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On Page 30 in the Art Of Electronics (2nd Edition).

The two images below describe the conservation of Sinusoidal signals to a complex representation and back. I am perfectly ok with the first image. The next image has me stumped.

In particular, I am stumped at the description of \$V(t)\$ and \$I(t)\$

For \$V(t)\$, I understand why they say, \$V(t) = \mathcal{Re}(\mathbf{V}e^{jwt})\$.

I do not understand the next line of the simplification, \$V(t) = \mathcal{Re}(\mathbf{V})\operatorname{cos}(wt) - \mathcal{Im}(\mathbf{V})\operatorname{sin}(wt)\$.

\$e^{jwt}\$ should simplify to \$\operatorname{cos}(wt) + j\operatorname{sin}(wt)\$.

I don't understand how \$\mathbf{V}\$ is distributed to the second equation. I guess \$I(t)\$ would follow similarly.

Can someone help with the factoring?


1 Answer 1


I'll just zero in on the last bit.

The authors are using bold, capitalized V and I to indicate that they are complex values. Not merely real values. That's all it is.

So, assume \$\mathbf{V}=5 j\$. Then:

$$ \begin{align*} V\left(t\right)&=\operatorname{Re}\left(\mathbf{V}\cdot e^{j \omega t}\right) \\ &= \operatorname{Re}\left(\mathbf{V}\cdot \left[\operatorname{cos}\left(\omega t\right)+i\operatorname{sin}\left(\omega t\right)\right]\right) \\ &= \operatorname{Re}\left(5 i\cdot \left[\operatorname{cos}\left(\omega t\right)+i\operatorname{sin}\left(\omega t\right)\right]\right) \\ &= \operatorname{Re}\left( 5 i\cdot\operatorname{cos}\left(\omega t\right)+5 i\cdot i\operatorname{sin}\left(\omega t\right)\right) \\ &=\operatorname{Re}\left( 5 i\cdot\operatorname{cos}\left(\omega t\right)-5 \operatorname{sin}\left(\omega t\right)\right) \\ &= -5 \operatorname{sin}\left(\omega t\right) \end{align*} $$

But perhaps \$\mathbf{V}=3-4 j\$? This just provides phase information where the polar magnitude is \$5\:\textrm{V}\$. But the result is arrived at in the same fashion as above. It's just that you now have sine and cosine terms mixed into the result.


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