The noise level in an electronic system is typically measured as an electrical power P in Watts or dBm, a root mean square (RMS) voltage (identical
to the noise standard deviation) in volts, dBµV or a mean squared error (MSE) in volts squared. Noise may also be characterized by its probability
distribution and noise spectral density N0(f) in watts per hertz.
Noise Power ` P = k*T*B ` where ` k = 1.3806504 * 10^-23 J/K ` (Boltzmann Constant).
T is the temperature in [K] (Kelvin) and B is the Bandwidth in [Hz].
` P = 10*log_10 (P/P_0) ` in [dBm] and P0= 1 mW
Noise Voltage ` U = sqrt(P*R) ` or ` U=sqrt(k*T*B*R) ` where R = 50 Ω. (usually)
Reference Temperature ` T = 273.15 K + theta_(a m b) ` usually T = 300 K.
Noise Figure may be given as a linear factor or in dB. It is a measure of degradation of the signal-to-noise ratio (SNR),
caused by one or more components in a signal chain.
F = SNRinput / SNRoutput whereas SNRinput` = S_(i n)/N_(i n) ` and SNRoutput` = S_(o u t)/N_(o u t) `
FdB `= 10*log_10 (F) `
SNRinput > SNRoutput as the 'device' adds noise, decribed by the Noise Figure :-(
⇒ You can compare receivers with the noise-figure. When using the 'sensitivity' a comparison is only
possible if this is done at the same bandwidth.
When using multi-stage-systems the noie figures add up using the 'de Friis-Formula'. calculator
The Noise Figure of the first device is decisive for the overall Noise Figure
The Noise Figures of the following stages loose influence with the Gain of the first stage.
Example : FM-Receiver
We have a Receiver with F = 2 dB operated at an Antenna. Between Receiver and Antenna we have an Coaxial cable of 25 m with
an Attenuation of 7 dB/100 m at 100 MHz. This means we have an attenuation of 1.75 dB which is equal to F = 1.75 dB.
F = 1.49 ( F = 10^(FdB/10)). This gives an overall Noise Figure of :
Ftotal = F1 `+ (F2 -1 )/ G1 = 1.49 + (1.58 - 1)/(1/1.49) = 2.37 ` ≡ 3.75 dB
The Attenuation of the Coaxial cable worsens the overall Noise Figure.
From a friend, we hear that adding an amplifier will improve the situation. Unfortunately we forgot to ask, where to put it.
Amplifier : Gain 20dB, F = 3 dB.
If we put it between Antenna and Coaxial cable, this results in F = 3.03 dB
If we put it between Coaxial cable and Receiver, this results in F = 4.76 dB
With the above formula (Noise Power), we may also calculate the Noisefloor of a Receiver. It depends only on the Noise Figure
and the Bandwidth :
Noise floor = -174 + NF + 10 log (Bandwidth)
Resistors are noisy. This noise depends on temperature. Therefore it is called 'thermal noise'. It further depends
on bandwidth. (See formula above). Using T = 300 K and B = 1 Hz we get :