Electronic – How does negative feedback reduce noise

I will try to describe the method used by the book.

First, we need to find the open-loop gain.

And I will use this circuit diagram

The Q2 voltage gain is

$$A_{V2} \approx \frac{R_5+R_4}{re2} \approx \frac{11\textrm{k}\Omega}{25\Omega} \approx 440 \: V/V $$

The first stage gain is:

$$ A_{V1} \approx \frac{R_3||(\beta +1)re2}{R_4||R_5} \approx \frac{500\Omega}{1\textrm{k} \Omega} \approx 0.5 \: V/V$$

Therefore the open-loop gain is:

\$ A_{OL} = A_{V1} * A_{V2} = 220 \:V/V \$

From the diagram we see that:

$$Z_{out(OL)} = R_4+R_5 = 11\textrm{k}\Omega$$

And

$$ Z_{out(CL)} = \frac{Z_{out(OL)}}{1+ A_{OL}\times\frac{R_4}{R_4+R_5} } = \frac{11k \Omega}{1 + 220 \times \frac{1}{11}} \approx 520\Omega $$

And now the closed-loop gain:

$$A_{CL} = \frac{A_{OL}}{1+ A_{OL}\times\frac{R_4}{R_4+R_5} } \approx 10.47 \:V/V$$

The LTspice is showing \$Z_{out(CL)} = 529\Omega\$ and \$A_{CL} = 10.47\: V/V\$ quite good.

Perhaps the shown block diagram can help to clarify the situation? (The last - most right - block should contain a minus sign).

In the diagram I have set Ie=Ic. Of course, we could include the base current into the diagram - however, this would not influence the feedback loop. Note the loop gain is Aloop=-gm*RE.

The circuit establishes current-controlled voltage feedback. (I do not like expressions like "series-parallel" or "parallel-parallel" or....because some authors are counting input-output and some others output-input. So it is somewhat confusing, I think).