Question

Q8) Consider that analog message signal is m (t) = a cos (21 fmt) + b cos (41 fmt) modulates a carrier c(t) = 2 cos (27 fot). The modulation type is the double-side band suppressed carrier modulation (y(t) = m(t) c(t)). The modulated signal is filtered with a band-pass filter with a following charac- teristics in the frequency domain. 1 f 5fm H(f) 1 fo +1, fo - 5fm< lf<fo 5 fm fo<\f|< fo +5fm fo +6, fo +5fm< lf<fo +6fm fm |f| > fo +6fm and 0 < lf<fo - 5fm 1 f+ fm The filter output is the vestigial side band modulation. Obtain the filter output. H(f) fo + 6fm N fo - 5 fm fo +15 fm

Answer 1

Oo olo 36) G o Ixxk 7,4) code ILuxu Pxn-x) PBpX3 nik Hamming code code [6] = information Generator matria [C] = [1110] I DOO 0 1 1 01001101 o olol ll 0 0 0 1 [c] Il lloOo C ti 10000] sum 14120 р (6) é 011 i = [11!!] [c] = [11] 0000 o 100 o 01 o 00 101 llo 0 0 1 ㅁ (C) [c] = [llll is [1010] [C] = iXG = [1010] OOO oll o loo oo lo 1 1 OOO CS Scanned withic [10 010101] CamScanner

(88) cos (21fmt) tb cos(unfmt) 2 coskafat), mt) = a cox (27 fmt) tbcos (unfint) Ct). E cos ceafat) d) = mt) [acor yk) c la cos(zatrt) con ez aft)+3 rn abcorchofimt) Com (20 fot) 2 COOA.cosB = [CORLA +B) +COR (A-B)] Yt) = gaxt [cos 2016 fmt falt +coser (fo-fo) +26X1,[compo(fot2 fm) + Cop 26{fo-fond] yt a cos en funtfoltta con call-fm) +6 cos20kfotz fm))+bcox [28(to-2fm)) House Coperoft -> 41864+)+${{5 fo)) ya) = f [ 864+64m] (8-64m ã [f (f +6fo-Pomo)+off-(o tom tz [(f+fotz fub+8[f- fot2 fm) + b2 [8(f+6o2fm) +86f-fo-2 from + pala YG ible 9/2 19/ able film freefom forff fotton fot 2fm Foton to foffen fot2fm 9 ed with CS Car anner

te) fo-5fm fo fotsfm to tofm. >f di g d2 J2 (fo-5.6m), o) (tot) slope = I₂ Y 2₂-X, fa-for-2fm Y= -fo-2fm 5 for H = =-to-%+ 56m to-fo+5fm = L y = -fo. +315 5fm 5fm ya amate c) intercept on yaxis. Cal. y = ti 5fm filter olp= Y(B) HP) ble al ble £+1) 3/2 1) alz £ 11/9/2 6Poti Astm NL f-2fm to-fm to fotfm futzfm & Scanned with tothm fot2fm 2 aritátmer