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Closed loop reduces internal disturbances

Short answer

Closed-loop reduces internal disturbances because feedback divides the effect of parameter variations and internal noise by (1+GH). For a disturbance D entering between G1 and G2, the output becomes G2D/(1+G1G2H), which is much smaller than the open-loop output G2D when loop gain is high.

Justify that the closed loop system reduces effect of internal disturbances in the system.

Internal disturbances are changes inside the system itself: variation of plant parameters (ageing, temperature, wear, supply changes) and internal noise. Feedback reduces their effect because the output is measured and the error is corrected.

1. Parameter variation

The sensitivity of the overall gain TT to a change in GG is

SGT=∂T/T∂G/G=∂T∂G⋅GTS_G^T = \frac{\partial T/T}{\partial G/G} = \frac{\partial T}{\partial G}\cdot\frac{G}{T}
  • Open loop: T=GT = G, so SGT=1S_G^T = 1 (a 10% change in GG gives a 10% change in output).
  • Closed loop: T=G1+GHT = \dfrac{G}{1+GH}, so ∂T∂G=1(1+GH)2\dfrac{\partial T}{\partial G} = \dfrac{1}{(1+GH)^2} and
SGT=1(1+GH)2⋅G(1+GH)G=11+GHS_G^T = \frac{1}{(1+GH)^2}\cdot\frac{G(1+GH)}{G} = \frac{1}{1+GH}

Since ∣1+GH∣≫1|1+GH| \gg 1, the effect of a change in GG is reduced by the factor (1+GH)(1+GH). (With GH=99GH = 99, a 10% change in GG changes the output by only 0.1%.)

2. Disturbance generated inside the plant

Let the disturbance D(s)D(s) enter between two forward blocks G1G_1 (controller) and G2G_2 (plant).

  • Open loop: CD(s)=G2(s)D(s)C_D(s) = G_2(s)D(s).
  • Closed loop (feedback HH), with R=0R = 0 (superposition):
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)

The disturbance output is divided by (1+G1G2H)(1+G_1G_2H). If ∣G1G2H∣≫1|G_1G_2H| \gg 1, CD≈DG1HC_D \approx \dfrac{D}{G_1H}, which is very small for large G1G_1.

Example

A DC motor's armature resistance rises as it heats. In open loop the speed falls in proportion. With tacho feedback and loop gain K=49K = 49, the change in speed is only 1/(1+49)=2%1/(1+49) = 2\% of the open-loop change.

Hence the closed-loop system is far less affected by internal disturbances, as long as the loop gain is high and the sensor itself is accurate (the sensitivity to HH is SHT=−GH/(1+GH)≈−1S_H^T = -GH/(1+GH) \approx -1).

✓ Answer checked by Claude Opus 5.5 (high effort)

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