0
4.6kviews
Prove Friss formula with refrence to noise factor in cascade.
1 Answer
0
574views
  • In practice the filters or amplifiers are not used in isolated manner, they are used in cascaded manner.

  • The overall noise factor of such cascade connection can be determined as follows:

  • Figure below shows two amplifiers connected in cascade.

enter image description here

  • Let the power gains of two amplifier be $G_1$ and $G_2$ respectively and le their noise factors be $f_1$ and $f_2$ respectively.

  • The total noise power at the i/p of the first amplifier is given as,

$P_{ni} \ (total) f_1 \ K \ To \ B$ - - - - (1)

  • The total noise power at the o/p of amplifier 1 will be addition of two terms.

$\therefore$ Noise e/p to amp2 = $G_1 f_1$ K To B + ($f_2 - 1)$ K To B - - - -(2)

The first term represents the amplified noise power (by $G_1$ ) and second terms represents the noise contributed by 2nd amp.

  • The noise power at the o/p of 2nd amp is $G_2$ times the i/p noise power to amp 2.

$\therefore$ $P_no = G_2 \times$ (Noise i/p to amp 2)

$\therefore$ $P_{no} = G_1 G_2 F_1 $ K To B + $G_2(f_2 - 1)$ K To B

  • The overall gain of the cascade connection is given by,

$G = G_1 G_2$

  • The overall noise factor F is defined as follows:

$F = \frac{P_{no}}{G_1 G_2 P_{ni}}$

Here, $P_{ni} = K \ To \ B$

Substituting the values of $P_{no}$ and $P_{ni}$ we get,

$F = \frac{G_1 G_2 F_1 K \ To \ B + G_2 (F_2 - 1) K \ To \ B}{G_1 G_2 K \ To \ B}$

$= F_1 + \frac{(f_2 - 1)}{G_1}$

The same logic can be extended for more number of amplifiers connected in cascade, then the overall noise factor F would be,

$F = F_1 + \frac{f_1 - 1}{G_1} + \frac{f_3 - 1}{G_1 G_2} + \frac{f_4 - 1}{G_1 G_2 G_3} + - - - $

This is Knows as Friss Formula.

Please log in to add an answer.