EECS 105 Spring 2004, Lecture 4
EECS 105 Spring 2004, Lecture 4
Prof. J. S. Smith
Bode plots
Since the majority of this lecture is on how to create
approximate Bode plots by hand, it is fair to ask
why we should do so when it can be done quickly
on a computer.
Lecture 4: Bode Plots
The answer is that a few features of transfer functions
that we will exploit for our graphs will appear often
in different contexts, and design of a circuit for a
particular purpose will often entail putting together
several of these features, and the language of the
circuit designer will use these constructs: poles,
zeros, resonances, etc.
Prof. J. S. Smith
Department of EECS
University of California, Berkeley
EECS 105 Spring 2004, Lecture 4
Prof. J. S. Smith
Department of EECS
EECS 105 Spring 2004, Lecture 4
Context
z
–
–
z
z
We discussed analyzing circuits with a sinusoidal input,
(in the frequency domain, a single frequency at a time)
How to simplify our notation with Phasors
and introduced Bode plots
In this lecture, we will:
–
–
–
–
Review how get a transfer function for a circuit
How to put the transfer function into a standard form
Find why magnitude and phase plots are a useful form.
How to create an approximate Bode plot for a circuit.
Department of EECS
Prof. J. S. Smith
Find the Transfer Function
In the last lecture:
–
University of California, Berkeley
University of California, Berkeley
z
z
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
This is found by using phasor notation to change
circuits into networks of complex resistors, then
applying Kirchoff’s laws repeatedly
Department of EECS
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1
EECS 105 Spring 2004, Lecture 4
Prof. J. S. Smith
EECS 105 Spring 2004, Lecture 4
Bode Plot Overview
z
Then put the transfer function into standard form:
ω
ω
ω
(1 + j
)(1 + j
) L (1 + j
)
ω z1
ωz2
ωz3
K
H (ω ) = G0 ( jω )
ω
ω
ω
(1 + j
)(1 + j
) L (1 + j
)
ω p1
ω p2
ω p3
1
Each of the frequencies: ωi =
correspond to
τi
z
z
z
time constants which are features of the circuit, and
are called break frequencies.
Those that appear in the numerator are called zeros,
and those that appear in the denominator are called
poles.
Department of EECS
University of California, Berkeley
EECS 105 Spring 2004, Lecture 4
Prof. J. S. Smith
Department of EECS
–
–
–
Real zeros
Real poles
Conjugate pairs of zeros
Conjugate pairs of holes
Each of these can appear in multiple orders (two poles
at the same frequency, for example)
Additional features of the function are the constant
G0, and the order of the overall term →K
z
0 dB
Simple Pole:
−1
z
Simple Zero: 0 dB
z
DC Zero:
ω
1+ j
ω zn
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0 dB
ω
j
ω0
z
ω = ωn
ω = ωn
⎛
⎞
⎜1 + j ω ⎟
⎜
⎟
ω
pn
⎝
⎠
DC1Pole:
j
Department of EECS
Prof. J. S. Smith
Summary of Individual Factors
Since the transfer function will always result in a
real voltage, the following features can appear:
–
University of California, Berkeley
EECS 105 Spring 2004, Lecture 4
Breakpoints
z
If we take the log magnitude of each factor (as in
db), they add to find the total magnitude.
The phase angle adds as well, each factor
contributes to the overall phase change.
So we just catalog each of the features, and then
add their magnitudes (in db) or contributions to the
phase angles
ω
ωo
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0 dB
− 90
+ 90
+ 90
Contribution to Phase angle
z
Adding them up
Magnitude in db
z
Prof. J. S. Smith
− 90
University of California, Berkeley
2
EECS 105 Spring 2004, Lecture 4
Prof. J. S. Smith
EECS 105 Spring 2004, Lecture 4
Prof. J. S. Smith
Example
z
Consider the following transfer function
H ( jω ) =
z
Phase
10 −5 jω (1 + jωτ 2 )
(1 + jωτ 1 )(1 + jωτ 3 )
z
τ 1 = 100 ns
τ 2 = 10 ns
τ 3 = 100 ps
∠H ( jω ) = ∠
ω2 = 100 Mrad/s
H ( jω ) =
ω3 = 10 Grad/s
− ∠{1 + j
jω
ω
(1 + j )
105
ω2
(1 + j
ω
ω
)(1 + j )
ω1
ω3
Department of EECS
z
University of California, Berkeley
EECS 105 Spring 2004, Lecture 4
Prof. J. S. Smith
= 20 log
ω
jω
(1 + j )
ω2
105
60
Prof. J. S. Smith
jω
105
40
ω
ω
)(1 + j )
ω1
ω3
0 dB
20
jω
ω
+ 20 log 1 + j
105
ω2
104
105
106
107
108
109
1010
1011
ω
-20
-40
Plot each factor separately and add them
graphically
Department of EECS
University of California, Berkeley
EECS 105 Spring 2004, Lecture 4
80
ω
ω
− 20 log 1 + j
− 20 log 1 + j
ω1
ω3
z
Plot each factor separately and add them
graphically
Magnitude Bode Plot: DC Zero
Recall log of products is sum of logs
(1 + j
ω
ω
} − ∠{1 + j }
ω1
ω3
Department of EECS
Magnitude
H ( jω ) dB = 20 log
10 −5 jω (1 + jωτ 2 )
(1 + jωτ 1 )(1 + jωτ 3 )
jω
ω
∠H ( jω ) = ∠{ 5 } + ∠{1 + j }
10
ω2
Break frequencies: invert time constants
ω1 = 10 Mrad/s
z
Since ∠a ⋅ b = ∠a + ∠b
-60
-80
University of California, Berkeley
Department of EECS
University of California, Berkeley
3
EECS 105 Spring 2004, Lecture 4
Prof. J. S. Smith
EECS 105 Spring 2004, Lecture 4
Phase Bode Plot: DC Zero
180
135
∠
Phase Bode Plot: Add First Pole
jω
105
180
135
90
90
45
45
104
105
106
107
108
109
1010
1011
ω
-45
-90
-90
-135
-135
-180
∠
jω
105
104
-45
105
106
107
108
109
1010
University of California, Berkeley
1+ j
EECS 105 Spring 2004, Lecture 4
Prof. J. S. Smith
Department of EECS
ω1 = 10 Mrad/s
jω
105
60
80
20
107
108
109
1010
1011
ω
104
-20
-20
-40
-40
-60
1
-80
1+ j
Department of EECS
ω2 = 100 Mrad/s
2nd
105
Zero
1+ j
dB
20
106
Prof. J. S. Smith
60
40
105
University of California, Berkeley
Magnitude Bode Plot: Add
40
104
ω
10 7
EECS 105 Spring 2004, Lecture 4
Magnitude Bode Plot: Add First Pole
ω
1011
1
∠
-180
Department of EECS
80
Prof. J. S. Smith
106
107
108
109
1010
ω
108
1011
dB
ω
-60
ω
10
7
-80
dB
University of California, Berkeley
Department of EECS
University of California, Berkeley
4
EECS 105 Spring 2004, Lecture 4
Prof. J. S. Smith
Phase Bode Plot: Add
2nd
EECS 105 Spring 2004, Lecture 4
Zero
Prof. J. S. Smith
Phase Bode Plot: Add
180
2nd
Pole
180
135
∠1 + j
90
ω
135
108
90
45
45
104
105
106
107
108
109
1010
1011
ω
104
-45
-45
-90
-90
-135
106
107
108
109
1010
− ∠1 + j
-135
-180
105
1011
ω
ω
1010
-180
Department of EECS
University of California, Berkeley
EECS 105 Spring 2004, Lecture 4
Prof. J. S. Smith
Magnitude Bode Plot: Add
2nd
Pole
1010
1011
Department of EECS
EECS 105 Spring 2004, Lecture 4
University of California, Berkeley
Prof. J. S. Smith
Comparison to “Actual” Mag Plot
80
60
ω3 = 10 Grad/s
40
20
104
105
106
107
108
109
ω
-20
1
-40
1+ j
-60
ω
1010
dB
-80
Department of EECS
University of California, Berkeley
Department of EECS
University of California, Berkeley
5
EECS 105 Spring 2004, Lecture 4
Prof. J. S. Smith
EECS 105 Spring 2004, Lecture 4
Prof. J. S. Smith
Series LCR Analysis
Comparison to “Actual” Phase Plot
z
With phasor analysis, this circuit is readily
analyzed
+
Vo
−
1
+I R
jω C
⎞
1
jωL +
+ R ⎟⎟
jω C
⎠
Vs = I jωL + I
⎛
Vs = I ⎜⎜
⎝
Vs
R
1
jωL +
+R
jω C
V0 = I R =
Department of EECS
University of California, Berkeley
EECS 105 Spring 2004, Lecture 4
Prof. J. S. Smith
Department of EECS
EECS 105 Spring 2004, Lecture 4
Second Order Transfer Function
z
The series resonant circuit is one of the most
important elementary circuits:
University of California, Berkeley
Prof. J. S. Smith
Second Order Transfer Function
z
So we have:
+
Vo
−
z
z
The physics describes not only physical LCR
circuits, but also approximates mechanical
resonance (mass-spring, pendulum, molecular
resonance, microwave cavities, transmission lines,
buildings, bridges, …)
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V0
R
=
1
Vs
jωL +
+R
jω C
To find the poles/zeros, let’s put the H in canonical
form:
H ( jω ) =
z
H ( jω ) =
V0
jω CR
=
Vs 1 − ω 2 LC + jω RC
One zero at DC frequency Æ can’t conduct DC due
to capacitor
Department of EECS
University of California, Berkeley
6
EECS 105 Spring 2004, Lecture 4
Prof. J. S. Smith
Poles of
z
2nd
jω
z
R
L
V0
jω CR
=
=
1
R
Vs 1 − ω 2 LC + jω RC
+ ( jω ) 2 + jω
LC
L
R
jω
1
L
ω02 ≡
H ( jω ) =
R
LC
2
2
ω 0 + ( jω ) + jω
L
H ( jω ) =
j
⎛ ω
⎞
ω02
0
±
− ω02 ⎟
= ± jω 0
2
⎜ 2Q
⎟
4Q
⎝
⎠ Q →∞
z
z
ω02 + ( jω ) 2 + j
ωω0
Q≡
Q
Department of EECS
ω0 L
R
z
University of California, Berkeley
EECS 105 Spring 2004, Lecture 4
The transfer function can parameterized in terms of
Im
loss. First, take the lossless case, R=0:
Prof. J. S. Smith
When the circuit is lossless, the poles are at real
frequencies, so the transfer function blows up!
At this resonance frequency, the circuit has zero
imaginary impedance
Even if we set the source equal to zero, the circuit
can have a steady-state response
Department of EECS
Prof. J. S. Smith
( jω ) 2 + j
ω=−
z
z
z
Magnitude Response
Let’s factor the denominator:
ω0
2Q
±
ω02
4Q 2
ωω0
Q
z
ω0
2Q
± jω 0 1 −
ω0 R
ω
jω 0
ω0 L
Q
H ( jω ) =
=
ω0 R
ω
2
2
2
2
ω 0 − ω + jω
ω 0 − ω + jω 0
ω0 L
Q
1
4Q 2
Poles are complex conjugate frequencies
The Q parameter is called the
“quality-factor” or Q-factor
This parameters is an important
→0
parameter:
Q ⎯R⎯
⎯→ ∞
Department of EECS
How strongly peaked the response is depends on Q
jω
+ ω02 = 0
− ω02 = −
University of California, Berkeley
EECS 105 Spring 2004, Lecture 4
Finding the poles…
z
Re
ω = ⎜−
ω ω0
Q
Prof. J. S. Smith
Resonance without Loss
Order Transfer Function
Denominator is a quadratic polynomial:
H ( jω ) =
EECS 105 Spring 2004, Lecture 4
H ( jω 0 ) = 1
Q =1
Im
j
Re
H (0) = 0
Q = 10
H ( jω0 ) =
ω02
Q
ω02 − ω02 + jω0
ω0
=1
Q
Q = 100
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Department of EECS
ω0
University of California, Berkeley
7
EECS 105 Spring 2004, Lecture 4
Prof. J. S. Smith
Get to know your logs!
dB
-20
-10
-5
-3
-2
-1
z
z
ratio
0.100
0.316
0.562
0.708
0.794
0.891
dB
20
10
5
3
2
1
ratio
10.000
3.162
1.778
1.413
1.259
1.122
Engineers are very conservative. A “margin” of
3dB is a factor of 2 (power)!
Knowing a few logs by memory can help you
calculate logs of different ratios by employing
properties of log. For instance, knowing that the
ratio of 2 is 3 dB, what’s the ratio of 4?
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8