I want to compare performance over AWGN channel of two systems that are exactly identical but their bandwidth. I will talk only about baseband system for sake of simplicity. For more details about notations, please take a look at this online course MIT DSP course.
Call $P$ the signal power (joules per second).
First, considering a system using bandwidth $W$. According to sampling theorem, the orthogonal basis set contains $2W$ vector. The single side power spectral density of white Gaussian noise $N(t)$ is defined as $N_0$ if its projection $N_k = <N(t), \Phi_k(t)>$ is set of i.i.d Gaussian variables with zero mean and variance $N_0 / 2$.
Sampling interval is $1/2W$, there is $2W$ signal symbol per second, then a symbol has power of $P/2W$. A noise sample has power of $N_0/2$ by definition, the $SNR = (P/2W)/(N_0/2) = P / N_0 W$. Well-known results !!!
Next, I wonder what happens if I increase the bandwidth, say double it, to $2W$ in the same physical environment.
Double bandwidth, half symbol duration. In one second, I have doubled the number of discrete symbol to $4W$. The signal power per discrete sample is now $P/4W$.
Question 1: But what happens to white Gaussian noise sample ?
If I understand $N_0/2$ as noise power per Hertz, does doubling bandwidth to $2W$ doubles noise power per sample ? If yes, SNR should be decreased by factor 4 (2 of reducing signal power per sample and 2 of increasing noise power per sample). I doubt this.
If I use the projection of the same $N(t)$ to the new orthogonal basis set. Number of vector in the set has increased twice from $2W$ to $4W$. Does projecting the same signal to more basis vectors reduce the power of each projected coordinate ? If yes, is it true that the noise sample $N_k$ now have variance $N_0/4$ instead of $N_0/2$ ? And SNR does not change because $SNR = (P/4W) / (N_0/4) = P/N_0 W$.
Question 2: Is $N_0/2$ watts per Hertz ?