# Disturbance Rejection in a Closed Loop Control System

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??

• Please don't write in capital letters, it's regarded as if you were shouting. – Enric Blanco Sep 20 '19 at 6:59
• 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. – Bimpelrekkie Sep 20 '19 at 7:30
• Also please remove the capital letters in the title. – Bimpelrekkie Sep 20 '19 at 7:33
• @Bimpelrekkie For a transfer function (where the text is about) that reduces to zero, is definitely means something. – Huisman Sep 20 '19 at 7:37
• @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 :-) – Bimpelrekkie Sep 20 '19 at 7:41

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.

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!)

• @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. – LvW Sep 20 '19 at 9:39
• @lvw , of course, nothing’s ideal. That’s life. – Chu Sep 20 '19 at 13:57