>Hi all, >I have my own numpty questions. When using Y-splitters such as the
010-0115 on this page: http://www.satcure.co.uk/accs/page15a.htm It says "Note: all 2-way splitters pass approximately 40% of the signal to each output." For the SkyLink compatible splitter, it says "As with all passive splitters, this unit will reduce the signal by around 4db (60%)".
It says several things on that site that are wrong. For instance: "Always put splitters/amplifiers as CLOSE to the aerial as possible (the loft is ideal)." It makes no difference where you put a splitter.
Say you split it again with a 2nd splitter, downstream on one half of the run, ie:
>Does splitter2 then pass 40% of the 40% of the original signal to each
half? Ie TV2 and TV3 get only 16% of the original signal? Yes
>Or do all 3 TVs
get the same "amount" of signal - if so how much? No. Depends on the splitter, really.
Forget the 40% figure. That's just a spuriously accurate sounding approximation. Decent splitters pass about half the signal less connector loss, but matching issues will make the response quite uneven.
>Also, does disconnecting any of the TVs, or whether they are on/off have
any effect to the amount of signal delivered to the others?
Not with decent splitters. Others might like to write you an essay about this.
Just for fun, think about this. A normal distribution amplifier as used on a head-end in a block of flats or wherever will have a maximum output of about 65dBmV (all figures are for analogue TV). Assuming that a splitter loses 3dB and that there are no cable losses, and that a telly needs +5dBmV, how many tellys could you run from the amp, and how many two way splitters would you use? But assuming that a splitter loses 4dB and that there is 8m of cable between every splitter and the next, losing 2dB, now how many tellys could you feed and how many splitters would you use?
On Mon, 9 Nov 2009 18:31:57 GMT, Bill <wrightsaeri...@f2s.com> wrote: > Just for fun, think about this. A normal distribution amplifier as used on a > head-end in a block of flats or wherever will have a maximum output of about > 65dBmV (all figures are for analogue TV). Assuming that a splitter loses 3dB > and that there are no cable losses, and that a telly needs +5dBmV, how many > tellys could you run from the amp, and how many two way splitters would you > use? But assuming that a splitter loses 4dB and that there is 8m of cable > between every splitter and the next, losing 2dB, now how many tellys could > you feed and how many splitters would you use?
Bill wrote: > Just for fun, think about this. A normal distribution amplifier as used on a > head-end in a block of flats or wherever will have a maximum output of about > 65dBmV (all figures are for analogue TV). Assuming that a splitter loses 3dB > and that there are no cable losses, and that a telly needs +5dBmV, how many > tellys could you run from the amp, and how many two way splitters would you > use?
8 tellys, 7 splitters
But assuming that a splitter loses 4dB and that there is 8m of cable
> between every splitter and the next, losing 2dB, now how many tellys could > you feed and how many splitters would you use?
Hundreds of tellys, zero splitters.
Just whack a huge high gain aerial on the output of the head-end and supply a load of cheapo set-back loop aerials purloined from the nearest sunday trash street market.
Then jump in the van and head north or south leaving no forwarding address.
> Quite a few?! 1048576 tellys and 1048575 splitters used?
>>But assuming that a splitter loses 4dB and that there is 8m of cable >>between every splitter and the next, losing 2dB, now how many tellys could >>you feed and how many splitters would you use?
> 1042 tellys and 1023 splitters used?
I'm with this man. Apart from the 1042 bit! Obviously a typo though & he meant 1024.
> I'm with this man. Apart from the 1042 bit! Obviously a typo though & he > meant 1024.
On a vaguely-related note, how much loss is incurred when using a UHF combiner, let's say to combine signals from an aerial with signals from an RF modulator?
And which is the best make/model of combiner to use?
Mike Henry <{$mrtickl...@nospam.demon.co.uk> wrote: >I have my own numpty questions. When using Y-splitters such as the >010-0115 on this page: >http://www.satcure.co.uk/accs/page15a.htm
See that cheap splitter? If you're using that, only put it right at the back of the set with no more than short flyleads between the splitter and the receivers. They typically contain this:
...which results in an impedance mismatch. And if you have long leads downstream of that splitter, reflections will cause destructive interference at some frequencies.
The "Deluxe UHF Y splitter" on that page, on the other hand, is *probably* a transformer splitter, which won't have that problem, but it'd be nice if its technical spec was available.
I got one of the following at Maplin, and it worked fine as a loft splitter. The F-plugs also have a lower insertion loss. If you're ordering online, try to find some place that sells it cheaper than Maplin:
Oh, and if you're splitting more than two ways, then use a 4-way splitter which will save on the insertion losses compared to stacked 2-way splitters, giving you, say, 7dB attenuation rather than 8dB.
>Oh, and if you're splitting more than two ways, then use a 4-way splitter >which will save on the insertion losses compared to stacked 2-way >splitters, giving you, say, 7dB attenuation rather than 8dB.
I couldn't resist the urge to buy a similar-looking 'Eagle' 2-way splitter in the local '99p Shop'. It claimed 5 to 2500MHz, and had power-passing to both output ports. I expected it to be rubbish. A quick measurement at around 570 MHz showed a definitely high insertion loss of some 6dB ("resistive?", I thought), but the isolation was surprisingly excellent 35dB. -- Ian
ON the fm question, are you saying they use a tv aerial not an fm one? If it works they must be in a very strong signal area.
Also if you did use an amp instead of passive splitters, then it could well be that this trick would not work any more depending on the amps bandwidth..I suspect all of this only work because the signal is strong on both tv and fm.
Brian
-- Brian Gaff....Note, this account does not accept Bcc: email. graphics are great, but the blind can't hear them Email: bria...@blueyonder.co.uk ___________________________________________________________________________ ___________________________________
"Mike Henry" <{$mrtickl...@nospam.demon.co.uk> wrote in message
> I have my own numpty questions. When using Y-splitters such as the > 010-0115 on this page: > http://www.satcure.co.uk/accs/page15a.htm > It says "Note: all 2-way splitters pass approximately 40% of the signal to > each output." > For the SkyLink compatible splitter, it says "As with all passive > splitters, this unit will reduce the signal by around 4db (60%)".
> Say you split it again with a 2nd splitter, downstream on one half of the > run, ie:
> Does splitter2 then pass 40% of the 40% of the original signal to each > half? Ie TV2 and TV3 get only 16% of the original signal? Or do all 3 TVs > get the same "amount" of signal - if so how much?
> Also, does disconnecting any of the TVs, or whether they are on/off have > any effect to the amount of signal delivered to the others?
> PS. I know you should use a distribution amplifier - this is to settle a > discussion :-). In the house in question, splitter1 is in the loft close > to the aerial, and splitter2 is down in the lounge. ("TV2" is actually a > receiver, with the TV aerial being used as an FM aerial - apparently it's > worked like a dream for years, and all his TV aerials have worked > excellently for FM)
In article <hdal46$q3...@aioe.org>, jamie_...@excite.com says...
> On a vaguely-related note, how much loss is incurred when using a UHF > combiner, let's say to combine signals from an aerial with signals from an RF > modulator?
Wow! I can't believe that you really asked that question, Jamie! (It definitely didn't come across as a joke.)
As I'm sure you already know, a combiner is just a splitter used backwards - in fact, in networks with a reverse channel, such as CATV, most splitters in the whole network function as splitters and combiners at the same time.
As for your aerial/modulator question, it depends on the relative levels. If both are similar, a normal 3dB splitter will be just the job. On the other hand, if the levels from your aerial are well below the output of the modulator (most probable) you would need a suitable directional coupler, the value of which would be the nearest available match to the level difference. The loss on the 'through' path will vary with coupler value (read the specification) but is minimal with the higher values (a 20dB coupler will probably have a through loss of less than 1dB.)
If you have a high output modulator - say 60dBmV - and much lower levels from the aerial system, I'd use a 20dB directional coupler with pads on the modulator output to make up the difference.
> And which is the best make/model of combiner to use?
A good one!
No doubt Bill will be along soon to advise what's currently available.
In article <slrnhfgqdc.qib.ab...@news.pr.network>, ab...@orac12.clara34.co56.uk78 says...
> On Mon, 9 Nov 2009 18:31:57 GMT, Bill <wrightsaeri...@f2s.com> wrote:
> > Just for fun, think about this. A normal distribution amplifier as used on a > > head-end in a block of flats or wherever will have a maximum output of about > > 65dBmV (all figures are for analogue TV). Assuming that a splitter loses 3dB > > and that there are no cable losses, and that a telly needs +5dBmV, how many > > tellys could you run from the amp, and how many two way splitters would you > > use? But assuming that a splitter loses 4dB and that there is 8m of cable > > between every splitter and the next, losing 2dB, now how many tellys could > > you feed and how many splitters would you use?
> 42.
That's the answer to the ultimate question of life, the universe and everything.
Now answer the penultimate question - the one set by Bill!
(I'm with the others on this one - 2^20 and (2^20)-1 for the lossless cable version and 2^10 and (2^10)-1 for the other.)
On Tue, 10 Nov 2009 16:35:51 -0000, Terry Casey <k.t...@example.invalid> wrote: >> > Just for fun, think about this. A normal distribution amplifier as used on a >> > head-end in a block of flats or wherever will have a maximum output of about >> > 65dBmV (all figures are for analogue TV). Assuming that a splitter loses 3dB >> > and that there are no cable losses, and that a telly needs +5dBmV, how many >> > tellys could you run from the amp, and how many two way splitters would you >> > use? But assuming that a splitter loses 4dB and that there is 8m of cable >> > between every splitter and the next, losing 2dB, now how many tellys could >> > you feed and how many splitters would you use?
>> 42.
> That's the answer to the ultimate question of life, the universe and > everything.
It is therefore correct.
> Now answer the penultimate question - the one set by Bill!
> (I'm with the others on this one - 2^20 and (2^20)-1 for the lossless > cable version and 2^10 and (2^10)-1 for the other.)
> In article <hdal46$q3...@aioe.org>, jamie_...@excite.com says...
>> On a vaguely-related note, how much loss is incurred when using a UHF >> combiner, let's say to combine signals from an aerial with signals from an >> RF >> modulator?
> Wow! I can't believe that you really asked that question, Jamie!
> And which is the best make/model of combiner to use?
>A good one! >No doubt Bill will be along soon to advise what's currently available.
Here I am, large as life and twice as 'orrible! I don't use combiners by themselves. I use a single channel pass filter, a single channel stop filter, and a Taylor TD2-4F splitter.
The message <C7SdndPXI9mC_WXXnZ2dnUVZ8s6dn...@pipex.net> from "Bill" <wrightsaeri...@f2s.com> contains these words:
====snip====
> Just for fun, think about this. A normal distribution amplifier as used on a > head-end in a block of flats or wherever will have a maximum output of about > 65dBmV (all figures are for analogue TV). Assuming that a splitter loses 3dB > and that there are no cable losses, and that a telly needs +5dBmV, how many > tellys could you run from the amp, and how many two way splitters would you > use? But assuming that a splitter loses 4dB and that there is 8m of cable > between every splitter and the next, losing 2dB, now how many tellys could > you feed and how many splitters would you use? > Bill
Off the top of my head, I'd say a thousand TVs and a total of 1023 splitters. It's a "binary question". Each 'split' involves (rather conveniently in this example) a 6 db reduction which, in this instance, can be applied up to ten times before the 65dbmV is reduced to the 5dbmV limit.
To simplify, assume a starting level of 15dbmV. This allows the signal to be split twice on its way to each TV set. The first split feeds two other splitters which feed a total of four TV sets.
The problem is rather analogous to determining the count range (or maximum number of unique bit patterns or codes) for any given number of binary digits in a binary word.
In this (simplified) case, a 2 digit word (the number of splitting stages the signal has to pass through to reach a TV set) resulting in the need to use 3 splitters (N-1 where N is the maximum number of TV sets that can be fed with the required minimum signal level, in this case 2 to the power of the number of splits that are applied (2)).
If the input level is raised another 6db this will allow another array of splitters to double up the total TV count to 8 (which requires another 4 splitters giving a total of (2 ^3) - 1 splitters).
HTH
PS This conundrum of Bill's is simply a reworked version of the classic chess board and grains of rice 'reward' asked of an ancient king for a service rendered wherein the hero simply asked for a single grain of rice on the first square and this amount to be doubled up on each successive square until all 64 squares had been so suitably filled. Said king, not having the necessary grasp of the mathematics involved soon came to regret his hasty decision to agree the proposed condition of the 'reward'.
-- Regards, John.
Please remove the "ohggcyht" before replying. The address has been munged to reject Spam-bots.
>May I ask what that page means by the term "Mutual isolation: 30dB" >please? Is a high number a good thing, (Ian Jackson's 35dB for example)?
It's the attenuation from one output to the other, which protects against the receivers interfering with each other, so the higher the better.
BTW, Maplin also sell a range of screw-fit F-plugs, which you select according to the thickness of the cable. So if you've already got cable in place, you measure each cable and select the appropriate plug for each one, which I usually find involves a bit of head scratching, but F-plugs *are* better than the old Belling-Lee connectors.
In article <w5qdnXUIL_emWGTXnZ2dnUVZ7rOdn...@pipex.net>, wrightsaeri...@f2s.com says...
> Here I am, large as life and twice as 'orrible! I don't use combiners by > themselves. I use a single channel pass filter, a single channel stop > filter, and a Taylor TD2-4F splitter.
Thanks Bill, all the modulators I've ever had any dealings with have had excellent output filtering built in. Obviously not the case in the real world, though.
> The message <C7SdndPXI9mC_WXXnZ2dnUVZ8s6dn...@pipex.net> > from "Bill" <wrightsaeri...@f2s.com> contains these words:
> ====snip====
> > Just for fun, think about this. A normal distribution amplifier as used on a > > head-end in a block of flats or wherever will have a maximum output of about > > 65dBmV (all figures are for analogue TV). Assuming that a splitter loses 3dB > > and that there are no cable losses, and that a telly needs +5dBmV, how many > > tellys could you run from the amp, and how many two way splitters would you > > use? But assuming that a splitter loses 4dB and that there is 8m of cable > > between every splitter and the next, losing 2dB, now how many tellys could > > you feed and how many splitters would you use?
> > Bill
> Off the top of my head, I'd say a thousand TVs and a total of 1023 > splitters. It's a "binary question". Each 'split' involves (rather > conveniently in this example) a 6 db reduction which, in this instance, > can be applied up to ten times before the 65dbmV is reduced to the 5dbmV > limit.
Having correctly stated that there are 1023 splitters, what's the point of 'simplifying' 1024 TVs to 1000?
> To simplify,
... at which point, your 'simplification' vanishes into a thick fog!
It is, as you say, a binary problem but try this for simplification:
We start with 65dBmV and want 5dBmV at the end, so 65-5 gives us a 60dB budget.
In the first scenario, with loss less splitters, each stage = -3dB so, dividing 60 by 3 gives us 20 stages, thus 2 raised to the power of 20 gives the number of TVs and the number of splitters is one less:
The message <31303030373730364AFA380...@plugzetnet.co.uk> from Johnny B Good <jcs.computersb...@plugzetnet.co.uk> contains these words:
====snip====
> To simplify, assume a starting level of 15dbmV. This allows the signal > to be split twice on its way to each TV set. The first split feeds two > other splitters which feed a total of four TV sets.
Oops! I meant to say 17dbmV ((2x6)+5). The rest is correct, though.
> The problem is rather analogous to determining the count range (or > maximum number of unique bit patterns or codes) for any given number of > binary digits in a binary word. > In this (simplified) case, a 2 digit word (the number of splitting > stages the signal has to pass through to reach a TV set) resulting in > the need to use 3 splitters (N-1 where N is the maximum number of TV > sets that can be fed with the required minimum signal level, in this > case 2 to the power of the number of splits that are applied (2)). > If the input level is raised another 6db this will allow another array > of splitters to double up the total TV count to 8 (which requires > another 4 splitters giving a total of (2 ^3) - 1 splitters).
-- Regards, John.
Please remove the "ohggcyht" before replying. The address has been munged to reject Spam-bots.
> In article <31303030373730364AFA380...@plugzetnet.co.uk>, > jcs.computersb...@plugzetnet.co.uk says...
> > The message <C7SdndPXI9mC_WXXnZ2dnUVZ8s6dn...@pipex.net> > > from "Bill" <wrightsaeri...@f2s.com> contains these words:
> > ====snip====
> > > Just for fun, think about this. A normal distribution amplifier as > > > used on a > > > head-end in a block of flats or wherever will have a maximum > > > output of about > > > 65dBmV (all figures are for analogue TV). Assuming that a splitter > > > loses 3dB > > > and that there are no cable losses, and that a telly needs +5dBmV, > > > how many > > > tellys could you run from the amp, and how many two way splitters > > > would you > > > use? But assuming that a splitter loses 4dB and that there is 8m > > > of cable > > > between every splitter and the next, losing 2dB, now how many > > > tellys could > > > you feed and how many splitters would you use?
> > > Bill
> > Off the top of my head, I'd say a thousand TVs and a total of 1023 > > splitters. It's a "binary question". Each 'split' involves (rather > > conveniently in this example) a 6 db reduction which, in this instance, > > can be applied up to ten times before the 65dbmV is reduced to the 5dbmV > > limit. > Having correctly stated that there are 1023 splitters, what's the point > of 'simplifying' 1024 TVs to 1000?
My bad, I _thought_ I _had_ said a thousand and twentyfour :-(
> > To simplify, > .... at which point, your 'simplification' vanishes into a thick fog!
Yes, well it _was_ 4 am and I had an "Early Morning Appointment" scheduled for 10:20am at the Health Centre. I'm a 'Night Owl' sort of person so find the concept of the very existance of a "10:00am" as a practical (rather than theoretical) idea, a little bit disturbing to my normal daily routine and was in a little bit of a rush (I had planned on an 'Early' retirement to my bed of no later than 3am ;).
> It is, as you say, a binary problem but try this for simplification: > We start with 65dBmV and want 5dBmV at the end, so 65-5 gives us a 60dB > budget. > In the first scenario, with loss less splitters, each stage = -3dB so, > dividing 60 by 3 gives us 20 stages, thus 2 raised to the power of 20 > gives the number of TVs and the number of splitters is one less:
That first example was based on a purely theoretical scenario, impossible to achieve in practice, so I ignored it and dealt solely with the practically realisable case that he posed where he had elected to use a nice and convenient 6db loss per stage.
Which, basically reflects the fact that the amplifier is providing a power level a million times stronger than the minimum acceptable level required by each TV where it possible to losslessly share it amongst that number of TVs.
> The second scenario has a loss of 6dB per stage so the number of stages > reduces to 60/6 - 10. > Answer(2) = 2^10 = 1024 TVs and (2^10)-1 = 1023 splitters.
This last figure, although about 3 orders of magnitude less than the one derived in the theoretical case, is, nevertheless, an impressive enough figure. More so due to it being actually possible to achieve as a 'practical' (if unwieldy) exercise which, I think, answers the OP's "Numpy question" much better by its very nature (i.e. a practical, rather than a purely theoretical solution).
-- Regards, John.
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>> Here I am, large as life and twice as 'orrible! I don't use combiners by >> themselves. I use a single channel pass filter, a single channel stop >> filter, and a Taylor TD2-4F splitter.
>Thanks Bill, all the modulators I've ever had any dealings with have had >excellent output filtering built in. Obviously not the case in the real >world, though.
>Which, basically reflects the fact that the amplifier is providing a
power level a million times stronger than the minimum acceptable level required by each TV where it possible to losslessly share it amongst that number of TVs.
The message <_ZqdncQ2C4qAgWbXnZ2dnUVZ8vudn...@pipex.net> from "Bill" <wrightsaeri...@f2s.com> contains these words:
> >Which, basically reflects the fact that the amplifier is providing a > power level a million times stronger than the minimum acceptable level > required by each TV where it possible to losslessly share it amongst > that number of TVs. > 60dB =1,000:1, not 1,000,000:1. > Bill
Could you please clarify how you arrived at this assertion?
-- Regards, John.
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