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How can I find the impulse response for the following system in time domain? I actually would like to find my mistake in my attempt. Below is what I have tried according to the answer given for this question: Why is particular solution zero for an impulse excitation signal?

The given system/circuit can be described by the following differential equation:

$$\frac{x(t)-V_c(t)}{R} - C \frac{\text{d}V_c(t)}{\text{d}t} = 0$$

Since we are looking for the impulse response, $x(t) = \delta(t)$ and $V_c(t) = h(t)$. By plugging in these values and rearranging the terms, we get:

$$\frac{\text{d}h(t)}{\text{d}t} + \frac{h(t)}{RC} = \frac{\delta(t)}{RC}$$

$h(t)$ is given to be $h_h(t) + A_0\delta(t)$ in the referred question and answer, where $h_h(t)$ is the homogeneous solution of the aforementioned differential equation. Here, $h(t)$ does not include any derivatives of $\delta(t)$ probably because the degree of the output is $\textbf{greater than}$ or equal to the degree of the input.

Homogeneous solution is found to be $h_h(t) = c_1e^{-\frac{t}{RC}}$. Therefore:

$$h'(t) = -\frac{c_1}{RC}e^{-\frac{t}{RC}} + A_0\delta'(t)$$

We can now find the unknowns by plugging $h(t)$ and $h'(t)$ into the differential equation.

$$\frac{\delta(t)}{RC} = -\frac{c_1}{RC}e^{-\frac{t}{RC}} + A_0\delta'(t) + \frac{c_1}{RC}e^{-\frac{t}{RC}} + \frac{A_0}{RC}\delta(t)$$

$$\frac{\delta(t)}{RC} = A_0\delta'(t) + \frac{A_0}{RC}\delta(t)$$

From $\frac{1}{RC}\delta(t) = \frac{A_0}{RC}\delta(t)$: $$A_0 = 1$$

From $0 = A_0\delta'(t)$: $$A_0 = 0$$

Why do I have such a contradiction and what is my mistake? Thank you in advance.

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    $\begingroup$ You don't wanna do derivatives of a dirac delta function. They be nasty. Consider what the current through $R$ is at the instant of $t=0$. Assume the voltage across $C$ is zero at $t=0$. What you want to determine is exactly what the capacitor voltage is immediately after the impulse, like when $t\approx 0$. $\endgroup$ Mar 19 at 21:59
  • $\begingroup$ @robertbristow-johnson Thank you. I cannot really visualize/think about it since dirac-delta function's magnitude is infinite at that point. It really confuses me to think about the physical system. Could you please explain that physical intuition shortly if possible? $\endgroup$
    – Lars Smith
    Mar 19 at 22:14
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    $\begingroup$ The voltage on $C$ is negligible compared to the infinite voltage of the dirac impulse. So then the current going through the resistor is $\frac{\delta(t)}{R}$. Then the capacitor integrates that current and it becomes a step discontinuity in the capacitor voltage. The size of that step is $\frac{1}{RC}$. Now what happens the instant of time after that? $\endgroup$ Mar 19 at 22:19
  • $\begingroup$ This really belongs in the electrical engineering SE. $\endgroup$ Mar 19 at 22:22
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    $\begingroup$ When $t>0$, then $x(t)=0$. But there is a non-zero voltage on the capacitor and you have a differential equation governing it. But at least now you know what the initial conditions are. $\endgroup$ Mar 19 at 22:27

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Since the system is causal, its impulse response has the form

$$h(t)=c_1e^{-t/RC}u(t)+A_0\delta(t)\tag{1}$$

Note the unit step function $u(t)$ in $(1)$.

From $(1)$ we obtain the derivative

$$h'(t)=-\frac{c_1}{RC}e^{-t/RC}u(t)+c_1\delta(t)+A_0\delta'(t)\tag{2}$$

Plugging $(1)$ and $(2)$ into the system's differential equation (with $\delta(t)$ as its input)

$$\frac{\delta(t)}{RC}=h'(t)+\frac{h(t)}{RC}\tag{3}$$

gives

\begin{align*} \frac{\delta(t)}{RC} &= -\frac{c_1}{RC}e^{-t/RC}u(t)+c_1\delta(t)+A_0\delta'(t) + \frac{c_1}{RC}e^{-t/RC}u(t)+\frac{A_0}{RC}\delta(t) \\ &= \left(c_1+\frac{A_0}{RC}\right)\delta(t)+A_0\delta'(t) \end{align*}

from which it follows that $A_0=0$ and $c_1=1/RC$.

Hence, the impulse response is given by

$$h(t)=\frac{1}{RC}e^{-t/RC}u(t)\tag{4}$$

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  • $\begingroup$ So $u(t)$ was the thing :) Thank you so much. $\endgroup$
    – Lars Smith
    Mar 19 at 22:11

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