EECS 105 SPRING 2004, Lecture 1
Lecture 2: Frequency domain
analysis, Phasors
Prof. J. Stephen Smith
Department of EECS
University of California, Berkeley
EECS 105 Fall 2004, Lecture 2
Prof. J. Stephen Smith
Announcements
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The course web site is
http://inst.eecs.berkeley.edu/~ee105
Today’s discussion section will meet
The Wednesday discussion section will move to
Tuesday, 5:00-6:00, 293 Cory
You can go to any or all discussions you like.
Labs will start Feb 3.
Reading assignments from the text will start next
week
There is a homework set due Wed. 1/28
Department of EECS
University of California, Berkeley
1
EECS 105 Fall 2004, Lecture 2
Prof. J. Stephen Smith
Context
In the last lecture we covered:
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how circuits can be modeled as linear circuits by
design or approximation.
z How to convert a linear circuit into a set of
differential equations.
In this lecture, we will cover:
z How to use complex analysis to solve circuits by
converting the differential equations in the time
domain into algebraic equations in the frequency
domain.
Department of EECS
University of California, Berkeley
Linear Circuit model↔ set of linear
differential equations
EECS 105 Fall 2004, Lecture 2
Prof. J. Stephen Smith
resistors
vr (t ) = ir (t ) ⋅ r
inductors
vL (t ) = L
capacitors
ic (t ) = C
diL (t )
dt
dvC (t )
dt
The wires convey the variables (voltages and currents) between
the equations (components) , by applying Kirchoff’s laws.
For the low pass example:
vout (t ) = vc (t )
vin = vr + vc
iin = ir = ic + iout
Department of EECS
University of California, Berkeley
2
EECS 105 Fall 2004, Lecture 2
Prof. J. Stephen Smith
Differential equation for low pass:
We are going to take the output current equal to zero, for simplicity, so:
vr = ir R = ic R = RC
dvC (t )
dt
vout (t ) = vc (t )
vin = vr + vout = vout + RC
dvout (t )
dt
Department of EECS
University of California, Berkeley
EECS 105 Fall 2004, Lecture 2
Prof. J. Stephen Smith
→Equation
For any linear circuit, you will be able to write:
L1{vout (t )} = L 2 {vin (t )} =
d
d2
avin (t ) + b1 vin (t ) + b2 2 vin (t ) + L
dt
dt
+ c1 ∫ vin (t ) + c2 ∫∫ vin (t ) + c3 ∫∫∫ vin (t ) + L
L {}, L {}
1
2
Here
represent Linear operators, that is, if you apply
it to a function, you get a new function (it maps functions to functions),
and linear operators also have the property that:
L{a ⋅ f (t ) + b ⋅ g (t )} = a ⋅ L{ f (t )} + b ⋅ L{g (t )}
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University of California, Berkeley
3
It’s now just mathematics, and
therefore easy!
EECS 105 Fall 2004, Lecture 2
Prof. J. Stephen Smith
Once we establish a linear model for a circuit, by
design or approximation:
z We can directly use the powerful methods of linear
analysis from mathematics.
z We can develop our intuition as to what will
happen, allowing us to design.
Department of EECS
University of California, Berkeley
EECS 105 Fall 2004, Lecture 2
Prof. J. Stephen Smith
Fourier Transform
One important linear analysis technique we will use is
the Fourier Transform:
The Fourier transform states:
F (ω ) =
+∞
∫ f (t )e
− jωt
dt
−∞
f (t ) =
1
2π
+∞
∫ F (ω )e
jωt
dω
−∞
Notice that what this says is that information that is
expressed as a function of time (voltage or current
for example) can be completely expressed as a
function of frequency: F (ω )
Department of EECS
University of California, Berkeley
4
EECS 105 Fall 2004, Lecture 2
Prof. J. Stephen Smith
Linearity
Let’s use functions of frequency F (ω ) (voltages and
currents) rather than functions of time as our new
variables. The Fourier relationship shows us that if
we can find these functions of frequency, we can
then convert them into the voltages and currents as
a function of time that we want. We can do this in
every equation for our linear components.
1
f (t ) =
2π
+∞
∫ F (ω )e
j ωt
dω
−∞
We just substitute this form into each of our equations
Department of EECS
University of California, Berkeley
EECS 105 Fall 2004, Lecture 2
Prof. J. Stephen Smith
Resistors:
vr (t ) = ir (t ) ⋅ r
1
2π
Department of EECS
+∞
1
v
(
ω
)
e
d
ω
=
r
∫
2π
−∞
jωt
+∞
∫ i (ω ) ⋅ re
r
jωt
dω
−∞
University of California, Berkeley
5
EECS 105 Fall 2004, Lecture 2
Prof. J. Stephen Smith
Capacitors:
ic (t ) = C
1
2π
1
2π
dvC (t )
dt
+∞
d 1
i
(
ω
)
e
d
ω
=
C
c
∫
dt 2π
−∞
+∞
jωt
1
∫−∞ic (ω )e dω = 2π
jωt
+∞
∫ v (ω )e
j ωt
c
dω
−∞
+∞
∫ C ( jω )v (ω )e
c
j ωt
dω
−∞
Department of EECS
University of California, Berkeley
EECS 105 Fall 2004, Lecture 2
Prof. J. Stephen Smith
Inductors:
vL (t ) = L
1
2π
1
2π
Department of EECS
+∞
diL (t )
dt
d 1
v
(
ω
)
e
d
ω
=
L
L
∫
dt 2π
−∞
+∞
jωt
1
v
(
ω
)
e
d
ω
=
L
∫
2π
−∞
jωt
+∞
∫i
L
(ω )e jωt dω
−∞
+∞
∫ L( jω )i
L
(ω )e jωt dω
−∞
University of California, Berkeley
6
EECS 105 Fall 2004, Lecture 2
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Prof. J. Stephen Smith
In each of these, we can eliminate the integration
over frequency, and the constant, to get:
vr (ω ) = ir (ω ) ⋅ r
ic (ω ) = vc (ω ) ⋅ C ( jω )
vc (ω ) =
1
ic (ω )
C ( jω )
vL (ω ) = iL (ω ) ⋅ L( jω )
Department of EECS
University of California, Berkeley
Conversion of linear circuits to
algebraic equations
EECS 105 Fall 2004, Lecture 2
Prof. J. Stephen Smith
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Something wonderful just happened: each of our
simultaneous linear differential equations were just
converted to algebraic equations (just
multiplication by a constant for these examples),
and the same thing happens to every linear circuit.
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Of course, the same thing happens to the
relationships derived from Kirchoff’s laws (they
are linear too; adding voltages, for example)
Department of EECS
University of California, Berkeley
7
EECS 105 Fall 2004, Lecture 2
Prof. J. Stephen Smith
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The advantage of changing differential equations
into algebraic equations comes at a small price: the
constants that we are multiplying by, and the
functions of frequency for both voltage and current,
are now complex numbers.
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No matter how complicated the circuit, if we drive
the circuit with a real function vin (t ) when we find
the output by using the inverse Fourier transform, it
will be real as well, as well as any voltage or
current at any node, at all times.
Department of EECS
University of California, Berkeley
EECS 105 Fall 2004, Lecture 2
Prof. J. Stephen Smith
Complex numbers
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It is important to think of complex numbers as just
an expansion over the definition of real numbers.
For example, if A, B, and C are complex:
A( B + C ) = AB + AC
AB = BA
A+ B = B + A
This seems trivial, but there is only one* other
definition for “numbers” which obeys these
properties**:
*The other one is “quaternions”
**finite, but not countable.
Department of EECS
University of California, Berkeley
8
EECS 105 Fall 2004, Lecture 2
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Prof. J. Stephen Smith
If you have a calculator which can handle complex numbers
you can just plug them in. Otherwise, you can use these
derived rules to get results for complex numbers from the
rules for real numbers*
A = ar + ai j
B = br + bi j
A + B = (ar + br ) + (ai + bi ) j
AB = (ar + ai j )(br + bi j ) = ar br − ai bi + a r b j j + ai br j
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j is used by electrical engineers for the imaginary constant
to avoid confusion with i for current which got there first.
*just as we handle reals by approximating with integer calculations and keeping
track of decimal points
Department of EECS
EECS 105 Fall 2004, Lecture 2
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University of California, Berkeley
Prof. J. Stephen Smith
Notice that unlike the reals, there is a complex
number that, when multiplied by itself, gives
negative one:
A = 0 +1 j
AA = 0 − 1 + 0 + 0 = −1
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Almost all functions have extensions over the
complex numbers, and in some ways complex
numbers seem more “complete” in that the inverses
of functions exist, roots can always be found, etc.
Department of EECS
University of California, Berkeley
9
EECS 105 Fall 2004, Lecture 2
Prof. J. Stephen Smith
http://mathworld.wolfram.com/
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A complex function is said to be analytic on a region R if it
is complex differentiable at every point in R. The terms
holomorphic function, differential function, complex
differentiable function, and regular function are sometimes
used interchangeably with "analytic function"
If a function is analytic in a region R, it is infinitely
differentiable in R. A function may fail to be analytic at one
or more points through the presence of singularities, or
along lines or line segments through the presence of branch
cuts.
A single-valued function that is analytic in all but possibly a
discrete subset of its domain, and at those singularities goes
to infinity like a polynomial (i.e., these exceptional points
must be poles and not essential singularities), is called a
meromorphic function.
Department of EECS
University of California, Berkeley
EECS 105 Fall 2004, Lecture 2
Prof. J. Stephen Smith
Why introduce complex numbers?
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They actually make things easier
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One insightful derivation of e
Consider a second order homogeneous DE
y '' + y = 0
⎧ sin x
y=⎨
⎩cos x
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Since sine and cosine are linearly independent, any
solution is a linear combination of the
“fundamental” solutions
Department of EECS
University of California, Berkeley
10
EECS 105 Fall 2004, Lecture 2
Prof. J. Stephen Smith
Insight into Complex Exponential
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But note that e jx is also a solution!
That means: e jx = a1 sin x + a2 cos x
e − jx = − a1 sin x + a2 cos x
e jx − e − jx = 2a1 sin x
But a number minus its complex conjugate gives an imaginary number so
is pure imaginary. Differentiating:
a1
je jx + je − jx = 2a1 cos x
a1 = ja2
And
a2
is real, so:
And since
e = a ( j sin x + cos x)
jx
e j 0 = 1 = a ( j sin 0 + cos 0) = a
Department of EECS
University of California, Berkeley
EECS 105 Fall 2004, Lecture 2
Prof. J. Stephen Smith
The Rotating Complex Exponential
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So the complex exponential is nothing but a point
tracing out a unit circle on the complex plane:
e ix = cos x + i sin x
e
Department of EECS
i ωt
e
− i ωt
e i ωt + e − i ωt
2
University of California, Berkeley
11
EECS 105 Fall 2004, Lecture 2
Prof. J. Stephen Smith
Magic: Turn Diff Eq into Algebraic Eq
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Integration and differentiation are trivial with
complex numbers:
d iωt
e = iωe iωt
dt
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∫e
iωτ
dτ =
1 i ωt
e
iω
Any ODE is now trivial algebraic manipulations …
in fact, we’ll show that you don’t even need to
directly derive the ODE by using phasors
The key is to observe that the current/voltage
relation for any element can be derived for complex
exponential excitation
Department of EECS
University of California, Berkeley
EECS 105 Fall 2004, Lecture 2
Prof. J. Stephen Smith
Complex Exponential is Powerful
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To find steady state response we can excite the system with
a complex exponential
Mag Response
e iωt
LTI System
H
H (ω ) ei (ωt +φ )
Phase Response
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At any frequency, the system response is characterized by a
single complex number H:
H (ω )
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This is not surprising since a sinusoid is a sum of complex
exponentials (and because of linearity!)
sin ωt =
z
φ =p H (ω )
eiωt − e − iωt
2i
cos ωt =
eiωt + e − iωt
2
From this perspective, the complex exponential is even
more fundamental
Department of EECS
University of California, Berkeley
12
EECS 105 Fall 2004, Lecture 2
Prof. J. Stephen Smith
Remember our low pass filter?
+
vout
vC (t )
vin
ic (t )
_
Department of EECS
University of California, Berkeley
EECS 105 Fall 2004, Lecture 2
Prof. J. Stephen Smith
LPF Example: frequency domain
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Let’s look at a single frequency from the source:
vs (t ) = v0 (t ) + RC
dv0
dt
vs (t ) = Vs e jωt
vo (t ) = V0 e j (ωt +φ ) = V0 e jωt
real
complex
Vs e jωt = V0 e jωt + RC ⋅ jω ⋅V0 e jωt
Vs = V0 (1 + jω ⋅τ )
V0
1
=
Vs (1 + jω ⋅τ )
Department of EECS
University of California, Berkeley
13
EECS 105 Fall 2004, Lecture 2
Prof. J. Stephen Smith
Magnitude and Phase Response
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The system is characterized by the complex
function
H (ω ) =
z
V0
1
=
Vs (1 + jω ⋅τ )
The magnitude and phase response:
H (ω ) =
V0
1
=
Vs
1 + (ωτ ) 2
p H (ω ) = − tan −1 ωτ
Department of EECS
University of California, Berkeley
EECS 105 Fall 2004, Lecture 2
Prof. J. Stephen Smith
Why did it work?
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The system is linear:
Re[ y ] = L(Re[ x]) = Re[L( x)]
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If we excite system with a sinusoid:
vs (t ) = Vs cos ωt = Vs Re[e jωt ]
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If we push the complex exp through the system first
and take the real part of the output, then that’s the
“real” sinusoidal response
vo (t ) = Vo cos(ωt + φ ) = Vo Re[e j (ωt +φ ) ]
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University of California, Berkeley
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