# Why is noise power not multiplied by spread gain?

In my intuitive consideration, noise power should be enlarged as spreading gain times. The reason why I think this is:

Let us consider that the information signal is of bandwidth W Hz. Then, the receiver suffers from the noise of bandwidth W Hz as well.

If we do the spreading work into the information signal with chip rate. Commonly it is much larger than W Hz. Then, (information signal X spread spectrum signal) will also have much larger bandwidth than W Hz. Thus, the receiver suffers from the noise of bandwidth much larger than W Hz.

Since noise is BW*N0/2, where BW=bandwidth and N0=noise density, I think much larger noise happens.

Why do people say DSSS makes much better SNR? (roughly, people say SNR will be increased around spreading gain times.)

$$C = BW \log_2 ({1+SNR}) \quad\longrightarrow\quad C = BW \log_2 ({1+G\times SNR})$$

Edit!!

If you measure the total background noise that comes in through the receiver bandwidth, then that's true. We can conceptually measure total noise by connecting a resistor to the IF (the Intermediate Frequency of the spectrum analyser we're using) and seeing how hot it gets, the total thermal power, and this increases with the bandwidth (assuming spectrally flat noise). As we increase the bandwidth 10 times, we get 10x more noise power in.

However, even though the spread signal looks noise-like on a spectrum analyser, as a result of being spread with a noise-like signal, we know something very important about it. We know what the spreading signal is. That means that once we have locked on to the spreading signal, we can average the signal coherently, which means the signal adds as voltage, not as power. What we do is correlate the incoming wideband signal with a locally-generated replica of the spreading signal, to extract the original narrow-band signal.

Now something quite important happens when we correlate and filter the wideband spread signal, because we are also correlating and filtering the wideband background noise as well. If the spreading signal is uncorrelated to the background noise (an assumption we can usually make safely) then this process reduces the power of the background noise by the ratio of the bandwidths. Because our local spreading signal is the same as the original, we retain all of the original signal power, while we reduce the total background noise. And there is your SNR improvement.

A corollary of this is, we need to generate a local replica of the signal to do the demodulation, if it's wrong, then we don't get the SNR improvement. So how do we lock-on in the first place? It doesn't happen by magic. But perhaps that's the answer to another question.

• May I ask a full word of IF in the sentence "We can conceptually measure total noise by connecting a resistor to the IF~"? – Danny_Kim Dec 30 '16 at 10:28
• I think this is a very nice answer although I do not understand well. Can you explain a bit easily the following sentence? "If the spreading signal is uncorrelated to the background noise (an assumption we can usually make safely) then this process reduces the power of the background noise by the ratio of the bandwidths." I understood a bit, i.e., you mean that signal power is not increased but noise power is decreased because uncorrelation between spreading signal and noise, so SNR is increased, right? I do not know why noise power is decreased. T..T sorry for my folly. – Danny_Kim Dec 30 '16 at 10:39
• It sounds like you're missing some of the necessary basics. Do you understand correlation? Do you understand sequences that are mutually orthogonal (ie do not correlate with each other)? Do you understand why doing an FFT splits the input noise into several bins? The FFT is easier, because the orthogonality of sines is easier to visualise. However, any bandwidth can be decomposed into orthogonal basis functions. The spread spectrum system is designed to use one basis function. When we corrlate both noise and sigal onto that basis, we reduce the noise power, but not the signal power, as in FFT! – Neil_UK Dec 30 '16 at 12:01

Let us consider that the information signal is of bandwidth W Hz. Then, the receiver suffers from the noise of bandwidth W Hz as well.

When you have (say) two spreading frequencies, the information signal adds coherently i.e. the signal doubles (6 dB increase) but, because the noise received in each spreading frequency is incoherent, the noise signal only increases by 3 dB ($\sqrt{N^2_1+N^2_2}$).

This means the signal increases by 6 dB whilst the noise only increases by 3 dB hence there is an improvement in SNR by 3 dB.

Take two more spreading frequencies (4 in total) and the SNR improves by 6 dB. With 8 spreading frequencies the SNR has improved by 9 dB.

This is just a simplified version of events.

• Thank you for good explanation. However, I do not understand why it happens. I drew a picture for better expression of my curiosity. – Danny_Kim Dec 30 '16 at 10:26
• Non-coherent signals (or noise sources) add as the square root of the two terms squared. Do you understand this? If not then you need to do some googling. The picture has no impact on what I'm saying. – Andy aka Dec 30 '16 at 10:42