# How to find impulse response for the given system?

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.

• 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 $t\approx 0$. Mar 19 at 21:59
• @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? Mar 19 at 22:14
• 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? Mar 19 at 22:19
• This really belongs in the electrical engineering SE. Mar 19 at 22:22
• 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. Mar 19 at 22:27

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

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

• So $u(t)$ was the thing :) Thank you so much. Mar 19 at 22:11