I have an $RC$ circuit and I calculate the differential equation as
$$RC \frac{dV_c}{ dt} + V_c = V_g$$ that is the same of
$$RCy'(t) + y(t) = x'(t).$$
If I assume $x(t) = u(t)$ I have that the equation is $$RCg’(t) + g(t) = 0$$ and the solution is $$g(t) = K_1 e^{-t/RC} + K_0$$ now for have the impulse response I have to calculate this for $t > 0$, $t < 0$ and $ t= 0$ but I have no idea how to do. Can you please help me ?
Dynamic electric circuits involving linear time-invariant resistors, capacitors, and inductors are described by linear constant coefficient differential equations (LCCDE). Mathematical solution of such LCCDE requires some physical (electrical circuit theoretic) insight too. This is especially true for solving circuits under impulse functions (such as finding impulse responses). Note, that the dirac impulse $\delta(t)$ is a non-differentiable, non-continuous function; therefore it cannot be the subject of a continuous-differential equation. Nevertheless, a procedure as I present below, is developed to solve LCCDEs involving such improper impulse functions.
Given the circuit assuming zero initial conditions, derive its DE by applying a KCL (Kirchhoff's current law) at the output ($y(t)$) node :
$$ \frac{y(t)}{R} = C \frac{d}{dt} \left( x(t) - y(t) \right) $$ $$ y(t) = RC x'(t) - RCy'(t) $$ $$ \boxed{ y'(t) +\frac{1}{RC} y = x'(t) } \tag{1} $$
Impulse reponse $h(t)$ is the solution of Eq.(1) when the input is an impulse: $$ x(t) = \delta(t) \implies y(t) = h(t) $$
Let's apply a two-stage procedure to find $h(t)$. First, modify RHS of Eq.(1) by replacing $x'(t)$ with $x(t)$ : $$ y'(t) +\frac{1}{RC} y = x(t) \tag{2}$$
and denote solution of Eq.(2) as $h_0(t)$ when $x(t) = \delta(t)$. Then, by the second stage, the overall impulse response $h(t)$ of Eq.(1) will be: $$\boxed{ h(t) = h_0'(t) .} \tag{3}$$
Solution of first stage, Eq.(2):
From the characteristic equation : $s + \frac{1}{RC} = 0 \implies s = - \frac{1}{RC}$.
The homogeneous solution is :
$$y_h(t) = K e^{-t/RC} u(t) \tag{4}$$
The particular solution $y_p(t) = 0$, as the input is an impulse; i.e., $$\lim_{t \to \infty} x(t) = 0 \implies y_p(t) = 0$$
And the complete solution: $$h_0(t) = y_h(t) + y_p(t) = K e^{-t/RC} u(t) $$
To find the unknown $K$, we can use $y(0^+) = h_0(0^+)$ ; and from the circuit physics we find $y(0^+) = h_0(0^+) = 1$.
Then $K = 1$, and the total solution of Eq.(2) is
$$ \boxed{ h_0(t) = e^{-t/RC} u(t) } \tag{5}$$
$u(t)$ function is used to indicate that the output is causal; i.e., zero for $t < 0$. (a realizable physical circuit will be a causal system.)
Finally, the impulse response $h(t)$ is found from Eq.(3), by the second stage, as: $$ h(t) = h_0(t)' = \left( e^{-t/RC} u(t) \right)' $$
$$ \boxed{ h(t) = \delta(t) - \frac{1}{RC} e^{-t/RC} u(t) } \tag{6}$$
An intuitive physical explanation of the circuit behaviour under impulse excitation is as follows. Due to the physical capacitor's continuity of voltage property, the initial impulse (physical spike) voltage applied at the input, appears right at the output as a result of KVL applied around the loop at $t=0$. This impulse voltage on the output resistor $R$ produces an impulse current $i(t)= \delta(t)/R$ at time $t=0$ looping clock-wise around the circuit. This impulse current passes through the capacitor and jumps its voltage from $V(0^-) = 0$ to $V(0^+) = 1/RC$. Then the input node is shorted as the impulse is gone at $t=0^+$. And the simple first-order capacitive discharge circuit is obtained. Note that the discharge current loops counter clock wise due to capacitor voltage polarity. Hence the output voltage becomes negative for $t>0^+$.