Alternating Current (AC) Circuits
•  We have been talking about DC circuits
–  Constant currents and voltages
–  Resistors
–  Linear equations
•  Now we introduce AC circuits
–  Time-varying currents and voltages
–  Resistors, capacitors, inductors (coils)
–  Linear differential equations
68
Recall water analogy for Ohm’s law…
(a) Battery
(b) Resistor
69
Now we add a steel tank with rubber sheet
(a) Battery
(b) Resistor
(c) Capacitor
70
Water enters one side of the tank and leaves the
other, distending but not crossing the sheet.
At first, water seems to flow through tank.
How to decrease capacitance of tank?
Make rubber sheet (a) smaller or (b) thicker.
71
Charge, like water is practically incompressible,
but within a small volume (closely spaced plates)
charge can enter one side and leave the other,
without flowing across the space between.
The apparent flow of
current through space
between the plates (the
“displacement current”) led
Maxwell to his equations
governing EM waves.
72
Basic Laws of Capacitance
•  Capacitance C relates charge Q to voltage V
Q
C=
V
•  Since Q = ∫ I dt ,
1
V = ∫ I dt
+
C
_
dV
I =C
dt
•  Capacitance has units of Farads, F = 1 A sec / V
73
Charging a Capacitor with Battery VB
•  Voltage across resistor to find current
VB − VC (t )
I (t ) =
R
•  Basic law of capacitor
dVC (t )
I (t ) = C
dt
•  Differential Equation yields exponential
dVC (t )
VC (t ) + RC
= VB
dt
t
−
⎛
⎞
RC
VC (t ) = VB ⎜ 1 − e ⎟
⎝
⎠
diminishing returns as
cap becomes charged
74
What determines capacitance C ?
•  Area A of the plates
•  Distance d between the plates
•  Permittivity ε of the dielectric between plates.
A
C=ε
d
Alignment of dipoles within dielectric between plates
increases capacitor’s ability to store charge (capacitance).
Permittivity of a vacuum ε0 ≈ 8.8541 × 10−12 F · m−1.
75
Types of Capacitors
•  Disk (Ceramic) Capacitor
–  Non-polarized
–  Low leakage
–  High breakdown voltage
–  ~ 5pF – 0.1μF
•  Electrolytic Capacitor
–  High leakage
–  Polarized
–  Low breakdown voltage
–  ~ 0.1μF – 10,000μF
•  Supercapacitor (Electrochemical Double Layer)
–  New. Effective spacing between plates in nanometers.
–  Many Farads! May power cars someday.
76
Inductor (coil)
•  Water Analogy
inductance is like
inertia/momentum
of water in pipe with
flywheel.
heavier flywheel
(coil wrapped around iron core)
adds to inertia/momentum.
77
Basic Laws of Inductance
•  Inductance L relates changes in the current to
voltages induced by changes in the magnetic
field produced by the current.
1
I = ∫ V dt
L
dI
V =L
dt
•  Inductance has units of Henries, H = 1 V sec / A.
78
What determines inductance L ?
•
•
•
•
•
Assume a solenoid (coil)
Area A of the coil
Number of turns N
Length  of the coil
Permeability μ of the core
2
N A
L=µ

Permeability of a vacuum μ0 ≈ 1.2566×10−6 H·m−1.
79
Joseph Henry
1797 – 1878
•  Invented insulation
•  Permitted construction
of much more powerful
electromagnets.
•  Derived mathematics for
“self-inductance”
•  Built early relays, used to
give telegraph range
•  Put Princeton Physics on
the map
80
Energy Stored in Capacitor
dV
I =C
dt
dV
P = VI = VC
dt
E = ∫ P dt
E = C ∫ V dV
1
2
E = CV
2
81
Energy Stored in Caps and Coils
•  Capacitors store “potential” energy in electric field
1
2
E = CV
2
independent of history
•  Inductors store “kinetic” energy in magnetic field
1
2
E= LI
2
independent of history
•  Resistors don’t store energy at all!
error in Scherz Eq. 2.42: missing “dt”
82
What is Magnetism?
•  Lorenz Contraction
 = 0 1 − v c
2
2
Length  of object
observed in relative
motion to the object is
shorter than the
object’s length  0 in its
own rest frame as
velocity v approaches
speed of light c.
Thus electrons in Wire 1
see Wire 2 as negatively
charged and repel it:
Magnetism!
83
Generating Sparks
•  What if you suddenly try to stop a current?
dI
V =L
dt
goes to - ∞ when
switch is opened.
use diode to shunt
current, protect switch.
•  Nothing changes instantly in Nature.
•  Spark coil used in early radio (Titanic).
•  Tesla patented the spark plug.
84
Symmetry of Electromagnetism
(from an electronics component point of view)
dV
I =C
dt
1
V = ∫ I dt
C
dI
V =L
dt
1
I = ∫ V dt
L
•  Only difference is no magnetic monopole.
85
Inductance adds like Resistance
Series
LS = L1 + L2
Parallel
1
LP =
1 L1 +1 L2
86
Capacitance adds like Conductance
Series
1
CS =
1 C1 +1 C2
Parallel
CP = C1 + C2
87
AC circuit analysis uses Sinusoids
88
Superposition of Sinusoids
•  Adding two sinusoids of the same frequency,
no matter what their amplitudes and phases,
yields a sinusoid of the same frequency.
•  Why? Trigonometry does not have an answer.
•  Linear systems change only phase and
amplitude
•  New frequencies do not appear.
89
Sinusoids are projections of
a unit vector spinning around the origin.
90
Derivative shifts 90° to the left
Taking a second derivative inverts a sinusoid.
91
Hooke’s Law
F = MA
⎛M⎞
⇒ d = −⎜ ⎟ a
⎝ k⎠
F = −kd
constant
Sinusoids result when a
function is proportional
to its own negative
second derivative.
92
acceleration
is negative of
displacement
also swings, flutes, guitar strings, electron
orbits, light waves, sound waves…
93
Complex numbers
•  Cartesian and Polar forms on complex plane.
•  Not vectors, though they add like vectors.
•  Can multiply two together (not so with vectors).
http://en.wikipedia.org/wiki/Phasor for nice animations
94
Complex Numbers
•  How to find r
95
Complex Exponentials – Phasor
•  All algebraic operations work with complex numbers.
•  What does it mean to raise something to an imaginary power?
Euler’s Identity
96
Cartesian and Polar forms
97
Cartesian and Polar forms (cont…)
98
Multiplying by a Complex Number
99
Dividing by a Complex Number
100
Rotating by + or - 90°
101
Examples
102
103
104
The “squiggly”
bracket: not an
algebraic expression.
y itself is real:
the coordinate on
the imaginary axis
105
106
2π, not π,
is the magic number
107
Now make the
phasor spin at
ω = 2π f
108
Multiplying by j shifts the phase by 90°
j=e
j
π
2
θ = 90°
solution to
Hooke’s Law
Just like a sinusoid: shifts 90° with each derivative.109
Complex Conjugates
110
Voltages and Currents are Real
111
Cosine is sum of 2 phasors
112
Sine is difference between 2 phasors
113
Trigonometry Revealed
114
115
Distribution of charge and voltage on multiple capacitors
•  To find the voltage on capacitors in series
–  Find total effective capacitance CTotal
–  Charge will be QTotal = CTotalV
–  Same charge will be on all caps (Kirchoff ’s Current Law)
–  Voltage will be distributed inversely to capacitance
These 2 slides
should have
gone earlier
QTotal = VCTotal
QTotal = Q1 = Q2
Q1 QTotal CTotal
V1 =
=
=
V
C1
C1
C1
Q2 QTotal CTotal
V2 =
=
=
V
C2
C2
C2
116
Distribution of charge and voltage on multiple capacitors
•  To find the charge in capacitors in parallel
–  Find total effective capacitance CTotal
–  Charge will be QTotal = CTotalV
–  Same voltage will be on all caps (Kirchoff ’s Voltage Law)
–  QTotal will be divided proportionally to capacitance
QTotal = VCTotal = Q1 + Q2
V = V1 = V2
Q1 = VC1
Q2 = VC2
117
Complex Impedance - Capacitor
represents
orthogonal
basis set
118
Complex Impedance - Inductor
119
Complex Impedance - Resistor
120
Impedance on the Complex Plane
121
Taxonomy of Impedance
122
Series Capacitors and Inductors
Two capacitors in series:
1
1
C1 + C2
Z=
+
=
=
jω C1 jω C2 jω C1C2
1
⎛
⎜
jω ⎜
⎜ 1
⎜
⎜⎝ C1
⎞
⎟
1
⎟
1 ⎟
⎟
+
C2 ⎟⎠
Two inductors in series:
Z = jω L1 + jω L2 = jω ( L1 + L2 )
123
Parallel Capacitors and Inductors
Two capacitors in parallel:
1
1
Z=
=
jω C1 + jω C2 jω (C1 + C2 )
Two inductors in parallel:
⎛
⎞
⎜ 1 ⎟
1
Z=
= jω ⎜
1
1
1 1 ⎟
⎜ + ⎟
+
jω L1 jω L2
⎝ L1 L2 ⎠
124
Same rules as DC circuits
now AC sources
125
Impedance of a Passive Branch – RC circuit
126
LC circuit - Resonance
At resonance,
impedances
add to zero
and cancel.
Analogous to spring
and weight system –
Energy in passed
between magnetic
and electric fields,
as in electromagnetic
wave.
127
Adding R to LC damps the ringing
Like dragging
your feet on the
swing. Energy
being passed
from magnetic
to electric field
eventually
dissipated by
resistor as heat.
128
“Tank” Circuit
permiability
and
permitivity
around loop
^
also pipe organs
and microwave
tubes.
129
current is now
through loop
rather than
around it
•  How can impedance be infinite through the parallel LC circuit
when each of the components can pass current?
•  At the resonant frequency the currents trying to pass from the
antenna to ground are shifted 90° in opposite directions and
thus are 180° out of phase and cancel.
•  This “null point” is an example of how lenses work with light.
130
Systems modeled as Filters
•
•
•
•
Has input and output signals (functions of time).
Most electronic systems use voltages as inputs and outputs.
Has a “transfer function” H(ω) in the frequency domain.
H(ω) determines what happens to e jωt, and thus any signal.
Vout (t )
H (ω ) =
Vin (t )
•  Consider system consisting of a voltage divider.
•  Divider can now use complex impedances.
H (ω ) =
Z1
Z1 + Z 2
131
Example: RC High-Pass Filter
Vout (t )
H (ω ) =
Vin (t )
H (ω ) =
R
1
R+
jω C
jω RC
=
1 + jω RC
At high frequencies, acts like a
piece of wire.
1
H (ω ) ≅ 1, ω >>
RC
At low frequencies, attenuates
and differentiates.
1
H (ω ) ≅ jω RC, ω <<
RC
Key frequency is reciprocal of time constant RC.
132
Example: LR Low-Pass Filter
Vout (t )
H (ω ) =
Vin (t )
H (ω ) =
R
1
=
R + jω L 1 + jω L
R
At low frequencies, acts like a
piece of wire.
R
H (ω ) ≅ 1, ω <<
L
At high frequencies,
attenuates and integrates.
R
R
H (ω ) ≅
, ω >>
jω L
L
Key frequency is reciprocal of time constant L/R.
133
Example: RC Low-Pass Filter
Vout (t )
H (ω ) =
Vin (t )
1
1
jω C
H (ω ) =
=
1
1 + jω RC
R+
jω C
At low frequencies, acts like a
piece of wire.
(assuming no current at output)
1
H (ω ) ≅ 1, ω <<
RC
At high frequencies,
attenuates and integrates.
1
1
H (ω ) ≅
, ω >>
jω RC
RC
Key frequency is reciprocal of time constant RC.
134
Decibels – ratio of gain (attenuation)
•  1 Bell = 10 dB = order of magnitude in power
⎛ Pout ⎞
1 dB ≡ 10 log10 ⎜
⎝ Pin ⎟⎠
so if Pin = 1 W, Pout = 100 W
•  Since power ∝ voltage
!
20 dB
2
⎛ Vout ⎞
1 dB ≡ 20 log10 ⎜
⎝ Vin ⎟⎠
so if Vin = 1 V, Vout = 10 V
Alexander
Graham Bell
!
20 dB
•  Decibels are pure ratios, no units
135
Magnitude and Phase of Low-Pass Filter
Recall low-pass filter:
Vout (t )
1
H (ω ) =
=
Vin (t ) 1 + jω RC
Vout (t ) = H (ω )Vin (t )
At corner (or “cut-off ”) frequency, ωC = 1/RC ,
1 1− j 1− j
H (ω ) =
⋅
=
1+ j 1− j
2
stationary phasor
changes magnitude
and phase of signal
assuming Vin(t) is a phasor
Magnitude (Gain/Attenuation)
Phase
1− j
1
H (ω ) =
=
2
2
⎛−1 ⎞
2⎟
∠H (ω ) = arctan ⎜
⎜⎝ 1 ⎟⎠
2
H (ω ) ≅ −3dB
∠H (ω ) = −45°
136
“Bode” Plot of Low Pass Filter (previous slide)
Simply a log/log
plot of
H (ω )
and
∠H (ω )
(this one vs. f , not ω)
http://www.electronics-tutorials.ws/filter/filter_2.html
137
Magnitude and Phase of High-Pass Filter
Recall high-pass filter:
Vout (t )
jω RC
H (ω ) =
=
Vin (t ) 1+ jω RC
Vout (t ) = H (ω )Vin (t )
At cut-off frequency, ωC = 1/RC ,
j 1− j 1+ j
H (ω ) =
⋅
=
1+ j 1− j
2
Magnitude (Attenuation)
again,
assuming Vin(t) is a phasor
Phase
1+ j
1
H (ω ) =
=
2
2
⎛1 ⎞
∠H (ω ) = arctan ⎜ 2 ⎟
⎜⎝ 1 ⎟⎠
2
H (ω ) ≅ −3dB
∠H (ω ) = 45°
138
“Bode” Plot of High Pass Filter (previous slide)
Simply a log/log
plot of
H (ω )
and
∠H (ω )
(this one vs. f , not ω)
http://www.electronics-tutorials.ws/filter/filter_2.html
139
Values for AC Voltage
“Peak” VP
or
“Max” VM
“Root Mean Square”
VRMS
“Peak-to Peak”
VPP
•  Any signal V(t) has all three values.
•  Since sin2 + cos2 = 1, and since sin2 and cos2 must each have the
same mean value, any sinusoid must have a mean value of ½.
1 + cos 2ω t
•  Or put another way:
2
cos (ω t ) =
•  Therefore, for a sinusoid VRMS
VP
=
2
(
2
)
mean = ½
140
AC Power in Resistor
When VRMS and IRMS are inphase (resistor), average
power is defined as in DC
Energy is not stored in the
resistor, but simply
dissipated as heat
For any signal:
2
V
(
2
RMS )
P = VRMS × I RMS =
= ( I RMS ) R
R
For sinusoids:
1
1 (VP )
1
2
P = VP × I P =
= (IP ) R
2
2 R
2
2
Power may be computed
from VP and IP for
sinusoids, or from VRMS and
IRMS for any signal.
because
VRMS
VP
=
2
and I RMS
IP
=
2
141
AC Power in Capacitor or Inductor
When VRMS and IRMS are 90° outof-phase (capacitor or inductor)
mean power is 0
sin ( 2ω t )
cos (ω t ) sin (ω t ) =
2
mean = 0
Thus no heat is dissipated, all
stored energy returned to circuit
Thus light propagates by passing
energy from magnetic to electric
fields without loss
sin and cos have 0 correlation
142
Transformer
iron core
•  Extremely efficient at preserving power:
•  Allows voltage (AC) to be changed:
V1I1 ≅ V2 I 2
N
V2 = V1
M
•  Voltages and currents assumed to be sinusoids, generally
reported as RMS
•  Permits efficient high-voltage power transmission, with small
current: thus little I 2R energy wasted in wire resistance.
143
World’s Fair
Chicago
1893
Tesla and
Westinghouse
(AC) beat
Edison (DC).
George
Westinghouse
Nikola
Tesla
Thomas
Edison
144
145
Phasor Notation
•  In BioE 1310, complex exponentials may be
described unambiguously with shorthand notation
jθ
re ⇒ "r∠θ "
•  Unfortunately when applied to real voltages and
currents, the same notation is widely used by
engineers to mean sin or cos, peak, or RMS.
Thus, A∠θ may mean (among other things)
A
v (t ) =
sin (ω t + θ )
2
or
v (t ) = A cos (ω t + θ )
146
Phasor Notation Ambiguity (cont…)
•  This ambiguity is allowed to continue because linear
systems change only magnitude and phase.
•  Thus a given network of coils, capacitors, and
resistors will cause the same relative change in
as it does in
A
v (t ) =
sin (ω t + θ )
2
v (t ) = A cos (ω t + θ )
so it doesn’t matter which definition of A∠θ they
use for real signals, so long as they are consistent.
See Phasor Notation Manifesto on website
147
Sample Problems with Phasor Notation
Using our unambiguous definition of phasor notation,
r∠θ = re jθ
Express the following as a complex number in Cartesian form
(x + jy):
( 4∠45°) ( 6∠45°) = 24∠90° = 0 + 24 j
•  In other words, multiply the magnitudes and add the
phases.
•  For division, divide the magnitudes and subtract the phases.
6∠30°
= 2∠ − 60° = 1 − j 3
3∠90°
148
Multiplying by j rotates any complex number 90°
z = a + jb
z ′ = jz = −b + ja
This can be recognized as
multiplication by a rotation
matrix, where
z′
im
z
b
re
a
z ′ = jz = a ′ + jb′
⎡ a ′ ⎤ ⎡ 0 −1 ⎤ ⎡ a ⎤
⎢
⎥=⎢
⎥⎢
⎥
⎣ b′ ⎦ ⎣ 1 0 ⎦ ⎣ b ⎦
Dot product projects
point (a,b) onto
rotated axes (circled).
149
Consider
e
jω t
graphically.
Its derivative
im
jω e
jω t
e jω t
jω t
de
jω t
= jω e
dt
re
is rotated by 90° and scaled by ω at all times.
Thus it must be spinning in a circle, and
jω t
since e = 1 when t = 0,
e
jω t
= cos ω t + j sin ω t
150
Superposition
Asin (ω t ) + B cos (ω t )
combine to form a sinusoid of frequency ω
amplitude
frequency phase
∑ A cos (ω t + θ ) is a sinusoid of frequency ω
i
i
i
any sinusoid of frequency ω
151
Sum of 2 sinusoids of same frequency…
cos (t )
sin (t )
cos (t ) + sin (t ) =
⎛ π⎞
2 cos ⎜ t − ⎟
⎝
4⎠
… is sinusoid of same frequency.
152
Superposition can be
illustrated by summing the
individual phasors
(without their complex
conjugates) within
sinusoids. For example,
just the positive frequency
phasor is shown here.
153
Recall complex conjugate pairs of phasors
e +e
cos (t ) =
2
jt
− jt
e −e
sin (t ) =
2j
jt
jt
− jt
e
_
2j
− jt
e jt
2j
e
2
e
2
− jt
(recall 1 = − j )
j
positive (solid) and negative (dashed) frequency.
154
Adding cos and sin conjugate pairs (black)…
sin (t )
cos (t )
…creates single conjugate pair (gray).
⎛ π⎞
cos (t ) + sin (t ) = 2 cos ⎜ t − ⎟
⎝
4⎠
hypotenuse
155
jω t
The complex exponential re forms
an orthogonal basis set for any signal.
Each phasor passes through a linear system
without interacting with any other.
To understand a system, all we need to know is
jω t
what it does to re
This is the system’s frequency response.
The system can only change the phase and
amplitude of a given phasor, not its frequency, by
multiplying it by stationary phasor H(ω).
156
Fourier Series
Applies only to periodic signals
Inverse Fourier Series
x (t ) =
+∞
∑ae
jnω 0t
n
n=−∞
fundamental
frequency
harmonic
number
Fourier coefficient: stationary phasor
(complex number)
determines magnitude and phase
of particular harmonic.
Any periodic signal x (t)
consists of a series of
sinusoidal harmonics of
a fundamental
frequency ω 0.
Each harmonic with n ≠ 0
is a pair of complex
conjugate phasors with
positive and negative
frequency.
The “DC” harmonic n = 0
has a constant value,.
157
Fourier Series: How to find coefficient an
Fourier Series
1
− jnω 0t
an = ∫ x (t ) e
dt
T0 T0
stationary
phasor for
harmonic
number n
Periodic signal x (t)
consists of phasors
forming the sinusoidal
harmonics of ω 0 .
Backward-spinning
phasor e − jnω 0t spins
fundamental the entire set of phasors
frequency in x(t), making the
jnω 0t
e
particular
phasor
backwards-spinning
stand still.
phasor.
2π
period T0 =
ω0
All other phasors
complete revolutions
integrating to 0.
158
The nth harmonic can be written as a
weighted sum of sin and cos at frequency nω 0 , (n ≥ 0)
An cos ( nω 0 t ) + Bn sin ( nω 0 t )
real coefficients
creating a single sinusoid of a given phase and amplitude.
(same as created by the conjugate phasors at n and –n).
The zero harmonic n = 0 (DC)
is a cosine of zero frequency
cos(0t)
159
Building a square wave by
adding the odd harmonics:
1, 3, 5, 7…
An infinite number of
harmonics are needed for a
theoretical square wave.
The harmonics account for
the harsher tone of the
square wave (buzzer),
compared to just the
fundamental 1rst harmonic
sinusoid (flute).
Wikipedia, Real Analysis
160
Fourier Transform
Applies to any finite signal (not just periodic)
Inverse Fourier Transform
1
x (t ) =
2π
∫
+∞
−∞
X (ω ) e
jω t
dω
Fourier Transform
X (ω ) = ∫
+∞
−∞
x (t ) e
− jω t
dt
Fourier coefficient an
has now become a
continues function
of frequency, X(ω).
X(ω) is a stationary
phasor for any
particular ω that
determines the
magnitude and phase of
the corresponding
jω t
phasor e in x(t).
As before, backwards-spinning phasor makes
corresponding component of x(t) stand still.
161
Transfer Function H(ω)
δ (t )
h (t )
t
system
t
For any input x(t), output y(t) = x(t) * h(t) (convolution).
H(ω) is the Fourier transform of the Impulse Response h(t).
A system is completely defined by h(t),
its response to the impulse function δ (t ) .
A system is completely defined by H(ω),
its response to each frequency.
162
Lumped Parameter Model - Loudspeaker
•
•
•
•
•
•
•
•
I
Electrical current in speaker voice-coil.
V
Mechanical force ‘voltage’ caused by electrical current.
Bl(x) Electro-dynamical force factor between I and V.
Re
Electrical resistance at DC.
ZL(x) Electrical inductance, varies with displacement x.
Cms(x) Compliance of mechanical suspension.
Mms Mechanical mass of loudspeaker diaphragm.
Rms
Mechanical friction and drag of loudspeaker (sound).
Energy goes both ways:
Speaker motion ‘current’ causes electrical voltage (neither shown).
163
Resulting Electrical Impedance
•  Electrical impedance results from electrical and mechanical parts.
•  Mechanical resonance generates a positive peak, even though
mechanically it is negative (L and C in series approach zero).
•  This is because electrical current causes mechanical ‘voltage’.
164