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I'm trying to learn how to correctly calculate (cascade==RF system cascading) the connection of 2-port networks, taking into account mismatch losses. For example There is a first 2-port network, its S parameters at 10GHz, enter image description here enter image description here And a second 2-port network. enter image description here enter image description here They are shown in the pictures below. If you try to add them up as an RF system (!!!not adding S parameters via ABCD or T matrices!!!) - you have to add their insertion loss ((-0.577)+ (- 0.895) = -1.472 ). So the total insertion loss should be 1.472 - only if the 2-port networks themselves are matched by exactly 50 ohms (perfect impedance matching). But in fact, as you can see in the picture in ADS and ADSimRF - the insertion loss is 1.508. enter image description here enter image description here enter image description here My question is this. If I know the input and output impedance of the 2-port networks (s11 and s22 impedance on SmithChart) how do I need to account for mismatch losses to get the same results in ADS and ADSImRF ? I think I need to calculate the mismatch loss between port 1(50 ohms) - and the first 2-port network (46.999+j16.888), then between network 1(46.999+j16.888) and network 2 (55.2+j2.594) and finally between network 2(55.2+j2.594) and port 2.

I tried to use the formula

MismatchLosses = -10log(1-Г^2)

but either I calculated something wrong, or it does not fit.

I will be grateful for any help or advice



@Jens I don't understand, did I dosmth wrong? enter image description here



Hi, Your calculation is almost correct. Note, in the above first GT -equation, there is a typo; it should read GT1 (not GT2 ). Your values (at the top right) of 0.895 and 0.577 are dB-Values; actually, they should be negative because 10log(|s21|2)=–0.895dB for NET1; and 10log(|s21|2)=–0.577dB for NET2. They represent the Power Transducer Gains of NET1 and NET2, respectively; and the values apply under the condition of perfect match, i.e., s22=0 of NET1 (or ΓS=0 in GT2 for NET2), and s11=0 of NET2 (or ΓL=0 in GT1 for NET1). So, the cascade (total) Power Transducer Gain is –0.895dB – 0.577dB = –1.472dB for this perfect matching, being a reference case or benchmark. Considering now the mismatch, your calculated linear values of 0.866 and 0.783 for GT1 and GT2, respectively, are correct; and the corresponding product of 0.678 is also ok. However, then we need to take 10log(0.678) which is –1.688dB. Note the Transducer Gain refers to Power, (not voltage or current). Therefore, the values |s21| appear as squared in the GT -equations. In conclusion, the mismatch between NET1 and NET2 imposes a penalty, or additional power loss of 0.22dB in your example case. It occurs that cascade programs (a) neglect the denominator in the GT -equations, and sometimes (b) assume implicitly the source as matched (ΓS=0 ). If we assume both (a) and (b), the resulting cascade (total) Transducer Gain becomes –1.49dB in the example. I hope the above clarifies the issue. Jens`

@Jens https://ham.stackexchange.com/review/suggested-edits/10545

Yes, I saw the typo about GT and about the negative dB(S21) values for network 1 and 2.

And still I can't understand you -

"However, then we need to take 10log(0.678) which is -1.688dB." - Yes it is, but where is the similarity with the values calculated in ADS? In ADS = -1.508dB. -1.688 is not equal to -1.508.

How do I get 1.508 ? or the difference is the interstage loss ? 1.508 (total of 2 nets) 1,472 (total value if adding up the values separately) 1.508 minus 1.472 = 0.036 dB

I do not see anywhere even close to a value of 0.036

"In conclusion, the mismatch between NET1 and NET2 imposes a penalty, or additional power loss of 0.22dB in your example case." - 0.036 does not equal 0.022 again

"It occurs that cascade programs (a) neglect the denominator in the GT -equations, and sometimes (b) assume implicitly the source as matched (ΓS=0). If we assume both (a) and (b), the resulting cascade (total) Transducer Gain becomes -1.49dB in the example." What is it mean? You mean EDA Keysight ADS or ADISimRF counts wrong or what? Then why do they both add up?

Let me explain what I need it for. If you try to calculate cascade P1dB with nets AS IF matched at 50 ohms each - you get one value. But, if you take mismatch losses into account - you will get the total P1dB higher, because mismatch losses - will increase the cell losses - accordingly if you look at the cascade P1dB formula - you will see that this leads to an increase in the total P1dB. I wrote in more detail here in a similar question and explained the reasons (https://www.edaboard.com/threads/how-to-take-to-account-impedance-mismatch-losses-when-cascading-2-port-networks-as-rf-system.398650/)

So - I still don't understand how do I calculate the interstage mismatch loss (get the total dB(S21) of the two networks = 1.508 as in ADS) AND/OR then - how much should I change the cell loss to get a more accurate total P1dB value.

Excuse me for my misunderstanding.

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    $\begingroup$ Hello and welcome to ham.stackexchange.com! $\endgroup$
    – rclocher3
    Jun 30 at 16:01
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Let’s assume a 50Ω Source feeding a 2-port network NET1, which is followed by a 2-port network NET2, which is terminated into 50Ω. We apply well-known equations from technical RF-literature. The (modified) input reflection coefficient $s’_{11}$ of the (output-mismatched) NET1 is

$$s’_{11} = s_{11} + \frac{s_{21}s_{12}\Gamma_{L}}{1-s_{22}\Gamma_{L}}$$

with (complex) s-Parameters of NET1, and $\Gamma_{L}$ for the reflection coefficient implied by the load that is seen by NET1 at its output. Because NET2 is terminated (matched) into 50Ω load, the $\Gamma_{L}$ in the above equation equals the $s_{11}$ of NET2. For the given two networks, the value of $s’_{11}$ is (0.211; 80.48deg) which corresponds to the value calculated by ADS for the $s_{11}$ of the cascade.

The cascade input mismatch is calculated by inserting $s’_{11}$ into $10log(1-|s’_{11}|^{2})$, assuming the source is 50Ω.

The approach for the output of the cascade is accordingly, i.e.,

$$s’_{22} = s_{22} + \frac{s_{21}s_{12}\Gamma_{S}}{1-s_{11}\Gamma_{S}}$$

with (complex) s-Parameters of NET2, and $\Gamma_{S}$ for the reflection coefficient implied by the source that is seen by NET2 at its input. Because NET1 is fed by a 50Ω source, the $\Gamma_{S}$ in the above equation equals the $s_{22}$ parameter of NET1. The value of $s’_{22}$ is (0.096; 161.036deg) which corresponds to the value calculated by ADS for the $s_{22}$ of the cascade.

The output mismatch is calculated by inserting $s’_{22}$ in $10log(1-|s’_{22}|^{2})$, assuming the load is 50Ω.

The following is about the results from ADSimRF and ADS, and the interstage mismatch.

In the considered example and for the ideal case of no mismatches along the cascade the (total) maximum Gain would be –0.577dB–0.895dB = –1.472dB.

ADISimRF applies a scalar approach; it obviously does not consider complex values. Therefore, it is no surprise when ADISimRF, in general, provides only approximative results when phase-dependent interstage mismatches exist. ADISimRF appears to tackle interstage mismatch in the following way. Let’s assume two stages NET1 and NET2 with individual Power Gains $|s_{NET1,21}|^{2}$ and $|s_{NET2,21}|^{2}$, respectively. NET1 has the real-valued output impedance $Z_{1,out}$, and NET2 has the real-valued input impedance $Z_{2,in}$. The resulting Power Gain $G$ of this 2-stage cascade is calculated in ADISimRF according to

$$G = |s_{NET1,21}|^{2} \cdot |s_{NET2,21}|^{2} \cdot (1-|\Gamma_{12}|^{2})$$

with

$$\Gamma_{12} = \frac{Z_{2,in}-Z_{1,out}}{Z_{2,in}+Z_{1,out}}$$

which considers the mismatch between NET1 and NET2. Performing the math (for the given $Z_{1,out}=46.99Ω$ and $Z_{2,in}=55.2Ω$), the resulting Power Gain $G$ is –0.577dB–0.895dB–0.028dB= –1.5dB which is exactly the result shown in ADISimRF. This explanation is an answer to the core of the Question.

ADISimRF applies the shown approach per each inter-stage interface along the cascade. Note, that $Z_{in}$ and $Z_{out}$ are in ADISimRF input values provided by the user for each stage independently. This is a simplification. (In general, the input impedance $Z_{in}$ of stage “i-1” depends on $Z_{in}$ of the next following stage “i”.)

The above mismatch loss (for real-valued $Z$) can be re-written in the form of

$$(1-|\Gamma_{12}|^{2}) = \frac{4Z_{1,out}Z_{2,in}}{(Z_{1,out}+Z_{2,in})^{2}}$$

For complex-valued impedances ($Z=R+jX$), the following equation applies

$$(1-|\Gamma_{12}|^{2}) = \frac{4R_{1,out}R_{2,in}}{(R_{1,out}+R_{2,in})^{2}+(X_{1,out}+X_{2,in})^{2}}$$

where NET1 is considered the source, and NET2 is the load. Note $(1-|\Gamma_{12}|^{2})=1$ if the two impedances $Z_{1,out}$ and $Z_{2,in}$ are complex conjugate ($Z_{1,out}=Z^{^{*}}_{2,in}$).

In the example, the relevant input and output impedances of the two stages do not much differ from each other. Therefore, from ADISimRF the resulting cascade Power Gain of –1.5dB is only slightly lower than the ideal –1.472dB for no mismatches.

From ADS, the obtained s-matrix for the cascade provides the power gain $|s_{21}|^{2}$ of –1.508dB which is similar to the cascade Power Gain result from ADISimRF. Although ADISimRF and ADS apply fundamentally different approaches, they provide here similar Power Gain results for the specific cascade because the relevant impedances do not much differ from 50Ω.

The cascade is terminated into 50Ω. When connecting this cascade to a 50Ω source, the $|s_{11}|=0.211$ of the cascade (result from ADS) determines the power mismatch between source and connected cascade. This mismatch is $(1-|s_{11}|^{2})=–0.198dB$. The mismatch at the cascade output is small ($|s_{22}|=0.096$) which implies a mismatch of only –0.04dB.

One can obtain the same mismatch results at cascade input and output by referring to the corresponding complex impedances (from ADS) and using the above equations.

In ADISimRF the mismatch between cascade input and source is not incorporated in the Power Gain result. Instead, in ADISimRF the cascade Input Power is a value (in dBm) that is provided by the user. The input impedance of the first stage and the output impedance of the last stage have no impact on the calculated cascade Power Gain (but they do influence voltage gains). Therefore, mismatches at input and output of the cascade must be considered separately also in ADISimRF for any adjustment of the calculated Power Gain if, for instance, a Transducer Gain value shall be obtained.

Now, about the Power Gain result from ADS.

The ratio of output-power to input-power, Gain $G_{1}$, of NET1 can be expressed as

$$G_{1} = \frac{|s_{21}|^{2}(1-|\Gamma_{L}|^{2})} {|1-s_{22}\Gamma_{L}|^{2}-|s_{11}-(s_{11}s_{22}-s_{12}s_{21})\Gamma_{L}|^{2}} = 0.9059\ \ (or\ \ –0.4292dB)$$

with $\Gamma_{L} = s_{11}$ of NET2 (terminated into 50Ω).

The ratio of output-power to input-power, Gain $G_{2}$, of NET2 can be expressed as

$$G_{2} = \frac{|s_{21}|^{2}} {1-|s_{11}|^{2}} = 0.8163\ \ (or\ \ –0.8815dB)$$

The gain $G_{C}$ (linear) of the cascade (terminated into 50Ω) is the product of $G_{1}$ and $G_{2}$,

$$G_{C}=G_{1}\cdot G_{2}=\frac{|s_{C,21}|^{2}}{(1-|s_{C,11}|^{2})}=0.7395\ \ (or\ \ –1.3108dB)$$

or

$$|s_{C,21}|^{2}= G_{1}\cdot G_{2}\cdot (1-|s_{C,11}|^{2})$$

where the s-Parameters now belong to the cascade. Inserting the values for $G_{1}$ and $G_{2}$ from above and $s_{C,11}$ for the cascade implies $|s_{C,21}|^{2}=–1.508dB$, which exactly corresponds to the value provided by ADS.

The above confirmed in every detail the values calculated by ADISimRF and ADS for the specific case of a cascade of only two stages considered in the Question. (Remark: ADS actually applies a method that is fundamentally different; the above approach is here doable for a result-verification and for the simple cascade of only two stages with 50Ω source and load.)

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    $\begingroup$ ham.stackexchange.com/questions/7333/… arrives at nearly the same equation, in case someone is wondering how it's derived. $\endgroup$ Jul 3 at 2:28
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    $\begingroup$ Moderator here — I saw your suggested edit of the other answer. Unfortunately, I had to delete that answer because the Q&A form of the site requires that "further questions" not be posted as answers — and edits should never be replies to the post being edited. But the text you wrote is preserved here so you can post it here or elsewhere as needed. $\endgroup$
    – Kevin Reid AG6YO
    Jul 16 at 22:10

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