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

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

Equation 2 can be written as

where

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

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

It is clear that when Ω≪βΩ≪β then M≈0M≈0. On the other hand if Ω≫βΩ≫β, M≈γ20log10Ωβ=γ20log10Ω-γ20log10βM≈γ20log10Ωβ=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.

Figure 1: Straight-line approximations to γ20log101-jΩβγ20log101-jΩβ.

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.

Figure 2: Straight-line approximations to γ20log101-jΩβγ20log101-jΩβ using logarithmic frequency axis.

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=γ20log102≈3γγ20log101-j=γ20log102≈3γ 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.

Figure 3: Straight-line approximations to -δ20log101-jΩα-δ20log101-jΩα.

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.

Figure 4: Straight-line approximations to γ20log10Ωγ20log10Ω.

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

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

The graph of this function is shown in Figure 5.

Figure 5: Graph of -arctanΩβ-arctanΩβ, 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 ΩΩ to positive values and plot -arctanΩβ-arctanΩβ on a logarithmic frequency scale, we get the graph shown in Figure 6.

Figure 6: Graph of -arctanΩβ-arctanΩβ, shown on a logarithmic frequency scale, for Ω>0Ω>0.

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.

Figure 7: Approximation (in red) to graph of -arctanΩβ-arctanΩβ on a logarithmic frequency scale.

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:

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

Lets find the magnitude response first:

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

Figure 8: Bode magnitude response plot derivation for Example 1.

The phase is given by

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

Figure 9: 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

Substituting s=jΩs=jΩ in Equation 13 gives

The magnitude response, expressed in decibels becomes

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.

Figure 10: Bode magnitude response plot derivation for Example 3.2.

The phase of the frequency response is found to be

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

Figure 11: Bode phase response plot derivation for Example 3.2.