Parallel Connection of Coaxcables
You are lucky, if have not one, I've two solutions for you:
- The first is an approximation and does not care much about how the system is build up
- The second is a much more convenient one, that you have a so called power splitter with input and output-matching.
1. Straight Forward Solution using equivalent circuit diagrams
Setting KVL and KCL of an infinitesimal section of the transmission line like the one depicted below
yields (as shown in more detail in this wiki link) to:
$$ \cfrac{\partial ^2 V(x)}{\partial z^2} + (R+j\omega L)\cdot(G+j\omega C)\, V(x) = 0 \\
\cfrac{\partial^2 I(x)}{\partial z^2} + (R+j\omega L)\cdot(G+j\omega C)\, I(x) = 0 \, . $$
Now igdl.y, in your case two transmission lines are connected in parallel! Looking at the equivalent circuit diagram of a transmission line (note that this circuit diagram is an infinite small section - it doesn't matter if you change R with L, etc.), you can easily conclude that:
$$R_\parallel = \cfrac{R}{2} \\
L_\parallel = \cfrac{L}{2}\\
G_\parallel = 2 \cdot G\\
C_\parallel = 2 \cdot C$$
Solving the partial differential equation leads to an characteristic wave impedance of
$$Z_{\parallel0} = \sqrt{\frac{\frac{R}{2} + j\omega\,\frac{L}{2}}{2 G + j\omega\, 2 C}\,} = \cfrac{1}{2} \cdot \sqrt{\cfrac{R + j\,\omega\, L}{ G + j\,\omega\, C}} = \cfrac{Z_o}{2}\,~\,.$$
Hence assuming equal lengths of the transmission lines, results in the following equivalent circuit:
simulate this circuit – Schematic created using CircuitLab
BTW: This model is only, and only valid if the length of both cables are equal! Otherwise you have different times until the wave reaches the resistor, and got partly reflected if the wave impedance and the load don't match.
In your special case the load and the wave impedance are matched, resulting in no reflection.
In case you are not interested in timing issues (how long does it take for the wave from port 2 to the load) you can simplify this circuit further: Since you have no reflection, the 37.5 transmission line and load simplifies to following equivalent circuit diagram (this can be rational if only interested in a power-analysis):
simulate this circuit
Discussion:
As you can see in this equivalent circuit diagram, due to different wave impedance you will have reflection at the joint connection where the 75 and 37.5 ohm transmission line will meet. This discussion has only academic character, since you have to take the geometry of the joint connection into account. To get a lumped equivalent circuit you have to apply Maxwell's equations on the geometry of the joint connection, that involves the materials and geometries. The result will be a frequency dependent impedance model; typical it's represented as S-matrix.
Parallel Connection of transmission lines have many practical issues:
- You can make an overall transmission line having another wave impedance
- With 2 cables the max. allowable power rating doubles in sum
2. A Much More Realistic Scenario - Using an Wilkinson Divider
Since RF engineer doesn't like to make their life's a hell, they make such joint connection in a much proper way:
They make the input and outputs of the divider (the thing that's split the input into to two outputs; it divides the input power into two output powers) and tries to match the (system) wave impedance.
Such a rf-device can be an Wilkinson divider as depicted in the picture below (taken from wikipedia). This device works only in a small frequency range, which is acceptable in most cases. Further: the 50 ohm has be replaced by 75 ohm, and the 71 ohm by 105 ohm in your question.
Using such a device, yields to the following equivalent circuit diagram:
simulate this circuit
There are a lot of devices which allow to maintain the wave impedance. So, the question you were given isn't clear enough to say what exactly happens. But I would prefer to take the second case.
