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

enter image description here

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:

enter image description here

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:

enter image description here

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.


2 Answers 2


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\$.

So your equation is:

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

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.

  • \$\begingroup\$ Thanks for your reply. But who says that 2(ac-bd) is negligible with respect to the rest? \$\endgroup\$
    – Stefanino
    Aug 23, 2021 at 12:22
  • \$\begingroup\$ @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. \$\endgroup\$
    – jonk
    Aug 23, 2021 at 17:16
  • \$\begingroup\$ @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. \$\endgroup\$
    – jonk
    Aug 23, 2021 at 20:54
  • \$\begingroup\$ 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. \$\endgroup\$
    – Stefanino
    Aug 23, 2021 at 21:10
  • \$\begingroup\$ @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. \$\endgroup\$
    – 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.


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