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Inside Collection (Textbook):

Textbook by: Carlos E. Davila. E-mail the author

# Bode Plots

Module by: Carlos E. Davila. E-mail the author

Summary: Describes how to manually sketch Bode amplitude and phase plots.

## Bode Plots

Plotting the magnitude and phase of the frequency response is most easily accomplished with a computer, provided you have the right software (for example, the Matlab function “freqs”). However if there is no computer handy, a classical method for quickly sketching the magnitude or phase response of a filter is using a Bode plot. Consider a general system function given by

H ( s ) = K 1 k = 1 q s - β k γ k k = 1 p s - α k δ k H ( s ) = K 1 k = 1 q s - β k γ k k = 1 p s - α k δ k
(1)

where we assume the γkγk and δkδk are integers. The corresponding frequency response is therefore

H ( j Ω ) = K 1 k = 1 q j Ω - β k γ k k = 1 p j Ω - α k δ k H ( j Ω ) = K 1 k = 1 q j Ω - β k γ k k = 1 p j Ω - α k δ k
(2)

Equation 2 can be written as

H ( j Ω ) = K k = 1 q 1 - j Ω β k γ k k = 1 p 1 - j Ω α k δ k H ( j Ω ) = K k = 1 q 1 - j Ω β k γ k k = 1 p 1 - j Ω α k δ k
(3)

where

K = K 1 k = 1 q β k γ k ( - 1 ) γ k k = 1 p α k δ k ( - 1 ) δ k K = K 1 k = 1 q β k γ k ( - 1 ) γ k k = 1 p α k δ k ( - 1 ) δ k
(4)

Since most frequency response plots are expressed in units of decibels, we have

20 log 10 H ( j Ω ) = 20 log 10 K k = 1 q 1 - j Ω β k γ k k = 1 p 1 - j Ω α k δ k = 20 log 10 K + k = 1 q γ k 20 log 10 1 - j Ω β k - k = 1 p δ k 20 log 10 1 - j Ω α k 20 log 10 H ( j Ω ) = 20 log 10 K k = 1 q 1 - j Ω β k γ k k = 1 p 1 - j Ω α k δ k = 20 log 10 K + k = 1 q γ k 20 log 10 1 - j Ω β k - k = 1 p δ k 20 log 10 1 - j Ω α k
(5)

where we have used basic properties of logarithms. The individual terms in the sums can be plotted, to within a reasonable approximation, with relative ease. So the Bode magnitude plot involves summing the graphs of each individual term in Equation 5. Lets consider first a single positive term having the form

M γ 20 log 10 1 - j Ω β M γ 20 log 10 1 - j Ω β
(6)

It is clear that when ΩβΩβ then M0M0. On the other hand if ΩβΩβ, Mγ20log10Ωβ=γ20log10Ω-γ20log10βMγ20log10Ωβ=γ20log10Ω-γ20log10β. If we plot this as a function of log10Ωlog10Ω, this represents a straight line having a slope of γ20γ20 that crosses the log10Ωlog10Ω-axis at log10Ω=log10βlog10Ω=log10β. These approximations are illustrated in Figure 1.

Instead of plotting MM as a function of log10Ωlog10Ω, it is more common to plot it as a function of ΩΩ with the frequency axis on a logarithmic scale. In this case the non-zero slope is 20γ20γ decibels per decade, where one decade represents an increase in ΩΩ by a factor of 10. The modified graph is shown in Figure 2.

We also note that when Ω=βΩ=β, the straight-line approximations are not valid, however the true value is easily found to be γ20log101-j=γ20log1023γγ20log101-j=γ20log1023γ dB. Negative terms in Equation 5 are approximated in a similar manner, but the non-zero slope is now -20γ-20γ dB/decade. The resulting approximation is shown in Figure 3.

Equation 5 did not take into account cases where the poles or zeros occur at s=0s=0. For example, if H(s)=sγH(s)=sγ, then M=20log10H(jΩ)=γ20log10ΩM=20log10H(jΩ)=γ20log10Ω, which is a line having a slope of 20γ20γ dB/decade passing through the ΩΩ-axis at Ω=1Ω=1 (see Figure 4). When there is a single or repeated pole at the origin, the graph appears just as in Figure 4 but with a negative slope.

Next we'll look at approximations to the phase response. Here we'll begin with H(jΩ)H(jΩ) as shown in Equation 3. Taking the phase of both sides gives

H ( j Ω ) = K k = 1 q 1 - j Ω β k γ k k = 1 p 1 - j Ω α k δ k = K + k = 1 q γ k 1 - j Ω β k - k = 1 p δ k 1 - j Ω α k H ( j Ω ) = K k = 1 q 1 - j Ω β k γ k k = 1 p 1 - j Ω α k δ k = K + k = 1 q γ k 1 - j Ω β k - k = 1 p δ k 1 - j Ω α k
(7)

where we have used the fact Z1Z2=Z1+Z2Z1Z2=Z1+Z2 and Zγ=γZZγ=γZ. Lets now consider how we approximate each of the terms in Equation 7. Consider the single term

1 - j Ω β = - arctan Ω β 1 - j Ω β = - arctan Ω β
(8)

The graph of this function is shown in Figure 5.

Since the magnitude response plots are on a logarithmic frequency axis, it would be desirable to do the same for the phase response plots. If we restrict ΩΩ to positive values and plot -arctanΩβ-arctanΩβ on a logarithmic frequency scale, we get the graph shown in Figure 6.

This graph can be approximated with straight lines as shown in Figure 7. The approximation is a straight line having a slope of -π/4-π/4 rad/decade passing through a phase of -π/4-π/4 at Ω=βΩ=β. The line then levels off at Ω=0.1βΩ=0.1β and Ω=10βΩ=10β. The resulting approximation is shown in Figure 7.

Constant terms in Equation 7 have a phase of 0 or ±π±π, depending on their sign, while poles or zeros of H(s)H(s) at the origin produce phase terms of γπ/2γπ/2 for zeros or order γγ or -δπ/2-δπ/2 for poles of order δδ. Next we'll illustrate these techniques with a few examples.

Example 3.1 Sketch the Bode magnitude and phase response plot for the following filter:

H ( s ) = 10 4 ( s + 10 ) ( s + 10 3 ) H ( s ) = 10 4 ( s + 10 ) ( s + 10 3 )
(9)

We begin by finding the corresponding frequency response by setting s=jΩs=jΩ:

H ( j Ω ) = 1 1 + j Ω 10 1 + j Ω 10 3 H ( j Ω ) = 1 1 + j Ω 10 1 + j Ω 10 3
(10)

Lets find the magnitude response first:

20 log 10 H ( j Ω ) = - 20 log 10 1 + j Ω 10 - 20 log 10 1 + j Ω 10 3 20 log 10 H ( j Ω ) = - 20 log 10 1 + j Ω 10 - 20 log 10 1 + j Ω 10 3
(11)

The resulting straight-line approximations to the two terms in Equation 11 along with their sum are shown in Figure 8.

The phase is given by

H ( j Ω ) = - 1 + j Ω 10 - 1 + j Ω 10 3 H ( j Ω ) = - 1 + j Ω 10 - 1 + j Ω 10 3
(12)

The straight-line approximations to the two phase terms, along with their sum are shown in Figure 9.

Next we'll look at an example of a second-order highpass filter.

Example 3.2 Find the Bode magnitude and phase response of the following filter

H ( s ) = s 2 s + 10 2 H ( s ) = s 2 s + 10 2
(13)

Substituting s=jΩs=jΩ in Equation 13 gives

H ( j Ω ) = - 10 - 2 Ω 2 1 + j Ω 10 2 H ( j Ω ) = - 10 - 2 Ω 2 1 + j Ω 10 2
(14)

The magnitude response, expressed in decibels becomes

20 log 10 H ( j Ω ) = 20 log 10 10 - 2 + 40 log 10 Ω - 40 log 10 1 + j Ω 10 20 log 10 H ( j Ω ) = 20 log 10 10 - 2 + 40 log 10 Ω - 40 log 10 1 + j Ω 10
(15)

The graphs of the straight-line approximations for the three terms in the right-hand side of Equation 15 along with their sum are shown in Figure 10.

The phase of the frequency response is found to be

H ( j Ω ) = - 10 - 2 Ω 2 - 1 + j Ω 10 2 = π - 2 1 + j Ω 10 H ( j Ω ) = - 10 - 2 Ω 2 - 1 + j Ω 10 2 = π - 2 1 + j Ω 10
(16)

The resulting Bode phase response plot is found in Figure 11.

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