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ECE405HW3Su2010Sol

# ECE405HW3Su2010Sol - ECE 405 HOMEWORK#3 Solutions SU 2010...

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ECE 405 HOMEWORK #3 Solutions SU 2010 ©DR. JAMES S. KANG 1 If ( ) ( / ) x t Arect t τ = , then 2 ˆ( ) ln 2 t A x t t τ τ π + = (a) 1 3 ˆ ( ) ln 3 t m t t π + = (b) ( ) 1,000 3 1,000 cos(2 1,000 ) ln sin(2 1,000 ) 6 3 USSB t t m t rect t t t π π π + = × × + t o t 2 ( ) 4sin(12 )cos(7 )cos(3 ) 2sin(12 )[cos(4 ) cos(10 )] m t t t t t t t = = sin(8 ) sin(16 ) sin(2 ) sin(22 ) t t t = + + + . ˆ ( ) sin(8 90 ) sin(16 90 ) sin(2 90 ) sin(22 90 ) o o o m t t t t t = + + + cos(8 ) cos(16 ) cos(2 ) cos(22 ) t t t = − 3 –10 10 t m(t) 4 (a) . ( ) 1 m t = ± ( ) ( ) 2 2 2 2 2 1 1 ( ) ( )cos ( ) 1 cos 2 2 c c c v t A m t t A m t t ω ω c = = + ( ) 2 2 2 1 ( ) ( )cos 2 2 c c v t A m t t ω = ( ) 2 1 ( ) cos 2 o c v t A t c ω = (b) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 1 ˆ ˆ ( )cos ( )sin 2 ( ) ( )sin cos c c c v t m t t m t t m t m t t t c ω ω ω = + + ω 1

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