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Now I needed to find the transfer function as well as impulse response $$\\ R=14\Omega \\ L=2H \\ \\ C=\frac{1}{12}F \\ In \ s-domain \\ R=14\Omega \\ L=sL=s\cdot 2=2s \\ \\ C=\frac{1}{sC}=\frac{1}{s\cdot \frac{1}{12}}=\frac{12}{s} \\ H(s)=\frac{Y(s)}{X(s)}=\frac{L+C}{R+L+C}=\frac{2s+\frac{12}{s}}{14+2s+\frac{12}{2}}=\frac{2s^{2}+12}{2s^{2}+14s+12} \\ L^{-1}[\frac{2s^{2}+12}{2s^{2}+14s+12}]=\delta (t)+\frac{7}{5}e^{-t}-\frac{42}{5}e^{-6t}$$

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  • \$\begingroup\$ Hi Stormy, homework questions require an attempt to solve or they will be closed. \$\endgroup\$
    – Drew
    Commented Sep 4, 2020 at 21:23
  • \$\begingroup\$ I think you should have simplified your transfer function. Both numerator and denominator are obviously divisible evenly by 2. Not sure about your inverse, off hand. \$\endgroup\$
    – jonk
    Commented Sep 5, 2020 at 4:11
  • \$\begingroup\$ Yes, the final answer is correct, but \$\frac{Y(s)}{X(s)}=\frac {L+C}{R+L+C}\$ Is wrong. \$\endgroup\$
    – Chu
    Commented Sep 5, 2020 at 5:58
  • \$\begingroup\$ what do you mean @Chu? Is the representation wrong or the transfer function itself \$\endgroup\$
    – stormy2020
    Commented Sep 5, 2020 at 11:16
  • \$\begingroup\$ \$(L+C)/(L+C+R)\$ is meaningless. \$\endgroup\$
    – Chu
    Commented Sep 5, 2020 at 15:16

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