# Why is particular solution zero for an impulse excitation signal?

We were being taught the impulse response for a series RC Circuit- consisting of simply one resistance, one capacitor, and an impulse excitation all in series.

I get that the homogeneous part of the solution will be of the form: $$x(t) = Ke^{\frac{-t}{RC}}$$

And we can find the value of $$K$$ using the initial conditions.

However, when it came to the particular solution part, we were straightaway told that since the excitation is an impulse, the particular solution will be equal to $$0$$. I'm not able to get why is this the case.

Also, I am not getting exactly how to calculate the particular solution for a given network, say RC network.

The correct form of the statement should be

the particular solution is zero for $$t>0$$

This is simply the case because the input $$\delta(t)$$ is zero for $$t>0$$. So for $$t>0$$, the impulse response equals the homogeneous solution. For $$t<0$$, the impulse response is zero because the system is causal. What remains is the moment $$t=0$$. The only thing that can happen there is an impulse (or its derivatives but let's not get into that here). So the complete form of the impulse response is

$$h(t)=A_0\delta(t)+y_h(t)\tag{1}$$

where $$y_h(t)$$ is the homogeneous solution of the differential equation. The constant $$A_0$$ can be determined by looking at the highest derivative of the input on both sides of the differential equation. For the given system we have

$$h'(t)+\frac{h(t)}{RC}=\frac{\delta(t)}{RC}\tag{2}$$

With $$h(t)$$ given by $$(1)$$ we see that on the left-hand side of $$(2)$$ we have a term $$A_0\delta'(t)$$ with nothing to match on the right-hand side. Consequently, for this specific differential equation we obtain $$A_0=0$$.

Note that it's easy to find a system for which $$A_0\neq 0$$. E.g., for a first-order RC highpass filter we have the following differential equation for the impulse response:

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

Using $$(1)$$ and matching the terms with $$\delta'(t)$$ on both sides of $$(3)$$ we obtain $$A_0=1$$, resulting in

$$h(t)=\delta(t)-\frac{1}{RC}e^{-t/RC},\qquad t\ge 0\tag{4}$$