A high pass circuit is a differentiator when used under the condition that input frequency of signal is much less than \$1/RC\$. But that region corresponds to stop band of the filter, so how do we get a differentiated wave at output?
Electronic – high pass circuit as differentiator
filter
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Best Answer
I believe I've answered this elsewhere but, regardless, here is the way to look at this. The transfer function for the RC HPF is:
\$\dfrac{j \omega RC}{1 + j \omega RC} \$
The "trick" is to look at the behaviour at low enough frequency such that \$ j \omega RC << 1\$. When this holds, the denominator is effectively just \$1\$ and the transfer function is effectively:
\$ j \omega RC\$
But, this is the transfer function for a differentiator with gain equal to \$RC\$. That's really all there is to it. For frequencies well below the corner frequency, the output is effectively proportional to frequency just as we would expect from a differentiator.
You're probably misunderstanding "stop band" in this context. A 1st order high pass filter has a gentle roll-off that is just about 20dB / decade. Signals aren't "stopped" below the corner frequency, they're increasingly attenuated.