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I'm studying Fourier transformations, and their relationship with electrical circuits. In the example below the capacitor is replaced by a resistance, in that way we can use the voltage-divider principle to find an expression for \$Y(j\omega t)\$, in terms of \$X(j\omega t)\$.

This is where I'm stuck, as the author leaves no reasonable explanation as to how the capacitor was turned into a resistance. Can someone provide an explanation as to how it was replaced by the resistance \$\frac{1}{j\omega100\cdot 10^{-9}}\$?

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I should add that I know about the relationship \$i(t) = C\frac{dv(t)}{dt}\$

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    \$\begingroup\$ He replaced the capacitor with a resistor whose value he determined from the definition of impedance. Wikipedia shows how to get from your last equation to the resistor value. (en.wikipedia.org/wiki/Electrical_impedance#Capacitor) \$\endgroup\$ – JRE Sep 16 '15 at 15:41
  • \$\begingroup\$ @JRE I'm a little confused about the jump from \$\frac{\sin{\omega t}}{\omega C \sin{(\omega t + \pi/2)}}\$ to \$-j\frac{1}{\omega C}\$. I understand the sinus functions gets replaced by \$-j\$. I don't see how, as that would mean \$\tan{\omega t} = -j\$, right? \$\endgroup\$ – B. Lee Sep 16 '15 at 16:03
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Okay, you already know that an ideal capacitor is defined in the time domain by the equation

$$i(t) = C\frac{\mathrm{d}v(t)}{\mathrm{d}t}$$

Now you should also know (or at least vaguely remember being taught once and be able to look up) the Fourier transform rule

$$\mathcal{F}\left[\frac{\mathrm{d}}{\mathrm{d}t}x(t)\right]=j\omega{}X(\omega)$$ when \$X(\omega)\$ is the transform of \$x(t)\$.

So we can rewrite the capacitor equation in the frequency domain as

$$I(\omega)=Cj\omega{}V(\omega)$$

or

$$\frac{V(\omega)}{I(\omega)}=\frac{1}{j\omega{}C}$$

which is the definition of the impedance of an ideal capacitor, and just what was substituted for the capacitor in your example.

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