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[SOLVED] [MOVED] calculating power or noise

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patel_123

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i know that power of noise is given by (noise voltage)^2 / 4R
where did this '4' come from?
shouldnt it be simply v^2/R?
WHY THE "4"

please answer my question.. i am banging my head over this..8-O
 

here's one guess --- don't take it as gospel, but it make put you on the right track :

noise voltage average is considered to be zero. However noise power is not. Thats where the v^2 comes from of course.
However note that its not absolute noise power, but rather mean noise power. That might imply a (V/2) ^2, and hence you get the 4.

EDIT : or you could look at this paper, where it is explained quite clearly in the first few pages itself.
 
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The power dissipated in a resistor ir Vr^2/R, where Vr ir the rms voltage value on the resistor and R is its resistance.
If you have a voltage source of rms voltage Vs and internal resistance Rs, then the available power (i.e. the maximum power you can get from it) is obtained when the load is matched to the source, i.e. when R=Rs. In this condition, R=Rs and Vr=Vs/2. Then the power (the available power) Vr^2/R is Vs^2/(4Rs).

Regards

Z
 
kripacharya said:
here's one guess --- don't take it as gospel, but it make put you on the right track :

noise voltage average is considered to be zero. However noise power is not. Thats where the v^2 comes from of course.
However note that its not absolute noise power, but rather mean noise power. That might imply a (V/2) ^2, and hence you get the 4.

EDIT : or you could look at this paper, where it is explained quite clearly in the first few pages itself.
Click to expand...

Thank you.. i find found it very helpfull

- - - Updated - - -

thank you for answering.. :)
 

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