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Im breaking my head over the solution of an old exam problem: Given is the following hertzian dipole:

Dipole

Given are also the equations: $I(t) = \Re\left(I_0 e^{j\omega t}\right)$, $Q(t) = \Re\left(\underline{Q} e^{j\omega t}\right)$ and $\frac{dQ}{dt} = I(t)$

The task is now "Calculate $\underline{Q}$"

The (very sparse) solution is as follows: $$ \begin{align} \frac{dQ}{dt} &= I(t) \tag 1 \\ \Re\left(j \omega \underline{Q} e^{j\omega t}\right) &= \Re\left(I_0 e^{j\omega t}\right) \tag 2\\ j \omega \underline{Q} &= I_0 \tag 3\\ \underline{Q} &= -j \frac{I_0}{\omega} \tag 4 \end{align} $$

Until step 2 everything is clear. But why can I leave away the $\Re()$ in step 3?

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  • $\begingroup$ I also find this a bit puzzling. I would start by expanding the complex exponential. $$j \omega e^{j \omega t} = j \omega (cos (\omega t) + j sin (\omega t) )$$ Then $$Re(j \omega e^{j \omega t}) = Re(j \omega cos (\omega t) - \omega sin (\omega t)) = - \omega sin(\omega t)$$ By the same treatment the RHS would be $I_{0} cos(\omega t)$. My only thought is that maybe $Re(j \omega e^{j \omega t})$ means $j \omega Re(e^{j \omega t})$ and it's just a notation thing. $\endgroup$ – Kyle Arean-Raines Jun 18 '15 at 15:35
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My buest guess: $\underline{Q}$, $j$, $\omega $ and $I_0$ are real constants. Then it would be trivial. But it's hard to tell in the current version of the question since not all symbols are defined.

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  • $\begingroup$ Okay let me define it a bit better: $j=\sqrt{-1}$ is the complex unit. $\omega$, $I_0$ and $t$ are real constants. $\underline{Q}$ is a complex number. $Q$ (not underlined) however is a real number (underlined = complex, not underlined=real) $\endgroup$ – Stefan D. Jun 18 '15 at 15:21
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Okay, I think I got it:

(In the following I'm using the short notation $Q_0 = \left|\underline{Q}\right|$ and $\phi = \arg{\underline{Q}}$, so that $\underline{Q}=Q_0 e^{j\phi }$)

$$ \begin{align} \Re{(I_0 e^{j \omega t})} &= \frac{d}{dt} Q(t) = \frac{d}{dt} \Re{(\underline{Q} e^{j \omega t})} = \frac{d}{dt} \Re{(Q_0 e^{j (\omega t + \phi)})} \\ I_0 \cos(\omega t) &= Q_0 \frac{d}{dt} \cos( \omega t + \phi) = -\omega Q_0 \sin( \omega t + \phi) = -\omega Q_0 \cos( \omega t + \phi -\pi/2)\\ I_0 \frac{e^{j \omega t}+e^{-j \omega t}}{2} &= -\omega Q_0 \frac{e^{j (\omega t +\phi -\pi/2)}+e^{-j (\omega t +\phi -\pi/2)}}{2} = -\omega Q_0 e^{\phi} e^{-j\pi/2} \frac{e^{j \omega t}+e^{-j \omega t}}{2} \\ I_0 = j\omega \underline{Q}\\ \underline{Q} = \frac{I_0}{j\omega} \end{align} $$

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