4.7 Bode 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

where we assume the ${\gamma }_k$ and ${\delta }_k$ are integers. The corresponding frequency response is therefore

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

It is clear that when $\Omega \ll \beta$ then $M\approx 0$ . On the other hand if $\Omega \gg \beta$ , $M\approx \gamma 20{log}_{10}\left|\frac{\Omega }{\beta }\right|=\gamma 20{log}_{10}\left|\Omega \right|-\gamma 20{log}_{10}\left|\beta \right|$ . If we plot this as a function of ${log}_{10}\left|\Omega \right|$ , this represents a straight line having a slope of $\gamma 20$ that crosses the ${log}_{10}\left|\Omega \right|$ -axis at ${log}_{10}\left|\Omega \right|={log}_{10}\left|\beta \right|$ . These approximations are illustrated in [link] .

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

We also note that when $\Omega =\beta$ , the straight-line approximations are not valid, however the true value is easily found to be $\gamma 20{log}_{10}\left|1,-,j\right|=\gamma 20{log}_{10}\sqrt{2}\approx 3\gamma$ dB. Negative terms in [link] are approximated in a similar manner, but the non-zero slope is now $-20\gamma$ dB/decade. The resulting approximation is shown in [link] .

[link] did not take into account cases where the poles or zeros occur at $s=0$ . For example, if $H\left(s\right)=s^{\gamma }$ , then $M=20{log}_{10}\left|H,\left(,j,\Omega ,\right)\right|=\gamma 20{log}_{10}\left|\Omega \right|$ , which is a line having a slope of $20\gamma$ dB/decade passing through the $\Omega$ -axis at $\Omega =1$ (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.

Next we'll look at approximations to the phase response. Here we'll begin with $H\left(j\Omega \right)$ as shown in [link] . Taking the phase of both sides gives

where we have used the fact $\angle \left\{Z_1,Z_2\right\}=\angle Z_1+\angle Z_2$ and $\angle \left\{Z^{\gamma }\right\}=\gamma \angle Z$ . Lets now consider how we approximate each of the terms in [link] . Consider the single term

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

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 $\Omega$ to positive values and plot $-arctan\left(\frac{\Omega }{\beta }\right)$ on a logarithmic frequency scale, we get the graph shown in [link] .

This graph can be approximated with straight lines as shown in [link] . The approximation is a straight line having a slope of $-\pi /4$ rad/decade passing through a phase of $-\pi /4$ at $\Omega =\beta$ . The line then levels off at $\Omega =0.1\beta$ and $\Omega =10\beta$ . The resulting approximation is shown in [link] .

Constant terms in [link] have a phase of 0 or $\pm \pi$ , depending on their sign, while poles or zeros of $H\left(s\right)$ at the origin produce phase terms of $\gamma \pi /2$ for zeros or order $\gamma$ or $-\delta \pi /2$ for poles of order $\delta$ . 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 $s=j\Omega$ :

Lets find the magnitude response first:

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

The phase is given by

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

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 $s=j\Omega$ in [link] 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 [link] along with their sum are shown in [link] .

The phase of the frequency response is found to be

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