EECS 105 Fall 2003, Lecture 2
Lecture 3: Bode Plots
Prof. Niknejad
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. A. Niknejad
Get to know your logs!
dB
-20
-10
-5
-3
-2
-1


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?
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. A. Niknejad
Bode Plot Overview

Technique for estimating a complicated transfer
function (several poles and zeros) quickly
(1  j z1 )(1  j z 2 )(1  j zn )
H ( )  G0 ( j )
(1  j p 2 )(1  j p 2 ) (1  j pm )
K

Break frequencies :
i 
Department of EECS
1
i
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. A. Niknejad
Summary of Individual Factors


Simple Pole: 0 dB
1
1  j

1


1

 90
Simple Zero: 0 dB
 90
1  j

DC Zero:
0 dB
 90
0 dB
 90
j

DC Pole:
1
j
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. A. Niknejad
Example

Consider the following transfer function
10 5 j (1  j 2 )
H ( j ) 
(1  j 1 )(1  j 3 )

 1  100 ns
 2  10 ns
 3  100 ps
Break frequencies: invert time constants
1  10 Mrad/s
H ( j ) 
Department of EECS
2  100 Mrad/s
3  10 Grad/s
j

(1  j )
5
10
2


(1  j )(1  j )
1
3
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. A. Niknejad
Breaking Down the Magnitude

Recall log of products is sum of logs
H ( j ) dB  20 log
 20 log
j

(
1

j
)
5
10
2


(1  j )(1  j )
1
3
j


20
log
1

j
105
2


 20 log 1  j
 20 log 1  j
1
3

Let’s plot each factor separately and add them
graphically
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. A. Niknejad
Breaking Down the Phase

Since a  b  a  b
10 5 j (1  j 2 )
H ( j )  
(1  j 1 )(1  j 3 )
j

H ( j )   5  1  j
10
2


 1  j  1  j
1
3

Let’s plot each factor separately and add them
graphically
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. A. Niknejad
Magnitude Bode Plot: DC Zero
80
60
j
5
10
40
0 dB
20
104
105
106
107
108
109
1010
1011

-20
-40
-60
-80
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. A. Niknejad
Phase Bode Plot: DC Zero
180
135
j
 5
10
90
45
104
105
106
107
108
109
1010
1011

-45
-90
-135
-180
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. A. Niknejad
Magnitude Bode Plot: Add First Pole
80
1  10 Mrad/s
j
105
60
dB
40
20
104
105
106
107
108
109
1010
1011

-20
-40
-60
1
-80
1 j
Department of EECS

107
dB
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. A. Niknejad
Phase Bode Plot: Add First Pole
180
135
90
j
 5
10
45
104
105
106
107
108
109
1010
1011

-45
-90

-135
1
1 j
-180
Department of EECS

10 7
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. A. Niknejad
Magnitude Bode Plot: Add 2nd Zero
80
2  100 Mrad/s
1 j
60

108
dB
40
20
104
105
106
107
108
109
1010
1011

-20
-40
-60
-80
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. A. Niknejad
Phase Bode Plot: Add 2nd Zero
180
135
1  j
90

108
45
104
105
106
107
108
109
1010
1011

-45
-90
-135
-180
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. A. Niknejad
Magnitude Bode Plot: Add 2nd Pole
80
60
3  10 Grad/s
40
20
104
105
106
107
108
109
1010
1011

-20
1
-40
1 j
-60

1010
dB
-80
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. A. Niknejad
Phase Bode Plot: Add 2nd Pole
180
135
90
45
104
105
106
107
108
109
1010
1011

-45
-90
 1  j
-135

1010
-180
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. A. Niknejad
Comparison to “Actual” Mag Plot
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. A. Niknejad
Comparison to “Actual” Phase Plot
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. A. Niknejad
Why do I say “actual”?



I plotted the transfer characteristics with
Mathematica
The range of frequency for the plot is 6 orders of
magnitude. The program has to find the “hot spots”
in order to plot the function. Near the hot spots,
more points are plotted. In between hot spots, the
function is interpolated. If you pick the wrong
points, you’ll end up with the wrong plot:
mag = LogLinearPlot[20*Log[10, Abs[H[x]]], {x, 10^4,
10^11},PlotPoints -> 10000, Frame -> True,PlotStyle ->
Thickness[.005], ImageSize -> 600,GridLines -> Automatic,
PlotRange -> {{10^4, 10^11}, {-20, 100}} ]
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. A. Niknejad
Don’t always believe a computer!
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. A. Niknejad
Second Order Transfer Function

The series resonant circuit is one of the most
important elementary circuits:

The physics describes not only physical LCR
circuits, but also approximates mechanical
resonance (mass-spring, pendulum, molecular
resonance, microwave cavities, transmission lines,
buildings, bridges, …)
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. A. Niknejad
Series LCR Analysis

With phasor analysis, this circuit is readily
analyzed
+
Vo
−
Vs  I jL  I

Vs  I 

1
IR
j C

1
j L 
 R 
j C

Vs
V0  I R 
R
1
j L 
R
j C
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. A. Niknejad
Second Order Transfer Function

So we have:
+
Vo
−

H ( j ) 
V0
R

1
Vs
jL 
R
j C
To find the poles/zeros, let’s put the H in canonical
form:
V0
j CR
H ( j ) 

Vs 1   2 LC  j RC

One zero at DC frequency  can’t conduct DC due
to capacitor
Department of EECS
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. A. Niknejad
Poles of 2nd Order Transfer Function

Denominator is a quadratic polynomial:
R
j
L
V0
j CR
H ( j ) 


2
1
R
Vs 1   LC  j RC
2
 ( j )  j
LC
L
R
j
1
2
L


H ( j ) 
0
R
LC
02  ( j ) 2  j
L
j
H ( j ) 
Q
  ( j )  j
2
0
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 0
2
0
Q
Q
0 L
R
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. A. Niknejad
Finding the poles…

Let’s factor the denominator:
( j )  j
2
0
02
0
Q
 02  0
0
1


  
 j0 1 

2Q
4Q
2Q
4Q 



2
0
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
Im
Re
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. A. Niknejad
Resonance without Loss

The transfer function can parameterized in terms of
Im
loss. First, take the lossless case, R=0:
2
 


2
0
0
  

 0 
  j0

 2Q

4
Q

 Q 



Re
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
University of California, Berkeley
EECS 105 Fall 2003, Lecture 3
Prof. A. Niknejad
Magnitude Response

The response peakiness depends on Q
0 R

j 0
0 L
Q
H ( j ) 

 R

02   2  j 0
02   2  j 0
0 L
Q
j
H ( j0 )  1
Q 1
j
H ( 0)  0
Q  10
H ( j 0 ) 
02
Q
02  02  j0
0
1
Q
Q  100
0
Department of EECS
University of California, Berkeley