"Dana" <raff242@yahoo.com> wrote in message
news:12q01dsdlua047d@corp.supernews.com...
| > > On Jan 6, 10:12 am, "Dana" <raff...@yahoo.com> wrote:
| > > > Everything produces noise.
| > > > And in the applications I was describing we had anywhere from 100 to
| 600
| > > > feet of cable to deal with.
| > >
| > > Wrong. Coax cable does not normally generate any noise by itself.
|
| Here is an article that describes noise in RF systems
|
http://telephonyonline.com/wireless/...luating_noise/
| All matter at temperatures above absolute zero (0K, about -460F) radiates
| electromagnetic energy. The amount of energy is related to temperature --
| the hotter the matter, the more energy is radiated. This energy is
described
| by Boltzmann's Constant, 'k' (k = -198.6dBm/degreesK-Hz). This constant,
| multiplied by the temperature of the matter a receiver views and the
system
| bandwidth, yields an irreducible background noise against which a desired
| signal must compete. This is thermal noise.
|
| In a cellular system, the receiving antennas are designed to view the
ground
| around the site because that's where the subscribers are. The ground
| temperature varies, but at 80F, it's about 300K (T=300K). RF engineers
| typically use this number as a rule of thumb.
|
| The receiver bandwidth varies depending on the technology, but the same
| principles hold for all technologies. EAMPS, for example, uses 30kHz-wide
| channels. Receiver bandwidth is a bit less than 30kHz, for rejection of
| adjacent channels. Assume the typical EAMPS receiver has a bandwidth of
| 25kHz (B=25,000Hz). By making this assumption, you can calculate the
amount
| of noise an EAMPS receiver will have in its passband if it contributes no
| noise of its own.
|
| This receiver thermal noise floor often is referred to as 'kTB.' In the
| example, assume consistent units:
|
| kTB = -198.6 + 10 Log(300)
|
| + 10 Log(25,000) in dBm
|
| kTB = -129.8dBm
|
| Thus, if you build a perfect EAMPS cellular receiver, it would have -129.
| 8dBm of noise in its passband competing with the wanted signal.
|
| CABLE LOSS Cable, filters and other passive elements exhibit a loss and
| produce thermal noise.
|
| If a cable (or other lossy element) has 10dB of loss, it will attenuate
the
| desired signal as well as the input noise by 10dB. But at the output of
the
| cable, you will see noise at least equal to kTB because the cable itself
| contributes it.
|
| If you put a signal into the cable at -100dBm over a thermal noise
of -129.
| 8dBm, you have a signal-to-noise ratio of 29.8dB at the cable input. At
the
| cable output, the signal has been attenuated by 10dB to -110dBm. The noise
| you put into the cable also has dropped the same amount, to -139.8dBm. But
| the cable contributes its noise floor of -129.8dBm, so the combined
| (uncorrelated) noise terms are -129.8dBm. The resulting signal-to-noise
| ratio is only 19.8dB at the cable output. You sacrifice 10dB of
| signal-to-noise ratio. This is why you spend money on 7/8", 15/8" or
larger
| coax at cell sites to reduce this loss.
In the real world such calculations mean little as with most wireless
systems the problem is not a noise but interference floor. Much of this is
in band interference but out of band can be significant especially in metro
areas.