# How do I replace a capacitor with a resistance?

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 $\frac{1}{j\omega100\cdot 10^{-9}}$?

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

• 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) – JRE Sep 16 '15 at 15:41
• @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? – B. Lee Sep 16 '15 at 16:03

## 1 Answer

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.