Question

Communication systems

d) The angle modulated signal is presented as = 1008 (2*10*4 +0.5sin(10*x+)) 1. Find and determine the frequency modulated signal x(C) with k, = 5. (5 marks) II. Calculate and find out the message signal m(t). (5 marks) e) A 40 MHz carrier is frequency-modulated by a sinusoidal signal with mascimum frequency deviation of 500 KHz. I. Find and calculate the approximate bandwidth of the FM signal the frequency of the message signal is 1 kHz. (5 marks) II. Find and calculate the approximate bandwidth of the FM signal if the frequency of the message signal is 500 KHz (5 marks)

Answer 1

Given that, 3 Angle modulated signal AC= 10 cos (srixi$t +o's Sin (1044)] frequency modulated signal with Kf=5 time Domain Equation of soft The General FM signal de s(t) = Accos [alfitt miks mt) at] T tone modulation message signal mt)= Am cos anfmt. for single Then XA = Ac cos [anfct +27111f (Am) siner fmt) 28fm As An sin (2.font)) = Ac cos [ 21 fet + fm Af sto(21 fmp] = AC COS[ all fet + X(t) - Accos Carlict + ß Siner-fmt] Kfm - modulation inden p = fm im AF = Frequency devotion.

Accos (27fet + 101 / mA) dt] - f-m forks as Coro ro cos [ an xiaft + os Sin (25x5x2 t)] T ß=0.521 Narrow Band F.M. mt) = 2 mt) = Am cosa fmt for st)= XcB)=Ą COS (911set + ß Sinattmt ]." Fm B = fc = 108 Hz = 100 MHZ fm fm= Sx03 = SKHZ. of = o.sfm = 1 ( 5x103) 2.58103 and Gress with it as Af=kf Am Am Af 25x103 = 0.5 103= soo. Kt 52 mt)= 500 cos(211 x 5x103) = 500g ollit) 1011s soo cos(104) NOW, X(t) = 10 cos [ 21 x 108t e pf.msignal for Kf=s.

e if fc= 40MHZ, Df= SOOKH. Sm=1 KHZ - first find the modulation index, Sook 800 fm x B> wide Band F.m. Bokl= 2017B) fm = 2 (501) & 1kHz) BH - 1002KHZ. = 1.002 MHz. ( if &m= SOOKHZ. Af ßa Soort al 5ook Im A ma Narrow Band Bel Ror wide Band im It He F.M then take as wide Band B. la= a01+B) fm a(1+1) fm = 4 fm - UX 500k 200 OK Hz B.W - 2MHZ

If you have any doubt, kindly comment below. I will be happy to help you