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Suppressing disturbance by increasing G(s) or H(s)

Short answer

Disturbance output is C_D = [G2/(1+G1G2H)]D; for large loop gain it approximates D/(G1H). Increasing G1 and/or H reduces this, suppressing disturbance, provided the system remains stable.

Effect of disturbance in case of feedback control system can be suppressed by increasing the gain G(s) and/or H(s). Justify.

The statement is justified, provided that the gain increased is the one in the loop before the point where the disturbance enters, and that the system stays stable.

Let the disturbance D(s)D(s) enter between G1(s)G_1(s) (controller) and G2(s)G_2(s) (plant), with feedback H(s)H(s).

 R  +      +----+  +     +----+  C
 -->(Σ)--->| G1 |-->(Σ)->| G2 |--+-->
     ^ -   +----+    ^ D +----+  |
     |       +----+              |
     +-------| H  |<-------------+
             +----+

With R=0R = 0 (superposition), the output due to the disturbance is

CD(s)=G2(s)1+G1(s)G2(s)H(s) D(s)C_D(s) = \frac{G_2(s)}{1+G_1(s)G_2(s)H(s)}\,D(s)

For large loop gain, ∣G1G2H∣≫1|G_1G_2H| \gg 1:

CD(s)≈G2G1G2HD(s)=D(s)G1(s)H(s)C_D(s) \approx \frac{G_2}{G_1G_2H}D(s) = \frac{D(s)}{G_1(s)H(s)}

So increasing G1G_1 and/or HH makes CDC_D smaller, and the disturbance is suppressed. In open loop, CD=G2DC_D = G_2D, with no reduction at all.

Output due to the reference: CR=G1G21+G1G2HR≈RHC_R = \dfrac{G_1G_2}{1+G_1G_2H}R \approx \dfrac{R}{H}, which is almost independent of the gains, so tracking is not harmed.

Limits

  • Raising G2G_2 alone does not help much: the result tends to D/(G1H)D/(G_1H), which does not depend on G2G_2.
  • Raising HH also lowers the closed-loop gain (≈1/H\approx 1/H), so it must be chosen with the required output level in mind.
  • Very high gain can reduce damping or cause instability in higher-order systems, and it amplifies sensor noise.

✓ Answer checked by Claude Opus 5.5 (high effort)

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