View Full Version : Audio Signal Question

pcancila

03-20-2008, 02:48 PM

I am going to be setting up a large multiamp system; and I need to find Coax Digital audio splitters. I'mt rying to split 1 coax signal into 2 or more. Does anyone know if this is possible and if there is a device available?

Thanks for help,

Phil

hermanv

03-20-2008, 09:06 PM

There are two possibilities:

Coax can be split with a conventional TV signal splitter. Like all passive splitters each leg will have about 2/3 the voltage, so it depends on the input compliance of the downstream device if it will recognize the signal or not with 2/3 the voltage. (The exact value should be 1 over the square root of 2 or .707)

The second way is with a signal buffer, I own a Monarchy DIP jitter reduction box that happens to have two outputs (most models have only one output). I bought it cheap years ago for about $100 if memory serves. My model handles only 2 channel digital, if you are talking about AC-3 I don't think this unit will work. There must be other active buffer/splitters on the market.

markw

03-21-2008, 03:39 AM

Coax can be split with a conventional TV signal splitter. Like all passive splitters each leg will have about 2/3 the voltage, so it depends on the input compliance of the downstream device if it will recognize the signal or not with 2/3 the voltage. (The exact value should be 1 over the square root of 2 or .707).It seems to me that for a two way splitter to send 2/3 the signal to each leg, it would have to been at least 4/3 the signal before the spli or have some inherent gain, which is not possible in a passive devicet.

Assuming balanced loads, the best a two-way splitter can do is to send 1/2 of the signal to each leg and this doesn't even account for any insertion loss.

hermanv

03-21-2008, 08:58 AM

It seems to me that for a two way splitter to send 2/3 the signal to each leg, it would have to been at least 4/3 the signal before the spli or have some inherent gain, which is not possible in a passive devicet.

Assuming balanced loads, the best a two-way splitter can do is to send 1/2 of the signal to each leg and this doesn't even account for any insertion loss.

You are correct for a resistive splitter. By using a pulse transformer 1/2 the power not 1/2 the voltage is sent to each leg giving each leg 0.707 the voltage for a loss of 3 dB, as you point out a resistive splitter has a loss of 6 dB.

A simple resistive splitter can be made by connecting 3 individual 27 Ohm resistors into a "Y" with one resistor lead going to each RCA jack center pin, the other resistor legs tied together and all the outer RCA shells (ground) connected together. When all three ports are connected, each port will end up with a 75 Ohm impedance.

I am unsure about minimum input signal levels for a SPDIF port so I didn't post the resistive solution.

Phill: If you decide to build and try this, keep the resistor leads reasonably short. Also have you considered using optical for one of the ports and coax for the other?

hermanv

03-21-2008, 09:10 AM

http://forums.audioreview.com/images/editor/insertimage.gif

Resistors really should be 25 Ohms, but 27 is the nearest standard 5% value.

markw

03-21-2008, 09:14 AM

It that what I refer to as a balun? In either case, you still lose at least 50% of the signal to each leg.

And, I dispute that 6 db loss figure. Assuming 100% efficiency, each leg of the split will be down only three db (or half the original signal) , not six (which would be one quarter the original signal).

hermanv

03-21-2008, 03:53 PM

It that what I refer to as a balun? In either case, you still lose at least 50% of the signal to each leg.

And, I dispute that 6 db loss figure. Assuming 100% efficiency, each leg of the split will be down only three db (or half the original signal) , not six (which would be one quarter the original signal).Using a resistive splitter each leg will be down to 1/2 the voltage. Since impedance is constant at 75 Ohms each leg will also have 1/2 the current (E over R is 1/2 of what it used to be). So 1/2 the voltage times 1/2 the current equals 1/4th the power or 6 dB.

dB for voltage = 20 * (log of ratio) for power it's 10 * (log of ratio). For a resistive splitter the voltage ratio is 1/2 so the dB equation is equal to 20 times (log of ratio). So, log of 2 is 0.3010299 times 20 equals 6.02 dB. You get an identical answer if you use 1/2 for the ratio instead of 2 (log of 0.5 is -0.3010299 times 20 equals -6.02 dB. Using the power equation it's the log of 1/4th or .602. 0.602 times 10 gives again 6 dB, in other words both voltage and power equations result in the same answer (as they should).

Using a transformer the ratio of each secondary to maintain a 75 Ohm input Z will be 1.414 to 1. On a transformer the impedance is the square of the turns ratio times the secondary load or 1.414^2 times 75 equals 150 Ohms. With two secondaries in parallel the Z returns to 75 Ohms (1/2 of 150). This means the voltage at each secondary is .707 times the input voltage not .5 times the input voltage. In a resistive splitter power is dissipated in the resistors, in the transformer design no power is lost (except for that power that is lost to imperfections which luckily are small in comparison) that's why the transformer solution results in less lost power, none of the power is converted to heat in any resistors. It is not practical to make a wide band transformer at home.

markw

03-22-2008, 10:25 AM

Using a resistive splitter each leg will be down to 1/2 the voltage. Since impedance is constant at 75 Ohms each leg will also have 1/2 the current (E over R is 1/2 of what it used to be). So 1/2 the voltage times 1/2 the current equals 1/4th the power or 6 dB.

dB for voltage = 20 * (log of ratio) for power it's 10 * (log of ratio). For a resistive splitter the voltage ratio is 1/2 so the dB equation is equal to 20 times (log of ratio). So, log of 2 is 0.3010299 times 20 equals 6.02 dB. You get an identical answer if you use 1/2 for the ratio instead of 2 (log of 0.5 is -0.3010299 times 20 equals -6.02 dB. Using the power equation it's the log of 1/4th or .602. 0.602 times 10 gives again 6 dB, in other words both voltage and power equations result in the same answer (as they should).

Using a transformer the ratio of each secondary to maintain a 75 Ohm input Z will be 1.414 to 1. On a transformer the impedance is the square of the turns ratio times the secondary load or 1.414^2 times 75 equals 150 Ohms. With two secondaries in parallel the Z returns to 75 Ohms (1/2 of 150). This means the voltage at each secondary is .707 times the input voltage not .5 times the input voltage. In a resistive splitter power is dissipated in the resistors, in the transformer design no power is lost (except for that power that is lost to imperfections which luckily are small in comparison) that's why the transformer solution results in less lost power, none of the power is converted to heat in any resistors. It is not practical to make a wide band transformer at home.

"The loss in a good splitter is around 4.0 dB per split. An average splitter has around 4.5 dB loss per split. (A two way splitter has one split, a four-way has 2, an eight way has 3.) Thus an average eight-way splitter should have about 12 to 13.5 dB insertion loss."

http://www.hometech.com/video/splitters.html

Here's one that has less loss, but it's still over 3 db.

http://www.smarthomeusa.com/Shop/cvc/CVC-Splitters/Item/C-0202/

Unless you've come up with a passive way to get more out than you put in to begin with, we're stuck with the fact that we're faces with that loss.

I really can't understand why you have to obfuscate this matter rather than simply state the facts. That tends to be a common trait with this hobby.

hermanv

03-22-2008, 06:02 PM

This has nothing to do with spin. You will find my posts have sound correct mathematics not opinion.

I never disagreed with 3dB, but I think you said 1/2 the voltage is lost and can not be less, with that I disagree. I do agree at least 1/2 the power is lost, that is 3dB. 1/2 the voltage is 6dB of loss - period as you said.

Anyway I didn't set out to hi-jack a thread, so I'm done.

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