Communication Systems
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New answer posted
4 months agoContributor-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 agoContributor-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 agoContributor-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
= = 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 agoContributor-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 agoContributor-Level 10
This is a Long Answer Type Questions as classified in NCERT Exemplar
as we know that I= I0
And I= 25%of I0=
I=I0/4
I0/4= I0
I0 cancel from both sides
¼=
Taking log on both sides log1 -log4= - loge
X= log4/
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