EECS 105 Fall 2004, Lecture 3
Lecture 3: Phasor notation,
Transfer Functions
Prof. J. S. Smith
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. J. S. Smith
Context
z
In the last lecture, we discussed:
–
–
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how to convert a linear circuit into a set of differential
equations,
How to convert the set of differential equations into the
frequency domain, a set of algebraic equations.
In this lecture we will cover:
–
–
–
–
Analyzing circuits with a sinusoidal input, (in the
frequency domain, a single frequency at a time)
How to simplify our notation with Phasors
Solving a couple of example circuits
How to present information about the circuit directly in
the frequency domain using diagrams of amplitude and
phase at different frequencies (Bode plots)
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University of California, Berkeley
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EECS 105 Fall 2003, Lecture 3
Prof. J. S. 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 )}
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. J. S. Smith
Single frequency approach
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Another way to look at the situation is that since the
circuit is linear, we can view any input as the sum
of sin waves at various amplitudes, frequencies,
and phases. (each piece of the Fourier transform)
If we can under the circuit for an arbitrary
sinusoidal input, we then can figure out what the
circuit will do for an arbitrary input, or inputs.
(Just break all the inputs up into sinusoids, put
them through one at a time, and then add the results
for each back up at the end, and that’s your
answer!)
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University of California, Berkeley
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EECS 105 Fall 2003, Lecture 3
Prof. J. S. Smith
Sinusoidal stimulus
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We are going to analyze circuits for a single
sinusoid at a time which we are going to write:
vin (t ) = Vi sin(ωt + φ )
But we are going to use exponential notation
vin (t ) = Vi sin(ωt + φ ) = Vi (e j (ωt +φ ) − e − j (ωt +φ ) ) / 2
vin (t ) = (Vi e j (φ ) e j (ωt ) − Vi e − j (φ ) e − j (ωt ) ) / 2
1
vin (t ) = [Vi e j (φ ) ]e j (ωt ) + C.C.
2
Complex conjugate
(same as first term, but with
(j)→(-j) wherever it occurs)
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. J. S. Smith
Sin in→Sin at every node!
It is especially interesting because any voltage or
current in our circuit, if this is the only input, must
also be sinusoidal with the same frequency, and so
can also be written in this form.
1
vany (t ) = [Vany e j (φ ) ]e j (ωt ) + C.C
2
1
iany (t ) = [ I any e j (φ ) ]e j (ωt ) + C.C
2
Because our equations will be linear, the same things
will happen to the complex conjugate terms as
happen to the first terms, so they will just tag along
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University of California, Berkeley
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EECS 105 Fall 2003, Lecture 3
Prof. J. S. Smith
Differentiating or integrating
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This form is particularly useful because it is easy to
differentiate or integrate with respect to time
1
vany (t ) = [Vany e j (φ ) ]e j (ωt ) + C.C.
2
d
1
vany (t ) = ( jω ) [Vany e j (φ ) ]e j (ωt ) + C.C.
dt
2
1 1
v
(
t
)
dt
=
[Vany e j (φ ) ]e j (ωt ) + C.C.
any
∫
( jω ) 2
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. J. S. Smith
Phasor notation
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As you may have noticed, we are going to write
expressions of this form a lot, so it is very common
to take the following shortcuts in notation:
Take the constants to include the phase, (they will
become complex constants)
Since all expressions will include the complex
constants, stop writing C. C. everywhere.
Since e-jωt appears everywhere, (or its C. C. in the
C. C. terms) stop writing it as well
Don’t write the ½ either
All of these, together, are called Phasor notation
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University of California, Berkeley
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EECS 105 Fall 2003, Lecture 3
Prof. J. S. Smith
Phasors
z
Each of the voltages between nodes, and each of
the currents, can then be represented by a single
complex number (remember, this is for a single
frequency input of a particular phase and
amplitude)
1
{
vany (t ) = [Vany e j (φ ) ]e j (ωt ) + C.C ⇒ Vˆany
2
Vˆany
{
1
iany (t ) = [ I any e j (φ ) ]e j (ωt ) + C.C ⇒ Iˆany
2
Iˆany
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. J. S. Smith
Tricky Bits:
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Phasor notation is very convenient, but there are
some tricky parts to look out for:
You can not use phasor notation (without added
precautions) if you need to multiply voltages and
currents (such as in a power calculation), because
that is not linear!
Another way to look at phasor notation is that
instead of adding the CC ( and dividing by 2), you
take the real part, which gives the same result.
However, you must not take the real part (or add
the complex conjugate) before you put back in the
time dependence e-jωt
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University of California, Berkeley
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EECS 105 Fall 2003, Lecture 3
Prof. J. S. Smith
Solving Linear Systems using Phasors
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Any linear circuit becomes a linear equation:
→ L1{vany (t )} = L 2 {vin (t )}& L1,2{} have the form
L{v(t )} = av(t ) + b1
z
d
d2
v(t ) + b2 2 v(t ) + L + c1 ∫ v(t )dt + c2 ∫∫ v(t )dt + L
dt
dt
For our complex exponential input Vejωt this is:
d jωt
d 2 jωt
L(Ve ) = aVe + b1V e + b2V 2 e + L + c1V ∫ e jωt + c2V ∫∫ e jωt + L
dt
dt
Ve jωt
Ve jωt
= aVe jωt + b1 jωVe jωt + b2 ( jω ) 2 Ve jωt + L + c1
+ c2
+L
( jω ) 2
jω
⎛
⎞
c
c
= H1Ve jωt = Ve jωt ⎜⎜ a + b1 jω + b2 ( jω ) 2 + L + 1 + 2 2 + L⎟⎟
jω ( jω )
⎝
⎠
jωt
z
jωt
Where H is just some complex number (at ω)
Department of EECS
EECS 105 Fall 2003, Lecture 3
z
University of California, Berkeley
Prof. J. S. Smith
Notice that linear operators acting on all of the
other voltages or currents are also a complex exp
times a complex number:
L2 {vany (t )} ⇒
⎞
⎛
c
c
H 2Vany e jωt = Vany e jωt ⎜⎜ a + b1 jω + b2 ( jω ) 2 + L + 1 + 2 2 + L⎟⎟
j ω ( jω )
⎠
⎝
So we are now prepared to calculate our circuits
response at any frequency using algebra, instead of
differential equations!
Department of EECS
University of California, Berkeley
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EECS 105 Fall 2003, Lecture 3
Prof. J. S. Smith
Complex Transfer Function
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Excite a system with an input voltage vin
Define the output voltage vany to be any node
voltage (branch current)
For a complex exponential input, the “transfer
function” from input to output( or any voltage or
current) can then be written:
H (ω ) =
n1 + n2 jω + n3 ( jω ) 2 + L
d1 + d 2 jω + d 3 ( jω ) 2 + L
( just multiply top and bottom by ejωt sufficient times)
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
z
Prof. J. S. Smith
The amplitude of the output is the magnitude of the
complex number and the phase of the output is the
phase of the complex number:
⎛
⎞
c
c
y = Hx = e jωt ⎜⎜ a + b1 jω + b2 ( jω ) 2 + L + 1 + 2 2 + L⎟⎟
jω ( j ω )
⎝
⎠
jωt
j p H (ω )
y = e H (ω ) e
Re[ y ] = H (ω ) cos(ωt + p H (ω ))
Department of EECS
University of California, Berkeley
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EECS 105 Fall 2003, Lecture 3
Prof. J. S. Smith
Impedance
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Suppose that the “input” is defined as the voltage of
a terminal pair (port) and the “output” is defined as
the current into the port:
+
v(t )
Arbitrary LTI
Circuit
i (t )
v(t ) = Ve jωt = V e j (ωt +φv )
i (t ) = Ie jωt = I e j (ωt +φi )
–
z
The impedance Z is defined as the ratio of the
phasor voltage to phasor current (“self” transfer
function)
Z (ω ) = H (ω ) =
V V j (φv −φi )
= e
I
I
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. J. S. Smith
Admitance
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Suppose that the “input” is defined as the current of
a terminal pair (port) and the “output” is defined as
the voltage into the port:
+
v(t )
Arbitrary LTI
Circuit
i (t )
v(t ) = Ve jωt = V e j (ωt +φv )
i (t ) = Ie jωt = I e j (ωt +φi )
–
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The admittance Z is defined as the ratio of the
phasor current to phasor voltage (“self” transfer
function)
Y (ω ) = H (ω ) =
Department of EECS
I
I
= e j (φi −φv )
V V
University of California, Berkeley
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EECS 105 Fall 2003, Lecture 3
Prof. J. S. Smith
Voltage and Current Gain
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The voltage (current) gain is just the voltage
(current) transfer function from one port to another
port:
+
+
v1 (t )
Arbitrary LTI
Circuit
i1 (t )
–
z
z
i2 (t )
v2 (t )
–
Gv (ω ) =
V2 V2 j (φ2 −φ1 )
=
e
V1 V1
Gi (ω ) =
I 2 I 2 j (φ2 −φ1 )
=
e
I1
I1
If G > 1, the circuit has voltage (current) gain
If G < 1, the circuit has loss or attenuation
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. J. S. Smith
Transimpedance/admittance
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Current/voltage gain are unitless quantities
Sometimes we are interested in the transfer of
voltage to current or vice versa
+
+
Arbitrary
linear Circuit
v1 (t ) or i1 (t )
–
Department of EECS
i2 (t )
or
v2 (t )
–
J (ω ) =
V2 V2 j (φ2 −φ1 )
=
e
I1
I1
[Ω]
K (ω ) =
I 2 I 2 j (φ2 −φ1 )
=
e
V1 V1
[S ]
University of California, Berkeley
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EECS 105 Fall 2003, Lecture 3
Prof. J. S. Smith
Direct Calculation of H (no DEs)
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To directly calculate the transfer function
(impedance, transimpedance, etc) we can
generalize the circuit analysis concept from the
“real” domain to the “phasor” domain
With the concept of impedance (admittance), we
can now directly analyze a circuit without explicitly
writing down any differential equations
Use KVL, KCL, mesh analysis, loop analysis, or
node analysis where inductors and capacitors are
treated as complex resistors
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. J. S. Smith
LPF Example: Again!
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Instead of setting up the DE in the time-domain,
let’s do it directly in the frequency domain
Treat the capacitor as an imaginary “resistance” or
impedance:
→
time domain “real” circuit
z
frequency domain “phasor” circuit
Last lecture we calculated the impedance:
ZR = R
Department of EECS
ZC =
1
j ωC
University of California, Berkeley
10
EECS 105 Fall 2003, Lecture 3
Prof. J. S. Smith
LPF … Voltage Divider
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Fast way to solve problem is to say that the LPF is
a voltage divider, using phasors:
1
V
ZC
1
j ωC
H (ω ) = o =
=
=
Vs Z C + Z R R + 1
1 + jωRC
j ωC
Department of EECS
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University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. J. S. Smith
Bigger Example (no problem!)
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Consider a more complicated example:
Z eff
H (ω ) =
Vo
ZC 2
=
Vs Z eff + Z C 2
H (ω ) =
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Z eff = R2 + R1 || Z C1
ZC 2
R2 + R1 || Z C1 + Z C 2
University of California, Berkeley
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EECS 105 Fall 2003, Lecture 3
Prof. J. S. Smith
Building Tents: Poles and Zeros
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For most circuits that we’ll deal with, the transfer
function can be shown to be a rational function
H (ω ) =
z
The behavior of the circuit can be extracted by
finding the roots of the numerator and denominator
H (ω ) =
z
n1 + n2 jω + n3 ( jω ) 2 + L
d1 + d 2 jω + d 3 ( jω ) 2 + L
( z1 − jω )( z 2 − jω ) L ∏ ( zi − jω )
=
( p1 − jω )( p2 − jω ) L ∏ ( pi − jω )
Or another form (DC gain explicit)
H (ω ) = G0 ( jω ) K
(1 − jωτ z1 )(1 − jωτ z 2 ) L
= G0 ( jω ) K
(1 − jωτ p 2 )(1 − jωτ p 2 ) L
∏ (1 − jωτ
∏ (1 − jωτ
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z ,i
)
p ,i
)
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. J. S. Smith
Poles and Zeros (cont)
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“poles”
The roots of the
numerator are called the
“zeros” since at these
frequencies, the transfer
function is zero
The roots of the
denominator are called
the “poles”, since at
these frequencies the
transfer function peaks
(like a pole in a tent)
Department of EECS
H (ω ) =
( z1 − jω )( z 2 − jω ) L
( p1 − jω )( p2 − jω ) L
University of California, Berkeley
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EECS 105 Fall 2003, Lecture 3
Prof. J. S. Smith
Finding the Magnitude (quickly)
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The magnitude of the response can be calculated
quickly by using the property of the mag operator:
H (ω ) = G0 ( jω ) K
= G0 ω K
z
(1 − jωτ z1 )(1 − jωτ z 2 ) L
(1 − jωτ p 2 )(1 − jωτ p 2 ) L
1 − jωτ z1 1 − jωτ z 2 L
1 − jωτ p 2 1 − jωτ p 2 L
The magnitude at DC depends on G0 and the
number of poles/zeros at DC. If K > 0, gain is zero.
If K < 0, DC gain is infinite. Otherwise if K=0,
then gain is simply G0
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. J. S. Smith
Finding the Phase (quickly)
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As proved in HW #1, the phase can be computed quickly
with the following formula:
(1 − jωτ z1 )(1 − jωτ z 2 ) L
p H (ω ) =p G0 ( jω ) K
(1 − jωτ p 2 )(1 − jωτ p 2 ) L
=p G0 + p ( jω ) K + p (1 − jωτ z1 )+ p (1 − jωτ z 2 ) + L
− p (1 − jωτ p1 )− p (1 − jωτ p 2 ) − L
z
Now the second term is simple to calculate for positive
frequencies:
p ( jω ) K = K
z
π
2
Interpret this as saying that multiplication by j is equivalent
to rotation by 90 degrees
Department of EECS
University of California, Berkeley
13
EECS 105 Fall 2003, Lecture 3
Prof. J. S. Smith
Bode Plots
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Simply the log-log plot of the magnitude and phase
response of a circuit (impedance, transimpedance, gain, …)
Gives insight into the behavior of a circuit as a function of
frequency
The “log” expands the scale so that breakpoints in the
transfer function are clearly delineated
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. J. S. Smith
Log ratios and definition of dB
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The frequency response can vary by orders of
magnitude rapidly
We can expand range by taking the log of the
magnitude response
dB = deciBel (deci = 10)
0
-10
-20
⎛ V0 ⎞
⎛V ⎞
⎜⎜ ⎟⎟ = 20 log⎜⎜ 0 ⎟⎟
⎝ Vs ⎠ dB
⎝ Vs ⎠
-30
-40
0.1
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1
10
100
University of California, Berkeley
14
EECS 105 Fall 2003, Lecture 3
Prof. J. S. Smith
Why 20? Power!
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Why multiply log by “20” rather than “10”?
Power is proportional to voltage squared to make
ratios in power and voltage come out the same, for
power ratios use 10×, for voltage ratios, use 20×
2
⎛V ⎞
⎛V ⎞
dB = 10 log⎜⎜ 0 ⎟⎟ = 20 log⎜⎜ 0 ⎟⎟
⎝ Vs ⎠
⎝ Vs ⎠
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. J. S. Smith
Example: High-Pass Filter
z
L
jω
j ωL
R
H (ω ) =
=
R + j ωL 1 + j ω L
R
jωτ
H (ω ) =
1 + jωτ
Using the voltage divider rule:
ω →∞⇒
H →
ω →0⇒
H →
ω=
Department of EECS
1
τ
⇒
jωτ
=1
jωτ
0
=0
1+ 0
j
1
H =
=
1+ j
2
University of California, Berkeley
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EECS 105 Fall 2003, Lecture 3
Prof. J. S. Smith
HPF Magnitude Bode Plot
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Recall that log of product is the sum of log
H (ω ) dB =
jωτ
dB
ωτ = 1 ⇒ jωτ
dB
jωτ
1 + jωτ
= jωτ
dB
+
dB
1
1 + jωτ
dB
Increase by 20 dB/decade
= 0 dB Equals unity at breakpoint
40 dB
20 dB
20 dB
0 dB
0.1
1
Department of EECS
10
100
-20 dB
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. J. S. Smith
HPF Bode dissection
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The second term can be further dissected:
1
1 + jωτ
0 dB
.1 / τ
-20 dB
1/τ
= 0 dB − 1 + jωτ
dB
dB
ω <<
10 / τ
20 dB
ω >>
-40 dB
ω=
-60 dB
ω=
Department of EECS
1
τ
1
τ
1
τ
1
τ
0 dB
-20 dB/dec
~ -3dB
- 3dB
University of California, Berkeley
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EECS 105 Fall 2003, Lecture 3
Prof. J. S. Smith
slope
z
At breakpoint:
⎛V ⎞
ω = 1 / τ → ⎜⎜ 0 ⎟⎟ = −3 dB
V
⎝
z
s
⎠ dB
Observe: slope of signal attenuation is 20
dB/decade in frequency
⎛V ⎞
ω = 100 / τ → ⎜⎜ 0 ⎟⎟ = −40 dB
⎝ Vs ⎠ dB
⎛V ⎞
ω = 1000 / τ → ⎜⎜ 0 ⎟⎟ = −60 dB
⎝ Vs ⎠ dB
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. J. S. Smith
Composite Plot
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Composite is simply the sum of each component:
jωτ
+
dB
1
1 + jωτ
jωτ
dB
0 dB
.1 / τ
1/τ
dB
High frequency ~ 0 dB Gain
10 / τ
Low frequency attenuation
-20 dB
1
1 + jωτ
dB
-40 dB
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17