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msbe
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If i wanted to make a 1:N coaxial RF switch (say ~100MHz, nothing too high), could you simply route N traces in a star pattern on a PCB and implement a mechanical switch to either connect to your measurement point, or terminate in the characteristic impedance (say 50 Ohm).
I'm getting thrown off a bit because I haven't come across much work on parallel transmission lines. The theory would be that if it's terminated in 50ohm it appears as an infinite length transmission line, i.e. has zero reflection. That would seem to imply that the signal would be split in power N ways. N-1 of them terminate in 50 ohms and have no reflection. The other one would go to the load and reflect back based on whatever the impedance seen is.
It seems like this is the design most RF switches use (absorptive switches that either connect to the output or terminate internally in 50ohm or 75ohm depending on the IC), albeit they use FET's typically for the actual switching.
Is this right? Or does connecting multiple transmission lines in parallel actually change the characteristic impedance seen going into the selected channel? I could see this since the power of the input signal is nessicarily lost due to the 50ohm terminations, but if the switches were open it would set up standing waves from the reflection.
Maybe the switch would simply present as an insertion loss by being terminated? So a 1:2 switch would reduce output power by 3dB since half of the power would go to a 50ohm termination?
I'm getting thrown off a bit because I haven't come across much work on parallel transmission lines. The theory would be that if it's terminated in 50ohm it appears as an infinite length transmission line, i.e. has zero reflection. That would seem to imply that the signal would be split in power N ways. N-1 of them terminate in 50 ohms and have no reflection. The other one would go to the load and reflect back based on whatever the impedance seen is.
It seems like this is the design most RF switches use (absorptive switches that either connect to the output or terminate internally in 50ohm or 75ohm depending on the IC), albeit they use FET's typically for the actual switching.
Is this right? Or does connecting multiple transmission lines in parallel actually change the characteristic impedance seen going into the selected channel? I could see this since the power of the input signal is nessicarily lost due to the 50ohm terminations, but if the switches were open it would set up standing waves from the reflection.
Maybe the switch would simply present as an insertion loss by being terminated? So a 1:2 switch would reduce output power by 3dB since half of the power would go to a 50ohm termination?
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