Plotting max gain and cutoff frequencies on a bode plot in MATLAB

Question

I want to highlight the differences between a simulated filter and real-life filter through a MATLAB plot. I am trying to show the max gain and the two cutoff frequencies for both filters. Ideally, each point on the bode plot would be shown by a circle with a dotted line down to the x-axis. I was able to get the max gain values by using the getPeakGain function and then converting to dB and Hz. I am not sure how to calculate the cutoff frequency. For both I am not sure how to plot them.

My code so far is below.

clear all;
s = tf('s');
G = (1.5791e08*s^2)/((s+1.257e04)^2*(s^2 + 62.83*s + 987))

options = bodeoptions;
options.Title.String = {["Bode Diagram For Designed Components"]}
options.FreqUnits = 'Hz'; % or 'rad/second', 'rpm', etc.
options.Grid = 'on';

figure(1)
subplot(1,2,1)
bode(G,options);
[gpeak,fpeak] = getPeakGain(G);
gpeak_dB = 20*log10(gpeak)
fpeak_hz = fpeak/(2*pi)

clear all;
s = tf('s');
G = (1.5039e08*s^2)/((s+1.257e04)*(s+1.197e04)*(s+41.18)*(s+21.65))

options = bodeoptions;
options.Title.String = {["Bode Diagram For Digikey Components"]}
options.FreqUnits = 'Hz'; % or 'rad/second', 'rpm', etc.
options.Grid = 'on';

figure(1)
subplot(1,2,2)
bode(G,options);
[gpeak,fpeak] = getPeakGain(G);
gpeak_dB = 20*log10(gpeak)
fpeak_hz = fpeak/(2*pi)

score 0 · Answer 1 · answered Feb 23 '23 at 19:25

To calculate the cutoff frequency of a filter, you can use the -3 dB point on the magnitude plot, which is the frequency where the gain is 3 dB below the maximum gain. Typically, the cutoff frequency is defined as the frequency at which the magnitude of the filter's transfer function is -3 dB (or 0.707 times the maximum gain in linear scale). You can find the cutoff frequency using the bode function and then looking for the frequency at which the magnitude of the transfer function is -3 dB.

clear all;

% Define the transfer function for the simulated filter
s = tf('s');
G_sim = (1.5791e08*s^2)/((s+1.257e04)^2*(s^2 + 62.83*s + 987));

% Define the transfer function for the real-life filter
G_real = (1.5039e08*s^2)/((s+1.257e04)*(s+1.197e04)*(s+41.18)*(s+21.65));

% Set up Bode plot options
options = bodeoptions;
options.Title.String = {["Bode Diagram For Designed Components"], ["Bode Diagram For Digikey Components"]};
options.FreqUnits = 'Hz'; % or 'rad/second', 'rpm', etc.
options.Grid = 'on';

% Plot the Bode diagrams for both filters
figure(1)
subplot(1,2,1)
[mag_sim,phase_sim,wout_sim] = bode(G_sim,options);
subplot(1,2,2)
[mag_real,phase_real,wout_real] = bode(G_real,options);

% Calculate and plot the maximum gain and cutoff frequency for the simulated filter
[max_gain_sim, max_gain_sim_idx] = max(mag_sim(:));
f_cutoff_sim = wout_sim(find(mag_sim>=max_gain_sim*0.707,1));
subplot(1,2,1)
hold on
plot(f_cutoff_sim, -3, 'ro')
plot(wout_sim(max_gain_sim_idx), 20*log10(max_gain_sim), 'ro')
hold off

% Calculate and plot the maximum gain and cutoff frequency for the real-life filter
[max_gain_real, max_gain_real_idx] = max(mag_real(:));
f_cutoff_real = wout_real(find(mag_real>=max_gain_real*0.707,1));
subplot(1,2,2)
hold on
plot(f_cutoff_real, -3, 'ro')
plot(wout_real(max_gain_real_idx), 20*log10(max_gain_real), 'ro')
hold off

score 0 · Answer 2 · answered Feb 24 '23 at 20:42

1.- No data, no way way to tell : CAPTURE TF SAMPLES

Mathworks is encouring the usage of handles, but direct modification of options in the input field is a MATLAB feature, so work has been put to keep both ways to do the same without conflict.

For this particular question, the following odd thing happens :

close all;clear all;clc

s = tf('s');
G = (1.5791e08*s^2)/((s+1.257e04)^2*(s^2 + 62.83*s + 987))

op1 = bodeoptions;
op1.Title.String = {["Bode Diagram For Designed Components"]}
op1.FreqUnits = 'Hz'; 
op1.Grid = 'on';
op1.PhaseVisible='off'

figure; bode(G) % same as bode(G,op1)

The previous 2 lines work ok.

However G are needed, so when one attempts to pull such values, like for instance with the following

figure
[Gmag,Gph,wout]=bode(G);

empty graph, not even the axes.

The following doesn't work either

figure
[Gmag,Gph,wout]=bode(G,op1);

again empty.

Don't ask me why but when calling bode twice, like this

figure
bode(G)
[Gmag,Gph,wout]=bode(G,op1);

Now it again works.

The bode plot is obtained AND the transfer function samples are also captured.

Yet bode is still ignoring both the option I added to keep phase off and your option to switch grid on.

Bode has been around since early versions of MATLAB. My guess would be bode bodeplot are mainly used in academia, and not that often comparing to directly working with G therefore no update has been done for a while.

Gmag Gph size check

size(Gmag)
size(Gph)

 =
     1     1    94
 =
     1     1    94

Again, another odd thing : Gmag Gph show up 3D despite both are actually 1D.

And the following doesn't work either.

op1.Grid = 'on';
op1.PhaseVisible='off'

2.- Not enough frequency resolution

Without G values the only way to find the 1dB bandwidth or 3dB, or whatever cut-off amplitude value you choose to decide how much pass band to be considered signal and what to be considered off band is using the 'markers' directly on the graph in the same way we do when working on measurement instruments like oscilloscopes, spectrum analysers and network analysers.

MATLAB allows multiple markersv on same graph curve, calling then DATA TIPS.

Let's say you are looking for the -1dB bandwidth, same as cut-off at -1dB from the pass band peak.

One can visually tell that the sought peak is not far from 0dB.

Look at the reading of 2 adjacent markers right around -1dB

If +17% -19% is ok to you just go ahead with the default input frequencies but the chances are you need better amplitude resolution.

To such purpose one has to use a higher density of frequencies, and the way do that is to define more input frequencies than the default amount, in this case 94

close all;

op1 = bodeoptions;
op1.Title.String = {['Bode Diagram For Designed Components']}
op1.FreqUnits = 'Hz'; 
op1.Grid = 'on';
op1.PhaseVisible='off';

fin=[1:1e2:1e5];  % [Hz]

figure
bode(G,op1)
ax=gca
hold(ax,'on')
[GmagdB,Gph,fout]=bode(G,fin,op1);

GmagdB=squeeze(GmagdB);
Gph=squeeze(Gph);

As asked there were 94 input frequencies, now you have 1e3 frequencies, but squeeze is needed to remove void dimensions because bode, for whatever reason, produces 1x1xN values, not just 1xN.

3.- Expected 1dB bandwidth

manually one can check expected results :

4.- Simplify : work directly on G

and despite having the graphic handle, plot cannot direct the new trace onto the generated graph.

fin=[0:10:1e7];
G2 = (1.5791e08*(2*pi*fin*1j).^2)./((2*pi*fin*1j+1.257e04).^2.*((2*pi*fin*1j).^2 + 62.83*(2*pi*fin*1j) + 987));

G2magdB=10*log10(abs(G2));
n=find(G2magdB>-10);

figure
ax2=gca
plot(ax2,fin,G2magdB)
hold(ax2,'on')
grid on
xlabel('f[Hz]');ylabel('|G| [dB]');
plot(ax2,fin(n),G2magdB(n),'r*')
axis([0 .5e6 -35 1.5])

5.- ** bode ** and ** bodeplot ** are slow

bode and bodeplot are useful functions but both are slow compared to working directly the the available data.

This G is not a really sharp filter.

Also the transfer function looks like a LPF but there's a notch when really close to 0Hz.

Do you really want DC through? Perhaps not the best choice of filter.

Plotting max gain and cutoff frequencies on a bode plot in MATLAB

2 Answers2