Value of power spectral density $N_0$ or effect of scaling bandwidth to SNR

Question

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 ?

  1. 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.

  2. 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 ?

Answer 1

The noise power continues to be $N_0/2$, independent of bandwidth. The reason is that the noise variance at the output of a filter with frequency response $H(f)$ is $$\sigma_n^2=\int_{-\infty}^\infty \frac{N_0}{2}|H(f)|^2\,df.$$ (This is a direct application of the Wiener-Khinchin theorem). Since you're assuming an orhonormal basis, the noise power is always $N_0/2$, independent of the bandwidth.

Then, doubling your signaling rate without a corresponding increase in spent energy results in halving the SNR.

Regarding your second question, $N_0/2$ is the PSD of the white Gaussian process assumed to exist at the input of your measuring devices. The PSD is indeed measured in W/Hz.

Note that it is a bit weird to define $P$ as watts per second. One watt is one joule spent per second; watts per second would be joules, or pure energy. Usually, you assume you have a source providing $P$ watts; this source operates for as long as necessary (even from $t=-\infty$ to $t=\infty$).