-1
\$\begingroup\$

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

\$\endgroup\$
  • \$\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
2
\$\begingroup\$

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?

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

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.