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