a voltage signal, V(t)=6*sin(8*pi*t+0.81)+3*cos(10*pi*t+0.4)+sin(12*pi*t+0.7), what are the amplitudes of the three waves?

Question

a voltage signal, V(t)=6*sin(8*pi*t+0.81)+3*cos(10*pi*t+0.4) +sin(12*pi*t+0.7), is composed of three waves; what are the amplitudes?

Answer 1

The amplitudes, mathematically determined, are the maxima and minima of the V function.

To find these we need the derivative of V:

dV/dt=48πcos(8πt+0.81)-30πsin(10πt+0.4)+12πtcos(12πt+0.7)=0 at extrema.

When this is solved for time t we get (approximately):

t=0.014364, 0.123466, 0.238676, 0.434563, 0.546765, 0.663799, 0.785952, 0.903143.

These values of t were derived using Newton's Method.

These correspond to voltage V:

V=8.450242, -6.244230, 2.606930, -5.172447, 7.001404, -7.525668, 8.051508, -8.815003.

These voltages were calculated by plugging t into V(t).

So we have 4 peaks and 4 troughs. To reduce these to three amplitudes we have:

8.4502, -7.5257, -8.8150 (approx).

The individual amplitudes for the three components of the voltage are 6, 3 and 1.