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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 ?

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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 $t=\infty$).

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    $\begingroup$ Thanks a lot. I have fixed my mistake of energy/power in $P$. Could I say, in discrete time model language, whatever the bandwidth, the noise sample always has power of $N_0/2$? Is the noise power your $\sigma^2_n$ ? If yes, the SNR is halved because of reducing energy per signal sample, nothing to do with noise ? $\endgroup$ – AlexTP Apr 4 '17 at 22:57
  • $\begingroup$ I am confused because your last sentence "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." So, should the energy of discrete noise sample depend on the bandwidth used by the signal ? $\endgroup$ – AlexTP Apr 4 '17 at 22:57
  • $\begingroup$ Regarding your first comment: yes to everything. $\endgroup$ – MBaz Apr 4 '17 at 23:12
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    $\begingroup$ Second comment: If you observe white gaussian noise through an ideal low pass filter with unity gain, then yes, doubling the bandwidth doubles the noise power at the filter output. But that is not what you're doing: you're observing the noise through a filter whose impulse response is part of an orthonormal set of signals, and which has energy one. In this case, the noise power is always $N_0/2$, irrespective of frequency. $\endgroup$ – MBaz Apr 4 '17 at 23:13
  • $\begingroup$ Re "it is a bit weird to define P as Watts per second." I think you meant watts per Hertz (which would be Watt seconds, i.e. Joules)? $\endgroup$ – SleuthEye Apr 4 '17 at 23:18

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