This is from my last year\'s paper but no soludtion i need help! I need help for

Question

This is from my last year's paper but no soludtion i need help!

I need help for the part a)b)c)d) and the other post asked for e)f)g)h) I divided the question by two and post it cuz i think its too long

please help me thank you

Fig. 2 shows a spectrum for the positive frequency of a modulated real signal Determine the following: Q3 0.5v 0.5v 30 kHz 35 kHz40 kHz FREQUENCY a)Modulation used Large carrier AM b)The frequency of the message signal 5 KHz The carrier frequency 35 KHz d) The bandwidth of the modulated signal e)The upper sideband frequency f)The total transmission power g) The percentage of power contained in the upper side band h)The modulation index 10 KHz 40 KHz 1/6 4

Explanation / Answer

a) The given diagram satisfies that AM equation x(t)= A2C/2 + ACm/2[ cos(Wc-Wm)+cos(Wc+Wm) ]

Whre m = modulation index

so thats why it is AMPLITUDE MODULATION BY LARGE CARRIER

b) the given diagram shows 3 frequcies ( ON THE FRQUIENCY AXIS )

1) WC-Wm = 30kHz

2)WC+Wm = 40kHz

and middle impulse represent Carrier frequency 35kHz with amplitude 1

so substitute above any equation  WC-Wm = 30kHz; YOU Get Wm =5kHz

c) middle impulse represent Carrier frequency WC= 35kHz with amplitude 1

d) Bandwidth = higher cuttoff freqency - lower cuttoff frequency = 40KHz - 30kHz = 10kHz;

e) THE Upper sideband freqency  WC+Wm = 40kHz;

h) the total transmitted power is Pt = Carrier power + Upper sideband frequency + Lower sideband frequency

Power = (RMS Value )2/R; RMS VALUE = PEAK VALUE / ROOT 2

Threfore Pt= A2C/2R +   A2Cm2/8R + A2Cm2/8R ; A2Cm2/2 = 0.5 ,; m= 1 & R=1 (default)

Pt = 2+ (0.5)2+ (0.5)2 = 3 watts (the given diagram only postive frequencies are given so double the value );

g) percentage of power contained in the upper side band  A2Cm2/8R = 1/6

h) modulation index A2Cm2/8R = 1/6 ; from this modulation index = 1

  

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