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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?

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  • \$\begingroup\$ What have you looked at so far? What information have you found that backs up the 1/RC condition. Have you looked at op-amp differentiators? Do you understand their limitations? \$\endgroup\$ – Andy aka Jun 19 '13 at 19:05
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    \$\begingroup\$ Note that stop band just refers to some frequency below which the signal is attenuated by at least a certain number of decibels. It doesn't refer to a "brick wall". What happens to the \$\omega\$ when you differentiate \$\sin(\omega t)\$? It goes out to the front: \$\omega\cos(\omega t)\$. I.e. low frequencies (small \$\omega\$) have small-amplitude derivatives. This is connected to high pass filtering. \$\endgroup\$ – Kaz Jun 19 '13 at 20:36
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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.

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  • \$\begingroup\$ So,can we say that for a signal whose frequency is above the corner frequency output will be proportional to the input.But for the similar signal whose frequency is much below corner frequency ,we get the differentiated output. \$\endgroup\$ – Abhishek Jun 20 '13 at 17:46
  • \$\begingroup\$ @Abhishek, that sounds correct to me. \$\endgroup\$ – Alfred Centauri Jun 20 '13 at 17:50
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If the transfer function of a high-pass system is:

$$\frac{Y\left ( s \right )}{X\left ( s \right )}= \frac{s}{s + a}$$

Then, $$\frac{Y\left ( s \right )}{sX\left ( s \right )}= \frac{1}{s + a}$$

The latter shows the transform of the output, Y, over \$ sX \left ( x \right ) \$, or the transform of the derivative of X. You'll notice that the right side of the equation now shows a low-pass filter! Thus, y will be a low-pass filtered version of the derivative of x.

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