For a Gaussian channel of 1 MHz bandwidth with the signal power to noise spectral density ratio of about 104 Hz, what would be the maximum information rate?

The correct answer is: C. 28000 bits/sec

The maximum information rate for a Gaussian channel is given by the Shannon-Hartley theorem:

$$C = B \log_2(1 + \frac{S}{N})$$

where $C$ is the maximum information rate in bits per second, $B$ is the bandwidth in hertz, $S$ is the signal power in watts, and $N$ is the noise power in watts.

In this case, we are given that $B = 1$ MHz and $S/N = 104$. Substituting these values into the Shannon-Hartley theorem, we get:

$$C = 10^6 \log_2(1 + \frac{10^4}{1}) = 28000\text{ bits/sec}$$

Therefore, the maximum information rate for this Gaussian channel is 28000 bits per second.

Option A is incorrect because it is the signal power in watts, not the maximum information rate.

Option B is incorrect because it is the noise power in watts, not the maximum information rate.

Option D is incorrect because it is the bandwidth in hertz, not the maximum information rate.