2006finalsolution

2006finalsolution - EE 279 Winter 2005-2006 Professor Cox...

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Unformatted text preview: EE 279 Winter 2005-2006 Professor Cox HO # Solution to Final 1. (12pt) a) ii) b) i) iii) c) i) iv) d) vi) 2. (35pt) In phase-acceleration modulation we have: ! + = t c d x K f t f " " ) ( ) ( . Therefore to recover the signal we should extract the phase as we did for PM and FM and then take the second derivative of the phase ) ( t ! . Therefore the demodulation process would be: ) ( 2 1 ) ( t t y v ! " & & = (5pt) First we calculate the R N S ! " # $ % & in the same way as we define it for PM and FM: 2 / 2 c R A S = (5pt) T R B N N = (5pt) T C R B N A N S 2 2 = ! " # $ % & As we have seen for PM and FM, when 10 > ! " # $ % & R N S then: ) ( 2 1 ) ( ) ( t n S t t q R v + = ! ! (5pt) So we have: ) ( ) ( ) ( t t Kx t y ! + = where ) ( ) 2 ( ) 2 ( 2 1 ) ( 2 4 f G f S f G n n R ! ! " = and therefore after the post detection filter we will have: x D S K S 2 = (5pt) ! " = = w w R D N w S df f G N 5 ) 2 ( ) ( 5 2 # $ (10pt) 3. (70pt) Recall from midterm that at B, we have s B ( t ) = cos 24 w o t + 24 k sin w m t ( ) = cos 24 w o t + 2.4048sin w m t ( ) (a) At J, we have s J ( t ) = 1 + g ( t ) ( ) " cos 24 # o t + 2.4048sin # m t ( ) = 1 + g ( t ) ( ) " J n (2.4048)cos (24 # o + n # m ) t ( ) n = $% % & We note that since " = 2.4048, J ( " ) = , we have no FM carrier. What about the switching sidebands from g ( t ) ? The switching sidebands are all AM modulated with carrier. The odd order FM sidebands are of opposite sign and so any superimposed AM sidebands from switching (if any) would cancel at 24f o . So, we have no output from these odd order FM sidebands. The even order FM sidebands are of same sign. What are the switching sideband amplitudes superimposed on the even order FM sidebands at 24f o ? g(t) is a square wave with even order sidebands ( kf s , k even) of 0 and odd order sidebands having amplitudes proportional to 1/k. Since f s = 200 Hz , even order sidebands are at k s " 200 Hz and even order FM sidebands are at k m f m = k m " 10 kHz . That is, k m f m for k m even = 20kHz, 40kHz, 60kHz Then, we have 20 kHz 200 Hz = spacingof FMevensideband...
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2006finalsolution - EE 279 Winter 2005-2006 Professor Cox...

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