Question

The output S/N at thereceiver must be greater than 40 dB. The audio signal has zero mean, maximum amplitude of 1, power of ½ Wand bandwidth of 15 kHz. The power spectral density of white noise N0/2 = 10-10W/Hz and the power loss in the channel is 50 dB. Determine the transmit power required and the bandwidth needed.

Answer 1

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

$\dfrac{N}{2}=10^{-10}\ W/Hz$

Power loss in channel= 50 dB

Suppose, Using DSB modulation

We need to calculate the power required

Using formula of power

$P_{L}_{dB}=10\log(P_{L})$

Put the value into the formula

$50=10\log(P_{L})$

$P_{L}=10^{5}\ W$

For DSB modulation,

Figure of merit = 1

We need to calculate the input signal

Using formula of FOM

$FOM=\dfrac{\dfrac{S_{o}}{N_{o}}}{\dfrac{S_{i}}{N_{i}}}$

$1=\dfrac{\dfrac{S_{o}}{N_{o}}}{\dfrac{S_{i}}{N_{i}}}$

$\dfrac{S_{i}}{N_{i}W}=\dfrac{S_{o}}{N_{o}}$

Put the value into the formula

$\dfrac{S_{i}}{2\times10^{-10}\times15\times10^{3}}$

$\dfrac{S_{i}}{30\times10^{-7}}$

$S_{i}$

$S_{i}=30\times10^{-3}$

We need to calculate the transmit power

Using formula of power transmit

$S_{i}=\dfrac{P_{t}}{P_{L}}$

$P_{t}=S_{i}\times P_{L}$

Put the value into the formula

$P_{t}=30\times10^{-3}\times10^{5}$

$P_{t}=3\ kW$

We need to calculate the needed bandwidth

Using formula of bandwidth for DSB modulation

$bandwidth=2W$

Put the value into the formula

$bandwidth =2\times15$

$bandwidth = 30\ kHz$

Hence, The transmit power is 3 kW.

The needed bandwidth is 30 kHz.