0
\$\begingroup\$

I can't intuitively believe that a signal could be demodulated whose voltage is below the noise floor. Can some one explain me how this demodulation of direct sequence spread spectrum happens without mathematics ?

\$\endgroup\$

closed as too broad by Marcus Müller, Dmitry Grigoryev, brhans, Voltage Spike, uint128_t Apr 4 '17 at 18:57

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • \$\begingroup\$ This is an easily researchable question. VTC telecomabc.com/d/dsss.html \$\endgroup\$ – Voltage Spike Apr 4 '17 at 15:19
  • \$\begingroup\$ "Please explain to me what is written in hundreds of textbooks." No, we can't write a textbook that is quicker to read then your textbook. Go read a textbook. \$\endgroup\$ – Marcus Müller Apr 4 '17 at 15:43
  • 1
    \$\begingroup\$ without mathematics won't work, by the way. that's really asking for "how does this mathematical algorithm work? Explain without math!". This is digital communication. The things we do are always very math-y. \$\endgroup\$ – Marcus Müller Apr 4 '17 at 15:44
1
\$\begingroup\$

If your symbols are much slower than the carrier or spreading rate, say 1000:1, then a precise correlation (aligned both in code bit and code edge phase) followed by lowpass to drop system bandwidth by 1,000X, will indeed drop the noise floor by 1,000X. Done it, liked it. Even injected a strong tone, to attempt to block; up to limit of linearity of the mixer in the correlator (I used a MC1596 multiplier) the blocking tone was spread out in spectral occupancy and as expected the desired energy was de-spread.

Thus for 1,000:1 despread, expect 30dB improvement in SNR.

\$\endgroup\$
1
\$\begingroup\$

Spread spectrum transmits the same information several times simultaneously but on separated carriers. That's a simple way of looking at it of course. Now imagine you had several receivers. Each receiver receives the desired signal and undesired noise.

Take two such receivers and add the outputs together. Your signal becomes twice as big but your noise only becomes \$\sqrt2\$ bigger. Therefore your signal to noise increases by 3 dB. The noises are uncorrelated hence they only become 3 dB bigger whilst your signals are totally correlated hence they become twice as big.

Two receivers give a 3 dB increase in signal to noise ratio and adding more receivers and channels improves this figure but there is a law of diminishing returns because to achieve 6 dB improvement you need 4 receivers/channels etc. 1024 channels is \$2^{10}\$ hence a 30 dB improvement can be realised.

\$\endgroup\$

Not the answer you're looking for? Browse other questions tagged or ask your own question.