Question 5.7: Figure 5–23 shows an oscilloscope display of a sinusoid. The......
Figure 5–23 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 5–23.
The maximum amplitude of the waveform is seen to be four vertical divisions; therefore,
V_{A} = (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
T_{0} = (8 div)(0.1 ms/div) = 0.8 ms
The two frequency parameters are f_{0} = 1/T_{0} = 1.25 kHz and ω_{0} = 2\pi f_{0} = 7854 rad/s. The parameters V_{A}, T_{0}, f_{0}, and ω_{0} do not depend on the location of the t = 0 axis.
To determine the time shift T_{S}, we need to define a time origin. The t = 0 axis is arbitrarily taken at the left edge of the display in Figure 5–23. 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 5–23 because it must lie beyond the left edge of the display. However, the positive peak shown in the display is located at t = T_{S} + T_{0} since it is one cycle after t = T_{S}. We can write
T_{S} + T_{0} = (5.5 div)(0.1 ms/div) = 0.55 ms
which yields T_{S} = 0.55 – T_{0} = -0.25 ms. As expected, T_{S} is negative because the nearest positive peak is to the left of t = 0.
Given T_{S}, we can calculate the remaining parameters of the sinusoid as follows:
\phi = -\frac{2\pi T_{S}}{T_{0}}= 1.96 rad or 112.5°
a = V_{A} \cos \phi =-7.65 V
b = -V_{A} \sin \phi -18.5 V
Finally, the three alternative expressions for the displayed sinusoid are
v(t) = 20 \cos [(7854 t) + 0.25 × 10^{-3}] V
= 20 \cos (7854 t + 112.5°) V
= -7.65 \cos 7854t – 18.5 \sin 7854t V