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# Noise Figure of cascaded stages

by on April 21, 2012

Following the discussion on thermal noise and it’s modeling and noise figure computation for a simple resistor network, in this article let us discuss the Noise Figure of cascaded stages.

## Definition of Noise Figure

From Section 2.3.2 of RF Microelectronics, Behzad Razavi, the noise figure is defined as

$N_F = \frac{SNR_{in}}{SNR_{out}}$, where

$SNR_{in}=\frac{S_{in}}{N_{in}}$ is the input signal to noise ratio and

$SNR_{out}=\frac{S_{out}}{N_{out}}$ is the output signal to noise ratio.

Consider that the system has a gain $G$ and $N_a$ is the noise added by the system.

With that,

$S_{out}=GS_{in}$

$N_{out}=N_a+ GN_{in}$

The Noise Figure is,

$\begin{array}{lll}N_F&=&\frac{S_{in}/N_{in}}{S_{out}/N_{out}}\\\\&=&\frac{S_{in}/N_{in}}{GS_{in}/$$N_a+GN_{in}$$}\\\\\\&=&\frac{N_a+GN_{in}}{GN_{in}}\end{array}$.

From the previous post on thermal noise, we know that noise power over bandwidth $B$ at temperature $T$ Kelvin is,

$N_{in} = kTB$.

With that the noise figure equation can be modified as,

$\Large{N_F = \frac{N_a+GkTB}{GkTB}}$.

Two key equations to remember:

a) Noise added by the noise sources in system is,

$\Large{N_a = kTBG(N_F-1)}$

b) Noise at the output of the system is,

$\Large{N_{out} = kTBGN_F}$

## Example

For example, consider a device of bandwidth $B=20MHz$ and having a gain $G=60dB$ with noise figure of $NF_{dB}=5dB$. What is the noise power seen at the output of the device?

Solution: Assuming temperature of 300K, the noise seen at the input of the device

$\begin{array}{lll}N_{in}&=&10\log_{10}$$kTB$$\\&=&10\log_{10}$$1.38*10^{-23}*300*20*10^6$$\\&=&-100.82dBm\end{array}$.

$N_{out}=N_a+GN_{in}=$$N_{in}G$$N_F$.

Expressing in deciBels, the noise power seen at the output of the device is,

$N_{out}=60+5+-100.82 = -35.82dBm$.

## Noise Figure of cascade of two stages

Consider a system with cascade of two gain stages having gains of $G_1$, $G_2$ and with noise figure of $N_{F1}$ and $N_{F2}$ respectively.

Figure : Cascaded gain stages with noise power output

The noise power at the output of the first stage is the sum of

a) Noise power at the input terminal of first gain stage $N_{in,1} = kTB$ which is amplified by gain $G_1$.

b) Noise power generated by the first gain stage $N_{a,1} = kTBG_1(N_{F1}-1)$

Summing them up, the noise seen at the output of first gain stage is,

$\begin{array}{lll}N_{out,1}&=&kTBG_1+kTBG_1(N_{F1}-1)\\&=&kTBG_1N_{F1}\end{array}$

* Note : The sum of powers is possible because it is assumed that the input noise and the noise added by the system are uncorrelated i.e.

$\begin{array}{lll}E\{(x+y)^2\}&=&E\{x^2\}+E\{y^2\}+2E\{xy\}\\&=&E\{x^2\}+E\{y^2\}\end{array}$, where

$E\{xy\}=0$ when $x$ and $y$ are uncorrelated.

Similarly the noise power seen at the output of the second stage is the sum of

a) Noise power at the input terminal of second gain stage $N_{out,1} = kTBG_1N_{F1}$ which is amplified by gain $G_2$.

b) Noise power generated by the second gain stage $N_{a,2} = kTBG_2(N_{F2}-1)$

Adding them up, the noise seen at the output of second gain stage is,

$\begin{array}{lll}N_{out,2}&=&kTBG_1G_2N_{F1}+kTBG_2(N_{F2}-1)\\&=&kTB\overbrace{G_1G_2}^{\mbox{gain}}\underbrace{$$N_{F1}+\frac{N_{F2}-1}{G_1}$$}_{\mbox{noise figure}}\end{array}$.

## Equivalent Noise Figure of 2 gain stages

Summarizing, the equivalent noise figure of the two cascaded stage is,

$\Large{N_{F12} = $$N_{F1}+\frac{N_{F2}-1}{G_1}$$}$

and the equivalent gain is,

$\Large{G_{12} = G_1G_2}$

Note : All the expressions are in linear terms

## Equivalent Noise Figure of n gain stages

Extending this to cascade of  $n$stages, the equivalent noise figure is,

$\Large{N_{F12\ldots n} = N_{F1}+\frac{N_{F2}-1}{G_1} + \frac{N_{F3}-1}{G_1G_2} + \ldots + \frac{N_{Fn}-1}{G_1G_2\ldots G_{n-1}}}$.

## References

a) Friis, H.T.; , “Noise Figures of Radio Receivers,” Proceedings of the IRE , vol.32, no.7, pp. 419- 422, July 1944

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{ 4 comments… read them below or add one }

Zhang Seven November 18, 2012 at 2:51 pm

Besides uncorrelation, another condition is required to make E[XY] = 0: the mean of noise E[X] = 0, so E[XY] = E[X]E[Y] = 0 * 0 = 0. Or if X and Y are orthogonal, then E[XY] = 0 for sure.

Reply

Krishna Sankar November 19, 2012 at 6:15 am

@Zhang: Yes, thanks for being specific.

Reply

Prem July 22, 2012 at 7:29 pm

Nice article.
This explains why any SMALL adjustments for gain should be attempted on amplifiers(VGA) far away from antenna. At the same time LNA should be the first candidate to adjust the gains in AGC process.

Would be good if you come up with another simple article explaining how the concept of cascaded systems NF is important( or applied) for AGC design.

Thank you
Prem

Reply

Krishna Sankar July 23, 2012 at 4:36 am

@Prem: Thanks, will do.

Reply

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