![]() ![]() Which can be algebraically solved for $Z_0$ as: You'll notice:Ī little more generally, a quarter-wave section of transmission line of impedance $Z_0$ converts between two impedances $Z_a$ and $Z_b$: This works because relative to 125Ω, 300Ω and 50Ω are (approximately) duals, that is they have the same magnitude reflection coefficient, but 180 degrees apart. More on that at the end.Īs an interesting example, VK6UU gives an example of a 300Ω folded dipole matched to 50Ω by way of a quarter-wave section of 125Ω coax. Of course you'd need to know the impedance of the thing you are trying to match. Then perform the same operation in reverse with the answer. Divide all the impedances by 75 to normalize them to 1, then multiply them by 50 to convert to what the calculator expects. You can also do the same with any transmission line calculator. The Smith chart has the advantage of not requiring any complicated math, and also facilitating intuition into what matching strategies are possible. Normally when using a Smith chart we divide all the impedances by 50 to normalize them to a 50Ω system, but you can divide by 75 (or any other number) and the results are just as valid. You can do this graphically, with a Smith chart.
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