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Given a simple series RC circuit where the output is measured across the resistor (i.e. a high-pass filter), how would one derive a general zero-input response for the system?

I can do it for a low-pass filter (output measured across the capacitor), but it almost seems to me like both responses will be the same!

I don't mind help with the solution in the time-domain, however I have a preference towards the complex-frequency domain personally.

Thank you!

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  • \$\begingroup\$ For zero-input the output should be zero too. \$\endgroup\$ – K. Rmth Nov 19 '15 at 5:57
  • \$\begingroup\$ That couldn't be more incorrect... \$\endgroup\$ – Connor Spangler Nov 19 '15 at 5:58
  • \$\begingroup\$ And why would that be so? \$\endgroup\$ – K. Rmth Nov 19 '15 at 6:02
  • \$\begingroup\$ Because the zero input response is resultant of the charge within the capacitor, I.e. Fully charged capacitor \$\endgroup\$ – Connor Spangler Nov 19 '15 at 6:15
  • \$\begingroup\$ A schematic of your circuit would be of much help here. \$\endgroup\$ – K. Rmth Nov 19 '15 at 6:22
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First of all zero input response doesn't make sense because after all circuit is meant to operate on the input in our case filter it based on where we take the output.

The zero input response is just the natural response of the circuit which is obtained as a differential equation using KVL ie;

RC dV/dt + V = 0 or R di/dt + i/C = 0

where V is the cap voltage and i is the current in the loop when there is no input.

Now, remember that the natural response of the system has nothing to do with where you take the output because there is no input.

For a High Pass Filter you are talking about,the capacitor comes in series with the input.

The capacitive impedance in frequency domain is given as 1/jwC where 'w' is the angular frequency of the input.

You can think of it as a simple impedance divider with the relation as follows:

Vout/Vin = R / (R + 1/jwC) = jwC/(R + jwC)

At very high frequencies, Vout/Vin = jwC / jwC = 1 ( approx) because the capacitor provides a low impedance path for the input.

However at low frequencies, the capacitor in series provides a high impedance path and the transfer function is :

Vout / Vin = jwC / R (approx)

where we can clearly see that this system is attenuating the input.

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One way to solve that problem is to use the (unilateral) Laplace transform, which allows to take initial conditions into account. For an RC high-pass circuit you have

$$U_1(s)=U_C(s)+U_2(s)\tag{1}$$

with \$U_1(s)\$ the input voltage, \$U_2(s)\$ the output voltage, and \$U_C(s)\$ the voltage across the capacitor. If \$u_C(0)\$ is the initial voltage across the capacitor, the current through the capacitor is

$$I(s)=C(sU_C(s)-u_C(0))\tag{2}$$

With \$(2)\$ and \$I(s)=U_2(s)/R\$ you can express \$U_C(s)\$ in terms of \$U_2(s)\$:

$$U_C(s)=\frac{U_2(s)}{sRC}+\frac{u_C(0)}{s}\tag{3}$$

For the zero-input response we set \$U_1(s)=0\$ and obtain from \$(1)\$ and \$(3)\$

$$U_2(s)=-U_C(s)=-\frac{U_2(s)}{sRC}-\frac{u_C(0)}{s}\tag{4}$$

which gives

$$U_2(s)=-\frac{u_C(0)}{s}\frac{sRC}{1+sRC}=-\frac{u_C(0)\tau}{1+\tau s}\tag{5}$$

with \$\tau=RC\$. In the time domain, \$(5)\$ corresponds to

$$u_2(t)=-u_C(0)e^{-t/\tau}\tag{6}$$

This zero-input response \$(6)\$ is indeed identical to the one of an RC low-pass filter (maybe apart from the sign, depending on the definition of \$u_c(t)\$), but this is no surprise because if you short-circuit the input, both circuits become equivalent.

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