## Q. 5.AE.6

An oscilloscope is a laboratory instrument that displays the instantaneous value of a waveform versus time. Figure 5–19 shows an oscilloscope display of a portion of an exponential waveform. In the figure, the vertical (amplitude) axis is calibrated at 2 V per division, and the horizontal (time) axis is calibrated at 1 ms per division. Find the time constant of the exponential. ## Verified Solution

For t > TS 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 VA or the time shift TS 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 TC 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

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