How to calculate impulse response of a RC circuit

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 ?

• Hi! I assume you want a time domain solution, instead of a Laplace based ? – Fat32 Oct 9 '18 at 9:39
• My book solve it as Cauchy problem, and the solution is having h(t) but I don’t know if it use Laplace. For now I solved the circuit for having the differential equation of it but now I have to solve it. Thank you – Elena Martini Oct 9 '18 at 10:03
• ok. a time domain solution. It's the classical approach and a little tricky. Laplace method works like simple algebra, but you will learn it later. – Fat32 Oct 9 '18 at 10:04

Linear and time invariant electric circuits involving capacitors and inductors are described via linear constant coefficient differential equations. The solution process greatly benefits from a physical (electrical) insight instead of relying purely on mathematical techniques.

This is especially true considering the fact that classical differential equations involving impulse functions (which is a generalised function) is understood to be an improper case. Nevertheless modification of the solution process for impulsive input / output functions, enables the computation of an impulse response which otherwise would not be possible by the classical treatment of differential equations.

Now, given the CR circuit (a differentiator) with zero initial conditions, I would like to obtain its differential equation by applying a KCL at the output ($$y(t)$$) node, equating the current entering and leaving as:

$$\frac{y(t)}{R} = C \frac{d}{dt} \left( x(t) - y(t) \right)$$ $$y(t) = RC x'(t) - RCy'(t)$$ $$y'(t) +\frac{1}{RC} y = x'(t)$$

The impulse reponse $$h(t)$$ is defined as the output $$y(t)$$ when the input $$x(t)$$ is an impulse: $$x(t) = \delta(t) \implies y(t) = h(t)$$

Now I will apply a two-stage solution to obtain $$h(t)$$. First I will treat the LHS equation as follows: $$y'(t) +\frac{1}{RC} y = x(t)$$

Denoting its output as $$h_0(t)$$ for $$x(t) = \delta(t)$$, then the resulting impulse response $$h(t)$$ will be: $$\boxed{ h(t) = h_0'(t) }$$

Solution of first stage: The characteristic equation yields $$s + 1/RC = 0 \implies s = - 1/RC$$. The homogeneous solution is $$y_h(t) = K e^{-t/RC}$$ for $$t > 0$$ The particular solution will be zero as the input is impulse. We only need an initial condition on $$y(0^+)$$ which is found to be $$y(0^+) = 1$$. Then $$K = 1$$ and the first stage solution is $$\boxed{ h_0(t) = e^{-t/RC} u(t) }$$ $$u(t)$$ function is used to indicate that the output is zero for $$t < 0$$.

Then the overall impulse response is simply found to be: $$h(t) = h_0'(t) = \left( e^{-t/RC} u(t) \right)' = \delta(t) - \frac{1}{RC} e^{-t/RC} u(t)$$

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

Electrically spleaking, this is the output $$y(t)$$ voltage when the input $$x(t)$$ voltage is an impulse $$x(t) = \delta(t)$$. Note that due to the capacitor's continuity of voltage (unless impulse current passes through) property, the initial impulse voltage appears at the output as a result of KVL applied around the loop at $$t=0$$. This impulse voltage on the output resistor $$R$$ induces an impulse current $$i(t)= \delta(t)/R$$ at time $$t=0$$. This current passes through the capacitor and forces its voltage from $$V(0^-) = 0$$ to $$V(0^+) = 1/RC$$. Then the input node is shorted as the impulse is gone. And the simple first order capacitive discharge circuit is obtained. Note that the discharge current is vertical up due to capacitor voltage polarity. Hence the output voltage becomes negative for $$t>0^+$$.

• Thank you so so much !!!!!!!! I ‘ only ‘ have two question.. maybe for my terrible English I didn’t understand why you apply the condition y ( 0+ ) = 1. Also, in the first steps I obtained $\ -1/RC x’(t) + 1/RC y’(t) = y(t)$ – Elena Martini Oct 9 '18 at 13:48
• @ElenaMartini I split the DE into two parts. Then solve the impulse response for the first. $y' + (1/RC) y = x'$ This first DE has an initial condition related to applied input $\delta(t)$. The initial condition for the DE $a_0 y^N + a_1 y^{N-1} + ...+ a_{N-1}y'(t) + a_N y(t) = x(t)$ is found to be $y^{N-1}(0^+) = 1/a_0$ where $N$ is the order of the DE... – Fat32 Oct 9 '18 at 20:03