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If the S parameters of a transistor given are S11=-0.811-j0.311 S12= 0.0306+j0.0048 S21=2.06+j3.717 S22=-0.230-j0.4517 Then ∆ for the given transistor is:

If the S parameters of a transistor given are S11=-0.811-j0.311 S12= 0.0306+j0.0048 S21=2.06+j3.717 S22=-0.230-j0.4517 Then ∆ for the given transistor is: Correct Answer 0.336

Given the S parameters of a transistor, the ∆ value of the transistor is given by │S11S22-S12S21│. Substituting the given values in the above equation, the ∆ of the transistor is 0.336.

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