I have to find the impulse response of an RC circuit (c up ). I have to find it from the step response $g(t)$.

I found that

$V_i - V_c - V_r = 0$ , with $V_r = V_u$


$V_u = V_r = R I_r$ , with $I_r = I_c$

$V_u= RC \frac{dV_c}{dt}$

$V_u = y(t) = RC \frac{d(x(t)-y(t))}{ dt}$

because $V_c = x(t)-y(t)$

From this i found $y(t) = RC ( x'(t) - '(t) )$ , i simply rewrote d/dt with '.

Now if i want to found the step response, i considered $x(t) = u(t)$ and $y(t) = g(t)$ that is the step response

So i wrote $$g(t) = RC u'(t) - RC g'(t)$$

but $u'(t) = 0$ so is $RC g’(t) + g(t) = 0$

Solving this differential equation i obtained that

$$g(t) = k_0 + k_1 \cdot e^{-t/RC }$$

Until now i obtained the same results of my professor but now that i have the step response i have to find the $h(t)$, impulsive response , but I don’t know how. Can someone please help me ? This is the same problem of my previous question but in this case i have to find the impulsive response using the step response. Thank you so much

  • $\begingroup$ Hi Elena... May I kindly ask why you didn't accept my answer (that I've already forgotten but thanks to @MattL. that I can see now!) to your previous very related question ? Furthermore, do you want me to add a solution for the step-response into that question or do you need a separate solution ? $\endgroup$
    – Fat32
    Aug 13 '19 at 13:47
  • $\begingroup$ Your response really help me but now I have to do the same thing but finding before the step response. I think I found it but now i’M blocked.. adding this response to my previous post should be perfect. When I posted This morning I didn’t think that posting this , as an alternative resolution method , to the previous post should be better. My fault, I’m sorry ! $\endgroup$ Aug 13 '19 at 14:05
  • $\begingroup$ Probably I thought I understood but , when I tryed to doing this exercise ( that is the same , with another method ) I stopped at the step response. $\endgroup$ Aug 13 '19 at 14:51

The answer (How to calculate the impulse response of an RC circuit using time-domain method) provides a direct time-domain solution of an RC circuit for the impulse reponse $h(t)$. Now this new answer modifies it to solve for the step-response $s(t)$ instead and then computes the impulse response according to :

$$h(t) = s(t)' $$

The differential equation of the first order circuit was derived as $$ \boxed{ y'(t) +\frac{1}{RC} y = x'(t) } \tag{1} $$

The step reponse $s(t)$ is defined as the output $y(t)$ of Eq.(1) when the input $x(t)$ is a unit-step function $$ x(t) = u(t) \implies y(t) = s(t) $$

Let's apply a one stage direct solution to obtain $s(t)$.

The homogeneous solution is found from $$ y'(t) +\frac{1}{RC} y = 0 \tag{2} $$

The characteristic equation : $s + \frac{1}{RC} = 0 \implies s = - \frac{1}{RC}$.

The (causal) homogeneous solution is :

$$y_h(t) = K e^{-t/RC} u(t) \tag{3}$$

Then, the particular solution $y_p(t)$ will be from the method of undetermined coefficients as follows:

For the particular input $x(t) = u(t)$ we may assume a particular solution as $y_p(t) = A u(t) + B \delta(t)$, then plug this assumed solution into Eq.(1) to find out the coefficients $A$ and $B$. This yields $A=0$ and $B=0$, hence the particular solution is found to be identically zero.

Then since the total solution is: $$s(t) = y_h(t) + y_p(t) = y_h(t)$$

to find the value of unknown $K$ in Eq.(4) we need an initial condition on the output $y(0^+) =s(0^+) $ ; found from the circuit physics: apply KVL around the loop of voltage source, capacitor and resistor, with the fact that the capacitor voltage is fixed at zero at time $t=0$. Then we find $y(0^+) = s(0^+) = 1$ which yields $K=1$

Hence the step-response is: $$ \boxed{ s(t) = e^{-t/RC} u(t) } \tag{4}$$

Then the impulse response of Eq.(1) is found to be $$ h(t) = s(t)' = \left( e^{-t/RC} u(t) \right)' $$

$$ \boxed{ h(t) = \delta(t) - \frac{1}{RC} e^{-t/RC} u(t) } \tag{5}$$

  • $\begingroup$ Now I obtained your same result but I don’t understand which condition you applied for obtain K. I mean that for obtain K=1 ( seeing the solution ) I should apply the condition g(t=0)=1 but I don’t know why.. $\endgroup$ Aug 30 '19 at 13:43
  • $\begingroup$ I've added further explanation of finding the initial condition from circuit topology and device physics by applying KVL... $\endgroup$
    – Fat32
    Aug 30 '19 at 22:55

Not the answer you're looking for? Browse other questions tagged or ask your own question.