SRC_stateplane_F_solution - SRC k=1 state‐plane diagrams...

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Unformatted text preview: SRC k=1 state‐plane diagrams for two special cases (a) F = 1, J = 2, hence M = 1 jl 4 π 2 β =π α =0 β =π A=π M=1 1+M=2 α =0 2 4 mc 1‐M=0 (b) F=0.5, M=0.5, hence J = 2/ jl M=.5 1 β =π ‐1 β =π ‐1 A=1+M=1.5 α = π ‐2 1+M=1.5 1 2 mc 1‐M=.5 A=M=0.5 α = π Thanks to Mark Norris for providing the diagrams ...
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This note was uploaded on 10/02/2010 for the course ECEN 5817 taught by Professor Maksimovic during the Spring '10 term at University of Colombo.

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