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I have solved a very long problem where I had to find \$|H(j\omega)|\$ for a given circuit. In the end

$$ |H(j\omega)|= \frac{90 }{\sqrt {(9-\omega^2)^2+ (10\omega^2)}} $$

In my book it says I also have to find the angular cutoff frequency \$\omega_c\$.

How do I find this?

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  • \$\begingroup\$ You have shown |H| as a magnitude. If you calculated it as a true transfer function somebody might be able to help you. Without knowing where the j's lie in the denominator this cannot be solved. \$\endgroup\$ – Andy aka Sep 10 '15 at 22:09
  • \$\begingroup\$ Magnitude is sufficient. \$\endgroup\$ – Houston Fortney Sep 11 '15 at 0:52
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The cut-off frequency is where the gain is 3dB below the passband gain, \$=K\$, say. Assuming your equation relates to voltage gain, -3dB equates to \$\dfrac{1}{\sqrt2}K\$

The equation represents a 2nd order low-pass filter and the passband gain is found when \$\omega=0\$. This gives a passband gain of \$K=10\$, hence the cut-off frequency can be determined by setting \$|H(j\omega)|=\dfrac{10}{\sqrt2}\$

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  • \$\begingroup\$ OK, got you now \$\endgroup\$ – Eugene Sh. Sep 10 '15 at 23:14

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