vanesch
Jul23-04, 07:41 AM
My professor told me there wasn't a proof, but I might have misunderstood him.
Of course there is a proof, it depends from what you start.
Let us consider a circuit containing components and sources, with two output wires. If we accept that the circuit is linear, meaning that the superposition principle holds (I'll come to that) then we can connect these two output wires to an external ideal voltage source. The current that will flow in that ideal voltage source can be written as:
I = (internal source 1) * A + (internal source 2) * B + ...
+ (internal source n) * C - (external source) * G
This is dictated by the superposition principle: there are n internal sources, and one external, so all voltages and currents anywhere in the circuit are a linear combination of these source values (if this is not the case, by definition, the circuit is non-linear, and indeed, thevenin's theorem doesn't hold). The coefficients do not, of course, depend on the sources (otherwise this is not a true linear combination).
Now the n first terms only depend on the internal workings of the circuit (coefficients and internal sources), so we can replace this, as the external source is concerned, by a constant, D.
So we have:
I = D - V_ext * G
Define V_th = D / G and G = 1/R and we have:
I = 1/R * (V_th - V_ext)
This is of course the equivalent relationship with a thevenin circuit with voltage V_th and resistance R, coupled to our external voltage source V_ext.
You could object that connecting an external voltage source is not the most general thing one could do, but this is not true. In ANY application, there will be A potential difference between the two wires ; it is evident that we can then replace the entire network by a voltage source with exactly this potential difference, and our circuit will not notice the difference.
The other facet is a bit more cumbersome: one has to show that any complicated circuit, only consisting of resistors and sources, constitutes a linear network. This can be done using Kirchhoff's laws.
cheers,
Patrick.
Recently this one for delta Y however it requires port variable theory :
For a T network the transfer matrix equation may be found by multiplying transfer matricies :
\[
\left[ \begin{array}{l}
1\,\,\,\,\,\,\,Z_1 \\
0\,\,\,\,\,\,\,1 \\
\end{array} \right]\left[ \begin{array}{l}
\,\,1\,\,\,\,\,\,\,\,\,\,\,0 \\
1/Z_3 \,\,\,\,\,1 \\
\end{array} \right]\left[ \begin{array}{l}
1\,\,\,\,\,\,\,\,Z_2 \\
0\,\,\,\,\,\,\,\,1 \\
\end{array} \right] = \left[ \begin{array}{l}
1 + \frac{{Z_1 }}{{Z_3 }}\,\,\,\,\,\,\,\,\,Z_1 + Z_2 + \frac{{Z_1 Z_2 }}{{Z_3 }} \\
\,\,\,\,\,\frac{1}{{Z_3 }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 + \frac{{Z_2 }}{{Z_3 }} \\
\end{array} \right]
\]
And for a pie network (of addmittances) the transfer matrix equation is :
\[
\left[ \begin{array}{l}
\,\,1\,\,\,\,\,\,\,\,0 \\
\,Y_1 \,\,\,\,\,\,\,1 \\
\end{array} \right]\left[ \begin{array}{l}
1\,\,\,\,\,\,\,1/Y_3 \\
0\,\,\,\,\,\,\,\,\,\,\,1 \\
\end{array} \right]\left[ \begin{array}{l}
\,\,1\,\,\,\,\,\,\,\,\,0 \\
\,\,Y_2 \,\,\,\,\,\,\,1 \\
\end{array} \right] = \left[ \begin{array}{l}
\,\,\,\,\,\,\,\,\,1 + \frac{{Y_2 }}{{Y_3 }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\frac{1}{{Y_3 }} \\
\,\,Y_1 + Y_2 + \frac{{Y_1 Y_2 }}{{Y_3 }}\,\,\,\,\,\,\,\,\,\,\,\,1 + \frac{{Y_1 }}{{Y_3 }} \\
\end{array} \right]
\]
Equate the transfer matrix components and you have the T-pi or Delta-Y transforms.
Best
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