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Old 4-Dec-2013, 2:48 AM   #6
mmbridges
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Join Date: Nov 2013
Posts: 8
So it looks like Frii's formula is what I was looking for. See

http://en.wikipedia.org/wiki/Noise_figure

I believe this is the formula behind the various online cascaded noise figure calculators ADTech mentioned. See

http://www.changpuak.ch/electronics/calc_01.php

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DISCLAIMER: What follows below will probably not be interesting to most but it helped me better understand how people were explaining noise figure accounting with respect to preamps, cable losses and tuner 1st circuit noise. What was confusing me was how various gain and noise figures were being added an subtracted sometimes but then later completely ignored using justifications like "the preamp gain is large enough". Hopefully I got it right. With luck there is maybe one other person that might find this was not a waste of time reading. I also suspect there is probably a much more succinct, easier and clearer way to explain it. If I made a mistake in my explanation please straighten me out.
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Here is my understanding from a mathematical point of view using a simplified system consiting of three cascaded stages consisting of preamp (pa), lumped cable losses (c) and set top box 1st transistor circuit (stb).


(1) Let F_sys = system noise factor = SNR_in/SNR_out

where SNR_in = signal to noise ratio at input to preamp immediately following antenna and
SNR_out = signal to noise ratio immediately following 1st transistor circuit of set top box

Since F_sys, SNR_in and SNR_out are linear ratios we can convert to dB and write

(2) NF_sys(dB) = System Noise Figure = 10 log(F_sys) = SNR_in(dB) - SNR_out(dB)

I believe ultimately, the SNR_out(dB) has to be at least > 15dB in order for the STB decoding circuitry to lock on/decode the signal.

Lets assume the antenna location and antenna gain are such that at the preamp input, SNR_in = 18dB. We can then write

(3) NM(dB) = Noise Margin = SNR_out(dB) -15dB= SNR_in(dB) - NF_sys(dB) - 15dB = 3dB - NF_sys(dB)


Now lets look at Frii's formula for the above three stage system. Basically used the Wikipedia link above to write the following equation:


(4) F_sys = F_pa + (F_c-1)/A_pa + (F_stb -1)/(A_pa*A_c)

where:

G_pa = preamp gain (dB) > = 0
L_c = cable loss (db) < = 0
NF_pa = preamp noise figure (dB) > = 0
NF_c = equivalent cable loss noise figure (dB) = -L_c > = 0
NF_stb = STB 1st circuit noise figure (dB) > = 0


F_pa = preamp noise factor (linear) = 10^NF_pa/10
F_c = cable loss equivalent noise factor (linear) = 10^NF_c/10 > = 1
F_stb = set top box 1st transistor circuit noise factor (linear) = 10^NF_stb/10 > = 1
A_pa = preamp gain ratio (linear) = 10^G_pa/10 > = 1
A_c = cable loss ratio (linear) = 10^L_c/10 < = 1

It is important to note that Frii's formula is specified in linear ratio terms even though component noise figures and gains are typically expressed in dB. We will see the significance of this later. So lets start with a perfect system.

Case 1: Perfect system with unity gain pre-amp
--------------------------------------------------

Defined as

F_pa = 1 noiseless preamp (NF_pa=0dB)
F_c = 1 ideal cables (NF_c=0dB)
F_stb = 1 noiseless stb (NF_stb=0dB)
A_pa = 1 unity gain preamp (G_pa=0dB)
A_c = unity gain cables (L_c=0dB)

Substituting into (4) we get

F_sys = F_pa + (F_c-1)/A_pa + (F_stb -1)/(A_pa*A_c)
= 1 + (1-1)/1 + (1-1)/(1*1) = 1 (linear) and substituting into (2) gives

NF_sys = 10log(1) = 0dB and substituting into (3) gives

NM = 3dB - NF_sys(db) = 3dB - 0dB = 3dB (the best an ideal system could ever get assuming the specified SNR_in). So what happens when we add in a noisy STB?


Case 2: lossless cables, ideal unity gain pre-amp and noisy STB
----------------------------------------------------------------

Defined as

F_pa = 1 noiseless preamp (NF_pa=0dB)
F_c = 1 ideal cables (NF_c=0dB)
F_stb = 10^(NF_stb/10) = 4 noisy stb (NF_stb=6dB)
A_pa = 1 unity gain preamp (G_pa=0dB)
A_c = unity gain cables (L_c=0dB)

Substituting into (4) we get

F_sys = F_pa + (F_c-1)/A_pa + (F_stb -1)/(A_pa*A_c)
= 1 + (1-1)/1 + (4-1)/(1*1) = 4 (linear) and substituting into (2) gives

NF_sys = 10log(4) = 6dB and substituting into (3) gives

NM = 3dB - NF_sys(db) = 3dB - 6dB = -3dB

A noisy STB 1st circuit has eaten up all of our noise margin. What happens when we add in a quiet unity gain preamp?


Case 3: lossless cables, unity gain but quiet pre-amp and noisy STB
----------------------------------------------------------------

Defined as

F_pa = 10^(N_pa/10) = 1.6 quiet preamp (NF_pa=2dB)
F_c = 1 ideal cables (NF_c=0dB)
F_stb = 10^(NF_stb/10) = 4 noisy stb (NF_stb=6dB)
A_pa = 1 unity gain preamp (G_pa=0dB)
A_c = 1 unity gain cables (L_c=0dB)

Substituting into (4) we get

F_sys = F_pa + (F_c-1)/A_pa + (F_stb -1)/(A_pa*A_c)
= 1.6 + (1-1)/1 + (4-1)/(1*1) = 4.6 (linear) and substituting into (2) gives

NF_sys = 10log(4.6) = 6.6dB and substituting into (3) gives

NM = 3dB - NF_sys(db) = 3dB - 6.6dB = -3.6dB

So before I understood Frii's formula I would have guessed that the NM would decrease by 6dB from the STB plus another 2dB from the preamp giving a net NF_sys of 8dB. Instead it only degrades the NM by 6.6dB. Stated another way, 2dB of preamp NF only increases the system noise figure by .6dB. The reason is that gains, losses and noise figures specified in dB don't add nicely when inserted into Frii's formula. log(a+b) does not equal log(a)*log(b)!


So lets add in cable losses.

Case 4: lossy cables, unity gain but quiet pre-amp and noisy STB
----------------------------------------------------------------

Defined as

F_pa = 10^(N_pa/10) = 1.6 quiet preamp (NF_pa=2dB)
F_c = 10^(NF_c/10) = 2.5 nonideal cables (NF_c=4dB)
F_stb = 10^(NF_stb/10) = 4 noisy stb (NF_stb=6dB)
A_pa = 1 unity gain preamp (G_pa=0dB)
A_c = 10^(L_c/10)=0.3981 lossy cables (L_c=-4dB)

Substituting into (4) we get

F_sys = F_pa + (F_c-1)/A_pa + (F_stb -1)/(A_pa*A_c)
= 1.6 + (2.5-1)/1 + (4-1)/(1*0.3981) = 10.65 (linear) and substituting into (2) gives

NF_sys = 10log(10.65) = 10.3dB and substituting into (3) gives

NM = 3dB - NF_sys(db) = 3dB - 10.3dB = -7.3dB


Adding an additional -4db of cable loss increases the system noise figure by 3.7dB. Interestingly this is closer to what I might have assumed if I simply added 4db of cable loss to the NF_sys from the previous case.


So lets look at how preamp gain affects things by looking at Frii's formula in the limit as G_pa->inf.

Case 5: lossy cables, quiet pre-amp with infinite gain and noisy STB
-----------------------------------------------------------------------

Defined as


F_pa = 10^(N_pa/10) = 1.6 quiet preamp (NF_pa=2dB)
F_c = 10^(NF_c/10) = 2.5 nonideal cables (NF_c=4dB)
F_stb = 10^(NF_stb/10) = 4 noisy stb (NF_stb=6dB)
A_pa = inf preamp with infinite gain (G_pa= inf dB)
A_c = 10^(L_c/10)=0.3981 lossy cables (L_c=-4dB)

Substituting into (4) we get

F_sys = F_pa + (F_c-1)/A_pa + (F_stb -1)/(A_pa*A_c)
= 1.6 + (2.5-1)/inf + (4-1)/(inf*0.3981)
= 1.6 + 0 +0
= 1.6 (linear) and substituting into (2) gives

NF_sys = 10log(1.6) = 2dB and substituting into (3) gives

NM = 3dB - NF_sys(db) = 3dB - 1dB = 2dB

So one can see how increasing the preamp gain allows one to ignore cable losses and STB noise. Large numbers in the denominator blows associated terms away in Frii's formula.

As G_pa--> inf then NF_sys --> N_pa.


Of course, as has been noted, you are stuck with the preamp noise figure and can get no more improvement in noise margin. Clearly infinite preamp gain is not practical so one might ask what is the minimum pre-amp gain to achieve positive NM?

Case 6: lossy cables, quiet pre-amp with minimum gain and noisy STB
-----------------------------------------------------------------------

Defined as


F_pa = 10^(N_pa/10) = 1.6 quiet preamp (NF_pa=2dB)
F_c = 10^(NF_c/10) = 2.5 nonideal cables (NF_c=4dB)
F_stb = 10^(NF_stb/10) = 4 noisy stb (NF_stb=6dB)
A_pa = to be solved for
A_c = 10^(L_c/10)=0.3981 lossy cables (L_c=-4dB)

To determine the minimum preamp gain set NM=0dB in (3) and solve to get

NF_sys (max) = 3dB and subsequently F_sys (max) = 10^(NF_sys/10) = 2


solving (4) for A_pa and substituting in F_sys (max) gives


A_pa_min = (F_c-1+(F_stb-1)/A_c)/(F_sys-F_pa)
= (2.5 - 1 + (4-1)/.3981)/(2-1.6)
= 22.6

Converting to dB gives the minimum preamp gain to reach the threshold of positive noise margin as

G_pa_min = 10log(A_pa) = 13.5 dB

So 13.5dB of preamp gain gets you back to 0db NM but you need an infinite amount of additional gain to get the last possible 1dB of remaining NM back.

So one question that still remains is what is the sweet spot selecting a preamp gain given the outlined diminishing returns. I suspect you could pick the maximum preamp gain hat would not overload the preamp or STB 1st circuit but I am not sure how one would determine that.

Another question is how can Frii's formula be used to analytically determine the preamp gain that would cause it to be dominant? I believe GroundUrMast outlined in his response to me something in terms of baseline receiver noise, baseline antenna noise and gain but I haven't been able to frame it in terms of Frii's formula.


Frii's formula for the above 3 stage system, explicitly written in terms of inputs expressed in dB's, is quite nonlinear and given as:

NF_sys = 10log[10^NF_pa/10 + (10^NF_c/10 -1)/10^G_pa/10 + (10^NF_stb/10 -1)/10^(G_pa+L_c)/10]

there is probably some way to rearrange the above equation so that it is clear how much bigger G_pa must be over L_c and NF_stb in order for everything down stream of the preamp to be ignored but I think I'll stop there for now.

Last edited by mmbridges; 4-Dec-2013 at 3:02 AM.
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