Impulse Response of a series RC filter where the output voltage is taken across the Capacitor
Take the equations of the relationship between voltage and current in a resister and capacitor

Ic = C dVc/dt     (1)
Ir = 1/R Vr         (2)


They are in series which indicates 2 things: 


The current in the resistor is the same as the capacitor (assuming no load on the output).  
The total Input voltage is the sum of the two.


Giving these equations:

Ic = Ir            (3)
Vi = Vc + Vr       (4)


putting (1) &amp; (2) into (3), use (4) to replace Vr

1/R Vr = C dVc/dt
1/RC (Vi - Vc) = dVc/dt    (5)


Before this can be integrated, the Vc must be with dVc.  Rearrange equation (5), and then integrate (assume Vi is constant and does not change with time)

1/RC dt = 1/(Vi-Vc) dVc

1/RC t = -ln(Vi-Vc) + K


Before time 0 Vi was stable for an limitless amount of time, Vc will match Vi as an initial condition; in this case 0V.  We evaluate the integral from time t0 and Vc=V0, to time t and output Vo.  (Vo is equal to Vc, and now is a good time to make that swap)

(1/RC) t |t=0-&gt;t = -ln(Vi-Vo) |Vc=0-&gt;Vo
(1/RC) t = -ln(Vi-Vo) -  -ln(Vi-V0) = -ln( (Vi-Vo)/(Vi-V0) )


A bit more rearranging

-t /RC = ln( (Vi-Vo)/(Vi-V0) )
exp(-t / RC) = (Vi-Vo)/(Vi-V0)            (6)


if V0 is 0 then:  

exp(-t / RC) = 1 - Vo/Vi
Vo/Vi = 1 - exp(-t (R/C))


The step response is:

Vo = Vi( 1 - exp(-t / RC) )                    (7a)


But you asked for the impulse response.  It might not be obvious, but I need to keep V0 in equation (6) to show what the impulse response is.

Vo = (V0-Vi) exp(-t / RC) + Vi                   (7b)


What if the system gets a pulse of infinite magnitude for a duration of 0 seconds.  A very theoretical question, as a pulse can not be 0 seconds long and infinity is not a value.  So we use a limit to see what equation (7a) equals when t-&gt;0 and Vi-&gt;1/t (The units of the '1' are Volt-seconds, such that no matter the duration the pules integrates to 1Vs, a unit pulse).

Lim t-&gt;0 of Vp = (1/t) ( 1 - exp(-t/RC) )                  (8)
Lim t-&gt;0 of Vp = (1/1) ( 0 - exp(-t/RC)*(-1/RC) )          (9)
Lim t-&gt;0 of Vp = 1/RC                                      (10)


Equation (8) results in a 'zero over zero' situation (exp(...) on top and 't' on bottom), which means that the limit of the quotient of the derivative of the 2 will equal the same value (derivative of top and bottom, and re-evaluate the limit); equation (9).  Equation (9) can be directly evaluated with `t=0, giving equation (10).  So immediately after the impulse (after... which is still at time zero), the Output voltage is suddenly at 1/RC. (who ever heard of a capacitor changing voltage in an instant! but remember this is theoretical with the 'impossible' impulse).

Bring back equation (7b).  The initial output V0 is 1/RC; the input Vi is zero (back to zero after that momentary lapse of reality that got the output suddenly to 1/RC!).  The result is a mathematical equation representing what would happen if an impulse function hit.

Vo = 1/RC exp(-t / RC)