Bode Plot and Corner Frequency

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I am trying to figure out how to get the corner frequency to show up on my bode plot or in the command window in matlab but everything i have tried doesn't work or gives me inf. is there any easy way to get the corner frequency (breakpoint) to show on the figure or the command window? here is what i have so far for my code:
w = logspace(-2,4,500);
num1 = [1 1];
den1 = [1];
num2 = [1 2 1];
den2 = [1];
sys1 = tf(num1,den1)
sys1 = s + 1 Continuous-time transfer function.
figure(1)
bode(sys1,w)
grid on
hold on
sys2 = tf(num2, den2)
sys2 = s^2 + 2 s + 1 Continuous-time transfer function.
figure(1)
bode(sys2, w, 'r')
grid on
  3 个评论
Star Strider
Star Strider 2022-11-17
@Adam Danz — Thank you for posting that reference!
Ethan Johnston
Ethan Johnston 2022-11-17
My question is how to add markers to the plot of those values. Sorry, I should have been more specific. I don't have any work that I calculated yet for it.

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采纳的回答

Star Strider
Star Strider 2022-11-17
The ‘sys1’ and ‘sys2’ transfer functions are actually improper, wince with a denominator of 1, they are infinite differentiators.
As such, they have no corner frequencies.
To get the frequencies at any points in the frequency response that you want, use freqresp and interp1
w = logspace(-2,4,500);
num1 = [1 1];
den1 = [1];
num2 = [1 2 1];
den2 = [1];
sys1 = tf(num1,den1)
sys1 = s + 1 Continuous-time transfer function.
figure(1)
bode(sys1,w)
grid on
H1 = freqresp(sys1,w);
H1 = squeeze(H1);
sys1_dB6_radfreq = interp1(mag2db(abs(H1)),w, 6) % +6dB Radian Frequency
sys1_dB6_radfreq = 1.7266
hold on
sys2 = tf(num2, den2)
sys2 = s^2 + 2 s + 1 Continuous-time transfer function.
figure(1)
bode(sys2, w, 'r')
grid on
[H2,wout2] = freqresp(sys2,w);
H2 = squeeze(H2);
sys2_dB6_radfreq = interp1(mag2db(abs(H2)),w, 6) % +6dB Radian Frequency
sys2_dB6_radfreq = 0.9976
It is also possible to calculate where the phase reaches a certain value, however it may be necessary to restrict ‘w’ to a monotonically-increasing or -decreasing part of the spectrum in order to avoid problems with non-unique values.
.
  2 个评论
Ethan Johnston
Ethan Johnston 2022-11-17
Thank you! I really appreicate it! I'm still learning this topic and it has been hard to grasp but this helps out so much.

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