Respuesta :
Given that,
The output signal at the receiver must be greater than 40 dB.
Maximum amplitude = 1
Bandwidth = 15 kHz
The power spectral density of white noise is
[tex]\dfrac{N}{2}=10^{-10}\ W/Hz[/tex]
Power loss in channel= 50 dB
Suppose, Using DSB modulation
We need to calculate the power required
Using formula of power
[tex]P_{L}_{dB}=10\log(P_{L})[/tex]
Put the value into the formula
[tex]50=10\log(P_{L})[/tex]
[tex]P_{L}=10^{5}\ W[/tex]
For DSB modulation,
Figure of merit = 1
We need to calculate the input signal
Using formula of FOM
[tex]FOM=\dfrac{\dfrac{S_{o}}{N_{o}}}{\dfrac{S_{i}}{N_{i}}}[/tex]
[tex]1=\dfrac{\dfrac{S_{o}}{N_{o}}}{\dfrac{S_{i}}{N_{i}}}[/tex]
[tex]\dfrac{S_{i}}{N_{i}W}=\dfrac{S_{o}}{N_{o}}[/tex]
Put the value into the formula
[tex]\dfrac{S_{i}}{2\times10^{-10}\times15\times10^{3}}<40\ dB[/tex]
[tex]\dfrac{S_{i}}{30\times10^{-7}}<10^{4}[/tex]
[tex]S_{i}<30\times10^{-3}[/tex]
[tex]S_{i}=30\times10^{-3}[/tex]
We need to calculate the transmit power
Using formula of power transmit
[tex]S_{i}=\dfrac{P_{t}}{P_{L}}[/tex]
[tex]P_{t}=S_{i}\times P_{L}[/tex]
Put the value into the formula
[tex]P_{t}=30\times10^{-3}\times10^{5}[/tex]
[tex]P_{t}=3\ kW[/tex]
We need to calculate the needed bandwidth
Using formula of bandwidth for DSB modulation
[tex]bandwidth=2W[/tex]
Put the value into the formula
[tex]bandwidth =2\times15[/tex]
[tex]bandwidth = 30\ kHz[/tex]
Hence, The transmit power is 3 kW.
The needed bandwidth is 30 kHz.
