20.5 Feedback Control Systems & Research Methods

20.5 Feedback Control Systems & Research Methods

Topic H of the Electrical Engineering subject in the PRC Enhanced Table of Specifications (PRBEE Resolution No. 40, s. 2024) is Instrumentation & Control, Feedback Control System and Research Methods, weighted 4.50% of the whole examination and 10 of the 100 Electrical Engineering items. Earlier sections of this chapter covered instrumentation, transducers and PLC-based industrial control; this section covers the feedback control theory and the research methodology that complete the topic. Governor speed control, automatic voltage regulation, load-frequency control and every process loop in a power plant are applications of exactly this material.

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1. Open Loop versus Closed Loop

Feedback buys accuracy and disturbance rejection at the price of possible instability — the central trade-off of the whole topic.

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2. Transfer Functions and Block Diagram Reduction

The transfer function is the Laplace-domain ratio of output to input at zero initial conditions:

$G(s) = \frac{C(s)}{R(s)}$

For a negative feedback loop with forward path $G(s)$ and feedback path $H(s)$:

$\frac{C(s)}{R(s)} = \frac{G(s)}{1 + G(s)H(s)}$

For positive feedback the denominator becomes $1 - G(s)H(s)$. When $H(s) = 1$ the system is unity feedback, and the loop reduces to $G/(1+G)$.

The characteristic equation is

$1 + G(s)H(s) = 0$

Its roots are the closed-loop poles, and every question about stability, overshoot and settling time is ultimately a question about where those roots lie. Poles in the left half of the s-plane give a stable system; any pole in the right half gives instability; poles on the imaginary axis give sustained oscillation.

Reduction rules: blocks in cascade multiply, blocks in parallel add, and a feedback loop collapses by the formula above. Mason's gain formula handles multiple-loop signal flow graphs.

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3. Transient Response

First-Order System

$G(s) = \frac{K}{\tau s + 1}$

Step response reaches 63.2% of final value at $t = \tau$ and 98.2% at $4\tau$. There is no overshoot and no oscillation.

Second-Order System

$\frac{C(s)}{R(s)} = \frac{\omega_n^{2}}{s^{2} + 2\zeta\omega_n s + \omega_n^{2}}$

Performance measures:

$\omega_d = \omega_n\sqrt{1-\zeta^{2}} \qquad\qquad \%OS = 100\,e^{-\pi\zeta/\sqrt{1-\zeta^{2}}}$

$t_p = \frac{\pi}{\omega_d} \qquad t_s \approx \frac{4}{\zeta\omega_n} \ (2\%\text{ band}) \qquad t_r \approx \frac{1.8}{\omega_n}$

Percent overshoot depends on $\zeta$ alone, not on $\omega_n$. Common values worth memorising: $\zeta = 0.5$ gives 16.3% overshoot, $\zeta = 0.707$ gives 4.3%, and $\zeta = 0.8$ gives 1.5%.

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4. Steady-State Error and System Type

System type is the number of pure integrators ($1/s$ terms) in the open-loop transfer function.

$K_p = \lim_{s\to 0}G(s)H(s), \quad K_v = \lim_{s\to 0}sG(s)H(s), \quad K_a = \lim_{s\to 0}s^{2}G(s)H(s)$

Adding an integrator removes steady-state error to that input class but adds 90° of phase lag, eroding stability margin. That trade-off is the entire reason PID exists.

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5. Stability Analysis

Routh-Hurwitz Criterion

Build the Routh array from the characteristic polynomial coefficients. The number of right-half-plane roots equals the number of sign changes in the first column; the system is stable only if there are none.

Two necessary conditions can be checked by inspection before building the array: all coefficients must be present and all must have the same sign. A missing or negative coefficient proves instability immediately, with no array required.

Root Locus

Plots closed-loop pole paths as gain $K$ varies from 0 to infinity. The locus starts at the open-loop poles and ends at the open-loop zeros or at infinity, and any branch crossing into the right half plane marks the gain at which the system goes unstable.

Frequency Response

The Nyquist criterion relates encirclements of the −1 point to right-half-plane poles. On a Bode plot, a system is stable when the gain crosses 0 dB before the phase reaches −180°, and phase margin correlates with damping — roughly $\zeta \approx PM/100$ for a well-behaved second-order system.

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6. PID Control

$u(t) = K_p e(t) + K_i\int e(t)\,dt + K_d\frac{de(t)}{dt} \qquad \Longrightarrow \qquad G_c(s) = K_p + \frac{K_i}{s} + K_d s$

Proportional-only control always leaves an offset — this is the classic examination point. Integral action removes it but adds phase lag and can cause integral windup when the actuator saturates, which is why practical controllers clamp the integrator. Derivative action improves damping but amplifies measurement noise, so it is filtered or omitted on noisy loops such as flow control.

Ziegler-Nichols ultimate-cycle tuning: increase $K_p$ until sustained oscillation begins at ultimate gain $K_u$ and period $T_u$, then set $K_p = 0.6K_u$, $T_i = 0.5T_u$, $T_d = 0.125T_u$ for a PID controller.

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7. Research Methods

The same TOS topic names Research Methods, reflecting the capstone design and research project required by CHED Memorandum Order No. 88, s. 2017.

An engineering capstone or technical report is judged on the same criteria the PRC applies to a Professional Electrical Engineer technical report: clarity of the problem, soundness of method, defensibility of results and originality of contribution.

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Solved Board Exam Examples

Example 1: Closed-Loop Transfer Function

A unity-feedback system has forward transfer function $G(s) = \dfrac{10}{s(s+4)}$. Find the closed-loop transfer function, $\omega_n$, $\zeta$ and the percent overshoot.

Solution.

$\frac{C}{R} = \frac{G}{1+G} = \frac{10/[s(s+4)]}{1 + 10/[s(s+4)]} = \frac{10}{s^{2} + 4s + 10}$

Comparing with the standard form $\omega_n^{2}/(s^{2}+2\zeta\omega_n s + \omega_n^{2})$:

$\omega_n = \sqrt{10} = 3.162 \text{ rad/s}, \qquad 2\zeta\omega_n = 4 \implies \zeta = \frac{4}{2(3.162)} = \boxed{0.632}$

$\%OS = 100\,e^{-\pi(0.632)/\sqrt{1-0.632^{2}}} = 100\,e^{-1.9855/0.7749} = 100\,e^{-2.5623} = \boxed{7.71\%}$

Settling time is $t_s \approx 4/(\zeta\omega_n) = 4/2.0 = 2.0$ s.

Example 2: Routh-Hurwitz Stability Limit

For what range of $K$ is the system with characteristic equation $s^{3} + 6s^{2} + 11s + K = 0$ stable?

Solution. Building the Routh array:

$\begin{array}{c|cc} s^{3} & 1 & 11 \\ s^{2} & 6 & K \\ s^{1} & \dfrac{66-K}{6} & 0 \\ s^{0} & K & \end{array}$

Stability requires no sign change in the first column, so both $\dfrac{66-K}{6} > 0$ and $K > 0$:

$\boxed{0 < K < 66}$

At exactly $K = 66$ the $s^1$ row vanishes and the system sustains oscillation at $\omega = \sqrt{11} = 3.317$ rad/s — the ultimate gain and frequency a Ziegler-Nichols tuning procedure would find experimentally.

Example 3: Steady-State Error and Controller Choice

A Type 0 system with $G(s)H(s) = \dfrac{20}{(s+2)(s+5)}$ is subjected to a unit step. Find the steady-state error, and state what controller action removes it.

Solution.

$K_p = \lim_{s\to 0}\frac{20}{(s+2)(s+5)} = \frac{20}{10} = 2.0$

$e_{ss} = \frac{1}{1+K_p} = \frac{1}{3} = \boxed{0.333 \text{ or } 33.3\%}$

Raising the proportional gain reduces this offset but can never eliminate it, because $e_{ss} \to 0$ only as $K_p \to \infty$. Adding integral action makes the system Type 1 and drives the step-input steady-state error to exactly zero, at the cost of 90° of added phase lag that must be paid for with reduced gain or added derivative action.