Plots of the frequency response magnitude and phase are most easily generated using a computer.
However, using the techniques of Bode, we can generate a quick sketch of the frequency response that can be very valuable in understanding the behavior of a system. The first step is to factor the transfer function so that it is composed of the following 7 types of terms:
1. k (constant)
When factored in this way, the transfer function is said to be in Bode form. We will describe how to plot the magnitude and phase for each type of term and then demonstrate how to generate composite plots for general transfer functions.
1. k
• magnitude plot is a horizontal line having magnitude k for allω
• phase plot isφ = 0 for all ω
The magnitude and phase plots for this case are shown in Fig. 7.1.
2. ( jω)n
• magnitude plot is a straight line with slope n passing through 1 atω = 1 rad/sec
• phase plot isφ = n ×90◦for allω
The magnitude and phase plots for this case are shown in Fig. 7.2.
100 7.4 Bode Plotting Techniques
3. ( jω)−n
• magnitude plot is a straight line with slope −n passing through 1 at ω = 1 rad/sec
• phase plot isφ = −n×90◦for allω
The magnitude and phase plots for this case are shown in Fig. 7.3.
4. jωτ + 1
These two expressions give asymptotes for small and large frequencies. Moreover, whenω = 1/τ, mag =√
2 = 1.414. This is the break point, or transition point, between the two asymptotes.
The magnitude and phase plots for this case are shown in Fig. 7.4.
5. 1
These two expressions give asymptotes for small and large frequencies. Whenω = 1/τ, mag = 1/√
2 = 0.707. This is the break point, or transition point, between the two asymptotes.
Frequency Response 101
The magnitude and phase plots for this case are shown in Fig. 7.5.
6. � jω
Again, these expressions give asymptotes for small and high frequencies. The break point between them occurs atω = ωn, at which
Note that the size of the magnitude ratio atω = ωn depends entirely on the damping ratioζ.
The transition between 0◦ and 180◦ depends onζ. Smaller ζ will have a more rapid transition, while largerζ will cause a slower transition. However, at ω = ωn, the phase angle will always be 90◦, halfway between the angles at low and high frequencies.
102 7.4 Bode Plotting Techniques
The magnitude and phase plots for this case are shown in Fig. 7.6.
7. 1
Again, these expressions give asymptotes for small and high frequencies. The break point between them occurs atω = ωn, at which
Again, the magnitude ratio atω = ωndepends entirely on the damping ratioζ.
• phase plot:
As before, the transition between 0◦ and −180◦ depends on ζ. Smaller ζ will have a more rapid transition, while larger ζ will cause a slower transition. However, at ω = ωn, the phase angle will always be −90◦, halfway between the angles at low and high frequencies.
The magnitude and phase plots for this case are shown in Fig. 7.7. In addition, Fig. 7.8 shows how the magnitude and phase plots change for varying values ofζ.
Frequency Response 103
10!2 10!1 100 101 102 103
10!2 10!1 100 101 102 103
Magnitude
Frequency (rad/s)
10!2 10!1 100 101 102 103
!60
!40
!20 0 20 40 60
Phase (deg)
Frequency (rad/s) k
Figure 7.1: Magnitude and phase plots for a constant k
104 7.4 Bode Plotting Techniques
10!2 10!1 100 101 102 103
10!2 10!1 100 101 102 103
Magnitude
Frequency (rad/s)
10!2 10!1 100 101 102 103
0 20 40 60 80 100 120 140 160 180
Phase (deg)
Frequency (rad/s)
Phase = 90° over all frequencies 1
1 +1 Slope
at ! = 1 rad/s, passes through 1
Figure 7.2: Magnitude and phase plots for ( jω)1
Frequency Response 105
10!2 10!1 100 101 102 103
10!3 10!2 10!1 100 101 102
Magnitude
Frequency (rad/s)
10!2 10!1 100 101 102 103
!180
!160
!140
!120
!100
!80
!60
!40
!20 0
Phase (deg)
Frequency (rad/s)
at ! = 1 rad/s, passes through 1
-1 1 -1 Slope
Phase = -90° over all frequencies
Figure 7.3: Magnitude and phase plots for ( jω)−1
106 7.4 Bode Plotting Techniques
10!2 10!1 100 101 102 103
10!2 10!1 100 101 102 103
Magnitude
Frequency (!")
10!2 10!1 100 101 102 103
0 10 20 30 40 50 60 70 80 90
Phase (deg)
Frequency (!") break frequency = 1/"
1.4
1 1 +1 Slope
.1/" .2/" 1/" 5/" 10/"
6°
11°
84°
79°
+45° at break frequency
Figure 7.4: Magnitude and phase plots for ( jωτ + 1)
Frequency Response 107 break frequency = 1/"
.7
-45° at break frequency
Figure 7.5: Magnitude and phase plots for 1/( jωτ + 1)
108 7.4 Bode Plotting Techniques
10!2 10!1 100 101 102 103
10!2 10!1 100 101 102 103 104 105
Magnitude
Frequency (!/!n)
10!2 10!1 100 101 102 103
0 20 40 60 80 100 120 140 160 180
Phase (deg)
Frequency (!/!
n)
1 2 +2 Slope
break frequency = !n
height of peak varies with "
2"
+90° at break frequency
Slope of phase plot varies with "
Figure 7.6: Magnitude and phase plots for ( jω/ωn)2+ 2ζ( jω/ωn) + 1;ζ = 0.1 for this case.
Frequency Response 109
10!2 10!1 100 101 102 103
10!4 10!3 10!2 10!1 100 101 102
edutingaM
Frequency (!/!n)
10!2 10!1 100 101 102 103
!180
!160
!140
!120
!100
!80
!60
!40
!20 0
)ged( esahP
Frequency (!/!n) break frequency = !n
height of peak varies with "
2"
-2 1
-2 Slope
-90° at break frequency
Slope of phase plot varies with "
1
Figure 7.7: Magnitude and phase plots for 1/[( jω/ωn)2+ 2ζ( jω/ωn) + 1];ζ = 0.1 for this case.
110 7.4 Bode Plotting Techniques
Figure 7.8: Illustration of variation of magnitude and phase withζ for a second-order system
Frequency Response 111 Given any transfer function G(s), we can always factor it into a form composed only of these seven terms. In other words, any transfer function can be factored to put it in Bode form. For example,
which is of the form
G( jω) = K ( jω)( jωτ1+ 1)
Sketching the Magnitude and Phase Plots
Once the transfer function has been factored, the magnitude and phase plots for each term can be plotted. The magnitude and phase plots for the whole transfer function are then found by simply adding the plots from each term. We can see this by considering an example transfer function,
G(s) = N1(s)N2(s)
Similarly, the phase of the total system comes from adding the phases of each individual compo-nent. Therefore, to generate the composite sketch, we:
1. Put the transfer function in Bode form, identifying terms according to the 7 types presented above.
2. Plot the magnitude for each individual term on a log-log scale.
3. Add the individual magnitude plots to generate the composite magnitude curve. Add the distances on the plot, not the values. Work from low to high frequencies.
112 7.4 Bode Plotting Techniques 4. Plot the phase associated with ( jω)n or ( jω)−n terms. This will be the low-frequency
asymptote for the phase plot.
5. Sketch the approximate phase curve by stepping ±90◦or ±180◦at the break points. Again, work from low to high frequencies.
6. Draw transition asymptotes.
7. Sketch the composite phase curve.
Example 1 On the log scale, each of these terms adds. Also,
∠G(s) = ∠k +∠(τ1s + 1) +∠�1
Blank magnitude and phase plots are found on the next pages for you to practice sketching; the correct answers are also shown.
Frequency Response 113
10!2 10!1 100 101 102 103
10!2 10!1 100 101 102 103
Magnitude
Frequency (rad/s)
G(s) = 2000(s + 0.5) s(s + 10)(s + 50)
114 7.4 Bode Plotting Techniques
10!2 10!1 100 101 102 103
10!2 10!1 100 101 102 103
Magnitude
Frequency (rad/s)
G(s) = 2000(s + 0.5) s(s + 10)(s + 50)
Frequency Response 115
10!2 10!1 100 101 102 103
!180
!160
!140
!120
!100
!80
!60
!40
!20 0
Phase (deg)
Frequency (rad/s) G(s) = 2000(s + 0.5)
s(s + 10)(s + 50)
116 7.4 Bode Plotting Techniques
10!2 10!1 100 101 102 103
!180
!160
!140
!120
!100
!80
!60
!40
!20 0
Phase (deg)
Frequency (rad/s) G(s) = 2000(s + 0.5)
s(s + 10)(s + 50)
Frequency Response 117
Example2
G(s) = 20,000s + 20,000 s3+ 40s2+ 10,000s We place the transfer function in Bode form:
G( jω) = 20,000( jω + 1) jω[( jω)2+ 40 jω + 10,000]
= 20,000( jω + 1) 10,000 jω�� jω
100
�2
+ 0.4100jω + 1
�
G( jω) = 2( jω)−1 jω + 1
�� jω
100
�2
+ 2(0.2)�jω
100
�+ 1�
This is in Bode form if
k = 2 n = −1 τ = 1rad/s ωn= 100rad/s ζ = 0.2
Blank magnitude and phase plots are located on the following pages for you to sketch these curves.
The correct answers are also given.
118 7.4 Bode Plotting Techniques
10!2 10!1 100 101 102 103
10!2 10!1 100 101 102
Magnitude
Frequency (rad/s) G(s) = 20,000s + 20,000
s3+ 40s2+ 10,000s
Frequency Response 119
10!2 10!1 100 101 102 103
10!2 10!1 100 101 102
Magnitude
Frequency (rad/s) G(s) = 20,000s + 20,000
s3+ 40s2+ 10,000s
120 7.4 Bode Plotting Techniques
10!2 10!1 100 101 102 103
!180
!160
!140
!120
!100
!80
!60
!40
!20 0
Phase (deg)
Frequency (rad/s)
G(s) = 20,000s + 20,000 s3+ 40s2+ 10,000s
Frequency Response 121
10!2 10!1 100 101 102 103
!180
!160
!140
!120
!100
!80
!60
!40
!20 0
Phase (deg)
Frequency (rad/s)
G(s) = 20,000s + 20,000 s3+ 40s2+ 10,000s