Quarter wave impedance transformer?

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Impedance transformation using a quarter wave transmission line (TLT) occurs due to the unique phase relationship of voltage and current along the line. At a quarter wavelength, the voltage experiences a 90-degree phase shift, causing an open circuit to appear as a short circuit and vice versa. This transformation is effective only at quarter wavelengths or odd multiples because of the specific phase cancellation that occurs at these points. When a signal encounters an open circuit, the reflected wave becomes increasingly out of phase, leading to cancellation at the quarter wavelength mark. To achieve impedance transformation, the transmission line impedance should be set to the geometric mean of the two impedances being matched.
hobbs125
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Hi everyone,

I keep trying to understand how impedance transformation works using a TLT but I'm stumped. There seems to be something I am just not understanding.

Can someone explain in different ways how a line that is a quarter wave long cause an open circuit to look like a dead short and vice versa?

How does the impedance transformation work, and why does it only work when the line is a quarter wavelength or one of it's odd multiples?
 
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You might be interested in my question for a previous post:
https://www.physicsforums.com/showthread.php?t=740203

Transmission Line Theory or even just visualizing a sine wave (at the desired frequency) traveling down the transmission line helps understand some of the aspect of what is happening. With a 1/4 wavelength (adjusted for velocity) length of coax, the voltage is a 90 degree phase shift. While one point the voltage on the center conductor is at the peak, 90 degrees down the line the voltage is at 0v. I now believe that the current in the outer conductor just rolls around and crosses directions so the currents balance. Anyway the impedance matching is directly related to the voltage changes of the lengths of the coax transformer matching section.

Others in this forum have a much better feel about the theory and can present a better explanation.
 
When a signal reaches an open on a transmission line it is reflected back towards the source. The reflected wave, while still increasing in phase is traveling in the opposite direction. This results in the reflected wave becoming more and more out of phase with distance up to 1/4 wavelength. At 1/4 wavelength from the open the forward wave and the reflected wave are 180 deg. out of phase and except for losses they cancel. If you calculate E/I at that 1/4 wave point, because E is very small and I is high, that point is essentially a short.

If you want to transform one impedance into another using a 1/4 wavelength transmission line, all you need to do is select a transmission line impedance equal to √(Z1 * Z2).
 
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