bodeplot.pdf

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 log10 of this magnitude and multiply it by 20 it takes on the form of
20 log10 (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 sket
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 ω
+ or
1
1
z
j ω
+
4) Complex Poles and Zeros (addressed later)

Effect of Constant Terms:
Constant terms such as K contribute a straight horizontal line of magnitude 20 log10(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 = | ω
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 =
ω
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? /zi) and (1+j? /pi). The values
zi and pi 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
ω
ω
+
+
1
1
20 log10(K)
?
0.1 1 10 100
(log scale)
20 log10(H)
?
0.1 1 10 100
(log10 scale)
20 log(H)
-20 db
dec.
dec.
+20 db
zi
pi
-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 superpositionall 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 howthe 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 p1 = 50
For the constant, K: 20 log10(0.01) = -40
For the pole, with critical frequency, p1:
Example 2:
Your turn. Find the Bode log magnitude plot for the transfer function,
4
2
5 10
505 2500
x s
TF
s s
=
+ +
Start by simplifying the transfer function form:
50
-40 db
0db ? (log scale)
20 log10(MF)

Example 2 Solution:
Your turn. Find the Bode log magnitude plot for the transfer function,
4
2
5 10
505 2500
x s
TF
s s
=
+ +
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
x s s
TF
s s s s
s s
= = =
+ + + + + +
Recognize: K = 20 à 20 log10(20) = 26.02
1 zero at the origin
2 poles: at p1 = 5 and p2=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 log10(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
100
80 db
-80 db
40 db
103
102
101
? (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
s s s
+
=
+ +
0 db
-40 db
100
80 db
-80 db
40 db
103
102
101
20log10(TF)
? (log scale)
0 db
-40 db
100
80 db
-80 db
40 db
103
102
101
? (log scale)

side of the graph.
Example 3 Solution: Find the Bode log magnitude plot for the transfer function,
200( 20)
(2 1)( 40)
s
TF
s s s
+
=
+ +
200*20
( 1) 100 ( 1)
200( 20) 40 20 20
(2 1)( 40) ( 1)( 1) ( 1)( 1)
0.5 40 0.5 40
s s
s
TF
s s s s
s s s s s
+ +
+
= = =
+ + + + + +
Recognize: K = 100 à 20 log10(100) = 40
1 pole at the origin
1 zero at z1 = 20
2 poles: at p1 = 0.5 and p2=40
0 db
-40 db
100
80 db
-80 db
40 db
103
102
101
20log10(TF)
40 db/dec
20 db/dec
20 db/dec
40 db/dec
? (log scale)

side of the graph.
The plot of the log magnitude vs. input frequency is only half of the story.
We also need to be able to plot the phase angle vs. input frequency on a log scale as well to complete the
full Bode diagram..
For our original transfer function,
)
1
(
)
1
(
)
(
1
1
1
1
+
+
=
p
jw
jwp
z
jw
Kz
jw
H
the cumulative phase angle associated with this function are given by
)
1
(
)
1
(
)
(
1
1
1
1
+
∠
∠
∠
+
∠
∠
∠
=
∠
p
jw
p
jw
z
jw
z
K
jw
H
Then the cumulative phase angle as a function of the input frequency may be written as





 +
−
−
−
+
+
+
∠
=
∠ )
1
(
)
(
)
1
(
)
(
1
1
1
1 p
jw
p
jw
z
jw
z
K
jw
H
Once again, to show the phase plot of the Bode diagram, lines can be drawn for each of the different terms.
Then the total effect may be found by superposition.
Effect of Constants on Phase:
A positive constant, K>0, has no effect on phase. A negative constant, K<0, will set up a phase shift of
±180o
. (Remember real vs imaginary plots – a negative real number is at ±180o
relative to the origin)
Effect of Zeros at the origin on Phase Angle:
Zeros at the origin, s, cause a constant +90 degree shift for each zero.
∠ TF
Effect of Poles at the origin on Phase Angle:
Poles at the origin, s -1
, cause a constant -90 degree shift for each pole.
∠ TF
? (log)
+90 deg
?
-90 deg

Effect of Zeros not at the origin on Phase Angle:
Zeros not at the origin, like
1
1
z
j ω
+ , have no phase shift for frequencies much lower than zi, have a +
45 deg shift at z1, and have a +90 deg shift for frequencies much higherthan z1.
∠ H
.
To draw the lines for this type of term, the transition from 0o
to +90o
is drawn over 2 decades, starting at
0.1z1 and ending at 10z1.
Effect of Poles not at the origin onPhase Angle:
Poles not at the origin, like
1
1
1
p
jω
+
, have no phase shift for frequencies much lower than pi, have a -
45 deg shift at p1, and have a -90 deg shift for frequencies much higherthan p1.
∠ TF
.
To draw the lines for this type of term, the transition from 0o
to -90o
is drawn over 2 decades, starting at
0.1p1 and ending at 10p1.
When drawing the phase angle shift for not-at-the-origin zeros and poles, first locate the criticalfrequency
of the zero or pole. Then start the transition 1 decade before, following a slope of
±45o
/decade. Continue the transition until reaching the frequency one decade past the critical frequency.
Now let’s complete the Bode Phase diagrams for the previous examples:
?
0.1z1 1z1 10z1 100z1
+90 deg
+45 deg
?
0.1p1 1p1 10p1 100p1
-90 deg
-45 deg

Example 1:
For the Transfer Function given, sketch the Bode diagram which shows how the phase of the system is
affected by changing input frequency.
1 (1/100)
2 100 ( 1)
50
TF
s
s
= =
+ +
20 log|TF|
TF
+90
-90
-40db
50
5 500 rad/s
?
?
0.5

Example 2:
Repeat for the transfer function,
20log|TF|
4
2
5 10 20
505 2500 ( 1)( 1)
5 500
x s s
TF
s s
s s
= =
+ + + +
0 db
-40 db
100
80 db
-80 db
40 db
103
102
101
? (log scale)
0o
-90o
100
180o
-180o
90o
103
102
101
? (log scale)
20 log10(MF)
Phase Angle

Example 2 Solution:
Repeat for the transfer function,
20log|TF|
4
2
5 10 20
505 2500 ( 1)( 1)
5 500
x s s
TF
s s
s s
= =
+ + + +
0 db
-40 db
100
80 db
-80 db
40 db
103
102
101
? (log scale)
0o
-90o
100
180o
-180o
90o
103
102
101
? (log scale)
20 log10(MF)
Phase Angle

Example 3: Find the Bode log magnitude and phase angle plot for the transfer function,
100 ( 1)
200( 20) 20
(2 1)( 40) ( 1)( 1)
0.5 40
s
s
TF
s s
s s s s
+
+
= =
+ + + +
0 db
-40 db
10-1
80 db
-80 db
40 db
102
101
100
? (log scale)
0o
-90o
10-1
180o
-180o
90o
102
101
100
? (log scale)
20 log10(MF)
Phase Angle

Example 4:
Sketch the Bode plot (Magnitude and Phase Angle) for
3
100 10 ( 1)
( 10)( 1000)
s
TF
s s
× +
=
+ +
=
20log10|TF|
Angle of TF

Example 4:
Sketch the Bode plot (Magnitude and Phase Angle) for
3
100 10 ( 1)
( 10)( 1000)
s
TF
s s
× +
=
+ +
=
10( 1)
1
( 1)( 1)
10 1000
s
s s
+
+ +
Therefore: K = 10 so 20log10(10) = 20 db
One zero: z1 = 1
Two poles: p1 = 10 and p2 = 1000
20log10|TF|
Angle of TF
40
20
0 db
-20
-40
10-2
10-1
100
101
102
103
104
10-2
10-1
100
101
102
103
104
180
90
0 deg
-90
-180

Matlab can also be used to draw Bode plots:
Matlab (with the sketched Bode Plot superimposed on the actual plot)
3
100 10 ( 1)
( 10)( 1000)
s
TF
s s
× +
=
+ +
w=logspace(-1,5,100); %setup for x-axis
MagH=100000*sqrt(w.^2+1^2)./(sqrt(w.^2+10^2).*sqrt(w.^2+1000^2));
%transfer function
MagHdb=20*log10(MagH); %transfer functionconverted to dB
PhaseHRad=atan(w/1)-atan(w/10)-atan(w/1000); %phase done in radians
PhaseHDeg=PhaseHRad*180/pi; %phase done in degrees
subplot(2,1,1)
semilogx(w,MagHdb,':b',x,y,'-b') %semilog plot
xlabel('frequency [rad/s]'),ylabel('20 log10(|TF|) [db]'),grid %xaxis label
subplot(2,1,2)
semilogx(w,PhaseHDeg,':b',xAng,yAngDeg,'-b')
xlabel('frequency [rad/s]'),ylabel('Phase Angle [deg]'),grid

10
-2
10
0
10
2
10
4
10
6
-20
-10
0
10
20
30
40
frequency [rad/s]
20
log10(|TF|)
[db]
10
-2
10
0
10
2
10
4
10
6
-100
-50
0
50
100
frequency [rad/s]
Phase
Angle
[deg]
Notice that the actual plot does not follow the sketched plot exactly. There is error between our sketched
method and the actual Bode plot. How much error is expected?
Let’s look at an example of a zero, )
10
1
(
s
TF +
= . Note, ? critical = 10 rad/s

The largest error that occurs on the Magnitude plot is right at the critical frequency. It is on the order of 3
db.
10
-1
10
0
10
1
10
2
10
3
0
10
20
30
40
50
frequency [rad/s]
20
log10(|TF|)
[db]
10
-1
10
0
10
1
10
2
10
3
0
20
40
60
80
100
frequency [rad/s]
Phase
Angle
[deg]
The largest error that is shown on the Phase plot occurs at 0.1? critical and 10? critical (one decade above and
below the critical frequency). Error at these points is about 6 degrees.
It’s understood that sketching the Bode diagrams will contain some error but this is generally considered
acceptable practice.

To quickly sketch the graphs:
1. Determine the starting value: |H(0)|
2. Determine all critical frequencies (break frequencies). Start from the lowest value and draw the graphs
as follows:
Magnitude Phase (create slope 1
decade below to 1 decade
above ωcritical)
Pole is negative -20dB/dec -45o
Pole is positive -20dB/dec +45o
Zero is negative +20dB/dec +45o
Zero is positive +20dB/dec -45o
Add each value to the previous value.
Examples:
1. H(s) = |H(0)| = |0+1/0+10 = 1/10 |= |0.1| => -20dB
Critical frequencies: zero@ -1 and pole @ -10
Magnitude Plot Phase Plot
The dotted line is a more accurate representation.
2. H(s) = |H(0)| = |10*(-1)/(-3)(-10) |= |-1/3| = 1/3 => -10dB
Note that the angle of (-1/3 real value) is 180o
critical frequencies: zero @ 1, pole@3 and 10

bodeplot.pdf

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bodeplot.pdf