In modern electrodynamics by Andrew Zangwill chapter 14, section 14.13.2 an analysis of RLC circuit is shown where Fourier transform of current, EMF, and impedance is used. And equation is $\hat{E}(\omega)=\hat{Z}(\omega)\hat{I}(\omega)$. Then it says in equation 14.148 that the physical current driven by a real electromotive force $\hat{E}(\omega)\cos(\omega t)$ is $$I(t)=\Re[\hat{I}(\omega)\exp(-i\omega t)]=\frac{\hat{E}(\omega)}{\sqrt{R^2+(1/\omega C-\omega L)^2}}\cos(\omega t+\phi)$$ I get that the middle implies the right side but I couldn't figure out why this equal to $I(t)$ which we should get by inverse Fourier transform from $\hat{I}(\omega)$. Any help from anyone is much appreciated.


1 Answer 1


Go through Phasors.

Phasors are used to represent a time-domain signal $f(t)$ as a frequency domain signal $\hat{f}(\omega)$ so as to solve the integro-differential equations quickly.

We have,

$\exp[\pm i(\omega t+ \phi)] = \cos(\omega t + \phi) \pm i\sin(\omega t + \phi). $


$\cos(\omega t + \phi) = \mathfrak{Re}\{\exp[\pm i(\omega t+ \phi)]\}$
$\sin(\omega t + \phi) = \pm\ \mathfrak{Im}\{\exp[\pm i(\omega t+ \phi)]\}$

If we have current, $I(t) = I\cos(\omega t+\phi)$ (where $I =$ amplitude of $I(t) $), we can write

$I(t) = \mathfrak{Re}\{I\exp[\pm i(\omega t+ \phi)]\}$

or, $I(t) = \mathfrak{Re}\{I\exp[\pm i(\omega t+ \phi)]\} = \mathfrak{Re}\{\hat{I}\exp(\pm i\omega t)\}$ where $\hat{I} = I\exp(\pm i\phi)$

Here as cosine function is even, you can consider either the $+$ or $-$ sign.

  • $\begingroup$ Thank you for your answer. I think I need to learn about Phasors real quick. $\endgroup$ Commented Oct 17, 2020 at 9:46
  • 1
    $\begingroup$ You're welcome! I missed $i$ in the first equation, edited now. $\endgroup$
    – 19aksh
    Commented Oct 17, 2020 at 9:49

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