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The question is to calculate the thevevin equivalent of the following two port network described by the transmission matrix [A B] [C D]

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The way I approached this was from the transmission matrix we know that $$ V_1 =AV_{th} - BI_2 $$ $$ I_1 =CV_{th} -DI_2 $$

Now, to find open circuit thevenin voltage source, we open circuit right side, making $$I_2 = 0$$ Subbing this back in to the first equations, $$ V_1 = AV_{th}$$ $$ I_1 = CV_{th}$$

I'm not sure where to go with this now?

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  • \$\begingroup\$ A HW problem with work shown!? A Christmas Miracle! \$\endgroup\$ – Bort Dec 12 '17 at 16:00
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Vth/V1 = 1/A and Vth/I1 = 1/C This means that the thevenin voltage is either 1/A or 1/C depending on the source your using, a voltage source (V1) or a current source (I1), at the input terminals. The network does not have any source term and that needs to be applied at the input, thus Vthev depends on the source impedance. These two case are a perfect voltage source (V1) or a perfect current source (I1).

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You have forgot that Vth finally comes from Vs and Rs also affects. Insert to your equations V1=Vs-I1*Rs and solve Vth.

Thevenin equivalent of the bare 2-port is quite useless. It has some Rth which depends on is the input open or shorted. Vth is zero if there's no signal source. To calculate the Rth of an inputless 2-port you must insert a temporary source to output and calculate V2/I2.

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