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$$ \frac{V_c}{V_s} = \frac{\frac{j}{\omega\cdot C}}{R+\frac{j}{\omega \cdot C}} $$

Given this starting formula, how to rationalize the denominator and find the gain?

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  • \$\begingroup\$ Multiply top and bottom by wC then multiply top and bottom by the complex conjugate of the denominator. That’s how you start to solve the problem. Does that make sense to you? \$\endgroup\$ – Andy aka Feb 23 '20 at 9:47
  • \$\begingroup\$ Moi, Are you sure you got that equation right? What circuit is that supposed to represent, specifically? It almost looks like an RC lowpass. But not quite, since \$\frac{-j}{\omega\,C}\$ is one of the usual forms for capacitance. \$\frac{j}{\omega\,C}\$ would be an interesting capacitor. \$\endgroup\$ – jonk Feb 23 '20 at 23:30
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Well, we have:

$$\frac{\text{V}_\text{c}}{\text{V}_\text{s}}=\frac{\frac{\text{j}}{\omega\text{C}}}{\text{R}+\frac{\text{j}}{\omega\text{C}}}=\frac{\omega\text{C}\cdot\frac{\text{j}}{\omega\text{C}}}{\omega\text{C}\cdot\text{R}+\omega\text{C}\cdot\frac{\text{j}}{\omega\text{C}}}=\frac{\text{j}}{\omega\text{CR}+\text{j}}=$$ $$\frac{\omega\text{CR}-\text{j}}{\omega\text{CR}-\text{j}}\cdot\frac{\text{j}}{\omega\text{CR}+\text{j}}=\frac{\omega\text{CR}\text{j}-\text{j}\text{j}}{\left(\omega\text{CR}\right)^2+1^2}=\frac{\omega\text{CR}\text{j}-\text{j}^2}{\left(\omega\text{CR}\right)^2+1}=$$ $$\frac{\omega\text{CR}\text{j}+1}{\left(\omega\text{CR}\right)^2+1}=\frac{1+\omega\text{CR}\text{j}}{1+\left(\omega\text{CR}\right)^2}\tag1$$

So:

$$\left|\frac{\text{V}_\text{c}}{\text{V}_\text{s}}\right|=\left|\frac{1+\omega\text{CR}\text{j}}{1+\left(\omega\text{CR}\right)^2}\right|=\frac{\left|1+\omega\text{CR}\text{j}\right|}{\left|1+\left(\omega\text{CR}\right)^2\right|}=$$ $$\frac{\sqrt{1+\left(\omega\text{CR}\right)^2}}{1+\left(\omega\text{CR}\right)^2}=\frac{1}{\sqrt{1+\left(\omega\text{CR}\right)^2}}\tag2$$

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