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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.

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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

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

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  • $\begingroup$ Hi! very similarly, let's solve a discrete-time LCCDE $\sum_0^N a_k y[n-k] = \sum_0^M b_k x[n-k]$ using homogenous $y_h[n]$ and particular $y_p[n]$ parts, where, $y_p[n]$ is determined by the method of undetermined coefficients. For a given particular input $x[n]=\alpha^n u[n]$, which form of particular output is assumed, that's plugged into the LCCDE to determine the coefficients of it? It seems it's not possible to find a valid form for $y_p$ for all $n \ge 0$ but only for $n \to \infty$ will work? What's your opinion? $\endgroup$
    – Fat32
    Commented Jul 18, 2023 at 10:59
  • $\begingroup$ Why is the system equation like that? I thought it was $\delta'(t) = h'(t) \cdot R + \frac{h(t)}{C}$. This way I ended up finding $A_0 = 0$ and $A_0 = \frac{1}{R}$ at the same time. Could you please help me? Here is my work: i.imgur.com/FC0Ocpq.png. Thank you. $\endgroup$
    – Lars Smith
    Commented Mar 19 at 3:14
  • $\begingroup$ @LarsSmith: It would be best to formulate a new question because it may be helpful for future users as well. Concerning the system equation, I'm not sure how you came up with it but make sure that the output is a voltage, not a current. $\endgroup$
    – Matt L.
    Commented Mar 19 at 9:21
  • $\begingroup$ Thank you for your reply. I understand that you have written that system equation as voltage being the output and I corrected my work accordingly. I still don't have a consistent value for $A_0$. Here is my work: i.imgur.com/rDKVLKx.png. I could create a new question but it might get closed and directed to this question back because the question will almost be the same. I will be grateful if you please answer my question. Thank you. $\endgroup$
    – Lars Smith
    Commented Mar 19 at 15:34
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    $\begingroup$ Thank you. I have just asked the question and here is the link for those who have similar questions: dsp.stackexchange.com/questions/93327/… $\endgroup$
    – Lars Smith
    Commented Mar 19 at 21:49

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