Assignment-II_A

# Assignment-II_A - EE3008 Solutions to Assignment 2 Q1(i m=...

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EE3008 Solutions to Assignment 2 Q1 (i) magnitude min. magnitude . max magnitude min. - magnitude . max + = m => 40 40 7 . 0 + = B B => B=226.7 (ii) % 6 . 19 % 100 7 . 0 2 7 . 0 % 100 2 2 2 2 2 = × + = × + = m m η (iii) t t m A t c m AM ω φ cos ) cos 1 ( ) ( + = => A (1+0.7)=226.7 => A =133.3 t t t c m AM cos ) cos 7 . 0 1 ( 3 . 133 ) ( + = (iv) t t t t c m c AM cos cos 3 . 93 cos 3 . 133 ) ( + = Q2 (i) (ii) Q3 (i) Now f ( t )= A cos(2 π f m t ). From φ FM ( t )=200cos(2 π 10 7 +10sin4000 π t ) we have f m =2kHz or ω m =4000 π rad/s, so θ (t)=2 π 10 7 t + 10sin4000 π t ω (t)=d /d t =2 π 10 7 +40000 π cos4000 π t Therefore, max. freq. deviation f =40000 π /2 π =20kHz −3 f c 0 f c AA c /4 3 f c f c Φ 2 ( f ) 2 f m 2 f m f 2 f m 2 f m 46.7 ∗0.5 133.3 ∗0.5 - f c f Φ( f ) 2 f m f c 2 f m 0 AA c /2 Φ 1 ( f ) 2 f m 2 f m f - f c f c

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(ii) W R t P FM t 400 50 2 200 ) ( 2 2 = = = φ (iii) ) sin cos( 200 ) ( t t t m c FM ω β + = where β =10; t n J t m c n n FM ) cos( ) ( 200 ) ( + = −∞ = carrier component, n =0, 200 J 0 (10)=200(-0.25) = -50 first side bands, n =1, 200 J 1 (10)=200(0.04) = 8 second side bands, n =2, 200 J 2 (10)=200(0.25) = 50 Fourier spectrum (iv) Carson’s Rule: B =2( f + f m )=44kHz. Q4 . (i) (ii) The peak frequency deviation 0 12 4, 6 , 6 8 t cc d tt d t t t dt θ ωπ π =+ = + = At t=8, 24 , 48 max max = = f Hz
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## This note was uploaded on 01/11/2011 for the course EE 3008 taught by Professor Pingli during the Fall '08 term at City University of Hong Kong.

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Assignment-II_A - EE3008 Solutions to Assignment 2 Q1(i m=...

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