Wave front retardation by terminal capacitance

An infinite rectangular wave $V$ arriving at a capacitance $C$ at the end of a line of surge impedance $Z$ sees the capacitor as a short circuit at first. The capacitor then charges towards $2V$ through $Z$:

$$V_C(t) = 2V\left(1 - e^{-t/ZC}\right)$$

So the vertical front of the incident wave becomes an exponential front: the wave is retarded (sloped). The amount of retardation is set by the time constant $ZC$.

Time constant

$$\tau = ZC = 500 \times 0.004\times10^{-6} = 2\times10^{-6}\ \text{s} = 2\ \mu\text{s}$$

Measures of retardation

• Time to reach the incident value $V$ (half the final $2V$):

$$2V(1 - e^{-t/\tau}) = V \;\Rightarrow\; t = \tau\ln 2 = 2 \times 0.693 = 1.386\ \mu\text{s}$$

• Time to reach 90% of the final value $2V$:

$$t = \tau\ln 10 = 2 \times 2.303 = 4.61\ \mu\text{s}$$

• Practically full value (about 99%) after $5\tau = 10\ \mu$s.

 V_C
 2V |            ___________
    |        __--
  V |    _--'
    |  /
    | /
    |/________________________ t
    0  1.39   4.61 us

A rectangular (zero rise time) wave thus becomes a wave with a front of a few microseconds. This reduces the steepness ($dV/dt$) of the surge, which protects the winding insulation of terminal equipment.

Answer: time constant $ZC$ = 2 μs. The front is retarded so that the voltage reaches the incident value after 1.39 μs and 90% of its final value ($2V$) after 4.61 μs.