Unitary property of scattering matrix
But how do we derive the above unitary property expression involving S22 and S21 ?
From S parameter’s definition.
Hi promach,
Mathematically speaking, a unitary matrix is one which satisfies the property [S]^* = [S]^{-1}. Re-arranging, we see that [S]^* [S] = [I], where [I] is the identity matrix.
Inserting the [S] matrix into this equation, we can then see that any column dotted with itself is equal to unity. Conversely, if any column is dotted with any other column, the product is equal to 0.
So if we take your 2x2 scattering matrix and look at some examples, we would find that |S12|^2 + |S22|^2 = 1. If the network is reciprocal, then S12 = S21. We would also have that |S11|^2 + |S21|^2 = 1 and (S11*)(S12) + (S21*)(S22) = 0.
Reciprocity is not required.
From nature of Unitary Matrix, [S]^* [S] = [S] [S]^* = [I]
So |S21|^2 + |S22|^2 = 1 is satisfied without reciprocity.
This could be explained by law of conservation of energy. The injected power must be equal to the reflected power + the transmitted power for a lossless network.
Wait, I do not understand why [S]^* [S] = [S] [S]^* = [I]
It is matrix nature of kindergarten level.
For [B][A]=[I], [B] has to be [A]^{-1}.
So [A][B]=[I] is satisfied.
This has no relation to S-parameter at all.
That?s all.
@pancho_hideboo
For ideal lossless case:
transmitted power + reflected power = source power
So, divide the whole expression just above by source power, we have the following:
|S21|^2 + |S22|^2 = 1
You can not understand things correctly.
Wrong.
We can not get this directly from "transmitted power + reflected power = source power".
We have followings from "transmitted power + reflected power = source power".
|S11|^2 + |S21|^2 = 1
|S12|^2 + |S22|^2 = 1
However from matrix nature of kindergarten level, |S21|=|S12| is satisfied.
So |S21|^2 + |S22|^2 = 1 is satisfied.
This does not require reciprocity at all.
Learn basic things surely.
Attached is an excerption from kindergarten level book.
Here "~" means conjugate complex and transpose.