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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