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I was reading Wireless Communications by Andrea Goldsmith in the book the author has stated that the Noise power for a wideband system with Bandwidth B is (B*N/2). Could someone help me understand why is how Noise Power is related to bandwidth? Does this mean that for a Narrowband system there would be negligible noise?

Also in statistics, we usually use the Noise power as the variance of AWGN which is No/2 why don't we include the bandwidth dependency here?

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    \$\begingroup\$ Not all noise is Gaussian. Also generally we call it Carrier/Noise or CNR then demodulation methods and type convert to SNR. But Signal and noise spectrum must be considered with matched filters to signal for best ratio and group delay error or linear phase adds degradation. Stats on AWGN all always random independent of BW but of course dependent on sampling rate. \$\endgroup\$ – Sunnyskyguy EE75 Jun 9 '18 at 1:55
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In the RF world one way to specify noise is as \$dBm/Hz\$ and where dBm is \$10*log_{10}(milliwatts)\$ this gives you a way to compare noise between systems with different bandwidths, just divide your noise power by the system's bandwidth.

E.g. Say I want to receive a 0dBm signal with a system that has a bandwidth of 1Hz and a noise power density on the order of \$-50dBm/Hz\$ (there is an absolute lower limit at \$-174dBm/Hz\$ because well... physics). Because my signal is \$0dBm\$ and my system has \$-50dBm/Hz\$ of noise power density and my bandwidth is \$1Hz\$, my SNR is going to be \$50dB = 0dBm - (-50dBm/Hz)*1Hz\$.

But what if I widen my bandwidth to \$1kHz\$ with the same \$0dBm\$ signal and \$-50dBm/Hz\$ noise power density? Well isn't that just 1000 lots of 1Hz blocks? So if each block contains -50dBm worth of noise power... then if I have 1000 of them I now have a receiver that has to deal with 1000 times as much noise as before, this brings my SNR from \$50dB\$ down to \$20dB\$ as \$0dBm - (-50dBm/Hz)*1kHz=20dB\$, what would happen if I increased the bandwidth again to 1MHz...? I'd have more noise than signal!

The more bandwidth a system has then the more noise power it is going to have to deal with, this is why there is a fundamental relationship between SNR and Bandwidth. Just as a room gets quieter as party goers are progressively removed... so too does system become quieter as you progressively reduce your system bandwidth, effectively blocking out all the unwanted stuff that's not part of your signal of interest.

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  • \$\begingroup\$ just a question, shouldnt the frequency or bandwidth multiplied ? in your equation I see dBm/Hz divided by bandwidth... \$\endgroup\$ – rsg1710 Aug 3 '18 at 11:11
  • \$\begingroup\$ Yes, you're right, it should be (dBm/Hz)*Hz \$\endgroup\$ – Sam Aug 4 '18 at 21:47

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