is periodic with period T x[(t+T)] = x(t )

T

t

Frequency, f, is a measure of how many periods (or cycles) the signal completes per second and is measured in Hertz (cycles per second )

fT

f

2

21

is the angular frequency and is measured in radians per second

Fall 2000 EE201 AC Steady-State Analysis 3

x(t) = XMsin(t + )

sin(120t)

sin(120t- /4)

Phase Shift - a sinusoid can be shifted right (left) by subtracting (adding) a phase angle.

Note that one function lags (leads) the other function of time.

Fall 2000 EE201 AC Steady-State Analysis 4

sin(120t+ /4)

Fall 2000 EE201 AC Steady-State Analysis 5

Some useful trigonometric identities:

)180sin()sin(sin

)180cos()cos(cos

90cos2

cossin

90sin2

sincos

ttt

ttt

ttt

ttt

sinsincoscos)cos(

cossincossin)sin(

Multiple-angle formulas:

sincos

cossin

sincoscossin

)sin()(

MM

MM

M

XBandXAwhere

tBtA

tXtX

tXtx

)(tansin)( 122A

BtBAtx

Fall 2000 EE201 AC Steady-State Analysis 6

Sinusoidal and Complex Forcing Functions

If we apply sinusoidal signals (voltages and/or currents) as inputs to linear electrical networks, all voltages and currents in the circuit will be sinusoidal also; only the amplitudes and phase angles will differ. The following example illustrates.

v(t)=VMcost

R

Li(t) tVtRidt

tdiL M cos)(

)(

Let us assume (this is a theorem from differential equations) that the solution to the differential equation, i(t), is also sinusoidal. That is, the forced response can be written

i(t) = Acos(t+)

The signal has the same frequency, but different magnitude and phase. However, this expression must satisfy the original differential equation.

Fall 2000 EE201 AC Steady-State Analysis 7

v(t)=VMcost

R

Li(t) tVtRidt

tdiL M cos)(

)(

i(t) = Acos(t+ )

tVALRAtRAALt

tVtRAtRAtALtAL

tVtAtARdt

tAtAdL

M

M

M

cossincos

cossincoscossin

cossincossincos

1212

2121

2121

Using the multiple angle formula: i(t) = Acostcos - Asin tsini(t) = A1 cost + A2 sint

Fall 2000 EE201 AC Steady-State Analysis 8

tVALRAtRAALt M cossincos 1212

Setting like terms on each side of the equation equal yields two simultaneous equations: MVRALA

LARA

12

12 0

Solving these two simultaneous equations yields:

2222

2221

LR

LVA

LR

RVA

M

M

Since i(t) = A1 cost + A2 sint tLR

LVt

LR

RVti MM

sincos)(

222222

Using trig identities we get:

R

Lt

LR

Vti M

1

222tancos)(

Conclusion: we do not want to use differential equations to analyze such circuits!

Fall 2000 EE201 AC Steady-State Analysis 9

Euler’s identity: sincos je j

te

te

tjte

tj

tj

tj

sinIm

cosRe

sincos

We can write our sinusoidal forcing functions in terms of the complex quantity e jt

v(t) = VMcost = Re[VMejt] = Re[VMcost + j VMsint ]

The currents and voltages produced can also be written in terms of e jt

i(t) = Imcos(t +) = Re[Imej(t + )] = Re[Imcos(t +) + j Imsin(t +) ]

Using complex forcing functions allows us to use (complex) algebra in place of differential equations.

Fall 2000 EE201 AC Steady-State Analysis 10

v(t)=VMejt

R

Li(t) tjM eVtRi

dt

tdiL )(

)(

i(t) = Imej(t +)

M

jM

tjM

jtjM

tjM

tjM

tjM

tjM

tjM

tjM

tjM

tjM

VeILjR

eVeeILjR

eVeILjR

eVeRIeILj

eVeRIdt

eIdL

)(

)()(

)()(

R

L

LR

VI

LR

RLV

LjR

VeI

MM

MMjM

122

22

1

tan

tan