Question

Q3 (a) Given the AM expression as below: VAM (t) = 20 sin(47 300 000t) + 4 cos (21 565 000t) - 4 cos (21 635 000t) Compute: (i) The modulation index. (1 marks) The modulating frequency of the signal. (2 marks) (iii) The bandwidth. (1 marks) (iv) The total transmitted power if R = 10 12. (3 marks)

Answer 1

detailed hand-written explanation is given after the solution

solution:

An AM wave can be represented as:

EAM Em E sin2 fet + bom cos2m (fc-fm) t-o cos27 (f. - fm) 2

where,

E_cis the maximum amplitude of carrier signal

Em is the maximum amplitude of modulating signal

f_cis the frequency of carrier signal

f_m is the frequency of modulating signal

Given that,

CAM 20sin (41 X 300000t) + 4cos (27 X 565000t) 4cos (27 X 635000t)

comparing the given AM wave equation with the equation in general form, we get:

Ec = 20 Em = 4Em = 8 2 fc = 600000Hz = 600kHz fm 635000 – 600000 = 35000 Hz = 35kHz

i.) modulating index is,

8 m Em Ec 0.4 20

ii.) modulating frequency is,

fm 35kHz

iii.) bandwidth is,

BW = 2 fm = 2 x 35 = 70kHz

iv.) total power transmitted is,

m Ptotal E2 2R 1+ 202 2 x 10 X 1+ 0.42 2 = 21.6W 2

PAGE-1 •The instantaneous Piven as: Value OF the amplitude modulated wave com be CAM (Ex + Eh. m 20m. E). Am cc 6 Where 'Em' is the maximum amplitude of modulating signal 'Ec' is the maximum amplitude of carrier signal fam' is the frequency of modulating signal the wa' is freuency of carrien signal CAM Ec. Si Cuct Eh. Sim . E . Am cc. + . Sin A. Sin B = CS (A-B) = Ec Sim weit + Em. Sim wcit sin Sin whit + Coo (478) 2 = Ec-Sin weit + Em. - . Cos (Weit- wmt)-a.cos(wc.t+Wmt) E+ und) 2 E. Sim c.t + En, TT Cos (wc-om).t „. Cos (wc +wm).t] = Ec. Sim. Wc.t + Em. Cos (Wc-wm).t Em. Cos (wc+wm).t 2 CU = 27 F CU Predluency in Tad I sec Prenuency in Hz Cam = Ec, sin arbeit. + bene. Cos (atfe - 2 sm).t - Em. Cos (atletañvent CAM = Ec. Sin an fc.t + Em En. Cos 27. Cfc- Vmot - Em. Cos 27. (1fc+fm).t

PAGE- Given that, Vam (t) = 20. Sim (4* . 300.000. t) + 4.CoS (22.565.0006) -4: Cos C&T : 635, Oops ) Vam (t) = 20. Sim (at * 600.000. t) +4.Cos (2* * 565,000 €) -4. Cos (QTX 635,000 €) Comparing 1 with the general forth of AM Wave derived in Page-1, we get : 4 Mazcimum amplitude of modulating signal, Em = 8 Em = 2x4 = 8 Maximum amplitude OF Carrier siynol, Ec = 20 Frenvuency of Carrier Sigma), Fc = 600,000 Hz = 600 KHZ .fc-ifm = 565,000 Freguency OF modulating signal, ifm = 35,000 HZ FC = 600,000 = 35 KHZ Fim = 600,000 - 56 57 000 = 35,000 Hz OR AM WAVE PARAMETERS fc +fm = 6 35,000 Em 8 fm = 635,000 - 600,000 Ec = 20 = 35,000 Hz fc = 600 KHz Ifm = 35 KHz

PAGE - 3 PART i of modulating • Modulation index is the ratio of maximum amplitude to maximum amplitude of arrier signal. Signal m Em Ес 8 20 m 0.4 زن - PART . Modulating frequency C derived eavier - Page 2) is : if m = 35 KHZ زنز - PART of the maximum freuency - Bandwidth Of an AM wave is twice OF modulating signal. BW = 2. Fm = 2x 35 BW .= 70 KHz

PAOE-4 PART- iv wave to a load resistor op'R' •The total Power delivered by an ohm is given by : Ptotal E 2R (1+ mode) R = 10 22 Given that , ... P total Eca 2R (it me:) 202 (1+0.42) 2x to = 20 (1+ 25) =.108 P. Total Power = Corrier Power + upper side band. Power + Lowey sidebord Power P total = 21.6 W Ptotal = Pc + PUSB + PLSB . + Ptotal = Ece 2R Ema 8R Ema 8R = Eca 2R + Ema 4R = Eca (m.Ec ) QR AR Eca AR + me. Eca 4R Ptotal Ес AR (i+ ha )