For t > T_S the general expression for an exponential in Eq. (5–17) becomes

υ(t) = [V_{A} e^{−(t  −  T_{S})/T_{C}}] u(t  −  T_{S}) V                        (5–17)

υ(t) =V_{A} e^{−(t  −  T_{S})/T_{C}}V

We have only a portion of the waveform, so we do not know the location of the t = 0 time origin; hence, we cannot find the amplitude V_A or the time shift T_S from the display. But, according to the decrement property, we should be able to determine the time constant since the decrement ratio is independent of amplitude and time. Specifically, Eq. (5–14) points out that

\frac{υ(t  +  Δt)}{υ(t)} = \frac{V_{A}e^{−t/T_{C}}e^{−Δt/T_{C}}}{V_{A}e^{−t/T_{C}}} = e^{−Δt/T_{C}}                      (5–14)

\frac{υ(t  +  Δt)}{υ(t)} = e^{−Δt/T_{C}}

Solving for the time constant T_C yields

T_{C} = \frac{Δt}{\ln \left[\frac{υ(t)}{υ(t  +  Δt)}\right]}

Taking the starting point at the left edge of the oscilloscope display yields

υ(t) = (3.6 div)(2 V/div) = 7.2 V

Next, defining Δt to be the full width of the display produces

Δt = (8 div)(1 ms/div) = 8 ms

υ(t + Δt) = (0.5 div)(2 V/div) = 1 V

As a result, the time constant of the waveform is found to be

T_{C} = \frac{Δt}{\ln \left[\frac{υ(t)}{υ(t  +  Δt)}\right]} = \frac{8  ×  10^{−3}}{\ln (7.2/1)} = 4.05  ms