Question

12. An FM signal has a deviation of 3 kHz and a modulating frequency of 1 kHz. Its total power Pris 5 W, developed across a 50 2 resistive load. The carrier frequency is 160 MHz (a) Calculate the RMS signal voltage VI (b) Calculate the RMS voltage at the carrier frequency and each of the first three sets of sidebands. (c) For the first three sideband pairs, calculate the frequency of each sideband. (d) Calculate the power at the carrier frequency and at each of the sideband frequencies found in part (C) (e) Determine what percentage of the total signal power is unaccounted for by the components described above (1) Sketch the signal in the frequency domain, as it would appear on a spectrum analyzer. The vertical scale should be power in dBm, and the horizontal scale should be frequency (g) Use Carlson's rule to calculate the bandwidth of the signal

Answer 1

Solution:- (a) The signal power will not change with modulation Pre V1² Vr = Verre . -15Wxson z 15.8V (RMS Value). (b) The modulation index must be calculated to use Bessel functions to find the corrien and side band voltages. » 3kHz fm J where x 20, 1, 2, 3 By Bessel function pairs are table the first three siele band Joz -0.26 J120.34 Sz20:49 Jg=0:31 | Vc = Jo Vt of Jo Thas negative sign which indicates a relationships between the components of the signal. Now Vol Jo I VT z Ve af on the comporta l Jol vt ? 0.26 X 15.8V 24.11v Now Viz J, VT 2 0.34 x 15.91 2 5.37V

V₂ z J₂ Vq z 0.49 X 15.3 V 77.74 V V32 J3 VT 20:31 X 15.8V z 4091 (C) The side band frequencies core tettm.n , fc-fm.n where n = 1,213 Ausbl = 160 MH3 + IkHz 2160.001 Metz fusB2 = 160 mitz + 2 kHz 160.002 MHz fusBz z 160m tz t3 Kitz • 160.003 MHz Fise, - 160mitz - IkHz - 159.999 mtz FLSB2 e 160 m 43 - 2kHz - 159.998 MHz Plsoy160 mtz - 3kitz e 159.997 mitz (d) All the components appear across the same 5012 load 4.112 50 z 0.338 W c Pezve a Paz Hill - 0.338 W Pre nego st. 0.576W Pasvalio r. 19 1.22 Bso vienuo 0.

e) Total powers Pi -> one cooolies + 6 sido bands 3 upper , 3 lower Prz Pc +21p 1 + P2 + P3) 20:333 +210.576 +1:2+0.43) W 24.85W This is not the total signal power which was given as 5w. The cremaining is in the additional sidebounds. To find how much is unaccounted by the Lawlies and first three side bounds, we can find by PL = 5-4.85 2015 W As a percentage of total power is Pe(%) 10:45) x100 23% Plasm). lolog limoms) i pel des m) - 10 log (333) - 25.3dBm Pi (dBm): 10 log (576) 10) 2 27.6 dBm Pz (dBm): 10 log (1200) - 30.8dBm Bildem) 10 log (450) - 2613 dem

30.8dBm 30.8dBm P(dBm) 27.6dBm - 27.6dBm 26.8 dBm 26.8 dBm 25.3dBm 199.993 159.998 159.999 160 160.00) 160.002 160.003 f(my) 19] Bandwidth By carson Rule. requen Band width = 2x1 peak Deviation + Highest modularing given If = 3K ltz fm = 1kHz. - 2x ( 3kHz +Ikitz) 2 2 x (UK13) [ Bw - 8k13]