ECE405HW4Su2010Sol

ECE405HW4Su2010Sol - ECE 405 HOMEWORK #4 Solutions SU 2010...

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ECE 405 HOMEWORK #4 Solutions SU 2010 ©DR. JAMES S. KANG 1 PM: The phase deviation is φ (t) = k p m(t) = 6 sin(2 π 5000t). m(t) = 2 sin(2 π 5000t). FM: The phase deviation is . Taking the derivatives on both sides, we obtain k f m(t) = 6 × 2 π× 5000 cos(2 π 5000t). m(t) = 10 cos(2 π 5000t). ( ) 6sin(2 500 ) t f kmd t λλ π −∞ = 2 () 2 T B Bf =+ Δ (a) . 220 2 80 , 30 kHz B kHz B kHz = (b) . 220 2 20 , 90 kHz kHz f f kHz Δ Δ = 3 Δω = k f m p = 2 π× 2000 × 6=2 π× 12000, Δ f = 12 kHz. ( ) ( ) 22 6 1 2 3 6 T B k H Δ = + = z 4 (a) 9 2 1,000 , 2 1,500 3,000 / / 6 fm m f mm kA kr A βω βπ ω ×× == = = × = a d s V (b) 2 32, 8 2 c c A A Power on the sideband at 60 kHz = 2 10 1 9 32 (0.1247) 0.4976 2 c AJ W = 5 The instantaneous phase is given by θ i (t) = 20 × 10 6 t + 20t + 5 sin(2 π× 10t). Since the instantaneous frequency is the time derivative of the instantaneous phase, we have ω i (t) = 20 × 10 6 + 20 + 2 π× 50 cos(2 π× 10t). 6 7 ( ) 2 10 8( 2 2,000)sin(2 2,000 ) i f tt ππ =×+ × × 7 2 10 2 16,000sin(2 2,000 ) t −× × Thus, 16 f kHz Δ= . 7 β = Δ f/f m = 99/11 = 9. (a) max 1% 2 2 13 11 286 Tm B nf k H z × × = (b) 2 2 11 99 220 T B k H Δ = + = z (c) ( ) 2 2 3 11 3 99 660 T B k H Δ = × + × = z 8 1
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This note was uploaded on 08/09/2010 for the course ECE ECE 405 taught by Professor Jameskang during the Summer '10 term at Cal Poly Pomona.

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ECE405HW4Su2010Sol - ECE 405 HOMEWORK #4 Solutions SU 2010...

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