Impulse Response - RLC Circuit

This is a very interesting one. There are some aspects to this that I don't understand.

Differential equations for the inductor and capacitor:

$V_S - R I - V_C = L \dot{I} \\ I = C \dot{V}_C$

Substituting the second equation into the first:

$V_S - R C \dot{V}_C - V_C = L C \ddot{V}_C$

Taking the Laplace transform (note that the Laplace transform of $\delta(t)$ is unity). I will refer to the transformed $V_C$ as simply $V$ . There are other terms in the transforms of the dot and double-dot quantities, but these are zero, per the stated initial conditions.

$1 - RC s V - V = s^2 LC V \\ V = \frac{1}{s^2 L C + s R C + 1}$

Doing the inverse transform gives:

$V_C (t) = 2 e^{-t} \sin(t)$

The current is therefore:

$I(t) = C \dot{V}_C = e^{-t} \Big(\cos(t) - \sin(t) \Big)$

The energy supplied by the source can be calculated using the sampling property of the Dirac delta function:

$E_S = \int_0^{\infty} V_S(t) I(t) \, dt = \int_0^{\infty} \delta(t) I(t) \, dt = I(0) = 1$

The energy dissipated in the resistor can be calculated by integrating the instantaneous resistor power:

$E_R = \int_0^{\infty} = I^2(t) R \, dt = \frac{1}{2}$

Here are some things that don't quite add up about this one:

1) The inductor current and capacitor voltage are zero at $t = \infty$ . Thus they both contain zero energy after infinite time elapses. I therefore expect the total energy dissipated by the resistor to equal the total energy supplied by the source. But this is apparently not the case

2) According to the initial conditions, the inductor current is supposed to be zero initially, but the derived current expression doesn't agree with this.

@Steven Chase has used the Laplace transform to solve this problem. That is the recommended approach because the problem greatly simplifies in the Laplace domain as the transformed Dirac function is simply unity. I am presenting an alternative approach to this problem.

The definition of the Dirac Delta function is:

$\delta(t)=\lim_{h \to 0}\begin{cases} 0 & t< 0 \\ \frac{1}{h} & 0\leq t\leq h \\ 0 & t>h \end{cases}$

Now, since we are working in the time-domain, $t<0$ becomes irrelevant, making: $\delta(t)=\lim_{h \to 0}\begin{cases} \frac{1}{h} & 0\leq t\leq h \\ 0 & t>h \end{cases}$

Now, if one were to plot this and compute the area under the curve, one would naturally arrive at the result:

$\int_{0}^{t} \delta(t) \ dt =1$

So, essentially, the integral of the Dirac delta function is the unit step function ( $H(t)$ ). In the case when the unit step occurs at $t=0$ , it can practically be treated as a constant function.

So now, let us consider the circuit equation:

$L\dot{I} + IR + \frac{Q}{C} = \delta(t)$ $\dot{Q} = I$

Integrate both sides of the above equation.

$Let \int_{0}^{t} Q \ dt = Q_s$ This makes the governing differential equation:

$L\ddot{Q}_s + R\dot{Q}_s + \frac{Q_s}{C} =\int_{0}^{t} \delta(t) \ dt$ $L\ddot{Q}_s + R\dot{Q}_s + \frac{Q_s}{C} =H(t)$ $L\ddot{Q}_s + R\dot{Q}_s + \frac{Q_s}{C} =1$

Now, one can easily solve for $Q_s(t)$ . Using that result:

$Q = \dot{Q}_s$ And finally:

$I = \dot{Q}$

So the conclusion here is that while dealing with impulse inputs, the way to tackle such a problem would be to compute the unit step response and then differentiate that with respect to time to obtain the impulse response. This makes the problem manageable numerically as well.