# I need help deriving the cutoff frequency of a negative feedback amplifier

I have the question and answer below for deriving the closed-loop cutoff frequency of a feedback amplifier circuit. I can follow the working out all the way until the last 2 lines, where the writer rearranges $$\A(f)\$$ and then forms the cutoff equation, $$\f_{cc}\$$.

I think they've missed out a couple of steps in between, so could anyone explain how to get to those last couple of lines?

1. Going from $$\A(f)=\frac{A_o}{1+j(\frac{f}{f_c})+A_o\beta}\$$ to $$\A(f)=\frac{A_o}{(1+\beta A_o)(1+j\frac{f}{(1+\beta A_o)f_c})}\$$
2. Going from $$\A(f)=\frac{A_o}{(1+\beta A_o)(1+j\frac{f}{(1+\beta A_o)f_c})}\$$ to $$\f_{cc}=(1+\beta A_o)f_c\$$  I think they've missed out a couple of steps in between, so could anyone explain how to get to those last couple of lines?

Just resolve the denominator: -

$$(1+\beta A_0)(1+ j\dfrac{f}{f_c(1+\beta A_0)}) = 1+\beta A_0 + j\dfrac{f}{f_c(1+\beta A_0)} +j\dfrac{\beta A_0f}{f_c(1+\beta A_0)}$$

$$= 1 + \beta A_0 + j\dfrac{f(1+\beta A_0)}{f_c(1+\beta A_0)}$$ $$= 1+\beta A_0 +j\dfrac{f}{f_c}$$

In other words it's correct.

For the 2nd equation, the only point of interest is this part of the denominator: -

$$1+ j\dfrac{f}{f_c(1+\beta A_0)}$$

This is the only point of interest because $$\1+\beta A_0\$$ (the bit of left out from the original equation) is just a constant gain modifier across the spectrum. And we know, that for a system like this, that to get the 3 dB point (aka $$\f_{cc}\$$), we equate real and imaginary terms thus: -

$$1 = \dfrac{f_{cc}}{f_c(1+\beta A_0)}$$

But, all of that could have been spotted from the very first equation you listed so I'm unsure why they drove round the hills to get there.