# Differentiator cut-off frequency

A practical differentiator has two cut-off frequency but can no longer perform differentiaiton after first cut-off. After first cut-off there is a 20db per decade gain reduction but why does this affect the mathematical function itself. The DC equation still remains $$V_{out}=-R_fC_f \frac{dV_{in}}{dt}.$$ In a low-pass filter the signal just attenuates and does not change mathematically, is something different here. Is it just that it is still differentiating but for some reason my output is distorted or is there a mathematical explanation behind this. I am using a standard practical differentiator as shown below and I am sure the values I have are correct.

• What is the purpose of Cf? You can design a stable (practical, non-ideal) diff. circuit even for Cf=0. – LvW Sep 2 '20 at 8:26

\begin{align} Z_f&=\frac{1}{\frac{1}{R_f}+sC_f} \\ Z_1&=R_1+\frac{1}{sC_1} \\ H(s)&=\frac{Z_f}{Z_1} \\ H(s)&=\frac{1}{\left(R_1+\frac{1}{sC_1}\right)\left(R_f+\frac{1}{sC_f}\right)} \\ H(s)&=\frac{\frac{s}{R_1C_f}}{s^2+\frac{R_fC_f+R_1C_1}{R_fR_1C_fC_1}s+\frac{1}{R_fR_1C_fC_1}} \end{align}
• @vikassajanani I can't see what's wrong and why it won't format it. At any rate, it's of the form s/(s^2+s+1). I've modifed my answer to include that you can no longer talk about it being a differentiator once a pole is reached. – a concerned citizen Sep 2 '20 at 7:01