The small-signal model (SSM) of a resistor is obviously the same resistor. Here I got a problem with understanding what SSM of inductor and capacitor would look like. Here's my analysis and please correct my if you see something wrong in my analysis:
To find SSM of the inductor we develop our model and write small-signal voltage as follows: \begin{eqnarray*} v_L=L\ \frac{d i}{dt} \\ v_{SSM}= \frac{\partial v_L}{\partial i_L}.i_L\\ v_{SSM}=\frac {\partial (L\ \frac{di_L}{dt})}{\partial i_L}.i_L\\ v_{SSM}=\frac {\partial }{\partial i_L}(L\ \frac{di_L}{dt}).i_L=L\frac{d}{dt}(\frac {\partial i_L}{\partial i_L}).i_L\\ v_{SSM}=L \ \frac{d}{dt}(1).i_L=0 \end{eqnarray*} So I infer that the SSM of the inductor is much the same as that of a voltage source, which is shorted out. Similarly for the SSM of the capacitor one can show, through the same kind of analysis, that \begin{eqnarray*} i_{SSM}=0,\ \end{eqnarray*} which says that the SSM of a capacitor behaves like an open to our circuit model. Do you think that my analysis is wrong? Where is the bug in my reasoning?
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