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Bode Plot Overview - Control Systems - Electrical Engineering (EE)
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Page 1
Introduction to Bode Plot
• 2 plots – both have logarithm of frequency on x -axis
o y-axis magnitude of transfer function, H(s), in dB
o y-axis phase angle
The plot can be used to interpret how the input affects the output in both magnitude and phase over
frequency.
Where do the Bode diagram lines comes from?
1) Determine the Transfer Function of the system:
) (
) (
) (
1
1
p s s
z s K
s H
+
+
=
2) Rewrite it by factoring both the numerator and denominator into the standard form
) 1 (
) 1 (
) (
1
1
1
1
+
+
=
p
s
sp
z
s
Kz
s H
where the z s are called zeros and the p
s are called poles.
3) Replace s with j? . Then find the Magnitude of the Transfer Function.
) 1 (
) 1 (
) (
1
1
1
1
+
+
=
p
jw
jwp
z
jw
Kz
jw H
If we take the log
10
of this magnitude and multiply it by 20 it takes on the form of
20 log
10
(H(jw)) =
?
?
?
?
?
?
?
?
?
?
+
+
) 1 (
) 1 (
log 20
1
1
1
1
10
p
jw
jwp
z
jw
Kz
=
) 1 ( log 20 log 20 log 20 ) 1 ( log 20 log 20 log 20
1
10 10 1 10
1
10 1 10 10
+ - - - + + +
z
jw
jw p
z
jw
z K
Each of these individual terms is very easy to show on a logarithmic plot. The entire Bode log magnitude plot is
the result of the superposition of all the straight line terms. This means with a little practice, we can quickly sketch
the effect of each term and quickly find the overall effect. To do this we have to understand the effect of the
different types of terms.
These include: 1) Constant terms K
2) Poles and Zeros at the origin | j? |
3) Poles and Zeros not at the origin
1
1
p
j w
+ or
1
1
z
j w
+
4) Complex Poles and Zeros (addressed later)
Page 2
Introduction to Bode Plot
• 2 plots – both have logarithm of frequency on x -axis
o y-axis magnitude of transfer function, H(s), in dB
o y-axis phase angle
The plot can be used to interpret how the input affects the output in both magnitude and phase over
frequency.
Where do the Bode diagram lines comes from?
1) Determine the Transfer Function of the system:
) (
) (
) (
1
1
p s s
z s K
s H
+
+
=
2) Rewrite it by factoring both the numerator and denominator into the standard form
) 1 (
) 1 (
) (
1
1
1
1
+
+
=
p
s
sp
z
s
Kz
s H
where the z s are called zeros and the p
s are called poles.
3) Replace s with j? . Then find the Magnitude of the Transfer Function.
) 1 (
) 1 (
) (
1
1
1
1
+
+
=
p
jw
jwp
z
jw
Kz
jw H
If we take the log
10
of this magnitude and multiply it by 20 it takes on the form of
20 log
10
(H(jw)) =
?
?
?
?
?
?
?
?
?
?
+
+
) 1 (
) 1 (
log 20
1
1
1
1
10
p
jw
jwp
z
jw
Kz
=
) 1 ( log 20 log 20 log 20 ) 1 ( log 20 log 20 log 20
1
10 10 1 10
1
10 1 10 10
+ - - - + + +
z
jw
jw p
z
jw
z K
Each of these individual terms is very easy to show on a logarithmic plot. The entire Bode log magnitude plot is
the result of the superposition of all the straight line terms. This means with a little practice, we can quickly sketch
the effect of each term and quickly find the overall effect. To do this we have to understand the effect of the
different types of terms.
These include: 1) Constant terms K
2) Poles and Zeros at the origin | j? |
3) Poles and Zeros not at the origin
1
1
p
j w
+ or
1
1
z
j w
+
4) Complex Poles and Zeros (addressed later)
Effect of Constant Terms:
Constant terms such as K contribute a straight horizontal line of magnitude 20 log
10
(K)
H = K
Effect of Individual Zeros and Poles at the origin:
A zero at the origin occurs when there is an s or j? multiplying the numerator. Each occurrence of this
causes a positively sloped line passing through ? = 1 with a rise of 20 db over a decade.
H = | w j |
A pole at the origin occurs when there are s or j? multiplying the denominator. Each occurrence of this
causes a negatively sloped line passing through ? = 1 with a drop of 20 db over a decade.
H =
w j
1
Effect of Individual Zeros and Poles Not at the Origin
Zeros and Poles not at the origin are indicated by the (1+j? /z
i
) and (1+j? /p
i
). The values
z
i
and p
i
in each of these expression is called a critical frequency (or break frequency). Below their critical
frequency these terms do not contribute to the log magnitude of the overall plot. Above the critical
frequency, they represent a ramp function of 20 db per decade. Zeros give a positive slope. Poles produce a
negative slope.
H =
i
i
p
j
z
j
w
w
+
+
1
1
20 log
10
(K)
?
0.1
1
10 100
(log scale)
20 log
10
(H)
?
0.1
1
10 100
(log
10
scale)
20 log(H)
-20 db
dec.
dec.
+20 db
z
i
p
i
-20 db
?
0.1
1
10 100
(log scale)
20 log(H)
dec
?
0.1
1
10 100
(log scale)
20 log(H)
20 db
dec
Page 3
Introduction to Bode Plot
• 2 plots – both have logarithm of frequency on x -axis
o y-axis magnitude of transfer function, H(s), in dB
o y-axis phase angle
The plot can be used to interpret how the input affects the output in both magnitude and phase over
frequency.
Where do the Bode diagram lines comes from?
1) Determine the Transfer Function of the system:
) (
) (
) (
1
1
p s s
z s K
s H
+
+
=
2) Rewrite it by factoring both the numerator and denominator into the standard form
) 1 (
) 1 (
) (
1
1
1
1
+
+
=
p
s
sp
z
s
Kz
s H
where the z s are called zeros and the p
s are called poles.
3) Replace s with j? . Then find the Magnitude of the Transfer Function.
) 1 (
) 1 (
) (
1
1
1
1
+
+
=
p
jw
jwp
z
jw
Kz
jw H
If we take the log
10
of this magnitude and multiply it by 20 it takes on the form of
20 log
10
(H(jw)) =
?
?
?
?
?
?
?
?
?
?
+
+
) 1 (
) 1 (
log 20
1
1
1
1
10
p
jw
jwp
z
jw
Kz
=
) 1 ( log 20 log 20 log 20 ) 1 ( log 20 log 20 log 20
1
10 10 1 10
1
10 1 10 10
+ - - - + + +
z
jw
jw p
z
jw
z K
Each of these individual terms is very easy to show on a logarithmic plot. The entire Bode log magnitude plot is
the result of the superposition of all the straight line terms. This means with a little practice, we can quickly sketch
the effect of each term and quickly find the overall effect. To do this we have to understand the effect of the
different types of terms.
These include: 1) Constant terms K
2) Poles and Zeros at the origin | j? |
3) Poles and Zeros not at the origin
1
1
p
j w
+ or
1
1
z
j w
+
4) Complex Poles and Zeros (addressed later)
Effect of Constant Terms:
Constant terms such as K contribute a straight horizontal line of magnitude 20 log
10
(K)
H = K
Effect of Individual Zeros and Poles at the origin:
A zero at the origin occurs when there is an s or j? multiplying the numerator. Each occurrence of this
causes a positively sloped line passing through ? = 1 with a rise of 20 db over a decade.
H = | w j |
A pole at the origin occurs when there are s or j? multiplying the denominator. Each occurrence of this
causes a negatively sloped line passing through ? = 1 with a drop of 20 db over a decade.
H =
w j
1
Effect of Individual Zeros and Poles Not at the Origin
Zeros and Poles not at the origin are indicated by the (1+j? /z
i
) and (1+j? /p
i
). The values
z
i
and p
i
in each of these expression is called a critical frequency (or break frequency). Below their critical
frequency these terms do not contribute to the log magnitude of the overall plot. Above the critical
frequency, they represent a ramp function of 20 db per decade. Zeros give a positive slope. Poles produce a
negative slope.
H =
i
i
p
j
z
j
w
w
+
+
1
1
20 log
10
(K)
?
0.1
1
10 100
(log scale)
20 log
10
(H)
?
0.1
1
10 100
(log
10
scale)
20 log(H)
-20 db
dec.
dec.
+20 db
z
i
p
i
-20 db
?
0.1
1
10 100
(log scale)
20 log(H)
dec
?
0.1
1
10 100
(log scale)
20 log(H)
20 db
dec
• To complete the log magnitude vs. frequency plot of a Bode diagram, we superposition all the lines
of the different terms on the same plot.
Example 1:
For the transfer function given, sketch the Bode log magnitude diagram which shows how the log
magnitude of the system is affected by changing input frequency. (TF=transfer function)
1
2 100
TF
s
=
+
Step 1: Repose the equation in Bode plot form:
1
100
1
50
TF
s
??
??
??
=
+
recognized as
1
1
1
K
TF
s
p
=
+
with K = 0.01 and p
1
= 50
For the constant, K: 20 log
10
(0.01) = -40
For the pole, with critical frequency, p
1
:
Example 2:
Your turn. Find the Bode log magnitude plot for the transfer function,
4
2
5 10
505 2500
xs
TF
ss
=
++
Start by simplifying the transfer function form:
50
-40 db
0db
? (log scale)
20 log
10
(MF)
Page 4
Introduction to Bode Plot
• 2 plots – both have logarithm of frequency on x -axis
o y-axis magnitude of transfer function, H(s), in dB
o y-axis phase angle
The plot can be used to interpret how the input affects the output in both magnitude and phase over
frequency.
Where do the Bode diagram lines comes from?
1) Determine the Transfer Function of the system:
) (
) (
) (
1
1
p s s
z s K
s H
+
+
=
2) Rewrite it by factoring both the numerator and denominator into the standard form
) 1 (
) 1 (
) (
1
1
1
1
+
+
=
p
s
sp
z
s
Kz
s H
where the z s are called zeros and the p
s are called poles.
3) Replace s with j? . Then find the Magnitude of the Transfer Function.
) 1 (
) 1 (
) (
1
1
1
1
+
+
=
p
jw
jwp
z
jw
Kz
jw H
If we take the log
10
of this magnitude and multiply it by 20 it takes on the form of
20 log
10
(H(jw)) =
?
?
?
?
?
?
?
?
?
?
+
+
) 1 (
) 1 (
log 20
1
1
1
1
10
p
jw
jwp
z
jw
Kz
=
) 1 ( log 20 log 20 log 20 ) 1 ( log 20 log 20 log 20
1
10 10 1 10
1
10 1 10 10
+ - - - + + +
z
jw
jw p
z
jw
z K
Each of these individual terms is very easy to show on a logarithmic plot. The entire Bode log magnitude plot is
the result of the superposition of all the straight line terms. This means with a little practice, we can quickly sketch
the effect of each term and quickly find the overall effect. To do this we have to understand the effect of the
different types of terms.
These include: 1) Constant terms K
2) Poles and Zeros at the origin | j? |
3) Poles and Zeros not at the origin
1
1
p
j w
+ or
1
1
z
j w
+
4) Complex Poles and Zeros (addressed later)
Effect of Constant Terms:
Constant terms such as K contribute a straight horizontal line of magnitude 20 log
10
(K)
H = K
Effect of Individual Zeros and Poles at the origin:
A zero at the origin occurs when there is an s or j? multiplying the numerator. Each occurrence of this
causes a positively sloped line passing through ? = 1 with a rise of 20 db over a decade.
H = | w j |
A pole at the origin occurs when there are s or j? multiplying the denominator. Each occurrence of this
causes a negatively sloped line passing through ? = 1 with a drop of 20 db over a decade.
H =
w j
1
Effect of Individual Zeros and Poles Not at the Origin
Zeros and Poles not at the origin are indicated by the (1+j? /z
i
) and (1+j? /p
i
). The values
z
i
and p
i
in each of these expression is called a critical frequency (or break frequency). Below their critical
frequency these terms do not contribute to the log magnitude of the overall plot. Above the critical
frequency, they represent a ramp function of 20 db per decade. Zeros give a positive slope. Poles produce a
negative slope.
H =
i
i
p
j
z
j
w
w
+
+
1
1
20 log
10
(K)
?
0.1
1
10 100
(log scale)
20 log
10
(H)
?
0.1
1
10 100
(log
10
scale)
20 log(H)
-20 db
dec.
dec.
+20 db
z
i
p
i
-20 db
?
0.1
1
10 100
(log scale)
20 log(H)
dec
?
0.1
1
10 100
(log scale)
20 log(H)
20 db
dec
• To complete the log magnitude vs. frequency plot of a Bode diagram, we superposition all the lines
of the different terms on the same plot.
Example 1:
For the transfer function given, sketch the Bode log magnitude diagram which shows how the log
magnitude of the system is affected by changing input frequency. (TF=transfer function)
1
2 100
TF
s
=
+
Step 1: Repose the equation in Bode plot form:
1
100
1
50
TF
s
??
??
??
=
+
recognized as
1
1
1
K
TF
s
p
=
+
with K = 0.01 and p
1
= 50
For the constant, K: 20 log
10
(0.01) = -40
For the pole, with critical frequency, p
1
:
Example 2:
Your turn. Find the Bode log magnitude plot for the transfer function,
4
2
5 10
505 2500
xs
TF
ss
=
++
Start by simplifying the transfer function form:
50
-40 db
0db
? (log scale)
20 log
10
(MF)
Example 2 Solution:
Your turn. Find the Bode log magnitude plot for the transfer function,
4
2
5 10
505 2500
xs
TF
ss
=
++
Simplify transfer function form:
4
4
5 10
5 10 20
5*500
( 5)( 500)
( 1)( 1) ( 1)( 1)
5 500 5 500
x
s
xss
TF
s s ss
ss
===
++
++++
Recognize: K = 20 à 20 log10(20) = 26.02
1 zero at the origin
2 poles: at p
1
= 5 and p
2
=500
Technique to get started:
1) Draw the line of each individual term on the graph
2) Follow the combined pole-zero at the origin line back to the left side of the graph.
3) Add the constant offset, 20 log
10
(K), to the value where the pole/zero at the origin line intersects the left
side of the graph.
4) Apply the effect of the poles/zeros not at the origin. working from left (low values) to right (higher
values) of the poles/zeros.
0 db
-40 db
10
0
80 db
-80 db
40 db
10
3
10
2
10
1
? (log scale)
Page 5
Introduction to Bode Plot
• 2 plots – both have logarithm of frequency on x -axis
o y-axis magnitude of transfer function, H(s), in dB
o y-axis phase angle
The plot can be used to interpret how the input affects the output in both magnitude and phase over
frequency.
Where do the Bode diagram lines comes from?
1) Determine the Transfer Function of the system:
) (
) (
) (
1
1
p s s
z s K
s H
+
+
=
2) Rewrite it by factoring both the numerator and denominator into the standard form
) 1 (
) 1 (
) (
1
1
1
1
+
+
=
p
s
sp
z
s
Kz
s H
where the z s are called zeros and the p
s are called poles.
3) Replace s with j? . Then find the Magnitude of the Transfer Function.
) 1 (
) 1 (
) (
1
1
1
1
+
+
=
p
jw
jwp
z
jw
Kz
jw H
If we take the log
10
of this magnitude and multiply it by 20 it takes on the form of
20 log
10
(H(jw)) =
?
?
?
?
?
?
?
?
?
?
+
+
) 1 (
) 1 (
log 20
1
1
1
1
10
p
jw
jwp
z
jw
Kz
=
) 1 ( log 20 log 20 log 20 ) 1 ( log 20 log 20 log 20
1
10 10 1 10
1
10 1 10 10
+ - - - + + +
z
jw
jw p
z
jw
z K
Each of these individual terms is very easy to show on a logarithmic plot. The entire Bode log magnitude plot is
the result of the superposition of all the straight line terms. This means with a little practice, we can quickly sketch
the effect of each term and quickly find the overall effect. To do this we have to understand the effect of the
different types of terms.
These include: 1) Constant terms K
2) Poles and Zeros at the origin | j? |
3) Poles and Zeros not at the origin
1
1
p
j w
+ or
1
1
z
j w
+
4) Complex Poles and Zeros (addressed later)
Effect of Constant Terms:
Constant terms such as K contribute a straight horizontal line of magnitude 20 log
10
(K)
H = K
Effect of Individual Zeros and Poles at the origin:
A zero at the origin occurs when there is an s or j? multiplying the numerator. Each occurrence of this
causes a positively sloped line passing through ? = 1 with a rise of 20 db over a decade.
H = | w j |
A pole at the origin occurs when there are s or j? multiplying the denominator. Each occurrence of this
causes a negatively sloped line passing through ? = 1 with a drop of 20 db over a decade.
H =
w j
1
Effect of Individual Zeros and Poles Not at the Origin
Zeros and Poles not at the origin are indicated by the (1+j? /z
i
) and (1+j? /p
i
). The values
z
i
and p
i
in each of these expression is called a critical frequency (or break frequency). Below their critical
frequency these terms do not contribute to the log magnitude of the overall plot. Above the critical
frequency, they represent a ramp function of 20 db per decade. Zeros give a positive slope. Poles produce a
negative slope.
H =
i
i
p
j
z
j
w
w
+
+
1
1
20 log
10
(K)
?
0.1
1
10 100
(log scale)
20 log
10
(H)
?
0.1
1
10 100
(log
10
scale)
20 log(H)
-20 db
dec.
dec.
+20 db
z
i
p
i
-20 db
?
0.1
1
10 100
(log scale)
20 log(H)
dec
?
0.1
1
10 100
(log scale)
20 log(H)
20 db
dec
• To complete the log magnitude vs. frequency plot of a Bode diagram, we superposition all the lines
of the different terms on the same plot.
Example 1:
For the transfer function given, sketch the Bode log magnitude diagram which shows how the log
magnitude of the system is affected by changing input frequency. (TF=transfer function)
1
2 100
TF
s
=
+
Step 1: Repose the equation in Bode plot form:
1
100
1
50
TF
s
??
??
??
=
+
recognized as
1
1
1
K
TF
s
p
=
+
with K = 0.01 and p
1
= 50
For the constant, K: 20 log
10
(0.01) = -40
For the pole, with critical frequency, p
1
:
Example 2:
Your turn. Find the Bode log magnitude plot for the transfer function,
4
2
5 10
505 2500
xs
TF
ss
=
++
Start by simplifying the transfer function form:
50
-40 db
0db
? (log scale)
20 log
10
(MF)
Example 2 Solution:
Your turn. Find the Bode log magnitude plot for the transfer function,
4
2
5 10
505 2500
xs
TF
ss
=
++
Simplify transfer function form:
4
4
5 10
5 10 20
5*500
( 5)( 500)
( 1)( 1) ( 1)( 1)
5 500 5 500
x
s
xss
TF
s s ss
ss
===
++
++++
Recognize: K = 20 à 20 log10(20) = 26.02
1 zero at the origin
2 poles: at p
1
= 5 and p
2
=500
Technique to get started:
1) Draw the line of each individual term on the graph
2) Follow the combined pole-zero at the origin line back to the left side of the graph.
3) Add the constant offset, 20 log
10
(K), to the value where the pole/zero at the origin line intersects the left
side of the graph.
4) Apply the effect of the poles/zeros not at the origin. working from left (low values) to right (higher
values) of the poles/zeros.
0 db
-40 db
10
0
80 db
-80 db
40 db
10
3
10
2
10
1
? (log scale)
Example 3: One more time. This one is harder. Find the Bode log magnitude plot for the transfer function,
200( 20)
(2 1)( 40)
s
TF
sss
+
=
++
Simplify transfer function form:
0 db
-40 db
10
0
80 db
-80 db
40 db
10
3
10
2
10
1
20log
10
(TF)
? (log scale)
0 db
-40 db
10
0
80 db
-80 db
40 db
10
3
10
2
10
1
? (log scale)
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