Question

# One of your commonly-used laboratory instantaneous test signal voltages (vs) is described by the equation... π...

One of your commonly-used laboratory instantaneous test signal voltages (vs) is described by the equation...
π vs=6sin2πft 4 −
where f = 1MHz and t represents time.
Make time (t) the subject of this formula, and hence determine the first point in time when the instantaneous signal voltage has a magnitude of +3V.
Note: A colleague has reminded you that you need to have your calculator in radians mode (RAD) for this calculation, because the angle is given in radians (i.e. π is featured).
Use suitable software to draw at least two cycles of this signal, and annotate the drawing so that non-technical colleagues may understand it.

Vs = 6*sin(2*pi*f*t - 4*pi)

Vs=3=6*sin(2*pi*f*t - 4*pi)

f=1MHz

pi=3.14

2*pi*f*t - 4*pi = sin-1(0.5) =30degree*3.14/180 = 0.5236

6.28*10^6*t = 0.5236 + 12.56 = 13.08

t = 13.08/(6.28*10^6) = 2.08*10^-6 seconds = 2.08 us.

The amplitude of Vs = 6V when tmax = 14.13/(6.28*10^6) = 2.25 us.

The amplitude of Vs is 1V when t1= 12.727/(6.28*10^6) = 2.026 us.

the amplitude is 0V when t0 = 12.56/(6.28*10^6) = 2 us.

Amplitude when Vs = -6V is t = 1.75 us

these values can be used to plot the signal in excel.

See that 2us - 1.75us = 0.25us and also 2.25 - 2 = 0.25 us

thats is the wave function is symmetric when voltage is positive and negative.

The wave is a symmetric sinusoidal function the period of the wave is the time between when voltage goes from -6V to 6V and that is 2*0.25 = 0.5 us.

The amplitude peak to peak is 6+6 = 12V.

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