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Insertion loss 0 db and return loss negative dB in low freq

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isla4018

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Hello everyone,

Why insertion loss is 0 dB and return loss is a very large negative dB value at low frequency ? I found an explanation in Eric Bogatin's book, but could not get the explanation. Someone please explain this?

Thanks!
 

isla4018 said:
Why insertion loss is 0 dB
and return loss is a very large negative dB value at low frequency ?
Click to expand...
Describe your situation in detail.
One possiblity is a error due to calibration of VNA(Vector Network Analyzer).
How do you calibrate VNA ?
 

Hello,

I simulated a four port board which is 0-5 GHz range. I compared my simulation result with the measurement. I got insertion loss (S21) is 0 dB and return loss (S11) is a very large negative dB value at low frequency (0 Hz). I found this is the characteristics of S21 and S11. But I do not know why this happens. Could you please help me with the explanation?

Thanks!
 

I understand that you got |S11|>1 (negative dB numbers) in simulation, not real measurement?

I'd say it's not possible with a passive network, might happen of course with an active device, e.g. an unstable amplifier. Unfortunately you didn't tell what the DUT is.

You may want to sketch the explanation you found in your book, along with a more detailed description of the test setup.
 

Thank you so much for all of your reply. I got the result for both measurement and simulation. The DUT is a two layer (Bottom and top) PCB board with FR4 dielectric (Just like Capacitor). The two layers are connected through two vias. There is no traces. For your convenience, I have attached the S11 and S21 figure herewith.

**broken link removed**


**broken link removed**


Thanks!
 

you have to understand wave theory in order to understand S11 and S21.

S11 = b1/a1 when a2=0
S21 = b2/a1 when a2=0

540px-TwoPortNetworkScatteringAmplitudes.svg.png

b1 & b2 reflected waves and a1 & a2 are incident waves
 

A negative return loss just means that the ratio of reflected power to incident power is very small - say 1:1000. 1:1000 would translate to 10LOG(1/1000), or -30dB. You would expect that from a system that has close to "0dB" insertion loss.
 
A negative return loss just means that the ratio of reflected power to incident power is very small - say 1:1000. 1:1000 would translate to 10LOG(1/1000), or -30dB. You would expect that from a system that has close to "0dB" insertion loss.
Click to expand...
Your explanation is of course correct. I guess the problem is a confusion of terms.

Return loss is commonly defined as

10 log (incident power/reflected power)

resulting in positive dB numbers for passive networks with |S11| <= 1. But I read that the quantity has been historically used with inverted sign, see https://en.wikipedia.org/wiki/Return_loss
 

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