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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

where we assume the γ k and δ k 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

[link] 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

where

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

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

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 [link] . Lets consider first a single positive term having the form

M γ 20 log 10 1 - j Ω β

It is clear that when Ω β then M 0 . On the other hand if Ω β , M γ 20 log 10 Ω β = γ 20 log 10 Ω - γ 20 log 10 β . If we plot this as a function of log 10 Ω , this represents a straight line having a slope of γ 20 that crosses the log 10 Ω -axis at log 10 Ω = log 10 β . These approximations are illustrated in [link] .

Straight-line approximations to γ 20 log 10 1 - j Ω β .

Instead of plotting M as a function of log 10 Ω , it is more common to plot it as a function of Ω with the frequency axis on a logarithmic scale. In this case the non-zero slope is 20 γ decibels per decade, where one decade represents an increase in Ω by a factor of 10. The modified graph is shown in [link] .

Straight-line approximations to γ 20 log 10 1 - j Ω β using logarithmic frequency axis.

We also note that when Ω = β , the straight-line approximations are not valid, however the true value is easily found to be γ 20 log 10 1 - j = γ 20 log 10 2 3 γ dB. Negative terms in [link] are approximated in a similar manner, but the non-zero slope is now - 20 γ dB/decade. The resulting approximation is shown in [link] .

Straight-line approximations to - δ 20 log 10 1 - j Ω α .

[link] did not take into account cases where the poles or zeros occur at s = 0 . For example, if H ( s ) = s γ , then M = 20 log 10 H ( j Ω ) = γ 20 log 10 Ω , which is a line having a slope of 20 γ dB/decade passing through the Ω -axis at Ω = 1 (see [link] ). When there is a single or repeated pole at the origin, the graph appears just as in [link] but with a negative slope.

Straight-line approximations to γ 20 log 10 Ω .

Next we'll look at approximations to the phase response. Here we'll begin with H ( j Ω ) as shown in [link] . 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

where we have used the fact Z 1 Z 2 = Z 1 + Z 2 and Z γ = γ Z . Lets now consider how we approximate each of the terms in [link] . Consider the single term

1 - j Ω β = - arctan Ω β

The graph of this function is shown in [link] .

Graph of - arctan Ω β , shown on a linear frequency scale.

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 Ω to positive values and plot - arctan Ω β on a logarithmic frequency scale, we get the graph shown in [link] .

Graph of - arctan Ω β , shown on a logarithmic frequency scale, for Ω > 0 .

This graph can be approximated with straight lines as shown in [link] . The approximation is a straight line having a slope of - π / 4 rad/decade passing through a phase of - π / 4 at Ω = β . The line then levels off at Ω = 0 . 1 β and Ω = 10 β . The resulting approximation is shown in [link] .

Approximation (in red) to graph of - arctan Ω β on a logarithmic frequency scale.

Constant terms in [link] have a phase of 0 or ± π , depending on their sign, while poles or zeros of H ( s ) at the origin produce phase terms of γ π / 2 for zeros or order γ or - δ π / 2 for poles of order δ . 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 )

We begin by finding the corresponding frequency response by setting s = j Ω :

H ( j Ω ) = 1 1 + j Ω 10 1 + j Ω 10 3

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

The resulting straight-line approximations to the two terms in [link] along with their sum are shown in [link] .

Bode magnitude response plot derivation for Example 1.

The phase is given by

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

The straight-line approximations to the two phase terms, along with their sum are shown in [link] .

Bode phase response plot derivation for Example 1.

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

Substituting s = j Ω in [link] gives

H ( j Ω ) = - 10 - 2 Ω 2 1 + j Ω 10 2

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

The graphs of the straight-line approximations for the three terms in the right-hand side of [link] along with their sum are shown in [link] .

Bode magnitude response plot derivation for Example 3.2.

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

The resulting Bode phase response plot is found in [link] .

Bode phase response plot derivation for Example 3.2.

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Source:  OpenStax, Signals, systems, and society. OpenStax CNX. Oct 07, 2012 Download for free at http://cnx.org/content/col10965/1.15
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