I am trying to understand how the loop gain of a system determines the stability of a complete system.

Loop gain is given by 1+GH where G is the forward transfer function (TF) and H is the feedback factor.

If we make the loop gain stable, then how it is possible that my transfer function is also stable? Because the poles of H become the zeros of the transfer function, so the complete system response changes.

How I can determine the transfer function shape from the loop gain/characteristic equation?

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  • 2
    \$\begingroup\$ If we make loop gain stable... It is not the loopgain that is stable/unstable. It is the system including feedback (for example an opamp + feedback resistors) that is stable or not. The loopgain needs to have certain properties (like loopgain < 1 when phase approaches 180 degrees) in order to make the system stable. All this is explained in all textbooks that discuss feedback. \$\endgroup\$ Nov 23, 2020 at 21:55
  • \$\begingroup\$ Also you might want to have a look at: allaboutcircuits.com/technical-articles/…. Also the ebook I always recommend about opamps, "Opamps for everyone" has a chapter on stability: web.mit.edu/6.101/www/reference/op_amps_everyone.pdf see chapter 5. \$\endgroup\$ Nov 23, 2020 at 21:58
  • 1
    \$\begingroup\$ Loop-gain is an open-loop measurement and hence is inherently stable. \$\endgroup\$
    – Andy aka
    Nov 23, 2020 at 22:42
  • \$\begingroup\$ Loop gain is equal to GH not (1 + GH). Or the purists might reason that loop gain is equal to -GH when the complete loop is taken into account. H is the feedback fraction which is often and very confusingly referred to as the feedback factor. I try to reserve the term "feedback factor" for the factor 1+GH. \$\endgroup\$
    – user173271
    Nov 24, 2020 at 6:59
  • \$\begingroup\$ @James...According to Harold Black the quantity H (he calls it "beta") is the "feedback circuit". And I think, in order to to mathematically treat this circuit we should identify this expression (H or beta) with its transfer function or transfer "factor" or simply feedback factor (because, in fact, it is the factor which defines the output portion that is fed back).. And for the denumerator (1+GH) some authors (Boris L. Lurie, in accordance with Bode) are using the term "return difference" . \$\endgroup\$
    – LvW
    Nov 24, 2020 at 9:38

4 Answers 4


It is incorrect to say that the loop-gain must be stable in order for the complete, closed loop system, to be stable.

Note: in the system \$ F = \frac{G}{1+GH} \$, loop gain is \$GH\$, not \$1+GH\$, as you stated.

It may be true that instability in the path \$GH\$ could cause closed-loop instability (for example, if \$H\$ itself is an ill-behaved entity), but as you noted, \$GH\$ appears in the denominator of the closed loop transfer function, and any notion of stability pertaining to \$GH\$ in and of itself is rendered moot, once you close the loop.

Rather than aiming for stability in \$GH\$, you are more concerned with obtaining a magnitude and phase response of \$GH\$ with frequency that does not cause feedback to become simultaneousy positive with a gain of one or more at any frequency. In response to your question about determining the shape of \$GH\$, that is the main purpose of Bode plots, graphs of gain and phase vs. frequency, from which you can visually identify such conditions.

I think it's fair to say (although I am not certain that it is formally correct to say this), that those required conditions for phase and magitude of \$GH\$ (to obtain stability) correspond to the requirement that the complex roots of \$1+GH\$ all reside in the left half of the complex plane (they all have negative real parts).

The fact that the denominator of the closed loop transfer function is \$1+GH\$, as opposed to simply \$GH\$, means that the poles and zeroes of \$GH\$ do not become zeroes and poles respectively of the closed loop transfer function, although I imagine that this could be a reasonable approximation for certain instances of \$G\$ and \$H\$.


Consider a G of 1,000,000,000 at 500 degrees, at 100Hz. (actually the frequency does not matter).

Implement an H of 10 (or 100, or 1,000)

This will be stable, despite the huge phase shift. Why?

Running the math on this, we find

  • The Numerator, of course, is 1,000,000,000 at 500 degrees phase shift

  • The Denominator, of course (1 + G * H) = (1 + 1,000,000,000/_ 500 * 0.1) which becomes 100,000,001 /_ 500

and the combined transfer function is stable, because the phaseshifts cancel.


Notice the mag/phase is nowhere near the Nyquist point.

  • \$\begingroup\$ H of 10 (or 100 or 1000)??? Or do you mean 1/10 or 1/100....? \$\endgroup\$
    – LvW
    Nov 24, 2020 at 13:20

Quote: If we make loop gain stable then how it is possible that my transfer function is also stable? because the poles of H becomes zeros of transfer function. So the complete systerm response changes? How I can determine the transfer function shape from loop gain/Characteristic equation?

In most cases, the loop gain (gain around the complete loop when it is open) is stable. But this does not mean that the closed-loop will also be stable. Stability properties (stability margin) must be checked using the stability criterion which can be applied in different ways: Bode diagram or Nyquist diagram or Nichols diagram).

It is to be noted that there are some rare cases where the lopp gain is NOT stable. In this case, the closed-loop can be made stable.

Concerning your last question:

  • Yes, the system response changes when the loop is closed. In general, it is not possible to guess how the "shape" of the closed-loop transfer function (magnitude) will look like.

  • However, in many cases this will be possible (approximately) when the gain of the forward function is pretty large (example: Operational amplifier) because in this case we have [G/(1+GH)]=1/[(1/G)+H] which equals 1/H for G>>1. Hence, the closed-loop function is just the inverse of the feedback function H.


Take, for example, a non-invering amplifier with a beta equal to 0.1 (R1 = 1k, R2 = 9k) operating at a frequency where its open loop gain is equal to 10. Its noise gain = 1/beta = 10 and it is operating at its closed loop -3dB frequency because the open loop gain is equal to the noise gain. Therefore the closed loop gain is equal to 10 * 0.707 = 7.07. If Vin is 1V pk to pk the output voltage will be 7.07V pk to pk. The voltage across the lower arm resistor is equal to 7.07*0.1 = 0.707V pk to pk. The voltage between the amplifier's inputs, Vdiff will be (output voltage)/(open loop gain) = 7.07/10 = 0.707V pk to pk. The loop gain, with out the inversion which occurs in the open loop loop gain measurement, is equal to the voltage across the lower arm resistor divided by Vdiff = 0.707/0.707 = 1.

The feedback fraction is 0.1 and the overall negative feedback factor (1 + beta*Aol) = Vin/Vdiff = 1.414 (Assuming open loop phase lag is -90 degrees).

Also of interest may be the closed loop phase lag which is -45 degrees.

The loop gain is equal to 1 (unity) and because the feedback network is purely resistive it means that the loop phase is equal to the open loop phase which is approx -90 degrees. The phase margin is 180-90 = 90 degrees and therefore the amp is very stable.

  • \$\begingroup\$ Quote: "...the overall negative feedback factor (1 + beta*Aol) .....How can you say that this expression (you call it "feedback factor") would be NEGATIVE ? Instead, see what wikipedia says (Quote): "...the voltage gain of the amplifier with feedback, the closed-loop gain AFB, is derived in terms of the gain of the amplifier without feedback, the open-loop gain AOL and the feedback factor β, which governs how much of the output signal is applied to the input (see Figure 1)". \$\endgroup\$
    – LvW
    Nov 25, 2020 at 10:42
  • \$\begingroup\$ @LvW That Wikipedia quote is perfectly accurate - beta does govern how much of the output signal is applied to the input although I would have been happier if they had referred to beta as the feedback fraction because beta is not the only factor that determines how much the input signal is being reduced by.The amount that the input signal is being reduced by also depends on how big Vdiff is. If the amplifier is being used over the frequency range where the output is fairly constant then the fraction of the output being fed back will be fairly constant but vdiff will be increasing with ..... \$\endgroup\$
    – user173271
    Nov 25, 2020 at 12:01
  • \$\begingroup\$ .... increasing frequency. The actual voltage into the amp Vdiff will become larger and larger fraction of Vin as frequency increases even though the magnitude of the voltage fed back is fairly constant. \$\endgroup\$
    – user173271
    Nov 25, 2020 at 12:13
  • \$\begingroup\$ @James...did you misunderstand my comments? Of course, Vdiff is not constant when the open-loop gain decreases. But this effect has nothing to do with my comment. It only concerns the definition of the feedback factor. I only have referenced the wikipedis definition for beta. There, the quantity "beta" is called "feedback factor" and the associated figure clearly shows what it is: It defines the feedback fraction that is fed back to the summing junction. Note that this factor must not necessarily be unitless. It can be V/I or I/V. \$\endgroup\$
    – LvW
    Nov 25, 2020 at 13:51
  • \$\begingroup\$ @LvW I was just trying to explain why I reserve the term "negative feedback factor" for (1+beta*Aol) as this factor best describes the negative feedback in the circuit whereas beta only describes the fraction of the output fed back. \$\endgroup\$
    – user173271
    Nov 25, 2020 at 14:31

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