Communication Systems

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New answer posted

4 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Frequency of career wave is 20MHz

Bandwidth require for modulation is 3kHz/2= 1.5kHz

For demodulating we need to reciprocal it

1/f= 1/20MHz= 0.5 * 10-7s

For modulation = 1/1.5KHz= 0.7 * 103s

According to first option =RC= 1 * 0.01= 10-5s

So it will demodulated

According to 2nd option- RC= 10-4s

So it can also be demodulated

According to third option – RC= 10-8s

So it cannot be modulated

New answer posted

4 months ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

in first case crowding spectrum is visible

If we want more wave to be modulated then more crowding will occur and more mixing up of signal.

But we want to accommodate this we use higher band width and frequency career wave

New answer posted

4 months ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Maximum voltage = 100/2 = 50V

And minimum voltage = 20/2 = 10V

Percentage modulation= max voltage -min voltage/ max voltage + min voltage * 100

= 50 - 10 50 + 10 * 100 = 66.67%

Peak career voltage= max voltage+ min voltage/2= 50+10/2=30V

Peak value of information voltage= 66.67/100 *  30= 20V

New answer posted

4 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

height of satellite hs= 600km
As we know velocity = distance/time

 2x/4.0410-3 = 38

So x=606 km after solving 
According to Pythagoras theorem d2=x2-h2= 6062-6002= 7236

So d= 85.06km so total distance will be double from receiver to transmitter = 170.1km
And d = √2Rh
 h=7236/2*6400 = 565m

New answer posted

4 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

as we know that I= I0 e - x

And I= 25%of I0=

I=I0/4

I0/4= I0 e - x

I0 cancel from both sides

¼= e - x

Taking log on both sides log1 -log4= - x loge

X= log4/

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