Timeline for 3 dB Power Divider Used as Power Combiner S Matrix Explanation
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| S Nov 13, 2023 at 17:15 | history | suggested | Jason | CC BY-SA 4.0 |
Formatted formulas.
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| Nov 13, 2023 at 16:08 | review | Suggested edits | |||
| S Nov 13, 2023 at 17:15 | |||||
| Nov 13, 2023 at 16:04 | vote | accept | Exclose | ||
| Nov 13, 2023 at 16:03 | comment | added | Jason | No worries - I thought that the first time I was thinking through the problem too. | |
| Nov 13, 2023 at 16:02 | answer | added | Jason | timeline score: 1 | |
| Nov 13, 2023 at 15:58 | comment | added | Exclose | Thank you @Jason I now understand what you are saying. I never stopped to think about the fact that it combines power not voltages directly. I thought we would get A+B in voltage as well. | |
| Nov 13, 2023 at 15:54 | comment | added | Exclose | Ok let me phrase it this way: With what you are saying then if i apply waves with power of P from both ports (in same phase) since we take the square of the scattering matrix we should get S21^2 = 1/2. Only half of the power would be transmitted from port 2 to port 1. If thats the case (S21^2=1/2) when we apply power of P from port 2 and 3 total input power is 2P and total output power is P. Which still doesnt make sense to me. I understand now what you meant though | |
| Nov 13, 2023 at 15:50 | comment | added | Jason |
Yes. Conservation of energy. |2*A/sqrt(2)|^2 = 2A^2 which is TWICE the power of A. If the answer were 2A then |2*A|^2 = 4A^2 which is 4x the power of A.
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| Nov 13, 2023 at 15:47 | comment | added | Exclose | @Jason If we apply waves in same phase both with amplitude of A volts. You are agreeing that the output will be 2A divided by sqrt of 2. Which equals to A times sqrt of 2. But our total applied input is 2*A which is not the same thing as A*sqrt2 Our input of 2*A does not equal to output of 1.414*A | |
| Nov 13, 2023 at 15:45 | comment | added | Jason |
I don't follow what you're saying. If you apply signal A to both port 2 and 3, port 1 will be 1/sqrt(2)*(A + B)=2A/sqrt(2) which is still twice the power of A.
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| Nov 13, 2023 at 15:42 | comment | added | Exclose | @Jason If we apply A from port 2 and 3 at the same time the total input is 2A in your case. Then we got A*sqrt2 at the output. 2A to 1.414*A therefore something is missing. | |
| Nov 13, 2023 at 15:39 | comment | added | Jason |
1/sqrt(2)*(A + B) is correct. If A = B, you'll have 2/sqrt(2)*A which is twice the power of A. It didn't go anywhere.
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| Nov 13, 2023 at 15:32 | history | asked | Exclose | CC BY-SA 4.0 |