Finding frequency corresponding to some gain in magnitude plot of bode

Hi Community members !

Can you please help me in getting a solution to my problem? I have a transfer function e.g. G(s) = 144000/(s*(s+36)*(s+100)), bode plot of which is given. The magnitude plot passes through -3.76 dB gain at some particular value of frequency. I want to access that particular frequency value using MATLAB program.

Thanks !

Answer 1

[Star Strider]

Try this:

s = tf('s');
G = 144000/(s*(s+36)*(s+100));

[mag,phase,wout] = bode(G);
mag = squeeze(mag);
phase = squeeze(phase);

wq = interp1(20*log10(mag), wout, -3.76);               % Find Desired Frequency

figure
semilogx(wout, 20*log10(mag), wq, -3.76, 'r+')
grid
text(wq, -3.76, sprintf('\\leftarrow -3.76 dB at %.2f rad/sec',wq), 'HorizontalAlignment','left', 'VerticalAlignment','middle')

The result is:

wq =
       39.034

so your transfer function will equal -3.76 dB at 39.034 rad/sec.

[Ari]

This works! Could you show us a code that would flip it, so I would be entering 10 rad/s and it would have to give me magnitude?

[Star Strider]

Sure!

To do that, the code changes (ever so slightly) to:

s = tf('s');
G = 144000/(s*(s+36)*(s+100));
[mag,phase,wout] = bode(G);
mag = squeeze(mag);
phase = squeeze(phase);
wq = 10;
magi = interp1(wout, 20*log10(mag),10);               % Find Desired Amplitude
figure
semilogx(wout, 20*log10(mag), wq, 10, 'r+')
grid
text(wq, magi, sprintf('\\leftarrow %.2f dB at %.2f rad/sec',magi,10), 'HorizontalAlignment','left', 'VerticalAlignment','middle')

producing:

.

[Star Strider]

Ari — Thank you for the vote!

[Ari]

Response time almost faster than the speed of light

[Star Strider]

I take that as a compliment!

[Lipan Ehmes]

can the code be modified for a bandstop filter and get both fl and fh from the bode plot? please help... transfer function is

s = tf('s');

Hs = (9.87e4*s^2+9.741e9)/(9.87e4*s^2+8.99e7*s+9.741e9)

[Star Strider]

@Lipan Ehmes — Probably. For a bandstop (or bandpass) filter the interpolation would need to be performed on both sides of the stopband (or passband) to get both values.

[Lipan Ehmes]

Thank you, i tried splitting the two halves and it worked. I was able to obtain the high and low frequencies at -3dB. thanks again

[Emmanuel Santiago Castillo Sánchez]

Gracias por su aporte!!

[parvathy ss]

Thank you so much....

[Star Strider]

A Vote is always appreciated!

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