# Finding the cut-off frequency

Find the cut-off frequency of this filter:

My attempt:

$$\text{H}\space_{\left(\omega\right)}=\frac{\text{R}_1+j\omega\text{L}}{\text{R}_1+\text{R}_2+j\omega\text{L}}=\frac{\text{R}_1+j\omega\text{L}}{\text{R}_1+\text{R}_2+j\omega\text{L}}\cdot\frac{\text{R}_1+\text{R}_2-j\omega\text{L}}{\text{R}_1+\text{R}_2-j\omega\text{L}}=$$ $$\frac{\text{R}_1^2+\text{R}_1\text{R}_2+\left(\omega\text{L}\right)^2+j\omega\text{L}\text{R}_2}{\left(\text{R}_1+\text{R}_2\right)^2+\left(\omega\text{L}\right)^2}$$

So when you're looking for the cut-off frequency you can say:

$$\Re\left(\text{H}\space_{\left(\omega\right)}\right)=\Im\left(\text{H}\space_{\left(\omega\right)}\right)$$

So we get:

$$\text{R}_1^2+\text{R}_1\text{R}_2+\left(\omega\text{L}\right)^2=\omega\text{L}\text{R}_2\Longleftrightarrow$$ $$\left(\omega\text{L}\right)^2-\omega\text{L}\text{R}_2+\text{R}_1^2+\text{R}_1\text{R}_2=0\Longleftrightarrow$$ $$\omega^2\text{L}^2-\omega\text{L}\text{R}_2+\text{R}_1^2+\text{R}_1\text{R}_2=0\Longleftrightarrow$$ $$\omega=\frac{\text{L}\text{R}_2\pm\sqrt{\left(-\text{L}\text{R}_2\right)^2-4\cdot \text{L}^2\cdot \left(\text{R}_1^2+\text{R}_1\text{R}_2\right)}}{2\cdot \text{L}^2}\Longleftrightarrow$$ $$\omega=\frac{\text{L}\text{R}_2\pm\sqrt{\left(\text{L}\text{R}_2\right)^2-4\cdot \text{L}^2\cdot \left(\text{R}_1^2+\text{R}_1\text{R}_2\right)}}{2\cdot \text{L}^2}$$

Am I doing something wrong?

• Possible duplicate of How do I calculate the cutoff frequency of a filter? Nov 10 '15 at 18:50
• What is your definition of the "cut-off frequency"? This filter has one zero and one pole so there is no frequency above or below which the output trends towards 0. To me it would make more sense to calculate either the zero frequency or the pole frequency, which can be done by considering the numerator and denominator of the transfer function seprately. Nov 10 '15 at 18:57
• Also, in general for this type of filter there's not always any frequency where the real and imaginary parts are equal, so your method of choosing the "cut off frequency" may not work at all. Nov 10 '15 at 19:00
• @Samuel: (et al.) Would you guys stop voting this as duplicate of that "How do I" question without reading it? It's by no means the same thing. This one is definitely more complicated. It is somewhat similar to the other "Cutoff frequency" question, but by no means the same. Also answers failed to calculate the frequency there.
– Fizz
Nov 20 '15 at 11:49
• @Respawned Sure, I stopped after the first vote. Nov 20 '15 at 15:31

The transfer function is (obviously):

$$H(s) = \frac{R_1 + L_1s}{(R_1+R_2)+L_1s}$$

The point of doing that was to plot

Adopting def of cut-off frequency as that of a high-pass filter, the eyeometer (or a numerical solver) will give us the -3dB roll-off around $10^6$ rad/s.

To calculate this symbolically just calculate the square of the gain (since $G(j\omega) = |H(j\omega)|$) as

$$(G(j\omega))^2 = H(j\omega)\overline{H(j\omega)} = \frac{R_1 + L_1j\omega}{(R_1+R_2) + L_1j\omega} \frac{R_1 - L_1j\omega}{(R_1+R_2) - L_1j\omega} = \\ = \frac{R_1^2 + (L\omega)^2}{(R_1+R_2)^2 + (L\omega)^2}$$

(Don't try to do the above in Wolfram Alpha as unlike Mathematica you can't tell it to assume those parameters are real, so it will give you some unsimplified complex number result.)

Now solve $(G(j\omega))^2 = \frac{1}{2}$ for $\omega$ positive, to get

$$\omega = \frac{\sqrt{-R_1^2+2 R_1 R_2+R_2^2}}{L_1} = \frac{R_2}{L_1} \sqrt {2-\Big(\frac{R_1}{R_2}-1\Big)^2}$$

This is certainly sensible (positive real) when $R_2>R_1$ (or more precisely when $R_2 > (\sqrt{2} - 1) R_1\ \approx 0.41 R_1$). If this doesn't hold (e.g. large $R_1$ say 100K in this example), then the gain doesn't ever drop below -3dB. Also note that for a simple high-pass RL filter you'd only have the $\frac{R_2}{L_1}$ part, which you can obtain as particular case by making $R_1 = 0$.

If you substitute the numerical values from your problem into this, you get basically the same result as via the graph/eyeometer: $1.09087×10^6$ rad/s (i.e. 175.3kHz).