## Chapter 5

## Q. 5.AE.7

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.

## Step-by-Step

## Verified Solution

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} = 2πf_{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 5–24. 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 = 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:

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

a = V_{A} cosϕ = −7.65 V

b = −V_{A} 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