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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 ?

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  • $\begingroup$ Hi! I assume you want a time domain solution, instead of a Laplace based ? $\endgroup$ – Fat32 Oct 9 '18 at 9:39
  • $\begingroup$ 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 $\endgroup$ – Elena Martini Oct 9 '18 at 10:03
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Dynamic electric circuits involving linear time-invariant resistors, capacitors, and inductors are described via linear constant coefficient differential equations (LCCDE). Solution of such LCCDE greatly benefits from physical (electrical circuit theoretic) insight. This is especially true considering the fact that classical differential equations involving generalised impulse functions are considered as improper cases.

Nevertheless, modification of the classical time-domain solution process for impulsive functions, enables the computation of an impulse response.

Now, given the RC circuit (a differentiator with R to ground) with zero initial conditions, I would like to derive its differential equation by any means possible such as applying a KCL at the output ($y(t)$) node, equating the current entering and leaving as:

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

The impulse reponse $h(t)$ is defined as the output $y(t)$ of Eq.(1) 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: $$ y'(t) +\frac{1}{RC} y = x(t) \tag{2}$$

Denoting the first stage output as $h_0(t)$ when $x(t) = \delta(t)$, then the resulting overall impulse response $h(t)$ will be found as: $$\boxed{ h(t) = h_0'(t) } \tag{3}$$

Solution of first stage: We first note that since this is a realizable physical circuit, it must denote a causal system, and the output must be in accordance with this.

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 $y_p(t)$ will be zero as the input is an impulse; i.e., $$\lim_{t \to \infty} x(t) = 0 \implies y_p(t) = 0$$

Then since $$h_0(t) = y_h(t) + y_p(t) = y_h(t)$$

to find the value of unknown $K$ in Eq.(4) we need an initial condition on the output $y(0^+) = h_0(0^+) $ ; found from the circuit physics as $y(0^+) = h_0(0^+) = 1$.

We find $K = 1$ and the first stage total solution 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 $t < 0$.

Then the overall impulse response of Eq.(1) is found to from Eq.(3) as: $$ h(t) = h_0(t)' = \left( e^{-t/RC} u(t) \right)' $$

$$ \boxed{ h(t) = \delta(t) - \frac{1}{RC} e^{-t/RC} u(t) } \tag{6}$$

From electrical circuit theory point of view, this is the output voltage $y(t)$ when the input $x(t)$ voltage is an impulse $x(t) = \delta(t)$.

The 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 around the circult. This 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. 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 $t>0^+$.

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  • $\begingroup$ 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) $ $\endgroup$ – Elena Martini Oct 9 '18 at 13:48
  • $\begingroup$ @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... $\endgroup$ – Fat32 Oct 9 '18 at 20:03

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