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

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

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?

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    \$\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\$ – Bimpelrekkie Nov 23 '20 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\$ – Bimpelrekkie Nov 23 '20 at 21:58
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    \$\begingroup\$ Loop-gain is an open-loop measurement and hence is inherently stable. \$\endgroup\$ – Andy aka Nov 23 '20 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\$ – James Nov 24 '20 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 '20 at 9:38
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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.

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

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

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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.

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  • \$\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 '20 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\$ – James Nov 25 '20 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\$ – James Nov 25 '20 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 '20 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\$ – James Nov 25 '20 at 14:31

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