Electronic – Calculating the cut off frequency of a electronic filter

componentscutoff frequencyfilteroperational-amplifiertransfer function

I've the following circuit:

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I know that the transfer function is given by:


Now, in this post they say that I can find the cut off frequency by finding:

$$\left|\mathscr{H}\left(\omega j\right)\right|=\frac{1}{\sqrt{2}}\cdot\left|\mathscr{H}\left(0 j\right)\right|\tag2$$

Question: so in my example I found that (when all the components have a value of \$1\$):


Is that correct?

Best Answer

If you replace all components values in the circuit by 1 (\$1\;\Omega\$ and \$1\;F\$) and use the equation I derived here, then the below graphs show the dynamic response and I can extract the cutoff frequency to 53 mHz or 0.335 rad/s as you correctly found.

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