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Consider a simple C-R 1-pole high pass filter. Apply a step input. The rising edge of the output rises up and then exponentially decays back to zero. Integrating the output over all times yields a positive number, so the output signal has a non-zero DC component. How is this reconciled with the notion that an HPF blocks DC with its zero at DC?

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    \$\begingroup\$ Why are you "integrating the output over all times" instead of, say, taking the limit as time goes to infinity? \$\endgroup\$ – Justin Nov 11 '20 at 19:33
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    \$\begingroup\$ Your thought experiment is mistaken, as that is not what happens. In actuality, the falling edge sends the output negative. \$\endgroup\$ – Chris Stratton Nov 11 '20 at 19:37
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    \$\begingroup\$ @ChrisStratton in my thought experiment, there is no falling edge. Just a single step input \$\endgroup\$ – alex Nov 11 '20 at 19:41
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    \$\begingroup\$ No conundrum. See what it does when you switch the DC source off! If you never switch off, consider this : a step input is not a DC signal. \$\endgroup\$ – Brian Drummond Nov 11 '20 at 19:45
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    \$\begingroup\$ To get the average DC output, you would have to integrate over all time then divide by the time interval. The integral is finite, but the denominator goes to infinity, so it tends to zero. \$\endgroup\$ – Austin Nov 12 '20 at 3:36
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You're confusing the steady-state response with the total response. The transfer function for that circuit is:

$$H(s)=\frac{s}{s+\frac{1}{RC}}$$

and if you solve for the step response:

$$h(t)=\text{e}^{-\frac{t}{RC}}$$

which shows that the steady-state response is zero for the step response, but the transient response is a decaying exponential, or what you saw.

What you're talking about, the integral of the area of the response, that's valid for the impulse response, where the response is:

$$h(t)=\delta(t)-\frac{\text{e}^{-\frac{t}{RC}}}{RC}$$

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The resolution to the conundrum is that there really is no conundrum.

Your input, a step function, has infinite dc component (its Laplace Transform, \$1/s\$, goes to infinity as \$s\rightarrow 0\$). Your output has a finite dc component. Therefore the transfer function of your filter, the ratio of the output to your input, has zero dc component because a finite number divided by infinity is zero. The transfer function has zero dc component, just as advertised.

If you put in any input with finite dc component, you will get zero dc component in your output. So, for example, as several comments have pointed out, if the input pulse steps back down to zero, you get the reverse spike that brings the dc component back to zero.

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The circuit you are describing and the output is shown below. Let me know if I got it correct.

There is no conundrum here. The HPF (capacitor) allows high frequencies to pass. The step voltage has a large number of high frequencies in it. A Laplace or Fourier series clearly shows this.

Integrating the area under the curve will result in showing you the amount of the signal that was allowed to pass through the filter. This is not a DC level type of value.

I agree that integral of the voltage plot will result in a scalar value of the voltage. This can be thought of as a DC level in some circuits. But in the context of a HPF, it can not be construed to mean that there is some continuous DC value being passed through the filter. It is instead a single scalar value that represents the amount of HF signal that did pass through the filter.

schematic

simulate this circuit – Schematic created using CircuitLab

enter image description here

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