ECE405HW5Su2010Sol

ECE405HW5Su2010Sol - ECE 405 HOMEWORK #5 Solutions SU 2010...

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ECE 405 HOMEWORK #5 Solutions SU 2010 ©DR. JAMES S. KANG 1 75,000/25 = 3,000. 3,000/60 = 50. multiplication factor = 50. 99.6/60 = 1.66 MHz, 50 × 0.2 = 10 MHz. f LO = 10 – 1.66 = 8.34 MHz or f LO = 10 – 1.66 = 8.34 MHz. 2 1 () T CC k m t =− 1 1 1( ) 2 ie kmt tk C LC ωω ⎡⎤ =+ = + ⎢⎥ ⎣⎦ ) f m t where 1 1 1 , 2 c cf k k C LC ω == (a) 6 31 2 1 11 10 10 / 51 0 21 0 c rad s LC −− = × ×× × (b) 15 6 3 12 1 3.351 10 10 10 8.3775 10 22 2 1 0 c f k k C × = × (c) 3 8.3775 10 6 8 2 1000 fm mm kA β π Δ× = = × × 3 200, 400, 600, . ......... , 2400 (a) 200 2 16 (1 ), 5.25 BW kHz =≥ × × + (b) 2 2 . 5 9 T B fB M H z + = + = . 4 25 550 1.08 ( 1) 5000 n +×× . . n = 9. ( 8.27542 n −≤ 2 (1 ) 2 25 (1 10) 550 B += × × + = 5 ( ) 2 4 1 2 T B M H z + = += . 6 Baseband Spectrum: 1
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-2 -1.5 -1 -0.5 0 0.5 1 1.5 2 x 10 4 0 0.2 0.4 0.6 0.8 1 f Sbb(f) FDM Baseband Spectrum Transmitted Signal Spectrum: -5 -4 -3 0 1 2 3 4 5 x 10 4 0 0.2 0.4 0.6 0.8 1 f FDM Baseband Spectrum Carrier frequency = 30 kHz. A c = 3V 0 1 2 3 4 5 x 10 4 0 0.5 1 1.5 2 f Sam(f) FDM Passband Spectrum 7 2 ( ) 3000 10 30 i SM PS f d f −∞ == × = 4 6 30 1.5 10 41.761 10 2000 i S o P SNR dB NB = × = × 2
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8 () 2 3 2 oi SNR SNR β = , 2 2 2 2 c ic i oo o A SA SNR NB NB Nf γ == = = m 2 10 3 40 10log 2 i SNR ⎛⎞ =+
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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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ECE405HW5Su2010Sol - ECE 405 HOMEWORK #5 Solutions SU 2010...

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