Coursify

Gain Reduction for Overshoot Specification in a Unity-Feedback System

Gain Reduction for Overshoot Specification in a Unity-Feedback System

Verified Sources
Sep 12, 2026

We study a unity-feedback control system with open-loop transfer function

G(s)=ks(1+sT),k>0,  T>0.G(s)=\frac{k}{s(1+sT)}, \quad k>0,\; T>0.

For a unit-step input, the peak (percent) overshoot depends on the closed-loop damping ratio of the resulting second-order dynamics. In this design problem, we will enforce that the overshoot decreases from 75%75\% to 25%25\% by scaling the amplifier gain kk accordingly.

Key relations we will use:

  • unity feedback implies the closed-loop characteristic equation is 1+G(s)=01+G(s)=0.
  • percent overshoot for underdamped second-order systems is PO=100 e−ζπ1−ζ2\mathrm{PO}=100\,e^{-\frac{\zeta\pi}{\sqrt{1-\zeta^2}}} where ζ\zeta is the damping ratio.
  • The given plant/feedback structure yields a standard second-order denominator, from which ζ\zeta can be expressed in terms of kk and TT.

Mermaid overview of the control loop:

Footnotes

  1. Percent overshoot and damping ratio relationship (including inverse formula) - Wikipedia “Damping” https://en.wikipedia.org/wiki/Damping - Provides PO = 100 exp(−ζπ/√(1−ζ^2)) and ζ inversion formula. ↩

Percent Overshoot & Damping Ratio (Second-Order Systems)

Step 1: Form the unity-feedback closed-loop transfer function

Unity feedback gives the characteristic equation:

1+G(s)=0  ⇒  1+ks(1+sT)=0.1+G(s)=0 \;\Rightarrow\; 1+\frac{k}{s(1+sT)}=0.

Multiply by s(1+sT)s(1+sT):

s(1+sT)+k=0.s(1+sT)+k=0.

Expand:

s+Ts2+k=0  ⇒  Ts2+s+k=0.s + Ts^2 + k = 0 \;\Rightarrow\; Ts^2 + s + k = 0.

Thus the closed-loop transfer function has a denominator proportional to Ts2+s+kTs^2+s+k.

To match the standard second-order form

s2+2ζωns+ωn2,s^2 + 2\zeta\omega_n s + \omega_n^2,

divide by TT:

s2+1Ts+kT=0.s^2 + \frac{1}{T}s + \frac{k}{T}=0.

Now compare coefficients:

  • 2ζωn=1T2\zeta\omega_n = \frac{1}{T}
  • ωn2=kT\omega_n^2 = \frac{k}{T}

From ωn2=kT\omega_n^2=\frac{k}{T}, we get ωn=kT\omega_n=\sqrt{\frac{k}{T}}. Then

2ζkT=1T  ⇒  2ζkT=1  ⇒  ζ=12kT.2\zeta\sqrt{\frac{k}{T}}=\frac{1}{T} \;\Rightarrow\; 2\zeta\sqrt{kT}=1 \;\Rightarrow\; \zeta=\frac{1}{2\sqrt{kT}}.

Key takeaway:

  • damping ratio ζ\zeta is inversely proportional to k\sqrt{k}: ζ∝1k.\zeta \propto \frac{1}{\sqrt{k}}.

This relation lets us compute the ratio k2/k1k_2/k_1 once we find ζ1\zeta_1 and ζ2\zeta_2 from the overshoot specs.

Compute damping ratios from the percent overshoot formula

  1. 1
    Step 1

    For an underdamped second-order step response, PO (decimal percent) satisfies PO = 100 e−ζπ/1−ζ2100\,e^{-\zeta\pi/\sqrt{1-\zeta^2}} .

    Footnotes

    1. Percent overshoot and damping ratio relationship (including inverse formula) - Wikipedia “Damping” https://en.wikipedia.org/wiki/Damping - Provides PO = 100 exp(−ζπ/√(1−ζ^2)) and ζ inversion formula. ↩

  2. 2
    Step 2

    For PO1_1 = 75%75\% and PO2_2 = 25%25\%: 75=100e−ζ1π/1−ζ1275 = 100e^{-\zeta_1\pi/\sqrt{1-\zeta_1^2}} and 25=100e−ζ2π/1−ζ2225 = 100e^{-\zeta_2\pi/\sqrt{1-\zeta_2^2}}.

  3. 3
    Step 3

    Rearranging gives ζ=−ln⁡(PO100)π2+(ln⁡(PO100))2\displaystyle \zeta=\frac{-\ln(\frac{PO}{100})}{\sqrt{\pi^2+\left(\ln(\frac{PO}{100})\right)^2}} .

    Footnotes

    1. Percent overshoot and damping ratio relationship (including inverse formula) - Wikipedia “Damping” https://en.wikipedia.org/wiki/Damping - Provides PO = 100 exp(−ζπ/√(1−ζ^2)) and ζ inversion formula. ↩

  4. 4
    Step 4

    Let PO/100=0.75PO/100=0.75. Then ζ1≈0.215\zeta_1\approx 0.215.

  5. 5
    Step 5

    Let PO/100=0.25PO/100=0.25. Then ζ2≈0.403\zeta_2\approx 0.403.

Step 2: Relate gain ratio to damping ratio

We derived:

ζ=12kT.\zeta=\frac{1}{2\sqrt{kT}}.

With TT fixed, this implies:

ζ∝1k⇒ζ2ζ1=k1k2.\zeta \propto \frac{1}{\sqrt{k}} \quad\Rightarrow\quad \frac{\zeta_2}{\zeta_1}=\sqrt{\frac{k_1}{k_2}}.

Therefore,

k2k1=(ζ1ζ2)2.\frac{k_2}{k_1}=\left(\frac{\zeta_1}{\zeta_2}\right)^2.

Insert the computed values:

k2k1=(0.2150.403)2≈(0.533)2≈0.284.\frac{k_2}{k_1}= \left(\frac{0.215}{0.403}\right)^2 \approx (0.533)^2 \approx 0.284.

So the amplifier gain should be reduced to about 0.2840.284 of its original value, i.e., reduced by a factor:

0.284 (approximately)\boxed{0.284 \text{ (approximately)}}

Equivalently, it is reduced by about:

1−0.284≈71.6%\boxed{1-0.284 \approx 71.6\%}

Why overshoot targets ζ

For standard underdamped second-order responses, peak overshoot depends only on the damping ratio ζ\zeta, via PO=100e−ζπ/1−ζ2=100e^{-\zeta\pi/\sqrt{1-\zeta^2}} . Frequency scale cancels out for PO, so we only need ζ(k)\zeta(k).

Footnotes

  1. Percent overshoot and damping ratio relationship (including inverse formula) - Wikipedia “Damping” https://en.wikipedia.org/wiki/Damping - Provides PO = 100 exp(−ζπ/√(1−ζ^2)) and ζ inversion formula. ↩

Underdamped requirement

The overshoot formula used is for underdamped systems (0<ζ<10<\zeta<1). Here both computed ζ1≈0.215\zeta_1\approx0.215 and ζ2≈0.403\zeta_2\approx0.403 satisfy ζ<1\zeta<1, so the model is consistent .

Footnotes

  1. Percent overshoot and damping ratio relationship (including inverse formula) - Wikipedia “Damping” https://en.wikipedia.org/wiki/Damping - Provides PO = 100 exp(−ζπ/√(1−ζ^2)) and ζ inversion formula. ↩

Damping ratio & gain scaling for overshoot reduction

From PO = 75% to PO = 25% (T fixed).

Common checks and derived formulas

Key Concepts Flashcards

1 / 5
Question · Term

Unity feedback characteristic equation

Click to reveal
Answer · Definition

For open-loop G(s), unity feedback gives 1+G(s)=0.

Knowledge Check

Question 1 of 4
Q1Single choice

For the given system with unity feedback, which characteristic equation is correct?

Explore Related Topics

1

Understanding Hysteresis in Sensors

Hysteresis is a sensor error where the output at a given input value changes depending on whether the input is approached from an increasing or decreasing direction.

  • It appears as a systematic up‑down difference (e.g., 49.95 kPa vs 50.05 kPa at 50 kPa).
  • Identification involves recording sensor readings while sweeping the input up and then down, then comparing values at identical points.
  • Unlike dead time, dead band, repeatability, or noise, hysteresis is a path‑dependent offset often expressed as a percentage of full scale.
  • Reducing it requires better material choice, mechanical design, and calibration that tests both directions.
2

Model Quantization from First Principles

Model quantization maps high‑precision tensors to low‑bit integer representations using affine scaling, enabling smaller models and faster integer‑only inference while preserving the network’s input‑output behavior.

  • Affine quantization: q=round⁡(r/s)+zq=\operatorname{round}(r/s)+z and r^=s(q−z)\hat{r}=s(q-z), with scale ss and zero‑point zz chosen to fit [rmin⁡,rmax⁡][r_{\min},r_{\max}] into an integer range (e.g., [−128,127][-128,127] for INT8).
  • Scale ss controls the trade‑off between rounding error (smaller ss → higher precision) and clipping error (larger ss → wider range).
  • Symmetric (r^=sq\hat{r}=sq) suits weight tensors; asymmetric (r^=s(q−z)\hat{r}=s(q-z)) handles shifted activation ranges, and per‑channel quantization assigns separate s,zs,z per output channel for improved accuracy.
  • Calibration (min‑max, histogram, percentile) estimates activation ranges from representative data; static, dynamic, and quantization‑aware training (QAT) are the three main deployment pipelines.
  • Practical success depends on bit‑width choice, granularity, hardware support for integer kernels, and evaluation of accuracy, latency, memory, and stability on the target platform.
3

Mastering Recurrence Relations: The Substitution Method

The lesson shows how to solve the classic recurrence T(n)=2T(n/2)+nT(n)=2T(n/2)+n with the substitution method, using a recursion‑tree intuition to guess an O(nlog⁡n)O(n\log n) bound and then proving it by induction.

  • Each level of the recursion tree costs nn, and there are log⁡2n\log_2 n levels → total Θ(nlog⁡n)Θ(n\log n).
  • Inductive hypothesis: T(n)≤c nlog⁡2nT(n)\le c\,n\log_2 n for n≥n0n\ge n_0, which simplifies to c≥1c\ge1 in the induction step.
  • The base case must start at n0>1n_0>1 (e.g., n0=2n_0=2) because T(1)≤c⋅1log⁡21T(1)\le c\cdot1\log_2 1 is false.
  • Substitution offers rigorous proof, while recursion trees give quick guesses; the Master Theorem confirms T(n)=Θ(nlog⁡n)T(n)=Θ(n\log n).
Browse all research articles