Re: Channel loading * CORRECTION *
by Buffalo Chips, 4/24/2002 10:35:14 PM , IP: 63.26.236.166
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I gave you the wrong formula by mistake. I will put the correct one on ASAP. The rest of the information is correct though. Sorry, I must have had a bad day that day...lol.
> With a standard DFB optical transmitter you would need to find out the manufacturer's reccomended input level at full (100%) channel loading.
>
> Then,
>
> input level = 10log(# actual channel's) - mfg. rec. level
>
> I hope that this helps. Also remember that whenever you use half (50%) of the maximum channel loading (100%), you have a 3dB difference in overall input. This means you could increase the input by 3dB.
>
> EXAMPLE: You have a DFB laser with a maximum input 15dBmV with 80 NTSC channels per the manufacturer spec's.
>
> In reality, you only have 40 active NTSC channels. Since you have half channel loading (50%), you should increase input by 3dB to 18dBmV for better c/n performance.
>
> If you have 20 channels, ditto. Another 3dB difference to 21dBmV input.
>
> Just like a 2-way splitter. You split the input in half, you lose 3.01dB in theory. Reality you lose 3.4 - 4.0dB depending on the quality and frequency of the splitter. In optical input power, you are much closer to that 3.01dB theory number than with coaxial splitters.
>
> Good luck!
>
>
> > Help! What is the formula for figuring a power level based on channel loading,e.g. I am allowed a maximum RF input level of +60dBmV on a device based on 1 NTSC channel loading and have 118 analog channels, what is my maximum allowable input level? THIS IS JUST AN EXAMPLE, I AM LOOKING FOR FORMULA.
> > Thanks for your help.
