Find the Transfer function of the given circuit in the frequency domain.The problem here is that the equations of this circuit become complicated by finding equal Z of R,C,L (right side of the circuit) in frequency domain which is ((Ls+R)/(LCs^2+RCs+1)). The next step is to apply voltage divider for Z and R which leads to Vc/VS.(as my friends mentioned in the answer part),then another voltage divider between R and L (Right side of circuit) which finally result in T.S.

firstly i am not sure that this approach is correct.

secondly ,if i calculate H(s) how should i use the given information (relation between Omega and other components) to get Vo(t). The given problem

  • \$\begingroup\$ Yes, they do become complicated so where are you stuck? \$\endgroup\$ – Andy aka Jul 2 at 8:32
  • \$\begingroup\$ why are they hard? It's hard to help you if we don't know what you're stuck with. To me, this looks like a pretty trivial linear network that with a single transformation can be simplified to a voltage divider, and then it's just plugging in numbers... \$\endgroup\$ – Marcus Müller Jul 2 at 8:33
  • \$\begingroup\$ Oh wait, this is a copy of electronics.stackexchange.com/questions/506234/… . Don't do that. \$\endgroup\$ – Marcus Müller Jul 2 at 8:33
  • \$\begingroup\$ In the process of simplifying the capacitor, inductor and resistor to \$Z\$, you have lost the node where the output was originally measured. \$V_0\$ was originally measured across the right most resistor. in the simplified figure, \$V_0\$ is effectively measured across the inductor-resistor combination. (You have noted it in your edited question). I think you shouldn't name the result you got as \$V_0 / V_s\$. You can name it as \$V_c / V_s\$. \$\endgroup\$ – AJN Jul 5 at 9:33
  • \$\begingroup\$ yes,that's correct .So it needs another voltage divider . \$\endgroup\$ – Rasoul Akbari Jul 5 at 12:34

until Z of right side of the circuit :(LS+R)/(LCS^2+RCS+1)

Well your almost there now that you have the parallel impedance of C, L and the right-hand R: -

$$Z = \dfrac{R+sL}{s^2LC+sRC+1}$$

Form the potential divider with the left-hand resistor to get: -

$$\dfrac{V_{C}}{V_{IN}} = \dfrac{\dfrac{R+sL}{s^2LC+sRC+1}}{\dfrac{R+sL}{s^2LC+sRC+1} + R}$$

Then multiply by top and bottom by \$(s^2LC+sRC+1)\$ to get this: -

$$\dfrac{V_{C}}{V_{IN}} = \dfrac{R + sL}{R+sL +R(s^2LC+sRC+1)}$$

$$\dfrac{V_{C}}{V_{IN}} = \dfrac{R + sL}{R+sL +s^2LCR+sR^2C+R}$$

$$\dfrac{V_{C}}{V_{IN}} = \dfrac{R + sL}{s^2LCR+s(R^2C+L)+2R}$$

To get \$V_{OUT}\$ we have this: -

$$\dfrac{V_{OUT}}{V_C} = \dfrac{R}{sL+R}$$

Therefore: -

$$\dfrac{V_{OUT}}{V_{IN}} = \dfrac{R}{s^2LCR+s(R^2C+L)+2R}$$

$$\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{s^2LC+s(RC+\frac{L}{R})+2}$$

Then drill down further if you want: -

$$\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{2}\cdot\dfrac{\frac{2}{LC}}{s^2 + s(\frac{1}{CR}+\frac{R}{L})+\frac{2}{LC}}$$

Can you take it from here?

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  • \$\begingroup\$ @RasoulAkbari OK I see what you mean...... \$\endgroup\$ – Andy aka Jul 5 at 12:51
  • \$\begingroup\$ I'm going to edit it. \$\endgroup\$ – Andy aka Jul 5 at 12:51
  • \$\begingroup\$ thanks my friend ,now that we have T.S how i should use the given information and calculate Vo(t) @Andy aka \$\endgroup\$ – Rasoul Akbari Jul 5 at 13:04
  • \$\begingroup\$ What is Vin(t)? If it is a step then multiply the TF by 1/s (the laplace of a step) and do the inverse laplace (with a few hurdles of partial fractions) or find a table of inverses. \$\endgroup\$ – Andy aka Jul 5 at 13:07
  • \$\begingroup\$ and when should i use info given by the question, \$\endgroup\$ – Rasoul Akbari Jul 5 at 13:10

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