While considering the value of S11 we always take -10dB as a reference for return loss. Why we consider -10 dB and what is the mathematical justification for the same.
Dear Samarth Agarwal , more accurately for a VSWR = 2 corresponds to a level of RL = -9.542 dB. In this regard, -10 dB is an approximation of -9.542 dB
Respected Vadym Slyusar Sir, Thank you for your response.
Return loss is always a positive number (in dB). As per the equation :
Return Loss = -20log(|reflection coefficient|)
For VSWR = 2; Reflection Coefficient will be = 1/3. Putting this value in above equation will yield 9.54 dB (positive value). How we are considering -10 dB even though we get a positive value of return loss.
In all my job interviews, interviewer ask me a mathematical derivation for the value of -10 dB as a reference for return loss. Earlier too, I explained by this way only but later got stuck on the fact that return loss will always be a positive number so why we consider a negative value and how do we evaluate that?
I hope you understand my concern.
Dear Samarth Agarwal , 20lg(1/3)= - 9.54. A negative value in dB means that the return loss is less than 1.
Respected Vadym Slyusar Sir, Thanks for your response but the formula for return loss is with minus sign as stated in my previous response. Please find in attach the image file for your reference.
Dear Samarth Agarwal I thijnk that this is a matter of taste, not nothing more.
a 10 dB return loss is not particularly good. As Vadym Slyusar says, it roughly corresponds to a VSWR of 2. It also limits the efficiency to no more than 90%, just due to the reflection of 10% of the power. However, the pressure to get wider bandwidths, and people and techniques from HF and VHF antennas (where 2:1 can be seen as good) moving up into microwave has led to poorer VSWR being acceptable. It is also cheaper to make up for the corresponding loss of power than it used to be. VSWR of 1.5:1, or 14 dB return loss, used to be a common target. However, in any system, the VSWR or return loss is set by overall cost and performance requirements. A better antenna is a once-off cost, but 10% extra power requirement goes on for ever. However, for rate of information transfer, bandwidth is more valuable than power (Shannon Hartley theorem), so if 10% extra bandwidth results in a loss of 10% power it is a trade worth making.
Vadym Slyusar
It isn't a matter of taste - but it is like the wild west out there, and people use + or - in a way that seems like it might be random, so you can never depend on it, and like you, we depend on the context to decide if it is really an increase or decrease, and if it matters I never depend on people understanding the convention. It gets complicated for reflection amplifiers, where the return loss can really be negative dB! (but S11 is positive dB, and you work outside the unit circle of the normal Smith chart).
The term loss implies a minus sign, but lots of people put an extra minus there anyway. S11 and return loss should be opposite sign. The return loss might be 20 dB, and the magnitude of S11 will be -20 dB, but most people seem to think S11 and return loss are the same thing.
It's different with attenuators. Almost nobody says a -20 dB attenuator.
Dear Malcolm White , I agree with you. I will stay forever on the position that Return loss in dB will be minus values. It is the easy mathematical formula:
20lg(1/3) = - 9,54.
Regarding taste, I said so to calm Samarth Agarwal down, and he did not torture himself because of someone's stupidity. There are many mistakes in science and in various scientific schools in the world.
Vadym Slyusar ..... and factors of 2 and + and - signs are sometimes the hardest to track down!
Yes, dear Malcolm White ! I had the same mistakes with factors of 2 and + and - signs as well.
Thank you John Colaco Sir.
I know that the return loss is a positive value but why do we consider as a -10dB of return loss as a reference. In any research paper as well as text book we can see that -10dB (return loss) is considered though the return loss will always be a positive value.
Respected Vadym Slyusar Sir, As stated the mathematical formula for the return loss
Return Loss = -20log(|reflection coefficient|)
will provide the positive value only for VSWR = 2. This formula is found everywhere either in text books or in research papers. So how we consider negative value of return loss.
Dear Malcolm White My question is that why do we consider -10 dB return loss as a reference for antennas? Further, if return loss is always a positive number as per the mathematical equation which can easily be found anywhere so how do we consider a negative value of return loss as a reference.
John Colaco RL can be positive in case of active elements only. In case of any passive element condition either load is complex or real the RL will be positive because RL = -20* log( mod( reflection coefficient)).
Samarth Agarwal
Return loss often has a negative sign for five reasons at least
1 People don't know the correct definition
2 People don't care
3 They made a mistake
4 They confused it with the magnitude of S11 or reflection coefficient.
5 There really was gain!
You can't depend on it being according to the definition, and you won't make a significant difference by pointing it out or complaining!
I think in my first post to this column I pointed out the reason why 10 dB, or vswr 2 as Vadym Slyusar helpfully added, is often chosen for the target.
John Colaco is wrong. The return loss does not have to be negative when the reflection coefficient is complex. It is only negative when the magnitude of the reflection coefficient is greater than 1, i.e. when there is gain.
You can always make any real reflection coefficient complex, if it is real, by measuring it a different distance along the transmission line.
As mentioned by Malcolm White for -10dB of reflection coefficient (S11) (10dB of return loss) the power reflected is 10%, which is mostly acceptable. In some of the works especially for mobile antennas the same is considered as -6dB.
Dear colleagues, Return loss every should be with "minus" sign. It is "LOSS". If you use the positive value that it is used with the word "loss". You can see the picture from Ansys HFSS.
In relative measurement such as dB, the mathematical representation of physical quantity like loss should always be negative.
For any antenna to radiate effectively forward incident power should be much greater than the reflected power. Ideally, for effective transmission VSWR should be 2, and thus return loss should be around -10 dB. But practically antenna starts to resonate and radiate in loaded condition at -8dB onwards.
Sovan Mohanty
If I was to say take minus 2 steps backwards, what would you understand by this?
In accounting, what is a loss of -$2?
Most of us, if the word loss is used, understand that it is instead of the minus sign, and S21 of -10 dB would be 10 dB loss. Similarly, it is usual to refer to attenuators as 3, 10 or 20 dB, for example, although -3, -10 or -20 would also be common, as there is never an attenuator with gain, so there can never be confusion with attenuators.
Dr Jugul Kishor
10 dB return loss does not mean 90% of the power is transferred to the load, it means 10% of the power came back. Less than 90% may have reached the load. Some may have been radiated, and some turned to heat in the cables, for instance.
Return loss is also known as reflective loss, is a parameter of the reflected signal performance. It indicates what portion of the incident power is reflected back to the signal source.
- Return Loss = Incident power / Reflected power, in dB
Greater Return Loss means the signal has a higher transmission efficiency - you’re losing less power and signal strength to your transmission gear.
Greater Return Loss potentially means the signal has a higher transmission efficiency.
One way of improving return loss is by introducing attenuation, which reduces transmission.
Just on the fun side: In ham radio: High SWR = power returns to transmitter and overheats it.
I had a colleague who told of seeing the insulation burning off the cables starting at the far end and coming back towards him! Time to turn off the power (if you can).
Dr Jugul Kishor Thank you so much sir for answering. Yes, I know that S11 of -10 dB represents the transfer of 90% power to the medium. But I want to know that is there any mathematical justification that we can prove with some sort of formula that we are getting 90% power at -10 dB S11
Not unless you are sure no power is going anywhere else.
Example
100% in, 10% back, 1% resistive and dielectric loss, 89% transmitted.
You could measure the loss using a calorimeter. You would need to be accurate.
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