If system 1 has 10 dB return loss and system 2 has 20 dB return loss then what is the overall return loss of cascaded system of 1&2 ?

  • 3
    $\begingroup$ Return loss is just 1 of the 4 S-parameters. You need them all for systems 1 & 2, and also the return loss for whatever you put after system 2. $\endgroup$
    – Juancho
    Commented Jun 11, 2016 at 11:34
  • $\begingroup$ @Juancho : I was asked this question in exam and only this information was given. $\endgroup$
    – Aparna B
    Commented Jun 12, 2016 at 6:10
  • $\begingroup$ If that was the only information you were given, you would first find the return loss looking at the load, then use system 2's S matrix to find the RL looking into it, then do the same with system 1 until you finally have the RL looking into the entire network. $\endgroup$
    – Andrew W.
    Commented Jun 13, 2016 at 17:26
  • $\begingroup$ @Andrew W. I tried your method n got answer as 9.5dB but actual answer is 9.63dB. can I approximate it to 9.63dB? $\endgroup$
    – Aparna B
    Commented Jun 15, 2016 at 6:34
  • $\begingroup$ Hum. In most practical calculation you probably could since that only about a 3% power difference.To get a more specific answer and account for all reflections you would need the actual S matrix, not just the insertion losses. (I prefer converting to the ABCD matrix which makes the cascading calculation simpler (although it can be annoying to convert back and forth). The link below has good examples of both methods and the pros/cons of each. ABCD method on page 40, S method using signal graphs on 54. .... ee.sc.edu/classes/Fall12/elct865/Class_Notes/… $\endgroup$
    – Andrew W.
    Commented Jun 20, 2016 at 18:23

2 Answers 2


Assuming the systems have no internal loss.

Reflected power ratio for system 1 is 0.1, for system 2 is 0.01. $ = \Gamma^2 = 10^{-RL/10}$

Power transmitted ratio for system 1 is 0.9, for system 2 is 0.99 $ = 1-\Gamma^2$

The first reflection back from each system will be the largest and subsequent reflections will be heavily attenuated, so we approximate by only considering the first reflections.

If we take A to be the power incident on system 1:

  • first reflection from system 1 will be $0.1*A$
  • incident power on system 2 will be $(1-0.1)*A = 0.9*A$
  • reflected power from system 2 will be $(0.01)*(0.9)*A$
  • reverse-transmission of system 2 reflected power through system 1 will be $(0.9)*(0.01)*(0.9)*A$

  • total power reflected (B) from both systems (first reflection only) $B = 0.1*A + (0.9)*(0.01)*(0.9)*A = 0.1081*A$

  • return loss of total system $RL = 10*log_{10}(0.1081) = -9.66dB$
  • 1
    $\begingroup$ Welcome to ham.stackexchange.com! $\endgroup$
    – rclocher3
    Commented Mar 18, 2020 at 22:42
  • $\begingroup$ Nice step by step explanation! $\endgroup$
    – Aparna B
    Commented Jan 3, 2021 at 2:12

Some time as passed since the original posting and I suppose the student has lost interest in this problem. But for the sake of completeness, I offer the following.

It must be said (as other commentators mentioned) that the problem as stated lacks sufficient information to give an exact result. Given more scattering parameters of each system, a conversion to T parameters to facilitate multiplying and then converting back to S parameters could be utilized. When only the combined S11 result is desired, this conversion process can deploy shortcuts to save on conversions.

Given the lack of required information , a statistical approach to the problem will yield the best available estimate. If we assume that the two stated Return Losses are unrelated, then applying RSS (Root of Sum of Squares) will yield a reasonable estimate (this is sometimes called "addition in quadrature").

First convert the RL figures to their respective reflection coefficient (Γ) or S11 scattering parameter:

$$ \Gamma_1 = 10^{10/-20} = 0.3162 $$

$$ \Gamma_2 = 10^{20/-20} = 0.1 $$

Then compute the RSS:

$$ \text{RSS} = \sqrt{\Gamma_1^2 + \Gamma_2^2} = 0.3316 $$

And finally, convert the RSS back to a combined RL:

$$ \text{RL} = -20 \log(0.3316) = 9.6 $$

It must again be stated that this results in a statistical estimate. If in fact there is significant gain, attenuation, or impedance transformation in these systems, this result could be substantially in error. In such a case, the order of the cascading could also have an effect on the results. But given the lack of information, it serves as a reasonable first order estimate. Meanwhile, I hope the good professor has removed or appropriately amended the question for future test pools.

It should be noted that this method can be used successfully when attempting to account for distributed reflections from connectors, etc. subject to the limitations noted above.

  • $\begingroup$ RL = 9.6 is indeed the correct answer. $\endgroup$
    – Aparna B
    Commented Jun 16, 2017 at 15:53
  • $\begingroup$ if system 1 has specified i/p and o/p return loss along with system 2 then cascase return loss will be same way calculated? $\endgroup$
    – user18725
    Commented Feb 4, 2021 at 11:56

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