Communication Systems

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

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

Range = $\sqrt[]{2hR}=\sqrt[]{2*20*6.4*{10}^6}=16km$

Area = $\pi R^2=3.14*16*16$ =803.84km²

When H= 25 m

Range = $\sqrt[]{2hR}+\sqrt[]{2HR}=\sqrt[]{2*20*6.4*{10}^6}+\sqrt[]{2*25*6.4*{10}^6}$ = 33.9km

Area= 3.14 $*33.9*33.9=3608.52km$ ²

percentage increase = $\frac{3608.52-803.84}{803.84}*100=348.9$ %

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

Total distance = 5km and loss is 2 dB/km

So total loss = 5 (2)= 10 dB

Total gain in amplifier  10+20= 30dB and gain in signal is 20dB

So by the formula 20= 10log_{10 $\frac{p_o}{p_i}$}

log_{10 $\frac{p_o}{p_i}$} = 2

so po/pi= 10²

so Po= Pi (100)= 101mW

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

In modulating signal we add career wave or noise signal to send it to receiver end and this wave is varied from time to time that is why more noise appear.

But in frequency modulation frequency is not varied so less noise appear.

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

Frequency tuned amplifier is $\frac{1}{2\pi \sqrt[]{LC}}=1MHz$

$\sqrt[]{LC}$ =1/2 $\pi *{10}^6$

$LC=2.54*{10}^{-14}s$

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

Maximum amplitude = Am+Ac =12 while minimum amplitude is Ac-Am=3

Using elimination method = 2Ac= 18

Ac=9 and Am= 6V

So modulating index m = 6/9= 2/3

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

As frequency of B is more than A so it has more refractive index also and if a wave have higher refractive index then it has less angle of refraction. So wave B travel more in ionosphere.

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

When we send a signal to be transmitted through sky waves must have a frequency range of 1710 kHz to 40 MHz.

But, the frequency of TV signals are 60 MHz which is beyond the required range.

So, sky waves will not be suitable for transmission of TV signals of 60 MHz frequency.

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 2^nd option- RC= 10^-4s

So it can also be demodulated

According to third option – RC= 10^-8s

So it cannot be modulated

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