2 added 230 characters in body edited Jun 19 '13 at 20:54 Alfred Centauri 24.3k11 gold badge1616 silver badges5454 bronze badges 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 $$\RC\$$. That's really all there is to it. For frequencies well below the corner frequency, the output is effectively proportional to frequency just as we would expect from a differentiator. You're probably misunderstanding "stop band" in this context. A 1st order high pass filter has a gentle roll-off that is just about 20dB / decade. Signals aren't "stopped" below the corner frequency, they're increasingly attenuated. I believe I've answered this elsewhere but, regardless, here is the way to look at this. The transfer function for the RC HPF is: $$\\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. 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 $$\RC\$$. That's really all there is to it. For frequencies well below the corner frequency, the output is effectively proportional to frequency just as we would expect from a differentiator. You're probably misunderstanding "stop band" in this context. A 1st order high pass filter has a gentle roll-off that is just about 20dB / decade. Signals aren't "stopped" below the corner frequency, they're increasingly attenuated. 1 answered Jun 19 '13 at 20:48 Alfred Centauri 24.3k11 gold badge1616 silver badges5454 bronze badges I believe I've answered this elsewhere but, regardless, here is the way to look at this. The transfer function for the RC HPF is: $$\\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.