# Question 5.AE.7: Figure 5–24 shows an oscilloscope display of a sinusoid. The......

Figure 5–24 shows an oscilloscope display of a sinusoid. The vertical axis (amplitude) is calibrated at 5 V per division, and the horizontal axis (time) is calibrated at 0.1 ms per division. Derive an expression for the sinusoid displayed in Figure 5–24.

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The maximum amplitude of the waveform is seen to be four vertical divisions; therefore,

VA = (4 div)(5 V/div) = 20 V

There are four horizontal divisions between successive zero crossings, which means there are a total of eight divisions in one cycle. The period of the waveform is

T0 = (8 div)(0.1 ms/div) = 0.8 ms

The two frequency parameters are f0 = 1/T0 = 1.25 kHz and ω0 = 2πf0 = 7854 rad/s. The parameters VA, T0, f0, and ω0 do not depend on the location of the t = 0 axis.
To determine the time shift TS, we need to define a time origin. The t = 0 axis is arbitrarily taken at the left edge of the display in Figure 5–24. The positive peak shown in the display is 5.5 divisions to the right of t = 0, which is more than half a cycle (four divisions). The positive peak closest to t = 0 is not shown in Figure 5–24 because it must lie beyond the left edge of the display. However, the positive peak shown in the display is located at t = TS + T0 since it is one cycle after t = TS. We can write

TS + T0 = (5.5 div)(0.1 ms/div) = 0.55 ms

which yields TS = 0.55 − T0 = −0.25 ms. As expected, TS is negative because the nearest positive peak is to the left of t = 0.
Given TS, we can calculate the remaining parameters of the sinusoid as follows:

$ϕ = -\frac{2πT_{S}}{T_{0}} = 1.96 rad or 112.5°$

a = VA cosϕ = −7.65 V

b = −VA sinϕ = −18.5 V

Finally, the three alternative expressions for the displayed sinusoid are

υ(t) = 20 cos [(7854t) + 0.25 × 10-3] V

= 20 cos(7854t + 112.5°) V

= −7.65 cos7854t − 18.5 sin7854t V

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