Question

Problem 1: An angle modulated waveform is given as s(t)=A_cos (1000 at +8071 +3204) where 0.=10001 , max|m(t)=3 , and m(t)=0,+<0. In a previous ICE, we determined that if the signal is phase modulated (PM), the phase sensitivity is k= 4001 - 133.31 , the message signal is m(t)=1 12 and the frequency deviation is Am=-80 tn 5-55.51 . Similarly, if the signal is frequency modulated (FM), the frequency sensitivity is kr , the message signal is m(t)==3(3), and the frequency deviation is Aw=- 80 In () – 55.51 (a) What additional information would you need in order to solve for the bandwidth for the FM and PM signals? (b) How would you determine the additional information needed? You do not need to solve for the information, but do need to explain how you would solve for it. (c) Do you expect the bandwidth to be different for the FM and PM signal? Why or why not?

Answer 1

An angle medulated wave form is given by SA? = Ac cos ( 100 ut + souf+) *+ 320 ] where Wca 1000 T max (mct) 1 =3 & mct) zo, tzo If the signal is phase modulated (P.M) Phase sensitivity, kp = 40 ou 33.3 message signal, met = 3 (2) + 2 u. frequency deviation, DW = – som dn(t) ~ 55.5 m

If the signal is frequency modulated (FM) fequenay Sensitruity, kp - Bouin () = 18$ message signal & met = -3 (2) + Frequency deviation, aw z-800 in (t) = 55.50 Band midth of FR = 26m if BLC) where B= kg Am fon Narracoband FM = 2(Af + fom] it Boi where 8 - ke Am of = reg Am wide band FM Band midth of por = 2 ( + 1] for & Where Ben = kp Am In order to calculate B.w for both For & pom we require & frequency of message Signal and the amplitude of the message Signal is also needed. man

PJ-3 (6) The amplitude of the message signal is can be determined by the graphical representation of the signal The maximum frequency of the message signal is obtained from frequency response. By using * onse Cena bee from we can determine the maximum frequency need not the FM and PM are Band midth of to be same. because, BeoM - 2(0f + fon) -2(xf Am + fm] = 2 xq Am + 2 fm Buwpor = {28pm + 1) m] - 2 { kp Am + 1] fm - 2 kp Am fm + 2 fm = 2fm [ kp Am + 1] - Bow required for pre increased by in- times when ever modulating signal frequency increased by ritirse - But it is not the case with Free so, usually FM is preferred in practice than PM.