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For the following question: $$\cos(\omega t)+2\cos\left(\omega t+\frac{\pi}{4}\right)+3\cos\left(\omega t+\frac{\pi}{2}\right)$$

We can add the three cosines using phasors / complex addition:

$$\begin{align} & 1\angle0^\circ + 2\angle45^\circ+3\angle90^\circ\\ =& 1+\sqrt2+j\sqrt2+3j\\ =& 1+\sqrt2 + j\left(3+\sqrt2\right) \end{align}$$

Which has

$$ A\approx5.03, \phi\approx1.07\text{rad}$$

Thus answer is $5\angle61^\circ$, or $5\cos(\omega t+1.07)$.

This is the correct answer, but what confuses me is that to use phasor form, don't we need both sine and cosine according to eulers identity? Why were we able to convert a cosine without a sine to a phasor and then convert it back to just a cosine without a sine? Eulers identity says that $Ae^{j\theta} = A\cos\theta - jA\sin\theta$

Thank you for anything you can provide :)

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  • $\begingroup$ I think I'm horribly missing a concept here. I tried reading through the section in the textbook but still don't get it. $\endgroup$ – user96193 Sep 22 '13 at 15:16
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Hint: Compare your original question, the real part of $($phasor$\times e^{j\omega t})$, and your answer.


More: Your original question is the real part of

$$\begin{align} e^{j\omega t} + 2e^{j\omega t+j\frac{\pi}{4}} + 3 e^{j\omega t+j\frac{\pi}{2}} =& e^{j\omega t}\left(1+2e^{j\frac{\pi}{4}}+3e^{j\frac{\pi}{2}}\right)\\ =& e^{j\omega t}\left(1+\sqrt2 + j\sqrt2+j3\right)\\ \approx& e^{j\omega t}\cdot5.03e^{j1.07}\\ =& 5.03 e^{j\omega t+j1.07}\\ \Re\left(e^{j\omega t} + 2e^{j\omega t+j\frac{\pi}{4}} + 3 e^{j\omega t+j\frac{\pi}{2}}\right) \approx& \Re\left(5.03 e^{j\omega t+j1.07}\right)\\ =& \Re\left[5.03 \cos(\omega t+1.07)+j5.03\sin(\omega t+1.07)\right]\\ =& 5.03 \cos(\omega t+1.07) \end{align}$$

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  • $\begingroup$ Thanks for the answer, but I'm still confused. The question only has cosines so everything in the question is a real number. However, the phasors at angle 45 and 90 have an imaginary component. The final answer is also not along the x axis so should it not have a sine component? $\endgroup$ – user96193 Sep 22 '13 at 16:08
  • $\begingroup$ OMG, you're awesome. Wow, that made it so simple to understand, I was banging my head on this for hours, woops. Thanks a bunch. $\endgroup$ – user96193 Sep 22 '13 at 16:12
  • $\begingroup$ Glad that you understand now. I have modified the hint a bit for a more accurate description. $\endgroup$ – peterwhy Sep 22 '13 at 16:30

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