# Finding impulse response of an RC circuit from its step response [duplicate]

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 $$V_r = V_u$$

So

$$V_u = V_r = R I_r$$ , 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

• 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 ? – Fat32 Aug 13 '19 at 13:47
• 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 ! – Elena Martini Aug 13 '19 at 14:05
• 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. – Elena Martini 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 $$h(t) = s(t)' = \left( e^{-t/RC} u(t) \right)'$$

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

• 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.. – Elena Martini Aug 30 '19 at 13:43
• I've added further explanation of finding the initial condition from circuit topology and device physics by applying KVL... – Fat32 Aug 30 '19 at 22:55