Electronic – Phasors: why is the book using sines

phasor

When I learned phasors I learned we passed from sinusoidal to phasor this way:

$$\sqrt{2} V\cos(\omega t+\phi) \rightarrow \sqrt{2} Ve^{j\phi}$$

However I'm now studying for a book that does this:

$$\sqrt{2} V\sin(\omega t+\phi) \rightarrow Ve^{j\phi}$$

Is this some kind of notation that I am not familiar with? Can I accept this as correct? I«m facing this notation studying three-phase systems, while the first notation was used in the context of circuit analysis.

Thank you!

Best Answer

I also saw the first definition in my education:

$$v(t) = \mathcal{R}\left\{\underline{V}\cdot e^{j\omega t}\right\} = \mathcal{R}\left\{(V\cdot e^{j\phi})\cdot e^{j\omega t}\right\} = V\cdot\cos(\omega t+\phi)$$

But as long as you are consistent with your formulation, both will lead to the same results when analyzing problems. The phasor amplitude will now represent the RMS value of the sinusoidal signal, and the phase will be relative to a sine wave (90 degrees shifted compared to the cosine).

[EDIT] If you're also interested in how exactly they're linked:

Both definitions are still strongly linked. Using the second definition is basically identical to

$$\begin{align} v(t) &= \mathcal{I}\left\{\underline{V}'\cdot \sqrt{2} e^{j\omega t}\right\}\\ &= \mathcal{I}\left\{(V'\cdot e^{j\phi'})\cdot \sqrt{2} e^{j\omega t}\right\}\\ &=\sqrt{2}V'\cdot \sin(\omega t + \phi')\\ &= (\sqrt{2}V')\cdot \cos\left(\omega t + \left(\phi' - \frac{\pi}{2}\right)\right) \end{align}$$

And you can find that

$$\begin{align} V &= \sqrt{2}V'\\ \phi &= \phi' - \frac{\pi}{2}\\ \underline{V} &= Ve^{j\phi} = (\sqrt{2}V')e^{j(\phi'-\frac{\pi}{2})} = -j\sqrt{2}\cdot \underline{V'} \end{align}$$

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