Question

When frequency division multiplexing (FDM) is used to aggregate several modulated channels together, no separation between channels (i.e., use of guard channels) is required.

True

False

Regarding amplitude modulation (AM), if the AM index, µ, is greater than 1, then the modulated passband signal will begin to show signs of signal distortion.

True

False

Given the following AM modulated signal, determine the occupied frequency bandwidth (note: assume double-sideband large carrier (DSB-LC):

s(t) = 4[1+0.95 cos(2π3000t)] cos(2π100E9t)

		a. Frequency bandwidth = 3000 Hz
		b. Frequency bandwidth = 6000 Hz
		c. Frequency bandwidth = 100E9 Hz
		d. Frequency bandwidth = 200E9 Hz

Given the following message and carrier equations, determine the AM modulated signal equation: m(t)=3cos(2π4kHz*t), c(t)=2cos(2π3MHz*t)

		a. s(t)=3[1 + 0.67cos(2π4000t)]cos(2π3E6t)
		b. s(t)=2[1 + 1.5cos(2π4000t)]cos(2π3E6t)
		c. s(t)=3[1 + 0.67cos(2π3E6t)]cos(2π4kHz*t)
		d. s(t)=2[1 + 1.5cos(2π3MHz*t)]cos(2π4000t)

Given the following FM modulated signal equation, determine the frequency bandwidth. s(t)=2cos(2π1E9t + 4sin(2π1E3t))

		a. 20kHz
		b. 10kHz
		c. 1kHz
		d. 106 kHz

Given the following message, m(t), carrier, c(t), and Kp=0.1(rad/v), determine the PM modulated signal equation and PM index. m(t)=2cos(2π4000t), c(t)=4cos(2π8MHzt).

		a. s(t)=4cos(2π4E6t + 0.75cos(2π3500t)), µPM=0.75
		b. s(t)=4cos(2π8E6t + 0.2cos(2π4E3t)), µPM=0.2
		c. s(t)=3cos(2π4E6t + 0.75cos(2π3500t)), µPM=0.75
		d. s(t)=4cos(2π4000t + 0.2cos(2π8E6t)), µPM=0.2

Given the following PM modulated signal equation, determine the original message equation, m(t). s(t)=3.2cos(2π2E9t + 0.99cos(2π5000t)), Kp=0.33 (Hz/v)

		a. m(t)= 4.4cos(2π5000t)
		b. m(t)= 4.4cos(2π2E9t)
		c. m(t)= 3cos(2π5000t)
		d. m(t)= 3.2cos(2π2E9t)

Given the following PM modulated signal equation, determine the frequency bandwidth. s(t)=3cos(2π13E9t + 2cos(2π13E3t))

		a. 22kHz
		b. 40kHz
		c. 78 kHz
		d. 116 kHz

Answer 1

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mt) = 3 ox (2 484) - Am Loo at tat M= Am = 3 =1.5 (4= 2 Cox (2113mt) - Ac cos anfct AC 84) Al II + M Colle en font] cop can fet 84)= 2 [ 1+195 lop (214K] con (@msmt) ® Part 5 SA= 2 Lop (27x19 t + 4 sin (24x10²t)] - A coplentet +ß sincan fint)] comporre, fm = 103, 824 B. W. a frequency Bandwidth = 2 [P+1) tm =21 4+1] xlo3 = lokHz ® part xpsod, mt=-2 lop (@ot fovo t) 4) = 4 op (255 Omlet) = Ac lop (eltet) 88 Ac lop anfit & kp mA = 4 com cenamitzt + 0.1*2 (or@2184000 t) 7 B Pent B- upm - kl Am tm - KP Am = 0.182=0.2 0 842 3.2 lop (2 2697 + 0.99 lop (2175000t)] de los califet & kl mail compass, 4ma - o. gg cop (alt soort) mq) = 0.99.coplast sooo t) 17142 0.33 O D.72

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