Engr228 - Chapter 9, Nilsson 11e 1 Chapter 9 Sinusoidal Steady-State Analysis Engr228 Circuit Analysis Dr Curtis Nelson Chapter 9 Objectives • Understand the concept of a phasor; • Be able to transform a circuit with a sinusoidal source into the frequency domain using phasor concepts; • Know how to use the following circuit analysis techniques to solve a circuit in the frequency domain: • Ohm’s Law; • Kirchhoff’s laws; • Series and parallel simplifications; • Voltage and current division; • Node-voltage method; • Mesh-current method; • Thévenin and Norton equivalents; • Maximum power theorem.
Transcript
Engr228 - Chapter 9, Nilsson 11e 1
Chapter 9Sinusoidal Steady-State
Analysis
Engr228
Circuit Analysis
Dr Curtis Nelson
Chapter 9 Objectives
• Understand the concept of a phasor;• Be able to transform a circuit with a sinusoidal source into the
frequency domain using phasor concepts;• Know how to use the following circuit analysis techniques to
solve a circuit in the frequency domain:• Ohm’s Law;• Kirchhoff’s laws;• Series and parallel simplifications;• Voltage and current division;• Node-voltage method;• Mesh-current method;• Thévenin and Norton equivalents;• Maximum power theorem.
Engr228 - Chapter 9, Nilsson 11e 2
Properties of a Sinusoidal Waveform
The general form of sinusoidal wave is
where:
• Vm is the amplitude in peak voltage;• ω is the angular frequency in radian/second, also 2πf;• q is the phase shift in degrees or radians.
v(t) =Vm sin(wt +q)
Frequency Review
0 1 2 3 4 5 6 7 8 9 10-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
sec
volts
fT 1=
Period ≈ 6.28 seconds, Frequency = 0.1592 Hz
period
Engr228 - Chapter 9, Nilsson 11e 3
Amplitude Review
Peak: Blue 1 volt, Red 0.8 voltsPeak-to-Peak: Blue 2 volts, Red 1.6 volts
Average: 0 volts
volts
0 1 2 3 4 5 6 7 8 9 10-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
sec
Phase Shift Review
)1sin(8.0)sin(
+==
tyty
red
blue
0 1 2 3 4 5 6 7 8 9 10-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
Period=6.28
Red leads Blue by 57.3 degrees (1 radian) !! 3.5736028.61
=´=f
Engr228 - Chapter 9, Nilsson 11e 4
More on Phase
• The red wave [VMsin(ωt + θ)] leads the wave in green by θ;• The green wave [VMsin(ωt)] lags the wave in red by θ;• The units of θ and ωt must be consistent.
Basic AC Circuit Components
• AC Voltage and Current Sources (active components)
• Resistors (R)
• Inductors (L) (passive components)
• Capacitors (C)
• Inductors and capacitors have limited energy storage capability.
Engr228 - Chapter 9, Nilsson 11e 5
AC Voltage and Current Sources
AC AC+
-
AC +-
AC
Voltage Sources Current Sources
AC+
-10sin(2πt + π/4)
Amplitude = 10Vpeak
ω = 2π so F = 1HzPhase shift = 45
Sinusoidal Steady State (SSS) Analysis
• SSS is important for circuits containing capacitors and inductors because these elements provide little value in circuits with only DC sources;
• Sinusoidal means that source excitations have the form VS cos(ωt + q) or VS sin(ωt + θ);
• Since VS sin(ωt + θ) can be written as VS cos(ωt + θ - p/2), we will use VS cos(ωt + θ) as the general form for our source excitation;
• Steady state means that all transient behavior in the circuit has decayed to zero.
Engr228 - Chapter 9, Nilsson 11e 6
Sinusoidal Steady State Response
The SSS response of a circuit to a sinusoidal input is also a sinusoidal signal with the same frequency but with possibly different amplitude and phase shift.
AC+
-5cos(3t+π/3)
i(t)
3H
2Ω
v2(t) cos wave
cos wave
v1(t) cos wave
vL(t) cos wave
• Complex numbers can be viewed as vectors where the X-axis represents the real part and the Y-axis represents the imaginary part.
• There are two common ways to represent complex numbers:
– Rectangular form: 4 + j3
– Polar form: 5 ∠ 37o
Review of Complex Numbers
jω
σ
3
4
Engr228 - Chapter 9, Nilsson 11e 7
qrqr
q
r
sincos
arctan
22
==
÷øö
çèæ=
+=
ba
abba
Complex Number Forms
Rectangular form: a + jb
Polar form: ρ ∠ θ
p = a + jb q = c + jd
• Addition and subtraction
x = p + q = (a + c) + j(b + d)
y = p – q = (a - c) + j(b - d)
• Example
p = 3 + j4 q = 1 - j2
x = p + q = (3 + 1) + j(4 - 2) = 4 + j2
y = p – q = (3 – 1) + j(4 – (-2)) = 2 + j6
Complex Math – Rectangular Form
Engr228 - Chapter 9, Nilsson 11e 8
p = a + jb q = c + jd
• Multiplication (easier in polar form)
x = p × q = ac + jad + jbc + j2bd = (ac – bd) + j(ad + bc)
• Example
p = 3 + j4 q = 1 - j2
x = p × q = [(3)(1) - (4)(-2)] + j[(3)(-2) + (4)(1)]
= 11 – j2
Complex Math – Rectangular Form
p = a + jb q = c + jd
• Division (easier in polar form)
• Example
p = 3 + j4 q = 1 - j2
Complex Math – Rectangular Form
( )( )( )( )
( ) ( )÷øö
çèæ
+-++
=÷÷ø
öççè
æ-+-+
=++
== 22 dcadbcjbdac
jdcjdcjdcjba
jdcjba
qp
x
( ) ( ) 215105
)2(1)2)(3()1)(4()2)(4()1)(3(
22 jjj
qp
x +-=+-
=-+
--+-+==
Engr228 - Chapter 9, Nilsson 11e 9
Euler’s Identity
• Euler’s identity states that e jθ= cos(θ) + jsin(θ)• A complex number can then be written as:
r = a + jb = ρcos(θ) + jρsin(θ) = ρ[cos(θ) + jsin(θ)] = ρe jθ
• Using shorthand notation, we write this as:
ρe jθ ≡ ρ ∠ θ
• Addition and subtraction - too hard in polar so convert to rectangular coordinates.
• Multiplication
• Example
Complex Math – Polar Form
qrr q Ð==+= jejbax 22 ba +=r ÷øö
çèæ= -
ab1tanq
( )11
qjemp = ( )22
qjemq =
( )2121
qq +=´= jemmqpz
÷øö
çèæ
= 66pj
ep÷øö
çèæ
= 22pj
eq ( )( )÷øö
çèæ
÷øö
çèæ +
==´= 32
26 1226ppp jj
eeqpz
°Ð=´=°Ð=°Ð= 12012902306 qpzqp
Engr228 - Chapter 9, Nilsson 11e 10
• Division
• Example
Complex Math – Polar Form
qrr q Ð==+= jejbax 22 ba +=r ÷øö
çèæ= -
ab1tanq
( )11
qjemp = ( )22
qjemq =
( )21
2
1 qq -=÷= jemmqpz
÷øö
çèæ
= 66pj
ep÷øö
çèæ
= 22pj
eq
!603326 326 -Ð===÷=
÷øö
çèæ -÷
øö
çèæ -
ppp jjeeqpz
More on Sinusoids
• Suppose you connect a function generator to any circuit containing resistors, inductors, and capacitors. If the function generator is set to produce a sinusoidal waveform, then every voltage drop and every current in the circuit will also be a sinusoid of the same frequency. Only the amplitudes and phase angles will (may) change.
• The same thing is not necessarily true for waveforms of other shapes like triangle or square waveforms.
• Fortunately, it turns out that sinusoids are not only the easiest waveforms to work with mathematically, they're also the most useful and occur quite frequently in real-world applications.
Engr228 - Chapter 9, Nilsson 11e 11
Phasors
• A phasor is a vector that represents an AC electrical quantity such as a voltage waveform or a current waveform;
• The phasor's length represents the peak value of the voltage or current;
• The phasor's angle represents the phase angle of the voltage or current;
• Phasors are used to represent the relationship between two or more waveforms with the same frequency.
• Phasors are complex numbers used to represent sinusoids of a fixed frequency;
• Their primary purpose is to simplify the analysis of circuits involving sinusoidal excitation by providing an algebraic alternative to differential equations;
• A typical phasor current is represented as I = IM ∠ f• For example, i(t) = 25cos(wt + 45º) has the phasor
representation I = 25∠45º• A phasor voltage is written as V = VM∠ f• For example, v(t) = 15cos(ωt + 120º) has the phasor
representation V = 15∠120º
More on Phasors
Engr228 - Chapter 9, Nilsson 11e 12
Phasor Equations
• The diagram at the right shows two phasors labeled v1 and v2;
• Phasor v1 is drawn at an angle of 0and has a length of 10 units;
• Phasor v2 is drawn at an angle of 45and is half as long as v1;
• In terms of the equations for sinusoidal waveforms, this diagram is a pictorial representation of the equations:
v1 = 10 cos(ωt)v2 = 5 cos(ωt + 45)
• The equations above and the diagram convey the same information.
Example Problems
Express each of the following currents as a phasor:1. 12sin(400t + 110º)A2. (-7sin800t – 3cos800t)A3. 4cos(200t – 30º) – 5cos(200t + 20º)A
• Now that we have defined phasor relationships for sinusoidal forcing functions, we need to define phasor relationships for the three basic circuit elements.
• The phasor relationship between voltage and current in a circuit is still defined by Ohm’s law with resistance replaced by impedance, a frequency dependent form of resistance denoted as Z(jω).
• In terms of phasors, Ohm’s law still applies so V = IZ(jω) where V is a phasor voltage, I is a phasor current, and Z(jω)is the impedance of the circuit element.- Note that Z is a real number for resistance and a complex number
for capacitance and inductance.• Since phasors are functions of frequency (ω), we often refer
to them as being in the frequency domain.
Phasors: The Resistor
In the frequency domain, Ohm’s Law takes the same form:
Engr228 - Chapter 9, Nilsson 11e 14
AC+
-
Asin(ωt)
i(t)
L
Finding the impedance (Z) of an inductor:
)2
(sin)cos(
cossin
sin1
)(1
)(
)()(
pww
ww
www
w
-=-=
÷øö
çèæ -==
==
=
ò
òò
tLAt
LA
tLAtdt
LA
tdtAL
dttvL
ti
dttdiLtv
Phasor Relationship for Inductors
Impedance of jωlPhase shift of -90º
Phasors: The Inductor
By dividing the phasor voltage by the phasor current, we derive an expression for the phasor impedance of an inductor shown in the figure below.Differentiation in time becomes multiplication in phasor form: (calculus becomes algebra).
Engr228 - Chapter 9, Nilsson 11e 15
AC+
-
Asin(ωt)
i(t)
C
)2
sin(1
)(cos
)sin()()(
pw
w
ww
w
+÷øö
çèæ
=
=
==
t
C
AtCA
dttAdC
dttdvCti
Impedance of 1/jωCPhase shift of +90º
Phasor Relationship for Capacitors
Phasors: The Capacitor
Differentiation in time becomes multiplication in phasor form: (calculus becomes algebra again).
Engr228 - Chapter 9, Nilsson 11e 16
Summary: Phasor Voltage/Current Relationships
Time Domain Frequency Domain
Calculus (real numbers) Algebra (complex numbers)
Impedance
• We define impedance as Z = V/I or V = IZ
ZR=R ZL=jωL ZC=1/jωC
• Impedance is a complex number (with units of ohms):– The real part of Z(jω) is called resistance;– The imaginary part of Z(jω) is called reactance.
• Impedances in series or parallel can be combined using the same “resistor rules” that you learned in Chapter 3.
Engr228 - Chapter 9, Nilsson 11e 17
1Ω
3H
Find the equivalent impedance, in polar form, for the circuit below if ω = 0.333 rad/sec.
!45213131 Ð=+=××+=+= jjLjRZEQ w
Impedance Example
Example: Equivalent Impedance
Find the impedance of the network at ω = 5 rad/s.
Answer: 4.255 + j4.929 Ω
Engr228 - Chapter 9, Nilsson 11e 18
Circuit Analysis Using Phasors
• Techniques that can be used in circuit analysis with phasors:– Ohm’s law;
– Kirchhoff’s voltage law (KVL);
– Kirchhoff’s current law (KCL);
– Source transformations;
– Nodal analysis;
– Mesh analysis;
– Thévenin's theorem;
– Norton’s theorem;
– Maximum power theorem.
Circuit Analysis Procedure Using Phasors
• Change the voltage/current sources into phasor form;• Change R, L, and C values into phasor impedances;
R L C R jωL 1/jωC
• Use normal DC circuit analysis techniques but the values of voltage, current, and impedance can be complex numbers;
• Change back to the time-domain form if required.
Engr228 - Chapter 9, Nilsson 11e 19
Example Problem 9.55 (Nilsson 11th)
Use the node-voltage method to find VO .
Answer: VO = 138.078 – j128.22V = 188.43∠-42.88º V
From the study of this chapter, you should:• Understand phasor concepts;• Be able to transform a circuit with a sinusoidal source into the
frequency domain using phasor concepts;• Know how to use the following circuit analysis techniques to
solve a circuit in the frequency domain:• Ohm’s Law;• Kirchhoff’s laws;• Series and parallel simplifications;• Voltage and current division;• Node-voltage method;• Mesh-current method;• Thévenin and Norton equivalents;• Maximum power theorem.
