SNR input, what is it?

If you know the gain of your unit/device/dut, then you can calculate the input-referred noise as the output-referred noise divided by the gain. And the signal powers at input and output you probably know.

Thanks jjx that makes sense.

So, for an example to put numbers in perspective:
Considering a 50-Ohm system, I got a 10 dB attenuator connected to a signal generator, -30dBm input power let's say.
Measurement is -40 dBm then. Gain -10 dB. So grouping this together I want to use the formula,
Noise factor F = (SNRinput)/(SNRoutput)= (Si/Ni)/(So/No)
Output side.
So=-40dBm or 0.0000001 W, known from measurement.
No= -156dBm or 2.5119e-19 W, known from measurement.
Input side.
Si=-30dBm or 0.000001 W, known because it is my input signal.
Ni is unknown, calculated based on output noise and gain( -10dB or 0.1) in linear
Ni= No/linear gain = 2.5119e-19/0.1=2.5119e-18

F=(Si/Ni)/(So/No)=(`1e-6/2.5119e-18)/(1e-7/2.5119e-19) = 1

NF = 10log10(F) = 0 dB

I expect NF to be 10 dB, equal to the attenuator value. What can be wrong with the variables above?
I still unsure about the calculation of the input noise, it seems that this is the only unknown in the noise factor calculation.

Thanks in advance for your inputs.

fgavin said:

NF = 10log10(F) = 0 dB

Click to expand...

How do you come to 0 dB?

I get 10 dB.

Klaus

KlausST said:

How do you come to 0 dB?

I get 10 dB.

Klaus

Click to expand...

I got noise factor F= 1 from F=(Si/Ni)/(So/No)=(`1e-6/2.5119e-18)/(1e-7/2.5119e-19) = 1

Hi,

makes sense. .. if F = 1
I mistakenly used F to be "10".

Thanks for clarifying.

Klaus

Have you payed attention that Output Noise is lower than Input Noise..!
No= -156dBm or 2.5119e-19 W
Ni=2.5119e-18 W=-146 dBm

Your assumption is wrong. You cannot know Input Noise unless a precise measurement is done.
Therefore Output Noise is divided by Gain and Input Referred Noise is found. It doesn't mean that the Noise is present at the Input and it's amplified by Gain, No..
Noise is a Distributed Random Process and any physical element contributes onto this Input Referred Noise in circuit.
So, sources are assumed as Noiseless in general cases. Antennas and some oscillators are defined with their Noise Characteristics.

BigBoss said:

Have you payed attention that Output Noise is lower than Input Noise..!
No= -156dBm or 2.5119e-19 W
Ni=2.5119e-18 W=-146 dBm

Click to expand...

Yes, output noise is lower than input because noise passes through the 10 dB attenuator.
Input noise is calculated based on the jjx's suggestion, if I undestood it correctly Ni is calcualated by No/0.1, 0.1 being the linear gain of -10dB attenuation.

BigBoss said:

Your assumption is wrong. You cannot know Input Noise unless a precise measurement is done.
Therefore Output Noise is divided by Gain and Input Referred Noise is found. It doesn't mean that the Noise is present at the Input and it's amplified by Gain, No..
Noise is a Distributed Random Process and any physical element contributes onto this Input Referred Noise in circuit.
So, sources are assumed as Noiseless in general cases. Antennas and some oscillators are defined with their Noise Characteristics.

Click to expand...

If we agree that all other variables are measurable, this brings back my original question, how does one measure/determine the input noise, Ni in the equation to find SNR at the input, to hence find the noise figure . In this example, I am using the formula to find a NF of a 10 dB attenuator, but I can apply it to amplifiers too. I just don't know how to find the input noise Ni.

fgavin said:

Yes, output noise is lower than input because noise passes through the 10 dB attenuator.
Input noise is calculated based on the jjx's suggestion, if I undestood it correctly Ni is calcualated by No/0.1, 0.1 being the linear gain of -10dB attenuation.

If we agree that all other variables are measurable, this brings back my original question, how does one measure/determine the input noise, Ni in the equation to find SNR at the input, to hence find the noise figure . In this example, I am using the formula to find a NF of a 10 dB attenuator, but I can apply it to amplifiers too. I just don't know how to find the input noise Ni.

Click to expand...

The Output Noise cannot be lower than Input Noise in ANY practical system. An Attenuator adds extra Noise into the System so that NF of an Attenuator=Attenuation Value

Input Noise of a system in other words Noise Density of a Source can be found by a Measurement and this is done by Phase Noise Measurement Techniques. This gives an idea how much Noise is carried by signal.

For more Information, refer to "Noise in Linear and Nonlinear Circuits. Stephen Maas" Artech House 2005

Hi,

BigBoss said:

The Output Noise cannot be lower than Input Noise

Click to expand...

For sure the output noise can be lower than the input noise....when we talk about noise, not SNR.

But I guess in this thread there is a confusion with noise vs SNR.
* Noise is an absolute value, given in V or A.
* SNR is a relative value given in dB.

SNR of output can not be lower than SNR of input.

Output_Noise = input_Noise × gain plus additional_noise
(Mind: I wrote "plus" instead of "+" because noise has to be added using square and squareroot)

Output_SNR = input_SNR plus additional_noise
(Not multiplied with gain, because "signal" becomes multiplied and "noise" becomes multiplied with gain, thus the ratio does not change)

Klaus