Great video that explains the simplicity and beauty of 1/4 wave transformers. It took me a while to understand why the reflected waves appear to behave differently (at Z0 & Z1, the reflection returns 'on top of' the incident wave, while at Z1 & ZL the reflection appears to invert and return as the negative of the incident wave). I now realise after thinking it through that the reflection continues the behaviour of the incident wave: at Z0 & Z1, it falls from high towards zero, yet at Z1 & ZL, BOTH incident and reflected waves fall from zero to negative. That is consistent and totally cool! Thanks for helping me visualise this - shall put it to work in a 15m HF antenna mismatch that I have. 73, Matt M6NJX
The key idea to understand that lies in the fact that the quarter wavelength accumulates 90 degrees of phase on the incident signal. So the reflected signal at the beginning of Z1 that comes from the end of the quarter wavelength is the same signal as the incident signal, but 180 degrees out of phase (90°+90°, or 2 times the quarter-wave phase accumulation ). It's that simple.
Thank you so much for sharing this content. I studied electrical engineering in the late 80's en beginning of 90's. Due to the "destiny" of life I am working as a networking engineer after my studies. It's good to see that after 30 years I still remember the concepts of this topic, although I have tend to forget the mathmatical details.
Great video. I understood how it works if the reflection coefficient is small. I don't understand how this works when the reflection coefficient is higher, for example, 0.5, because the reflected waves would have different amplitudes. I appreciate if someone could elaborate on this question.
hi thanks for your video, since there is some cancelation although no standing waves appear on Zo, isn't that there is effectively standing waves in perfect antiphase so that they cancel and there is still power loss going into the loss? i've read people described the quarter wave transformer as 'perfect' but mustn't it be lossy?
We need the spatial impedance relationship for Z1, so in 3:10, Z0 is actually Z1. Also in 3:46, we need the Γ=0 between Z0 & Z1, for Z1's (-λ/4). This is the way to solve the eqs.
How the reflected wave at i/p terminal of lambda by 4 transformer(Z1) will get reflected at the same phase toward i/p(Z0)?? The reflected wave i/p terminal of lambda by 4 transformer(Z1) should have 180 deg. phase shift with (Z0).If it has 180 deg.phase shift reflected wave then both the reflected waves will add.
Sort of... The point is that any time you have a sudden impedance mismatch there will be a reflection back. The overall signal consists of what is going towards the load from your transmitter and what is reflected back. Since ideally you don't want anything going back to the transmitter you put in the matching section so that there are two reflections going back that are the same (small) amplitude and 180 degrees out of phase with each other so what is bouncing back sums up to zero. Remember that amplitude isn't conserved so this allows full power (which is conserved) transfer to the load.
so does that mean that quarter wave matching is less than ideal, because there is still energy being reflected back even though the energy cancels itself out
Hey, if anyone is around to answer... do the two reflected waves exactly cancel out even though some of the 2nd reflected wave will get re-reflected back towards the load? (Is it just that this 2nd reflection (and third and fourth etc. etc. as this wave bounces around inside the quarter wavelength area) is small enough to ignore?
Short and efficient! Thank you so much, I've been trying to understand the significance of quarter wavelengths and you demonstrated it perfectly
Great video that explains the simplicity and beauty of 1/4 wave transformers. It took me a while to understand why the reflected waves appear to behave differently (at Z0 & Z1, the reflection returns 'on top of' the incident wave, while at Z1 & ZL the reflection appears to invert and return as the negative of the incident wave). I now realise after thinking it through that the reflection continues the behaviour of the incident wave: at Z0 & Z1, it falls from high towards zero, yet at Z1 & ZL, BOTH incident and reflected waves fall from zero to negative. That is consistent and totally cool! Thanks for helping me visualise this - shall put it to work in a 15m HF antenna mismatch that I have. 73, Matt M6NJX
Queuerious Guy that helped me a lot, thank you for the explanation
Thanks for that.......that question was bothering me as well
The key idea to understand that lies in the fact that the quarter wavelength accumulates 90 degrees of phase on the incident signal. So the reflected signal at the beginning of Z1 that comes from the end of the quarter wavelength is the same signal as the incident signal, but 180 degrees out of phase (90°+90°, or 2 times the quarter-wave phase accumulation ). It's that simple.
Thank you so much for sharing this content. I studied electrical engineering in the late 80's en beginning of 90's. Due to the "destiny" of life I am working as a networking engineer after my studies. It's good to see that after 30 years I still remember the concepts of this topic, although I have tend to forget the mathmatical details.
Thanks for your videos. I did learn a lot.
Fantastic! My electromagnetics professor, as good as he is, left some of his explanations a little empty. This filled them right in!
Thanks for such a good content.
Great explanation. I was wondering why quarter of a wavelength is taken. Now it's CRYSTAL CLEAR
Hi there is an error at 3:13. The Z0 in the left equation should be Z1. Otherwise we can not have Zin=Z1^2/ZL
Great video. I understood how it works if the reflection coefficient is small. I don't understand how this works when the reflection coefficient is higher, for example, 0.5, because the reflected waves would have different amplitudes. I appreciate if someone could elaborate on this question.
hi thanks for your video, since there is some cancelation although no standing waves appear on Zo, isn't that there is effectively standing waves in perfect antiphase so that they cancel and there is still power loss going into the loss? i've read people described the quarter wave transformer as 'perfect' but mustn't it be lossy?
We need the spatial impedance relationship for Z1, so in 3:10, Z0 is actually Z1. Also in 3:46, we need the Γ=0 between Z0 & Z1, for Z1's (-λ/4). This is the way to solve the eqs.
thx you sooo much!!!
How the reflected wave at i/p terminal of lambda by 4 transformer(Z1) will get reflected at the same phase toward i/p(Z0)?? The reflected wave i/p terminal of lambda by 4 transformer(Z1) should have 180 deg. phase shift with (Z0).If it has 180 deg.phase shift reflected wave then both the reflected waves will add.
amazing thanks!!
if the reflected signal is 180 out of phase from the 1/4 length matching section, wouldn't that then destroy your transmission signal ?
Sort of... The point is that any time you have a sudden impedance mismatch there will be a reflection back. The overall signal consists of what is going towards the load from your transmitter and what is reflected back. Since ideally you don't want anything going back to the transmitter you put in the matching section so that there are two reflections going back that are the same (small) amplitude and 180 degrees out of phase with each other so what is bouncing back sums up to zero. Remember that amplitude isn't conserved so this allows full power (which is conserved) transfer to the load.
so does that mean that quarter wave matching is less than ideal, because there is still energy being reflected back even though the energy cancels itself out
Hey, if anyone is around to answer... do the two reflected waves exactly cancel out even though some of the 2nd reflected wave will get re-reflected back towards the load?
(Is it just that this 2nd reflection (and third and fourth etc. etc. as this wave bounces around inside the quarter wavelength area) is small enough to ignore?
BEST!!!!!!
....what about complex loads...
You sound like Tom Cruise