Question

2. A 3 KHz sinusoidal signal Phase Modulates a carrier at 245.3 MHz. The modulating tone has an amplitude of 4 volts and the phase sensitivity of the modulator is 0.5 radians/volt. a. (15 points) Evaluate the required transmission bandwidth using Carson's Rule. b. (15 points) If the resulting PM signal is passed through an IBPF with bandwidth given by Carson's Rule (plus a few Hz), what percentage of the signal power is lost?

Answer 1

Given that, frequency of message signal fm = 3KHZ frequency of carster Signal fe = 245-3 MHz Amplitude of message signal Am = 4 volta Phase sensitivity of the moduladoo kg = 0.5 sadions holt

PM >5lt m(t) 1 Modulator PM modulated wave : 51t) - Accoo 2TT.ft + kpmct) ] instantanious phase 9.() = 277fct + Kymct) Phax deviation AQ;t) = Kymct). instantancouz frequency f:(!) 0C) - filt) - r [ 21 dt + Kp & mct) ] & fc + Kp doct) ke units sadians/wolts at dt f (t) = (245.3x106) + 0.5xd (4 cos afmt) = 245.3x100 + + x4 ( 21 fm) (-sin at fmt) Af = 25 fm Sina7 fmt Maximum foequency deviation Afmax 12fm sinatifunt = axfm - 2x 3 KHZ Afmax = 6kHz 2T dt ATT I max

Bandwidth = 2 (p+1) from according to carson's Pp: Modulation index rule - 1 Bp = Kp Am - 0.5 x 4 = 2 BW = 2 (2+1) 3KHZ BW = 18 KHZ b BPF center frequency fo=18kHZ SC) y(t) 5(t) = AC COS (2Tfct + B 5inatifmt] This can be written in terms of Bersel's function s(t) - Ac ¿ Jn (B) COB 2T (fc+nfm) + - n=-o costi fct. - Ac Jo (Pp) + AJ. (B) Cosat (fct fom) tt AcJ_, CPp) cos at (fc-fm) + + ..

Spectrum of PM Modulated have АТ. А.Т.Р) 2.1 AJ; (B) 50f) f- 3 fm fc 2 fm fc fm fc fet fmi ft 2 fm fc + 3 for fet lufm 18kHz Bandwidth an After passing through BPF of Bandwidth 18kHz output is : Y(f) AcJelp Ac Il AJ 2(B) 2 1 AJ, (B) 12 AJ (B) L ? TACJE (B? f-2fm fc 2 fm fer fm fe fct for fat2fm fc + 3 fm of The power of the output of BPF ylt) is e Ac JILB Ac Sa Te +2 2 Pages = AE 5P +2 APRILE 2 AFP RAO JE UP T: JLPJ), TCP - ICP) ] 22 Ac e Jo (F) 2 Ac Jol + J. (B ) + & 2 2 Ac Jz CF ² + 2 Ad

Ac=1V Bp=2 overall power of PM modulated wave Sct) i Po(t) = Power of [ AC Cos(alifet + kp m(t)}] Pock). Ź Pylt) 25(2)+ J; 2) + 5, 12) + *(2) Power loss = Post) - Pyet) - (25 (2) +57C2) +5.22) +5%)]