Impulse response of RC filter

        R
 x(t) o--/\/\/--+--------o
  +             |        +
               === C    y(t)
  -             |        -
 ---o-----------+--------o

Impulse response

By KVL, with $i(t) = C\,\dfrac{dy}{dt}$:

$$x(t) = RC\frac{dy(t)}{dt} + y(t)$$

Take the Fourier transform:

$$\begin{aligned} X(j\omega) &= j\omega RC\,Y(j\omega) + Y(j\omega) \\ H(j\omega) &= \frac{Y(j\omega)}{X(j\omega)} = \frac{1}{1 + j\omega RC} \end{aligned}$$

Write it as $H(j\omega) = \dfrac{1/RC}{1/RC + j\omega}$ and use the pair $e^{-at}u(t) \leftrightarrow \dfrac{1}{a + j\omega}$ with $a = 1/RC$:

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

(Physically: an impulse charges the capacitor instantly to $1/RC$, which then discharges through $R$ with time constant $RC$.)

h(t)
1/RC|.
    | `.
    |   `-.
    |      `--.___
    +-------+-------`----> t
    0      RC

$h(t) = 0$ for $t < 0$, so the filter is causal; $\int |h|\,dt = 1$, so it is stable.

Magnitude and phase response

$$|H(j\omega)| = \frac{1}{\sqrt{1 + (\omega RC)^2}}, \qquad \angle H(j\omega) = -\tan^{-1}(\omega RC)$$

|H|                         angle H
 1 |----.                  0 |----.
   |     `.                   |     `.
.707|- - - -*                -45|- - - -*
   |         `-.              |         `-.___
   |             `---.    -90 |- - - - - - - -
   +------+-----------> w     +------+--------> w
        1/RC                       1/RC

The cutoff (half-power, $-3$ dB) frequency is $\omega_c = 1/RC$; the circuit is a first-order low-pass filter.