Communication Systems

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2 months ago

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V
Vishal Baghel

Contributor-Level 10

Modulation Index,  μ=AmAC

Variation = 2Am=8Am=4v

Am+Ac=9

AC=9Am=5v

μ=45=0.8

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Payal Gupta

Contributor-Level 10

For high wavelength, size of antenna should be high.

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4 months ago

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A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Analog and digital signals are used to transmit information, usually through electric signals. In both these technologies, the information such as any audio or video is

transformed into electric signals.

The difference between analog and digital technologies is that in analog technology, information is translated into electric pulses of varying amplitude. In digital technology, translation of information is into binary formal (zero or one) where each bit is representative of two distinct amplitudes.

Thus, (a) and (b) would produce analog signal and (c) and (d) would produc

...more

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4 months ago

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A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- b, d

Explanation- modulation index m= Am/Ac

If m>1 then Am>Ac

maximum modulation frequency mf= frequency deviation / maximum frequency of modulating wave

here if m>1 then it means overlapping of both sides resulting loss of information.

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4 months ago

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A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- a, b, c

Explanation- production of modulated wave is given by frequency of upper side band – frequency of lower side band so in first three cases it is clearly visible.

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4 months ago

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A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-b, c, d

Explanation- height of tower= 240m

For line of sight communication maximum distance would be d= 2 R h

So d= 2 * 6.4 * 10 6 * 240  = 55.4km

So distance under this are communicable

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4 months ago

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A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- b, d

Explanation- wm=3KHz

And wc= 1.5MHz= 1500KHz

By using these two 1500 ? 3= 1503 and 1497

Bandwidth = 2wm= 2 * 3 = 6KHz

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