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I was analyzing the following RLC circuit: RLC circuit

And applying the Fourier transform on the circuit, I ended up with the following transfer function:

$$ H(\omega)=\frac{V_c(\omega)}{V_i(\omega)}=\frac{1}{1+CR(j\omega)+LC(j\omega)^2} $$ Now I want to do the inverse Fourier transform in that function: $$ h(t)=F^{-1}\{H(\omega)\} \\ h(t)=f(R,L,C,t) $$ but I can't find any table that has an equation similar to this one. I also know that I can find a similar equation by doing partial fractions, but I can't find a way to solve this function by partial fractions.

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    \$\begingroup\$ Have you tried a table of Laplace/inverse Laplace transforms? \$\endgroup\$ Commented Dec 19, 2021 at 7:57
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    \$\begingroup\$ You need to find the Laplacian transfer function and not the Fourier TF. Then you need to do an inverse Laplace to get an answer in the time domain. You might also choose to apply an impulse or a step to the Laplace TF. Not knowing how to find something doesn't constitute a question on this site. \$\endgroup\$
    – Andy aka
    Commented Dec 19, 2021 at 11:09
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    \$\begingroup\$ Jjeico, I believe the choice as to whether or not to use partial fraction decomposition prior to applying the inverse Laplace transform methods will depend a bit on whether or not the system is underdamped or overdamped (or critically.) You could do use the approach Syed mentions to generate the partial fractions, easily enough. Each of which relates directly to the quadratic solution pair. But if the system is underdamped, directly solve with the higher order inverse. \$\endgroup\$
    – jonk
    Commented Dec 19, 2021 at 21:13

3 Answers 3

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I'm curious why didn't anybody mention partial fraction decomposition yet.

First of all, you have to keep in mind that \$\mathcal{F\{}\}\neq\mathcal{L\{}\}\$. The transfer function is defined as \$H(s):=\mathcal{L}\{h(t)\}=\mathcal{L}\{Y\{\delta(t)\}\}=\frac{\mathcal{L}\{y(t)\}}{\mathcal{L}\{u(t)\}}\$. What you have derived using the F-transform is another beast, I'm not sure about the correct English terminology of it, but it was called "transfer characteristic" in my (not English) course. You have \$H(j\omega)=\frac{\mathcal{F}\{y(t)\}}{\mathcal{F}\{u(t)\}}\$ In most well-behaved cases, the two are very similar, and could be used so, if you do a \$j\omega\Leftrightarrow s\$ substitution - but that only applies if the system is casual (so \$h(t<0)\equiv0\$), and the system is BIBO-stable (\$\int{h(t)dt}<\infty\$). Even then, you have to remember that \$\mathcal{F}\$ is used in periodic steady state analysis, and \$\mathcal{L}\$ is used with casual systems and if \$u(t<0)\equiv0\$ (again I can't find the correct English terminology, it was called "entering" signal on my course).

So all of that in mind, since your system can be considered as BIBO-stable and casual (BIBO-stable because it's only built from passives), you can swap \$H(j\omega)\$ for \$H(s)\$.

Your task is to find the impulse response. That can be done by inverse Laplace transforming \$H(s)\$. Your transfer function is not something you could inverse transform by hand (but tools like matlab, octave or mathematica can do it for you anyway), so in practical problems we usually do a partial fraction decomposition (waaay easier when we have numbers for \$R\$,\$L\$ or \$C\$, but still possible). For that, you could use the Heaviside cover-up method, but not necessary (and again, it works best when you have numbers).

From your transfer function, you'd get two fractional parts: \$\frac{A}{s-B}+\frac{A^*}{s-B^*}\$. From the table of \$\mathcal{L}\$, you can now inverse-transform them to be \$\varepsilon(t)(Ae^{(Bt)}+A^*e^{(B^* t)})\$. You can leave that in that form, but usually it's better to write \$A=|A|e^{j\phi}\$ and then separating the real and imaginary parts of \$B\$ to bring the result into a form of an entering damped sinusoid signal.

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  • \$\begingroup\$ I really like all the packing of information you provided. You could have written more in order to help unpack some of what you said. But +1 anyway. It covers the essentials. \$\endgroup\$
    – jonk
    Commented Dec 19, 2021 at 21:07
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\$\mathcal{L^{-1}}\left(\frac{k}{a s^2+b s+c}\right)\$ would give:

$$-\frac{k \left(e^{t \left(-\frac{\sqrt{b^2-4 a c}}{2 a}-\frac{b}{2 a}\right)}-e^{t \left(\frac{\sqrt{b^2-4 a c}}{2 a}-\frac{b}{2 a}\right)}\right)}{\sqrt{b^2-4 a c}}$$

You can spot the familiar components of the quadratic formula.


Using Mathematica the code would be,

InverseLaplaceTransform[k/( a s^2 + b s + c), s, t]
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Well, the transfer function of the circuit is given by:

$$\mathscr{H}\left(\text{s}\right):=\frac{\text{V}_\text{o}\left(\text{s}\right)}{\text{V}_\text{i}\left(\text{s}\right)}=\frac{\frac{1}{\text{sC}}}{\frac{1}{\text{sC}}+\text{sL}+\text{R}}=\frac{1}{\text{CLs}^2+\text{CRs}+1}\tag1$$

Where I used the well-known Laplace transform.

The impulse response of a system is given by the output response when a the input function is a dirac delta function, so:

$$\text{V}_\text{i}\left(\text{s}\right)=\mathscr{L}_t\left[\delta\left(t\right)\right]_{\left(\text{s}\right)}=1\tag2$$

The Laplace transform can be found using the definition of the Laplace transform or by examining a table of selected Laplace transforms.

Now, the output time response of this given by:

\begin{equation} \begin{split} \text{v}_\text{o}\left(t\right)&=\mathscr{L}_\text{s}^{-1}\left[\underbrace{\text{V}_\text{i}\left(\text{s}\right)}_{=\space1}\cdot\frac{1}{\text{CLs}^2+\text{CRs}+1}\right]_{\left(t\right)}\\ \\ &=\mathscr{L}_\text{s}^{-1}\left[\frac{1}{\text{CLs}^2+\text{CRs}+1}\right]_{\left(t\right)}\\ \\ &=\frac{2\exp\left(-\frac{\text{R}t}{2\text{L}}\right)\sinh\left(\frac{t\sqrt{\text{CR}^2-4\text{L}}}{2\text{L}\sqrt{\text{C}}}\right)}{\sqrt{\text{C}\left(\text{CR}^2-4\text{L}\right)}} \end{split}\tag3 \end{equation}


We can find the general output by using the convolution property of the Laplace transform:

\begin{equation} \begin{split} \text{v}_\text{o}\left(t\right)&=\mathscr{L}_\text{s}^{-1}\left[\text{V}_\text{i}\left(\text{s}\right)\cdot\frac{1}{\text{CLs}^2+\text{CRs}+1}\right]_{\left(t\right)}\\ \\ &=\int_0^t\mathscr{L}_\text{s}^{-1}\left[\text{V}_\text{i}\left(\text{s}\right)\right]_{\left(\tau\right)}\cdot\mathscr{L}_\text{s}^{-1}\left[\frac{1}{\text{CLs}^2+\text{CRs}+1}\right]_{\left(t-\tau\right)}\space\text{d}\tau\\ \\ &=\int_0^t\text{v}_\text{i}\left(\tau\right)\cdot\left(\frac{2\exp\left(-\frac{\text{R}\left(t-\tau\right)}{2\text{L}}\right)\sinh\left(\frac{\left(t-\tau\right)\sqrt{\text{CR}^2-4\text{L}}}{2\text{L}\sqrt{\text{C}}}\right)}{\sqrt{\text{C}\left(\text{CR}^2-4\text{L}\right)}}\right)\space\text{d}\tau\\ \\ &=\frac{2}{\sqrt{\text{C}\left(\text{CR}^2-4\text{L}\right)}}\int_0^t\text{v}_\text{i}\left(\tau\right)\exp\left(-\frac{\text{R}\left(t-\tau\right)}{2\text{L}}\right)\sinh\left(\frac{\left(t-\tau\right)\sqrt{\text{CR}^2-4\text{L}}}{2\text{L}\sqrt{\text{C}}}\right)\space\text{d}\tau \end{split}\tag4 \end{equation}

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