Question

1) TWO bose band signals Mice) Malt) each hos Bandwidth B ore Simultaneously applied to the following Am systems @ Find the output at points 1,2,3,4. . X.): M.ct) sin(WE) exalt): m2(t) sin Cuct) COS(wt) . *3(t): m.(t) cos Cuct) X4(6), 32(), Xalt) malt) ma(t) sin(at) + M. (t) cos ( 6) - mayo 121 sketch the spectrum at points 2,3,4 . 22(£): 3 [m2(fo) - M2 (E-fe)] . 33.(). [M. Oft fe) MCF-fo) ] . X4 (A). 5 [M2 (f+fo) - M2 (6-fo) ] + ! [m, (ftfa) + M, CA-fo)]

Answer 1

Summary:

1) Take care of -90 degree in multiplication of message signal with cos function and use trigonometry properties

2)Apply multiplication property of Fourier transform

m.lt cos (wat) malt) Let the output at o is x,ct) 2, (t) = milt) cos(wet. I ) = molt) cos (I-wet)) = moct) cos(I wat) (since cos function is even, x.lt) = molt) sin (wet) let the output at @ is dalt) blz (t) = m(t) cos (Wet- .) (t) = m₂ (t) sin wat let the output at point (t) = m.lt) coswat is xght) Let the output at point o is xact) xact) = m.ct) coswit + m₂ (t) sincet X417) = Hz(t) +X3 (t)

b xict = mict) sin(Wet) mict), Fourier Troms form Mact) (scf-fc) - 8 (ft fc)) milt) a sin(wet) & Fil i using multiplication molt) sin (wit) ft By property of Mo(f) * i Fourier Transform sef-fc-8cft fc)] y convolution symbol Hlt), m.ct) sin(wet) <1> rict) M. Cf) * 86f-fe) - Micf) * 8cftf) [Milf-fc)-Mi (ft fc)] = x, (f) we know that the Bandwidth of milt of mact) is B Milf) M₂(f) + Bf -B +B f Magnitude spectrum of Xo(f) Plocf 7/2 -fe-B - fc -fctB lo fuß ft B

phase spectrum of xilt) A. (f) KIN f

xact) = malt) sin(wet) malt) ft M₂(f) sin wat & (olf-fc) – 8 (fx fc)) R2(t) = m₂ (t) sinwet > (M2Cf- fe) – Mz ft fc)) excf) Magnitude spectrum of selt) F.T 1 xz () - - - 1/2 - - - - -fc-B -fc - fctB fc-B fc fctB af phase spectrum of salt) A 1X2(f) kid

it FT X3 (t) = milt coswet molt stil milf coswet & FT 2 [def-fc) + dlftfu) 8g(t)= moct coswet & Xz (f) = IT Molf) & (f-fc) + Mocf 8 (ft fc) X3<f) = [ Mo f=fc) + M, cf+f0] mag spectrum of x=(f) ------ ---- - fc-B - - fitB fc-B fc Pagt phase spectrum is nil 3 (f) > F Note: magnitude are some spectrum but of phase Both X (f) and X3 (f) spectrum is different.

xa(t) = x₂ lt) +xz(t). X4 (F) = x2(6) +X3(f) X4(f) = { [M, (f-fc) - Melft fc)] + [mo(ff) IM, (ft fe)] , X264)) ----- ------- - fct FEB & fctß Filters of – fc fet B fc-B X3(f) N Xe f(-B fc fet B fc-B magnitude spectrum of Xa(f) 1 1X4(f)) 1 ------ -- - PC-B fet B fc-B – fc fc tB fc & since Fourier Linear property. Transform pLX2(f) follows I TL fc f 1/X3 (f) CC 1 *ZX44) →f N12 f