# How to derive cutoff frequency equations from the transfer function

I am working on a problem from the textbook Electric Circuits 10th edition

simulate this circuit – Schematic created using CircuitLab

I am trying to figure out first how to go from the transfer function to the magnitude. It's almost as if the denominator gets a $$\ w^2 \$$ out of nowhere. Then, I can't seem to figure out how to move from the magnitude in order to solve for the cutoff frequencies.

$$\ H(s)=\frac{\frac{R}{L}\times s}{s^2+(\frac{R+R_i}{L})s+\frac{1}{LC}} \$$ - using voltage divider

$$\ |H(jw)|=\frac{\frac{R}{L}w}{\sqrt{(\frac{1}{LC}-w^2)^2+(w\frac{R+R_1}{L})^2}}\$$

We can find the cutoff frequencies:

$$\\frac{R}{R_1+R}(\frac{1}{\sqrt{(2)}}) = \frac{\frac{R}{L}w}{\sqrt{(\frac{1}{LC}-w^2)^2+(w\frac{R+R_1}{L})^2}}\$$

The cutoff frequencies are then defined as:

$$\wc_1 = -\frac{R+R_1}{2L}+\sqrt{(\frac{R+R_1}{2L})^2+\frac{1}{LC}}\$$

$$\wc_2 = \frac{R+R_1}{2L}+\sqrt{(\frac{R+R_1}{2L})^2+\frac{1}{LC}}\$$

Well, we have:

$$\mathcal{H}\left(\text{s}\right):=\frac{180}{180+20+40\cdot10^{-3}\cdot\text{s}+\frac{1}{40\cdot10^{-9}\cdot\text{s}}}=$$ $$\frac{4500\cdot\text{s}}{625000000+\text{s}\cdot\left(5000+\text{s}\right)}\tag1$$

Now, for sinuscoidal signals we can write:

$$\text{s}=\omega\text{j}\tag2$$

So, we get:

$$\left|\mathcal{H}\left(\omega\text{j}\right)\right|=\frac{4500\cdot\omega}{\sqrt{\left(625000000-\omega^2\right)^2+\left(5000\cdot\omega\right)^2}}\tag3$$

And the maximum is given at:

$$\frac{\text{d}}{\text{d}\hat{\omega}}\left(\frac{4500\cdot\hat{\omega}}{\sqrt{\left(625000000-\hat{\omega}^2\right)^2+\left(5000\cdot\hat{\omega}\right)^2}}\right)=0\space\Longleftrightarrow\space\hat{\omega}=25000\tag4$$

By the definition of the cutoff frequency we have to find:

$$\frac{4500\cdot\omega}{\sqrt{\left(625000000-\omega^2\right)^2+\left(5000\cdot\omega\right)^2}}=\frac{1}{\sqrt{2}}\cdot\left|\mathcal{H}\left(\hat{\omega}\text{j}\right)\right|=\frac{1}{\sqrt{2}}\cdot\frac{9}{10}\space\Longleftrightarrow\space$$ $$\omega=2500\cdot\left(\sqrt{101}\pm1\right)\tag5$$

• How did you get $w^2$ in the denominator for step 3? Commented Nov 10, 2018 at 21:55
• @ArthurGreen Try to expand the factor $500000000+\text{s}(4000+\text{s})$ from equation $(1)$ Commented Nov 10, 2018 at 21:57
• I put your cutoff $$\omega=400\cdot\sqrt{3156\pm\sqrt{19711}}\tag4$$ into the calculator and got 21.965k and 22.965k The textbook solution is 22.625k and 27.625k imgur.com/a/05ojdoz I will try and go through the math again Commented Nov 10, 2018 at 22:01
• @ArthurGreen I verified mine using Mathematica. Commented Nov 10, 2018 at 22:05
• Did you get the same values I did? Commented Nov 10, 2018 at 22:10