2
\$\begingroup\$

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\$.

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
  • \$\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
1
\$\begingroup\$

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, $$ \lim _{s\to0}sH(s)=\lim _{s\to0}(s.\frac{V_1}{s}.\frac{5}{\frac{100}{10s+10}+5} )$$ $$=\frac{V_1}{3}$$

|improve this answer|||||
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.