Finding the angular cutoff frequency

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?

• 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. – Andy aka Sep 10 '15 at 22:09
• Magnitude is sufficient. – Houston Fortney Sep 11 '15 at 0:52

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 $|H(j\omega)|=\dfrac{10}{\sqrt2}$