How does Impedance matching equation work out from original link?

Question

From this link: L-Pad Impedance Matching Equation

How do you get those two equations? They do not look like a voltage divider formula, do they?

Questions repeated here:

First answer from Adil Malik:

Say R1=475,R2=56.2,Rload=50,Rsource=500

You simply want from the source side: R1+(R2∗Rload)/(R2+Rload)=Rsource

So lets try that: 475+(50∗56.5)/(50+56.5)=501.5

So this shows that it works.

But as you can see there are 2 unknowns, ie R1 and R2 . So you need atleast two equations. The one I did above was looking in from the source side. You can form a similar equation from the load side and solve simultaneously:

R2||(R1+Rsource)=Rload

Answer 1

They're basic series and parallel impedance combinations, based on the fact that the impedance looking into a transmission line terminated with its characteristic impedance is equal to the characteristic impedance.

Answer 2

The answer is: "Resistances in series and parallel."

Consider the 56.2Ω resistor. It is in parallel with the 50Ω impedance presented by the transmission line. Resistances in parallel are equivalent to Requiv = R1R2/(R1+R2). Thus the 56.2Ω and the 50Ω together act as (56.2)(50)/(56.2+50) = 26.46Ω. Thus the input sees the input resistor 475Ω in series with this 26.46Ω which is 501.46Ω which I suppose is close enough for the purposes of that post.

To derive the formulas, replace "56.2" with (variable) "R1" and "50" with "Z" (where I am using "Z" to represent the impedance of the transmission line). The 475Ω resistor -- call it R2. Introduce a 4th variable (say Zi) to represent the final input impedance (in that case, 500Ω).

The rest is algebra; solve for R1 and R2 in terms of Z and Zi. Two equations, two unknowns.

Also be aware that the value [ab / (a+b)] is equal to [ 1 / (1/a + 1/b) ] as you can easily verify.