In the KD9PDP's answer in the post below why can't you calculate the transfer function in time by taking the inverse Laplace transform of the transfer function in Laplace domain?
The original post: https://electronics.stackexchange.com/q/551541
$$
\frac{1}{1+sRC}V_i(s)=V_o(s)\\
\frac{V_o(s)}{V_i(s)} = \frac{1}{1+sRC}\\
\frac{V_o(t)}{V_i(t)} = \frac{e^{\frac{-t}{RC}}}{RC}
$$
\$\begingroup\$
\$\endgroup\$
Add a comment
|
2 Answers
\$\begingroup\$
\$\endgroup\$
Because the Laplace transform of \$V_o(t)/V_i(t)\$ is not the ratio of the Laplace transforms of \$V_o(t)\$ and \$V_i(t)\$.
A little deeper dive...
Everything is correct up to this point:
$$ \frac{V_o(s)}{V_i(s)} = \frac{1}{1+sRC} = \mathcal{L}\left( \frac{e^{\frac{-t}{RC}}}{RC} \right ) $$
and taking \$\mathcal{L}^{-1}\$ of both sides yields:
$$ \mathcal{L}^{-1}\left( \frac{V_o(s)}{V_i(s)} \right ) = \frac{e^{\frac{-t}{RC}}}{RC} $$
but now you're stuck because you can't express the lefthand side in terms of \$\mathcal{L}^{-1}V_o\$ and \$\mathcal{L}^{-1}V_i\$
\$\begingroup\$
\$\endgroup\$
Well, a general transfer function is given by:
$$\mathscr{H}\left(\text{s}\right):=\frac{\text{v}_\text{o}\left(\text{s}\right)}{\text{v}_\text{i}\left(\text{s}\right)}=\frac{\text{k}}{\text{n}+\text{ms}}\tag1$$
This implies that:
$$\text{v}_\text{o}\left(\text{s}\right)=\frac{\text{k}}{\text{n}+\text{ms}}\cdot\text{v}_\text{i}\left(\text{s}\right)\tag2$$
Using the convolution property of the Laplace transform, we can write:
$$\text{V}_\text{o}\left(t\right)=\int_0^t\mathscr{L}_\text{s}^{-1}\left[\frac{\text{k}}{\text{n}+\text{ms}}\right]_{\left(t-\tau\right)}\cdot\mathscr{L}_\text{s}^{-1}\left[\text{v}_\text{i}\left(\text{s}\right)\right]_{\left(\tau\right)}\space\text{d}\tau=$$ $$\int_0^t\frac{\text{k}\exp\left(\frac{\text{n}\left(\tau-t\right)}{\text{m}}\right)}{\text{m}}\cdot\text{V}_\text{i}\left(\tau\right)\space\text{d}\tau=\frac{\text{k}}{\text{m}}\int_0^t\exp\left(\frac{\text{n}\left(\tau-t\right)}{\text{m}}\right)\cdot\text{V}_\text{i}\left(\tau\right)\space\text{d}\tau\tag3$$
