D/A Communications - Minimum bandwidth of a channel

[stn0091]

Homework Statement

Homework Equations

The given channel transfer function and PSDs. Note that the capital Pi symbol in the PSD of the message symbolizes the rectangle function so Sm(f) = 8rect(f/8000)

The Attempt at a Solution

From my understanding, channel noise is not filtered at this point since it's additive to the signal in the channel.

I get an answer of B = 0.000164 Hz, which doesn't seem reasonable. Including the negative frequencies, a bandwidth of 2B = 0.000328 Hz still doesn't make much sense to me.

 

[Valkarie]

I'm not sure if I'm missing something here.I also get power spectral densities for the noise and message symbol, but I don't understand how to interpret them. Any help would be greatly appreciated.The channel transfer function is given by H(f) = 0.8e^-j2πf/7000 and the PSDs are given by Sx(f) = 4rect(f/4000) and Sn(f) = 8rect(f/8000).From this, we can calculate the bandwidth of the channel as B = 1/(2π/7000) = 0.000164 Hz. This is the bandwidth of the signal at the output of the channel, which is the sum of the input signal and the channel noise. The power spectral densities of the message signal and the noise can then be calculated using Parseval's theorem:Sx(f) = 4rect(f/4000) Sn(f) = 8rect(f/8000)These PSDs represent the power per unit of frequency (in hertz) of the message signal and the noise respectively. From these PSDs, we can determine how much power is contained in each frequency range. For example, at f = 4000 Hz, Sx(f) = 4, meaning that there is 4 units of power in the frequency range from 4000 Hz to 8000 Hz. Similarly, at f = 8000 Hz, Sn(f) = 8, meaning that there is 8 units of power in the frequency range from 8000 Hz to 16000 Hz.