I am confused in some basic terminologies of feedback system and its requirements

1.what is difference between actual input ,reference input and disturbance input ?

2.is reference input setup by designer or its unwanted input ?

3.is output of of the feedback settle at value equal to reference input after infinite time?

4.I know feedback minimize the effect of disturbance input ,but what about other two inputs i.e actual input and reference input ,how does feedback system effect these two inputs ?

I am ok with rigorous mathematics ,if anyone wants to explain by maths by taking any particular example and even a theoretical answer would also be appreciated!

I'll add paragraph of book by Bernard Friedland for more context about question enter image description hereenter image description here!

  • \$\begingroup\$ I'd like to see a diagram illustrating the question better. I can read words. I sometimes even imagine I know what someone else is thinking when using them. But I suspect all this would just go away if you'd just sit down with paper and pencil and see what feedback does to equations. By the way, are you only talking about NFB? (It looks that way to me, but I have to ask.) Or are you thinking about some very generalized analysis of both PFB and NFB? (I don't think anyone will address the entire theoretical space for that. But who knows?) \$\endgroup\$
    – jonk
    Sep 2 '20 at 20:18
  • \$\begingroup\$ Thanks jonk for your concern ! I'll add paragraph of book that I'm reading and figure associated with it , \$\endgroup\$
    – user215805
    Sep 2 '20 at 20:30
  • \$\begingroup\$ I'll sit with pen paper from last 2days didn't come to any conclusions ,I know all the mathematical stuffs in control system but rather I wanted to understand how it actually works and even if you consider only NFB that's enough for me to understand it \$\endgroup\$
    – user215805
    Sep 2 '20 at 20:33
  • \$\begingroup\$ Okay. I'll provide a few examples to point in the right direction. I don't have time today to do more than that. \$\endgroup\$
    – jonk
    Sep 2 '20 at 20:41

I'm not interested in spending a lot of time on this, right now. Other work needs doing today. But I can offer a thought or two on the basis that you are discussing NFB.

NFB linearizes. It's amazing at it. You should play around with it, using paper and pencil. You'll gain a lot from taking some arbitrary equation, applying NFB, and then seeing the new closed equation.

For example, start with:


Apply 10% NFB to that to get:

$$\begin{align*} y &= e^{x-10\%\,\cdot\, y}\\\\ &= 10\cdot\operatorname{LambertW}\left(10\%\cdot e^x\right) \end{align*}$$

The original and resulting curves are:

enter image description here

Well, one looks more linear than the other to me. (See LambertW Appendix below for derivation of the closed equation.)

Suppose we do something trickier:

$$y = e^{\sin(5\,x)+1}$$

(The above equation isn't too far from what you might get from a BJT operated as a grounded-AC CE arrangement.)

And apply 30% NFB to it? We get this:

enter image description here

Notice the nice recovery of the sine wave? This is why global NFB is critical when you are using a grounded-AC BJT amplifier to get high voltage gain!

I've taken a few examples based upon what I consider to be kind of crazy -- functions that are exponentials of other functions -- just to exaggerate the problem a bit.

But don't stop there. You can do this to polynomials, too. Or anything, really.

Suppose this polynomial:

$$y = 4\,x^2 - 6\,x + 3$$

Apply 10% feedback to it (there will be two solutions) and get:

enter image description here

And so it goes.

NFB is very, very powerful.

LambertW Appendix

The LambertW function is also known as the product-log function. In the first case I listed above, I followed this derivation:

$$\begin{align*} y &= e^{x-10\%\cdot y}\\\\ y &= e^{x}\cdot e^{-10\%\cdot y}\\\\ y\cdot e^{10\%\cdot y} &= e^{x} \\\\ 10\%\cdot y\cdot e^{10\%\cdot y} &= 10\%\cdot e^{x} \\\\ \operatorname{LambertW}\left(\left[10\%\cdot y\right]\cdot e^{\left[10\%\cdot y\right]}\right) &= \operatorname{LambertW}\left(10\%\cdot e^{x}\right) \\\\ 10\%\cdot y &= \operatorname{LambertW}\left(10\%\cdot e^{x}\right) \\\\ y &= \frac1{10\%}\cdot \operatorname{LambertW}\left(10\%\cdot e^{x}\right) \\\\ y &= 10\cdot \operatorname{LambertW}\left(10\%\cdot e^{x}\right) \end{align*}$$

As exponentials (and logarithms) appear frequently, it's a function worth learning about. Anytime a variable appears both inside and outside of a logarithm or an exponential, you should think about it.

There's an odd history to it. Leonhard Euler investigated it. But in that paper he refers to Johann Lambert and as there were already lots of stuff named after Euler. So the function got Lambert's name assigned to it.

Footnote: Images above were generated using Wolfram Alpha.

  • \$\begingroup\$ +1, even if only for "NFB linearizes". I would embolden it, make it blink, sounds, maybe some confetti, too. \$\endgroup\$ Sep 2 '20 at 20:45
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    \$\begingroup\$ @jDAQ I didn't have the time today to do more. If I get a chance, later, I may improve it. Regardless, I don't disagree with your choice and I thank you for taking a moment to explain your reasoning. It goes a long way with me and too few seem willing. Your criticism is constructive. Thanks, again. \$\endgroup\$
    – jonk
    Sep 2 '20 at 21:56
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    \$\begingroup\$ @jDAQ I should add, though, that I added my answer without seeing any of the OP's additions. (You can check it my looking at the "edited N hours ago" link there.) So I was working with very little, as well. That said, it is really good to hear from you about this. I very much enjoy constructive feedback. \$\endgroup\$
    – jonk
    Sep 2 '20 at 22:09
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    \$\begingroup\$ @jDAQ While you are right about OP's 4 points, I feel obliged to add -- maybe in a bit of an objective manner -- that OP has selected the answer, which might suggest that this answer addresses what OP meant to ask. But it could also be that OP was hasty. \$\endgroup\$ Sep 3 '20 at 7:27
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    \$\begingroup\$ +1 for you @jonk: an excellent and very original response to show the benefit of feedback through a simple equation. Bravo ! Perhaps some further light shed on the LambertW function would be appreciated when time permits. Thanks. \$\endgroup\$ Sep 3 '20 at 7:55

I will try to give you a short answer to the 4 questions (with reference to the shown block diagram):

(1) Actual input is the most left signal (xr). At the same time, this signal serves as a reference for the output x (both should be as close to each other as possible. Therefore, the signal "x" is fed back to the summing block (100% negative feedback).

(2) Of course, "xr" is a "wanted" input (externally applied reference) because it will be compared at the most left summing block with the output "x".

(3) No, both signals (x resp xr) cannot be equal because in this case the loop gain (gain of the open loop) must be infinite (difference signal at the output of the summing block zero). In your case, the loop gain is LG=G*Hp).

(4) To answer this question it is best to simply compare the two closed-loop transfer functions. For this principle comparison the feedforward block Gr is neglected (for the sake of simplicity). The transfer function of the process is designated as Hp.

  • Forward transfer function (set xd=0): x/xr=Hr=GHp/(1+GHp)

  • Disturbance transfer function (set xr=0): x/xd=Hd=-H/(1+GHp)

As you can see, only for a very large product GHp the function Hr approaches unity (both signals nearly identical). Comparing both transfer functions, we see that Hr is larger by a factor G. Hence, the output portion caused by the disturbance is suppressed with this factor. Therefore, the block G should be as large as possible.

  • \$\begingroup\$ Thanks for answer and for addressing all my concerns ! I am confused in point 1 and 3 of your answer hopefully you'll clear my doubt . Isn't x is reference and output of process block is actual output which becomes equal to x at steady State for unity negative feedback as in figure ? \$\endgroup\$
    – user215805
    Sep 3 '20 at 12:22
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    \$\begingroup\$ Well...perhaps you are right. There are two meanings of the term "reference": Either the output should follow the reference applied to the input (your interpretation) or the input is to be compared with a signal - called "reference" (at the summing block). However, now I see that the input is called "xr" - this is an indication that my asumption was not the correct one. I will revise my answer. Thank you. \$\endgroup\$
    – LvW
    Sep 3 '20 at 14:46

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