ECE405HW8Su2010Sol

ECE405HW8Su2010Sol - ECE 405 1 For 0 T, HOMEWORK #8...

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ECE 405 HOMEWORK #8 Solutions SU 2010 ©DR. JAMES S. KANG 1 For 0 τ T, () 2 2 ( ) 00 1 2 TT a at a t a at a T H e Re e d t e e d t e a ττ τ −− −−+ − − == = ∫∫ Since R H (- τ ) = R H ( τ ), 2( ) 1, 2 0, a aT H e eT R a elsewhere ⎡⎤ −≤ ⎣⎦ = ) 1 2 0, a XH e RR aT T elsewhere Since [ ] 1 cos( ) sin( ) 11 aT aj T j ee H aj ωω ω −+ =− = = ++ + + T The power spectral density is given by 2 2 22 1c o s ( ) s i n ( (2 ) aT aT X Hf Sf e T e T a f π + ⎢⎥ + ) 2 12 c o s ) aT aT e Ta f + + 2 (a) ()( ) H R htht d t −∞ =+ (0) 4 1 2.5 H R T =× +× = 21 H R T ⎛⎞ =×× = ⎜⎟ ⎝⎠ For 0 τ T/2, 56 4 1 2 H T R T 5 × + × +− × = For T/2 τ T, () ( ) 22 1 H RT T T × = Since R H (- τ ) = R H ( τ ), we have 1
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() 56 1, 25 21 , 2 0, H TT T T 2 R T elsewhere ττ τ ⎛⎞ −≤ ⎜⎟ ⎝⎠ =− which is shown below. 2.5T 0 T/2 –T/2 T –T τ R H ( τ ) Notice that () 2 2 2 H T RT t r i t r i T T =+ (b) 3 22 44 ( ) sinc sinc 2 jf fT T fT Hf T e e ππ −− 2 24 1 sinc fT Te e e π 4 T (c) 1 , 2 0, MH T T T 2 R elsewhere == which is shown below.
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ECE405HW8Su2010Sol - ECE 405 1 For 0 T, HOMEWORK #8...

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