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

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




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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