Steady-State Sinusoidal
Analysis
Introduction
Previously we have analyzed circuits with time-independent
sources – voltage and current that do not change with time
DC circuit analysis
In this section we will analyze circuits containing time-dependent
sources – voltage and current vary with time
One of the important classes of time-dependent signal is the
periodic signals
x(t) = x(t +nT),
where n = 1,2 3, … and T is the period of the signal
Typical periodic signals in electrical engineering:
Sawtooth wave
Square wave
t
t
Triangle wave
pulse wave
t
t
Vm
ωt
i
Im
-Vm
ωt
-2π
T
T
T
2π
Vm
-Im
ωt
sinusoidal signals
-Vm
T
In this course we will deal with one of the most important periodic signal
of all :- sinusoidal signals
Signals that has the form of sine or cosine function
t
Circuit containing sources with sinusoidal signals is called an AC circuit. Our
analysis will be restricted to the steady state behavior of AC circuit.
Study of sinusoidal AC circuit are so important due to:
•
It is widely used in the electric power industries worldwide – household and
industrial applications
•
ALL periodic waveforms (e.g. square, triangular, sawtooth, etc) can be
represented by sinusoids
Objectives for studying AC circuit
1. Identify the frequency, angular frequency, peak
value, rms value, and phase of a sinusoidal signal.
2. Solve steady-state ac circuits using phasors and
complex impedances.
3. Compute power for steady-state sinusoidal ac
circuits.
4. Apply circuit theorems for steady-state ac voltages.
5. Solve balanced/unbalance three-phase circuits.
Sinusoidal waveform
Let a sinusoidal signal of a voltage is given by:
v(ωt) = Vm sin (ωt)
v(ωt)
Vm
π
2π
3π
4π
Vm – the amplitude or maximum value
ω – the angular frequency (radian/second)
ωt – the argument of the sine function
ωt
Sinusoidal waveform
Let a sinusoidal signal of a voltage is given by:
v(ωt) = Vm sin (ωt)
The voltage can also be written as function of time: v(t) = Vm sin (ωt)
T is known as the period of the waveform. In T seconds, the
voltage goes through 1 cycle.
Thus in one second there are 1/T cycles of waveform
v(t)
• The number of cycles per second is the frequency f
measured in Hertz
1
f=
T
Vm
T/2
T
(3/2)T
2T
t
Sinusoidal waveform
A more general expression of a sinusoidal signal is
v1(ωt) = Vm sin (ωt + φ)
φ is called the phase angle, normally written in degrees
Let a second voltage waveform is given by: v2(ω
ωt) = Vm sin (ω
ωt - θ)
θ
v(ωt)
v1(ω
ωt) = Vm sin (ω
ωt + φ)
v2(ω
ωt) = Vm sin (ω
ωt - θ)
Vm
θ
φ
ωt
Sinusoidal waveform
v(ωt)
v1(ω
ωt) = Vm sin (ω
ωt + φ)
v2(ω
ωt) = Vm sin (ω
ωt - θ)
Vm
θ
ωt
φ
v1 and v2 are said to be out of phase
v1 is said to be leading v2 by φ − (-θ) or (φ + θ)
alternatively,
v2 is said to be lagging v1 by φ − (-θ) or (φ + θ)
Sinusoidal waveform
Some important relationships in sinusoidals
Vmsin (ω
ωt) = Vmsin (ω
ωt − 360o)
⇒ Vmsin (ωt + θ) = Vmsin (ωt − (360o − θ))
e.g., Vmsin (ωt + 250o) = Vmsin (ωt − (360o − 250o)) = Vm sin (ωt − 110o)
v(ωt)
Vm
ωt
250o
110o
Vmsin (ωt + 90o) = Vmcos (ωt )
Vmcos (ωt - 90o) = Vmsin (ωt )
-Vm sin (ω
ωt)
v(ωt)
180o
ωt
Vmsin (ωt ± 180o) = -Vmsin (ωt )
Parameters of Sinusoidal Currents and Voltages
•Vm: is the peak value, unit is volt
•ω : is the angular frequency, unit is radians per second
•f : is the frequency，
，unit is Hertz (Hz) or inverse second.
•θ : is the phase angle, unit is radian or degree.
ω = 2πf
1
f =
T
2π
ω=
T
sin(z ) = cos(z − 90o )
Average and Effective value of a sinusoidal waveform
An average value of a periodic waveform is defined as:
X ave
1
=
T
∫
x( t )dt
t
Vave
For a sinusoidal voltage:
For half a period we can
calculate the average
value as:
t +T
V ave
=
1
π
1
=
2π
2
=
T
π
∫V
0
m
θ + 2π
∫θ
T /2
∫V
m
Vm sin(ωt )d (ωt ) = 0
sin ω t dt
0
sin ω t d (ω t ) =
2
π
V m = 0 .636 V m
Average and effective value of a sinusoidal waveform
An effective value or Root-Mean-Square (RMS) of a periodic current (or
voltage) is defined as:
The equivalent value of the DC current that when flowing through a resistance
of R-ohm delivers the same average power as does the periodic current.
v(t)
i(t)
R
Average power:
(absorbed)
Ieffec
Vdc
R
Average power:
(absorbed)
P=
1
T
T
2
∫ i R dt
0
P = I2effecR
Power to be equal:
1
I2effecR =
T
∴ Ieffec =
1
T
∫
T
2
i R dt
0
T
2
∫ i dt
0
• Root-Mean-Square Value or Effective Value
Vrms
1
=
T
T
∫
v (t )dt
2
0
I rms
1
=
T
T
∫
i 2 (t )dt
0
If the AC current is given by: i=Imsinωt
T
I rms
1
Im
2
(
)
=
I
sin
ω
t
dt
=
= 0.707 I m
m
∫
T 0
2
Vrms
1 2
Vm
=
v
dt
=
= 0.707Vm
∫
T 0
2
T
The rms value of a sinusoid is the peak value divided by the square root of two.
Pavg = I
2
rms
R
Pavg
2
Vrms
=
R
v(t ) = 100 cos(100πt )
A voltage given by v(t ) = 100 cos(100πt ) is applied to a 50Ω
Ω resistance. Find
the rms value of the voltage and the average power delivered to the
resistance.
Vrms =
Vm
= 70.71V
2
V 2 rms 70.712
Pavg =
=
= 100 W
R
50
v 2 (t )
p (t ) =
= 200 cos 2 (100π t ) W
R
complex numbers and Phasors
OUTLINE
–
–
–
–
Sinusoidal Voltages
Complex impedances
Phasors
Circuit analysis using complex impedances
Complex Numbers
imaginary
axis
z
y
θ
real
axis
x
• Rectangular Coordinates
Z = x + jy
j = ( − 1)
• Polar Coordinates: Z = z ∠ θ
• Exponential Form: Z = Z e jθ = ze jθ
Euler’s Identities
•
•
•
•
x is the real part
y is the imaginary part
z is the magnitude
θ is the phase
z= x +y
2
2
θ = tan
Z = z (co s θ + j sin θ )
e jθ + e − jθ
co s θ =
2
jθ
e − e − jθ
sin θ =
2j
e jθ = co s θ + j sin θ
e jθ =
y = z sin θ
x = z cos θ
co s 2 θ + sin 2 θ = 1
1 = 1e j 0 = 1∠0°
j = 1e
j
π
2
= 1∠90°
−1
y
x
Operations of Complex Numbers
• Add and Subtract: it is easiest to do this in rectangular
format
– Add/subtract the real and imaginary parts separately
• Multiply and Divide: it is easiest to do this in
exponential/polar format
– Multiply (divide) the magnitudes
– Add (subtract) the phases
Z 1 = z1 e jθ 1 = z1 ∠ θ 1 = z1 co s θ 1 + jz1 sin θ 1
Z 2 = z 2 e jθ 2 = z 2 ∠ θ 2 = z 2 co s θ 2 + jz 2 sin θ 2
Z 1 + Z 2 = ( z1 co s θ 1 + z 2 co s θ 2 ) + j ( z1 sin θ 1 + z 2 sin θ 2 )
Z 1 − Z 2 = ( z1 co s θ 1 − z 2 co s θ 2 ) + j ( z1 sin θ 1 − z 2 sin θ 2 )
Z 1 × Z 2 = ( z1 × z 2 ) e j ( θ 1 + θ 2 ) = ( z1 × z 2 ) ∠ (θ 1 + θ 2 )
Z 1 / Z 2 = ( z1 / z 2 ) e j ( θ 1 − θ 2 ) = ( z1 / z 2 ) ∠ (θ 1 − θ 2 )
Complex Exponentials & Phasors
• We represent a real-valued sinusoid as the real part of a
complex exponential.
• Complex exponentials
– provide the link between time functions and phasors.
– make solving for AC steady state an algebraic
problem.
• Phasors allow us to express current-voltage
relationships for inductors and capacitors much like we
express the current-voltage relationship for a resistor.
• A complex exponential is the mathematical tool needed
to obtain this relationship.
Phasors
A phasor:
phasor is a complex number that used to represent a sinusoidal
waveform. It contains the information about the amplitude and phase
angle of the sinusoidal wave.
In steady state condition, the sinusoidal voltage or current will have the
same frequency. The differences between sinusoidal waveforms are
only in the magnitudes and phase angles
Analysis of AC circuit will be much more easier using phasors
A phasor diagram is a graph of several phasors on the
complex plane (using real and imaginary axes).
A phasor diagram helps to visualize the relationships
between currents and voltages
Phasors: transformation of sinusoidal waveforms to phasors
Phasor is rooted in Euler’s identity:
e jθ = cos θ + j sin θ
Real
{ }
sin θ = ℑ{e }
⇒ cos θ = ℜ e
⇒
jθ
jθ
Supposed v(t) = Vm cos (ωt + θ)
Imaginary
cos θ is the real part of e
jθ
sin θ is the imaginary part of e jθ
j( ωt + θ )
}
This can be written as v(t) = Vmℜ{e
Phasors
j( ωt + θ )
}
v(t) = Vmℜ{e
j( ωt + θ )
}
= ℜ{Vm e
jθ jω
= ℜ{Vm e e }
{
jθ j ω
v(t) = ℜ Vm e e
}
VVVV
Vm e jθ is the phasor transform of v(t)
v(t) = Vmcos (ωt +θ)
phasor transform
= Vm e jθ
VVVV
Phasors
VVVV
= Vm e jθ
= Vm ∠θ
VVVV
consider these
notations
Polar forms
o
= Vm cos θ + Vm j sin θ
Rectangular forms
VVVV
Im
= Vm∠θo
Phasors can be graphically
represented using Phasor Diagrams
Vm sin θ
θo
VVVVaaaa
Some examples ….
= Vm∠ − φ
IIII
va(t) = Vmcos(ωt-φ)
Re
o
Vm cos θ
VVVVxxxx
= Im∠θo
i(t) = Imcos(ωt+θ)
= Vm ∠(βo − 90o )
vx(t) = Vmsin (ωt +β) vx(t) = Vmcos (ωt +β - 90o)
VVVV3333
VVVV2222
Draw the phasor diagram for the following phasors:
= −40∠100 o
= −5 + j5
•Adding Sinusoids Using Phasors
•Example: v ( t ) = 20 cos (ωt − 45 ) V
o
1
v2 ( t ) = 10sin (ωt + 60o ) V
Find vs = v1 + v2 = ?
Solution: take v=Vmcosωt as reference, i.e. v=Vm<0
V1 = 20∠ − 45o V
V2 = 10∠ − 30o V
V&s = V&1 + V&2
= 20∠ − 45o + 10∠ − 30o
= 14.14 − j14.14 + 8.660 − j5
= 23.06 − j19.14
= 29.97∠ − 39.7o V
∴ vs ( t ) = 29.97 cos (ωt − 39.7o ) V