C1 has an impedance of 1/0.1s Ohm in S-domain. Its impedance together with R1's is 100/(10s+10) Ohm

The transfer function is: $$\frac{V_2}{V_1}=\frac{5}{\frac{100}{10s+10}+5}=\frac{50s+50}{50s+150}$$

Just looking at this transfer function, because the only pole is has a negative real part, in time domain, it's going to 0. But that is not correct according to the schematic. At time=infinity, C1 is open circuit, so \$V2=(1/3) * V1\$.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ The FVT is speaking of \$sF(s)\$, not just \$F(s)\$. \$\endgroup\$ – Eugene Sh. Sep 23 '19 at 14:21
  • \$\begingroup\$ How do you model a DC voltage source in the Laplace domain? \$\endgroup\$ – TimWescott Sep 23 '19 at 16:48

Here is the FVT in Laplace domain \$-\$ $$\lim _{t\to\infty} H(t) = \lim _{s\to0}sH(s)$$ Transfer function is \$-\$ $$H(s)=V_2(s) = V_1(s).\frac{5}{\frac{100}{10s+10}+5} $$ \$V_1\$ is a DC voltage source. $$\implies V_2(s)=\frac{V_1}{s}.\frac{5}{\frac{100}{10s+10}+5}$$ @time = infinity, $$ \lim _{s\to0}sH(s)=\lim _{s\to0}(s.\frac{V_1}{s}.\frac{5}{\frac{100}{10s+10}+5} )$$ $$=\frac{V_1}{3}$$

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