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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?
- Determine the Transfer Function of the system:
1
1 ss p
K s z H s
- Rewrite it by factoring both the numerator and denominator into the standard form
1 1
(^11)
p sp s
z Kz s H s
where the z s are called zeros and the p s are called poles.
- Replace s with j?. Then find the Magnitude of the Transfer Function.
(^11)
(^11)
p jwp jw
z Kz jw H jw
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) ) =
20 log (^11)
(^11) 10 p jwp jw
z Kz jw =
20 log 20 log 20 log ( 1 ) 20 log 20 log 20 log ( 1 ) (^10 ) K + z + jwz + − p − jw − jwz +
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
Poles and Zeros at the origin | j? |
Poles and Zeros not at the origin 1
p
j ω
z
j ω
- 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 = | j ω |
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 =
jω
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 frequenc y, 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
20 log 10 ( K )
0.1^?
(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
Example 2 Solution: Your turn. Find the Bode log magnitude plot for the transfer function,
4 2
x s TF s s
Simplify transfer function form: 4 4
x s x s s TF s s s^ s^ s^ s
Recognize: K = 20 ‡ 20 log10(20) = 26.
1 zero at the origin
2 poles: at p 1 = 5 and p 2 =
Technique to get started:
- Draw the line of each individual term on the graph
- Follow the combined pole-zero at the origin line back to the left side of the graph.
- 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.
- 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
80 db
-80 db
40 db
? (log scale)
Example 3: One more time. This one is harder. Find the Bode log magnitude plo t for the transfer function, 200( 20) (2 1)( 40)
s TF s s s
Simplify transfer function form:
0 db
8 0 db
40 db
20 log 10 (TF)
? (log scale)
0 db
-40 db
80 db
-80 db
40 db
? (log scale)
Technique to get started:
- Draw the line of each individual term on the graph
- Follow the combined pole-zero at the origin line back to the left side of the graph.
- 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.
- Apply the effect of the poles/zeros not at the origin. working from left (low values) to right (higher values) of the poles/zeros. 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,
(^11)
(^11)
p jwp jw
z Kz jw H jw
the cumulative phase angle associated with this function are given by
(^11)
(^11)
∠ ∠ ∠ +
p jw p jw
z K z jw H jw
Then the cumulative phase angle as a function of the input frequency may be written as
∠ ( )=∠^ + +( + 1 )−( )− −( + 1 )
(^1 11) p 1 jw p jw z H jw K z jw
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
z
j ω
- , have no phase shift for frequencies much lower than z i, have a +
45 deg shift at z 1 , and have a +90 deg shift for frequenc ies much higher than z 1.
∠ H
To draw the lines for this type of term, the transition from 0o^ to +90o^ is drawn over 2 decades, starting at 0.1 z 1 and ending at 10 z 1.
Effect of Poles not at the origin on Phase Angle:
Poles not at the origin, like
1
p
jω
, have no phase shift for frequencies much lower than p i, have a -
45 deg shift at p 1 , and have a -90 deg shift for frequenc ies much higher than p 1.
∠ TF
To draw the lines for this type of term, the transition from 0o^ to -90o^ is drawn over 2 decades, starting at 0.1 p 1 and ending at 10 p 1.
When drawing the phase angle shift for not-at-the-origin zeros and poles , first locate the critical frequency 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:
0.1z 1 1 z 1 10 z 1 100 z 1
+90 deg +45 deg
0.1 p 1 1 p 1 10p 1 100 p 1 -90 deg
-45 deg
Example 2: Repeat for the transfer function,
20log| TF |
4 2
x s s TF s s s^ s
0 db
-40 db
80 db
-80 db
40 db
? (log scale)
0 o
-90o
180 o
-180o
90 o
? (log scale)
20 log 10 ( MF )
Phase Angle
Example 2 Solution: Repeat for the transfer function,
20log| TF |
4 2
x s s TF s s s^ s
0 db
80 db
4 0 db
? (log scale)
0 o
-90o
180 o
-180o
90 o
? (log scale)
20 log 10 ( 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
10 -^1
8 0 db
4 0 db
? (log scale)
0 o
-90o
10 -^1
180 o
-180o
90 o
? (log scale)
20 log 10 ( MF )
Phase Angle
Example 4: Sketch the Bode plot (Magnitude and Phase Angle) for 100 10 (^3 1) ( 10)( 1000)
s TF s s
× +
20log 10 | TF |
Angle of TF
Matlab can also be used to draw Bode plots:
Matlab (with the sketched Bode Plot superimposed on the actual plot)
100 10 (^3 1) ( 10)( 1000)
s TF s s
× +
w=logspace(-1,5,100); %setup for x-axis MagH=100000sqrt(w.^2+1^2)./(sqrt(w.^2+10^2).sqrt(w.^2+1000^2)); %transfer function MagHdb=20log10(MagH); %transfer function converted to dB PhaseHRad=atan(w/1)-atan(w/10)-atan(w/1000); %phase done in radians PhaseHDeg=PhaseHRad180/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 -2^100 102 104
0
10
20
30
40
frequency [rad/s]
20 log10(|TF|) [db]
10 -2^100 102 104
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
s TF = +. Note,? (^) critical = 10 rad/s
To quickly sketch the graphs:
- Determine the starting value: |H(0)|
- 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:
- H(s) = |H(0)| = |0+1/0+10 = 1/10 |= |0.1| => -20dB Critical frequencies: zero@ -1 and pole @ -
Magnitude Plot Phase Plot
The dotted line is a more accurate representation.
- 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
