50 Ohm Wave Terminates in Purely Reactive Load

[chiques]

I have a load which resonates at my operating frequency (S11 ~ -40dB). If I have a 50 Ohm system and I transmit and terminate using this load, does the impedance go infinity?

If so, this would greatly increase the voltage across the load being:


P=V²/R

 

[abdulwahababid]

I have a load which resonates at my operating frequency (S11 ~ -40dB). If I have a 50 Ohm system and I transmit and terminate using this load, does the impedance go infinity?

If so, this would greatly increase the voltage across the load being:


P=V²/R

Are you trying to remove resonance from the circuit??
you may use resistance having less power dissipation; I think so it will help you

 

[chiques]

Are you trying to remove resonance from the circuit??
...

No, I'm not trying to remove resonance. I just need to know how the energy of the wave propagating around a 50 ohm load translates into e.g. an antenna; which has no real impedance but is reactivley matched.

 

[E Kafeman]

No real impedance => no radiation. If it not radiates is it a rather useless antenna.

 

[biff44]

If the load were purely reactive, S11 would be 1. Since it is near zero, the load impedance is obviously 50 ohms with no reactive part. And yes, if this is a resonant circuit, the voltage across this load will be much higher than you would normally see.

 

[chiques]

If the load were purely reactive, S11 would be 1. Since it is near zero, the load impedance is obviously 50 ohms with no reactive part. And yes, if this is a resonant circuit, the voltage across this load will be much higher than you would normally see.

Referring to this diagram:



So the voltage across C2 would be much larger than the voltage across C in series? Is there a theory I can to refer for this?

 

[E Kafeman]

Your diagram is opposite to a purely reactive load, at resonance frequency. Reactance X is then infinite and total load impedance Z is: 1/R + 1/X = 1/Z => R=Z , a purely resistive load.

I have a load which resonates at my operating frequency (S11 ~ -40dB).

I just need to know how the energy of the wave propagating around a 50 ohm load translates into e.g. an antenna; which has no real impedance but is reactivley matched.

First sentence is at least possible and reactive load is almost none. Second sentence is the opposite, now is it a purely reactive load, but it is not is possible to reactivly match something that have no real impedance to become a matched real load in a 50 Ohm system.
Theory is: Z=R+jX
What ever you do with X, if R is none from beginning, it will remain that way as long as it only is pure reactive matching.

 

[biff44]

In your diagram on the right side "impedance 50 ohm", the L and C2 tell where in frequency a resonance will occur. As stated above at resonance, the Yl and Yc susceptances will cancel out, since at that one frequency Yl = - Yc. That means all you have left are the coupling capacitor C1 and the equivalent R. You can vary both to get just about any |S11| you want.

 

[chiques]

...Yl and Yc susceptances will cancel out, since at that one frequency Yl = - Yc.

Thanks for the variation of explanations.

So is it safe to to say: when Yl or Yc begin to drift (due to heat, climate change or environmental reasons) then the total impedance is changed which causes a mismatch in the system?