Question
Download Solution PDFA short-circuit admittance matrix of a two-port network is
\(\left[ {\begin{array}{*{20}{c}} 0\\ {\frac{1}{2}} \end{array}\begin{array}{*{20}{c}} { - \frac{1}{2}}\\ 0 \end{array}} \right]\)
The two-port network is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Admittance Matrix:
- It is also known as a short circuit matrix or Y matrix.
- Y matrix is represented as:
\(\begin{bmatrix} Y_{11} & Y_{12}\\ Y_{21}& Y_{22} \end{bmatrix}\)= \(\begin{bmatrix} Y_{A}+ Y_{C}& -Y_{C}\\ -Y_{C}& Y_{B}+ Y_{C} \end{bmatrix}\)
- The condition of symmetry and reciprocity in Y parameters are given by:
Symmetry: \(Y_{11}= Y_{22}\)
Reciprocity: \(Y_{12}= Y_{21}\)
Explanation:
Given, that the Y matrix is \(\left[ {\begin{array}{*{20}{c}} 0\\ {\frac{1}{2}} \end{array}\begin{array}{*{20}{c}} { - \frac{1}{2}}\\ 0 \end{array}} \right]\)
Here, \(Y_{12}\neq Y_{21}\)
Hence Y matrix is not reciprocal.
The shunt admittance dissipates energy, hence it is a passive element.
Therefore, option 1 is correct.
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