We have the following circuit:

enter image description here

I am trying to get the transfer function using the Mesh Method. The first equation I've got is:

(s+1) I1(s) - sI2(s) = Vi(s)

The second one is:

-sI1(s) + (s + 2) I2(s) = 0

By that, the transfer function is:

2/(3s + 2).

The teacher used the nodal method:

and the result was:

1/(s + 2).

I know that using both methods, should give the same solution. Any help ?

  • \$\begingroup\$ 2nd equation is wrong. \$\endgroup\$ – Chu Jul 15 '18 at 6:22
  • \$\begingroup\$ Yeah. Its should be -sI1(s) + (s+1) I2(s)=0 but still giving a wrong end result of V0/Vi \$\endgroup\$ – alim1990 Jul 15 '18 at 7:02
  • 3
    \$\begingroup\$ Answer is \$\frac{s}{1+2s}\$; teacher is wrong. Clearly DC gain = 0, since L is a short circuit at DC. \$\endgroup\$ – Chu Jul 15 '18 at 7:47
  • \$\begingroup\$ You can see that when \$s = 0\$, the inductor is a short circuit so your equation you should return 0 when \$s = 0\$ while for \$s\$ approaching infinity, the inductor is open-circuited and you are left with a divider exhibiting an attenuation of 0.5. This is the simple engineering judgment you must do to check your results. \$\endgroup\$ – Verbal Kint Jul 15 '18 at 7:51
  • \$\begingroup\$ Verbal kint can you explain more about what you said so I always check it on all transfer functions. Mr. Chu can you explain the process in an aswer ? \$\endgroup\$ – alim1990 Jul 15 '18 at 8:44

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