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How does DSSS bleed-over occur?

If a wavelength is cycling 2.432 billion times per second (2.432GHz, channel 5)...how can it also be cycling 2.437 billion times per second (2.437 GHz, channel 6) at the same time?

Wouldn't this be similar to the analogy of a light switch being turned on & off 4 times per second and also being turned on & off 5 times per second at the same time?

How is it possible for something to be moving at 2 (or more) different speeds at the same time?

Thanks!

1) The ceter frequencies of the channels are separated by 5 mHz

2) The channels are 22 mHz wide ( each channel is 5 times as wide as the space between channels )

3) The channels overlap. Except 1, 6 and 11.

http://www.wirelessnetworkproducts.com/index.asp?ID=73&PageAction=Custom#Channels%20For%20FHSS

That is some defining characteristics of channels. It doesn't explain the reason bleed-over occurs in the first place.

Channels are simply labels that have been assigned to certain ranges of frequencies, therefore making it easier for people to verbally communicate what frequency ranges are being discussed.

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Let me restate the question a little differently...

What property(ies) of electromagnetic energy causes bleed-over to occur (from one frequency to the neighboring frequency)? (or, is it due to inaccurate tuning capabilities of the RF radios?)

If I understand your question correctly, the answer to your question is that DSSS uses Spread Spectrum technology to communicate on multiple frequencies. In comparison, FHSS uses narrowband technology and communicates on one channel at high power. Then it hops to other channels and transmits briefly there, one frequency center at a time. DSSS transmits on multiple frequency centers (22) at lower powers at the same time, which is why it interferes with other channels. For example, if you put a DSSS device on channel 6, it is communicating 11MHz above and 11MHz below the center frequency of channel 6 (2.437GHz, thus using 2.426GHz through 2.448GHz to transmit, if my math is correct). That is why DSSS devices suffer from co-channel interference and only 1, 6, and 11 do not overlap. I hope this helps.

Thanks...

OK, then why is the signal strength stronger on the center frequency than the surrounding frequencies (in the same channel)? Does the RF radio transmit at a lower power level on the neighboring frequencies (within the same channel)?

Also, why in DSSS is "there still some level of overlap" even if using channels 1, 6, & 11 (CWNA Official Study Guide, pg 140-141)? Surely that overlap isn't the result of intentional transmissions at lower power...

Let me step back for a second and try to answer the question. I should clarify/correct what I said previously about FHSS. It is actually a spread spectrum technology, but each transmission, in isolation, is similar to a narrowband transmission, without the high power. It probably made my answer less accurate.

Because the unlicensed frequency space that is used in wireless networks is surrounded by licensed frequency space (2.400GHz is next to 2.399GHz), the FCC regulates the way in which unlicensed frequency transmissions can occur. When manufacturers design equipment, they must use technologies that abide by the regulations. Where DSSS transmits on several frequencies at a time, the transmission is filtered (regulated) so that the transmission's power is not overly strong on neighboring frequencies.

So, DSSS signals are filtered in such a way as to be strongest on the center frequency (farthest from other channels, and farthest from licensed frequency space), but not excessively strong on neighboring frequencies (they must meet a certain level of loss/noise). So, where a 2.412GHz (channel 1) transmission is strongest, the 2.406 transmission will be weaker so that the RF signal will have reached a certain level of loss as heard at 2.399GHz. If you look at the specral mask of a DSSS transmission, you can see this feature more readily. It is still using subcarriers on the 22MHz of frequency space, but it is using modified power for transmissions, not equal power across the entire 22MHz. The signal will still be heard by other frequencies, but it will be weak enough to be background noise.

I hope that clarifies. Good questions.

The Fourier Transform of a pulse (RECT) results in a SINC function. This is the reason why you have sidelobes that extend beyond the 22 MHz wide carrier. There is filtering done for 802.11b DSSS that minimizes sidelobes, however they are nearly impossible to eliminate.

In the frequency domain, 802.11g OFDM signals appear much more square due to the multiple sub-carriers, which also decrease interference for adjacent (1,6,11) channels.