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I want to make equation that describe relation between vi and vc. I did it like this. Is this right way?

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2 Answers 2

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EE&O ...
To help a little (check results of differential equations), here are pictures of the inductor current and the voltage across the capacitor ...
Values : L:= 1e-3 H: C:= 1e-6 F: r1:= 1 ohm : r2:= 100 Ohm: Vin:=step 1 V.

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Well, using Laplace transform we can see that:

$$\mathscr{H}\left(\text{s}\right):=\frac{\text{V}_\text{out}\left(\text{s}\right)}{\text{V}_\text{in}\left(\text{s}\right)}=\frac{\text{R}_2\space\text{||}\space\frac{1}{\text{s}}}{\text{R}_1+\text{sL}+\left(\text{R}_2\space\text{||}\space\frac{1}{\text{s}}\right)}\tag1$$

Where \$\alpha\space\text{||}\space\beta:=\frac{\alpha\beta}{\alpha+\beta}\$.

So, when we simplify we can write:

$$\text{R}_2\space\text{||}\space\frac{1}{\text{s}}=\frac{\text{R}_2\cdot\frac{1}{\text{s}}}{\text{R}_2+\frac{1}{\text{s}}}=\frac{\text{s}}{\text{s}}\cdot\frac{\text{R}_2\cdot\frac{1}{\text{s}}}{\text{R}_2+\frac{1}{\text{s}}}=\frac{\text{R}_2\cdot\frac{\text{s}}{\text{s}}}{\text{sR}_2+\frac{\text{s}}{\text{s}}}=\frac{\text{R}_2}{1+\text{sR}_2}\tag2$$

So, we get:

$$\mathscr{H}\left(\text{s}\right)=\frac{\text{R}_2}{1+\text{sR}_2}\cdot\frac{1}{\text{R}_1+\text{sL}+\frac{\text{R}_2}{1+\text{sR}_2}}=\frac{\text{R}_2}{\text{R}_2+\left(\text{R}_1+\text{sL}\right)\left(1+\text{sR}_2\right)}\tag3$$

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  • \$\begingroup\$ thx for fast answer but i don't know Laplace transform... what is that jw thing?? \$\endgroup\$
    – moonjy1120
    May 27 at 10:11
  • \$\begingroup\$ j is i, sqrt(-1). It's an engineer thing. w is lowercase omega, angular frequency, f/2pi, basically f but more convenient to avoid pis popping up everywhere. Laplace transform is a way of solving many differential equ'ns by transforming your functions into others where doing differentiation maps to a simple algerbraic operation . A tool a bit like the way it makes more sense to think of response in terms of frequency than time-axis response in AC circuits (which is effectively looking at the Fourier transform). Doing the similar but different Laplace is useful for analysing filters. \$\endgroup\$
    – Dannie
    May 27 at 10:39
  • \$\begingroup\$ thanks for nice explanation @Dannie, in this circuit the input is not AC but DC square wave.... is it apply to both ac and dc? \$\endgroup\$
    – moonjy1120
    May 27 at 11:26
  • \$\begingroup\$ "what is that jw thing??" Somewhat out of place here as this is a Laplace domain expression. It should be s, the complex frequency variable used in place of f or ω in Laplace expressions, so should read \$\frac{1}{sC}\$ \$\endgroup\$
    – Graham Nye
    May 27 at 14:27

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