0
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

I calculated the transfer function $$\frac{I_L}{In}=\frac{R_1R_2}{s^2L^2+sL(R_1+R_2)}$$

starting from the node analysis at the nodes V1 and V2, so: $$\frac{V_1}{R_1}+\frac{V_1}{Ls}-\frac{V_2}{Ls}=In $$ $$-\frac{V_1}{Ls}+\frac{V_2}{Ls}+\frac{V_2}{R_2}=0$$

Because I obtain a second order system, this seems quite wrong.

Have I consider the admittances in the node analysis? So: $$\frac{V_1}{G_1}+V_1Ls-V_2Ls=In$$ $$-V_1Ls+V_2Ls\frac{V_2}{G_2}=0$$ In this case I obtain a more likely: $$\frac{I_L}{In}=\frac{1}{sL(R_1+R_2)+R_1R_2}$$

But I'am not sure about the node analysis.

What do you think is the correct solution?

Note: In both cases I consider $$I_L=\frac{V_2}{Ls}$$

\$\endgroup\$
2
  • 1
    \$\begingroup\$ inspection of the transfer function at dc (s at zero) tells you it is wrong. \$\endgroup\$
    – Andy aka
    Feb 8, 2014 at 17:27
  • \$\begingroup\$ @Andy aka: Thanks for the advice. I think that I found the problem, it's just that I(L) must be equal to (V1-V2)/Ls. At least, I think :D \$\endgroup\$
    – FdT
    Feb 8, 2014 at 18:14

1 Answer 1

Reset to default
1
\$\begingroup\$

The simplest path to the transfer function is to use current division which gives you the answer by inspection:

$$I_L = I_n \frac{R_1}{R_1 + R_2 + sL} $$

Also, the final formula in your question is incorrect; the voltage across the inductor is not \$V_2\$

\$\endgroup\$
1
  • \$\begingroup\$ Indeed Alfred, that's the solution that I was looking for :D I found that in two different ways: Thevenin and simply calculate I as (V1-V2)/Ls with the first system. Thanks. \$\endgroup\$
    – FdT
    Feb 8, 2014 at 19:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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