# Ideal closed-loop gain derivation for negative feedback

Considering a negative feedback system, the closed-loop transfer function Af is given by:

where A is the open-loop transfer function and AB is the loop gain.

Every text I read says that if the magnitude of the loop gain is much larger than one, then the closed-loop gain becomes:

whereas if the loop gain is much smaller than one, the closed-loop gain becomes equal to the open-loop gain A.

I would perfectly agree with these approximations if all the quantities were real. But in general both the open-loop gain as well as the loop gain are complex quantities.

Question: how can I prove that if the magnitude of the loop gain is much larger than one, then the closed-loop transfer function is approximately equal to 1/B? Here my calculations:

In the last expression, I cannot go on: I would like to get rid of the term 2Re(BA), so that the square root would become exactly the magnitude of the loop gain and would simplify with the numerator.

If you think more geometrically, then it's not terribly complex (sorry for the pun.)

Multiplication of complex numbers is counter-clockwise rotation and scaling. Division is clockwise rotation and inverse scaling.

In polar, you have $$\A=r_a e^{I\theta_a}=r_a\angle\theta_a\$$ and $$\\beta=r_b e^{I\theta_b}=r_b\angle\theta_b\$$. Then multiplication just produces $$\r_ar_b\:\angle\: \theta_a+\theta_b\$$.

$$\frac{r_a\angle\theta_a}{1+r_ar_b\:\angle\: \theta_a+\theta_b}$$

So long as $$\r_ar_b\$$ is very much larger than one, this reduces (via clockwise rotation due to division) to:

$$\frac{r_a\angle\theta_a}{r_ar_b\:\angle\: \theta_a+\theta_b}=\frac{r_a}{r_ar_b}\angle \theta_a-\left(\theta_a+\theta_b\right)=\frac1{r_b}\angle -\theta_b=\frac1{r_b\angle \theta_b}=\frac1{\beta}$$

You can keep this in Euler form instead as:

$$\frac{r_a e^{I\theta_a}}{1+r_ar_b e^{I\left(\theta_a+\theta_b\right)}}$$

And again, so long as $$\r_ar_b\$$ is very much larger than one:

$$\frac{r_a e^{I\theta_a}}{r_ar_b e^{I\left(\theta_a+\theta_b\right)}}=\frac{ e^{I\theta_a}}{r_b e^{I\left(\theta_a+\theta_b\right)}}=\frac1{r_b e^{I\left(\theta_a+\theta_b\right)}e^{-I\theta_a}}=\frac1{r_b e^{I\theta_b}}=\frac1{\beta}$$

If you insist, you can $$\A=a +bI\$$ and $$\\beta=c+dI\$$. Then $$\\mid A\mid=\sqrt{a^2 +b^2}\$$ and $$\\mid \beta\mid=\sqrt{c^2 +d^2}\$$. Then your absolute value approach yields:

\begin{align*} &\frac{\mid A\mid}{\sqrt{1+a^2c^2+a^2d^2+b^2c^2+b^2d^2+2\left(ac-bd\right)}}\\\\&=\frac{\mid A\mid}{\sqrt{1+\mid \beta\mid^2\cdot\mid A\mid^2+2\left(ac-bd\right)}} \end{align*}

And I think you can readily see the reduction here, too.

• Thanks for your reply. But who says that 2(ac-bd) is negligible with respect to the rest? Aug 23, 2021 at 12:22
• @Stefanino 1st, you should be able to already see that it must be true from the polar proofs. 2nd, I do. And you should be able to easily see why. Think about how you might try to set up a counter-case. (You only need one.) Then see where that inevitably takes you. You must be able to do that much algebra.
– jonk
Aug 23, 2021 at 17:16
• @Stefanino It's been 3 hours so here's a hint: Try to imagine how you'd construct a large value for $2\left(ac-bd\right)$. For example, let $a\to\infty$, $c\to\infty$, $b=0$ and $d=0$. That's got to make $2\left(ac-bd\right)$ pretty big, right? Now see where that necessarily takes you in the denominator.
– jonk
Aug 23, 2021 at 20:54
• Maybe I got it... Since b=d=0 before I calculate the limit, then all terms containing b or d are 0. Then (1+ac)^2 is under the square root. But in this case, A=a and beta=c. If betaxA is much larger than 1, then 1/beta is obtained. Aug 23, 2021 at 21:10
• @Stefanino Suppose $b=0$ and $d=0$ and also $\mid \beta\mid^2\cdot\mid A\mid^2\gg 1$, then this means $\left(ac\right)^2\gg 1$ and that is a heck of a lot bigger than $2ac$. Suppose $a=0$ and $c=0$ and also $\mid \beta\mid^2\cdot\mid A\mid^2\gg 1$, then this means $\left(bd\right)^2\gg 1$ and that is a heck of a lot bigger than $-2bd$. And, taken to extremes, with $a\to\infty$ and $c\to\infty$, it's darn sure that $\left(\infty\cdot\infty\right)^2$ is a lot bigger than $\infty\cdot\infty$. The limit is clear and consistent with polar results.
– jonk
Aug 23, 2021 at 21:58

I think you have assumed in your calculations that A is a real number? that way: Real(AB) = AReal(B) and Im(AB) = AIm(B), and thinking intuitively, that if A is much larger than 1, A^2 is that much larger than A, thus {A/A^2 -> 0}.

Having known that, the expression:

2Re(BA) + Re(BA)^2 + Im(BA)^2

becomes BA^2

inserting that back in again, you will again get 1/B

Note: if it's strictly AB>>1, then again you can think that:

1. (BA)^2 >> (BA), so in the expression

2Re(BA) + Re(BA)^2 + Im(BA)^2

you can see again that

BA^2 >> 2Re(BA)

giving in the same results again.