Impulse response of a two-stage RC network

The impulse response is the inverse Laplace transform of the transfer function, $h(t) = \mathcal{L}^{-1}\{E_o(s)/E_i(s)\}$, because $\mathcal{L}\{\delta(t)\} = 1$.

Circuit in the s-domain

Impedances: $C_1 = 1$ F $\to \dfrac{1}{s}$, $R_1 = 2\ \Omega$, $R_2 = 1\ \Omega$, $C_2 = 2$ F $\to \dfrac{1}{2s}$.

 ei --||-----+----/\/\/----+------ eo
     1/s     |     1 ohm   |
            2 ohm        1/(2s)
             |             |
 ----------- +-------------+------

Let $V_1$ be the voltage at the junction of $C_1$ and $R_1$.

Node equations

At the output node (current through $R_2$ flows into $C_2$):

$$\frac{V_1 - E_o}{1} = 2sE_o \;\Rightarrow\; V_1 = (1 + 2s)E_o$$

At node $V_1$ (KCL):

$$(E_i - V_1)s = \frac{V_1}{2} + (V_1 - E_o)$$ $$sE_i = V_1\left(s + \tfrac{3}{2}\right) - E_o$$

Substituting $V_1$:

$$\begin{aligned} sE_i &= E_o\left[(1+2s)\left(s+\tfrac{3}{2}\right) - 1\right] = E_o\left[2s^2 + 4s + \tfrac{1}{2}\right] \\ \frac{E_o(s)}{E_i(s)} &= \frac{2s}{4s^2 + 8s + 1} = \frac{0.5\,s}{s^2 + 2s + 0.25} \end{aligned}$$

Poles

$$s = -1 \pm \frac{\sqrt{3}}{2} \;\Rightarrow\; s_1 = -0.134,\quad s_2 = -1.866$$

Partial fractions

$$\frac{0.5s}{(s+0.134)(s+1.866)} = \frac{k_1}{s+0.134} + \frac{k_2}{s+1.866}$$ $$\begin{aligned} k_1 &= \frac{0.5(-0.134)}{-0.134+1.866} = \frac{-0.067}{1.732} = -0.0387 \\ k_2 &= \frac{0.5(-1.866)}{-1.866+0.134} = \frac{-0.933}{-1.732} = 0.5387 \end{aligned}$$

Impulse response

$$h(t) = 0.5387\,e^{-1.866t} - 0.0387\,e^{-0.134t}, \quad t \ge 0$$

Exact form:

$$h(t) = e^{-t}\left[\tfrac{1}{2}\cosh\tfrac{\sqrt{3}}{2}t - \tfrac{1}{\sqrt{3}}\sinh\tfrac{\sqrt{3}}{2}t\right]$$

Answer: $e_o(t) = 0.5387e^{-1.866t} - 0.0387e^{-0.134t}$ V for a unit impulse input. Check: $h(0) = 0.5$, which matches $\lim_{s\to\infty} sT(s) = 2/4$.