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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 $$I_L=\frac{V_2}{Ls}$$

  • 1
    \$\begingroup\$ inspection of the transfer function at dc (s at zero) tells you it is wrong. \$\endgroup\$
    – Andy aka
    Feb 8 '14 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 '14 at 18:14

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\$

  • \$\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 '14 at 19:38

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