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I was reading an article about disturbance rejection of closed loop control system and I came across the sentence "becomes almost zero, and the effect of the disturbance is suppressed."

I have also attached snapshot of article and highlighted the confusing part of sentence. I am not able to understand how closed loop transfer function of the disturbance accounted in the system becomes zero??

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    \$\begingroup\$ Please don't write in capital letters, it's regarded as if you were shouting. \$\endgroup\$ – Enric Blanco Sep 20 at 6:59
  • \$\begingroup\$ Writing "reducing something to almost zero" doesn't mean much in my opinion. Example: the noise is 1 uV which is almost zero Sure, 1 uV is a small voltage but what if my signal is also 1 uV? In feedback systems usually the disturbances are reduced by the excess loopgain. I would just ignore the "almost zero" statement as it is meaningless. \$\endgroup\$ – Bimpelrekkie Sep 20 at 7:30
  • \$\begingroup\$ Also please remove the capital letters in the title. \$\endgroup\$ – Bimpelrekkie Sep 20 at 7:33
  • \$\begingroup\$ @Bimpelrekkie For a transfer function (where the text is about) that reduces to zero, is definitely means something. \$\endgroup\$ – Huisman Sep 20 at 7:37
  • \$\begingroup\$ @Huisman Note "consider the case... the >> 1" that means that the transfer function will be zero only if the gain approaches infinity. It is more like a limit function. It never really becomes zero. But feel free to prove me wrong :-) \$\endgroup\$ – Bimpelrekkie Sep 20 at 7:41
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In control systems the loop gain LG (product of all transfer functions within the closed loop) is a very important parameter.

In your case, the loop gain is LG(s)=G1(s)*G2(s)*H(s).

As you can see, the closed-loop transfer function for the disturbed input

Hd(s)=Cd(s)/D(s)=G2(s)/[1+LG(s)]

will be rather small ("almost zero" in the text) for a large loop gain LG(s)>>1.

For the reference input the situation is different because the product G1(s)*G2(s) appears also in the numerator of the closed-loop function Hr(s)=Cr(s)/R(s).

With other words: Both closed-loop functions have the same denominator (1+LG) - however, the numerator for the closed-loop reference function (G1*G2) is larger than for the closed-loop disturbance function (G2). Hence, the influence of the disturbance signal is smaller if compared with the reference signal.

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Assume there's an integrator somewhere in the system, so that \$ G_1(s)G_2(s)H(s)=\large\frac{A(s)}{sB(s)}\$ where \$A(s)\$ and \$B(s)\$ are polynomials without a free \$s\$ term.

Now Set \$\small R(s)=0\$, and work out the CLTF: \$G(s)=\frac{C(s)}{D(s)}=\frac{sB(s)}{A(s)+sB(s)}\$. The DC gain is found by setting \$s=0\$, thus \$G(0)=0\$.

This means that disturbance will have no effect on the response once once the transients have decayed to zero. We say that the disturbance is rejected (but not instantaneously!)

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  • \$\begingroup\$ @Chu....of course, I agree (theoretically). However, this is true only if there would be an IDEAL integrator which - however - cannot be realized. Hence, the disturbance cannot be fully rejected. \$\endgroup\$ – LvW Sep 20 at 9:39
  • \$\begingroup\$ @lvw , of course, nothing’s ideal. That’s life. \$\endgroup\$ – Chu Sep 20 at 13:57

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