Electronic – Why can’t phasors be equated to a sinusoid

phasorsine

I'm having trouble understanding why a phasor for an AC sinusoid, let's say voltage (represented as Vrms/_50° for Vsin(ωt + 50°) to a current, for example) cannot be equated to the sinusoid in question. Sources such as my textbook say that it is wrong to do so, as the phasor is a complex constant, but should it still not represent the sinusoid?

Best Answer

Phasors are just a convenient way of doing certain math with sinusoidal signals.

Basically, phasors are a good way to do bookkeeping on sinusoidal signals. Let's approach this like so:

First of all, note that adding two sinusoids isn't trivial:

$$ A_1 \cos(\omega t + \phi_1) + A_2\cos(\omega t+ \phi_2) = ? $$

One nice way to do it is expressing a sinusoid as a linear combination of two basis vectors: $$ A \cos(\omega t + \phi) = I \cos(\omega t) + Q\sin(\omega t) \\ A = \sqrt {I^2 + Q^2} \\ \tan \phi = \frac I Q $$

When you do that, you can just add the I's from one sinusoid and Q's from another sinusoid to get your resultant sinusoids.

If you go another step further, you could just come up with this brilliant idea:

Let's just forget about \$A \cos(\omega t + \phi)\$ and just use \$I\$ and \$Q\$ from now on, as a representation of the original signal.

That's basically what a phasor is: the phasor is just \$I + jQ\$. You're just "bookkeeping" by using a phasor.

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