Control Engineering: free practice, theory and problems

1. Laplace and transfer functions

What is it about?

In control engineering we describe how a system (a motor, a temperature process, a robot arm) responds to an input by transforming the differential equation into the Laplace domain. There, differential equations become algebra, and the system can be described by one compact fraction: the transfer function $G(s)$. The poles and zeros of $G(s)$ tell you almost everything you need to know about the system's behavior – whether it is stable, how fast it responds, and whether it oscillates.

The goal of this unit is to be able to read a transfer function and say something qualitative and quantitative about the step response without solving the differential equation in the time domain every time.

Concepts and formulas

• Transfer function: $G(s) = Y(s)/U(s)$, the ratio of the Laplace-transformed output to the input with zero initial conditions.
• Poles: the values of $s$ that make the denominator (the characteristic equation) zero. Zeros: the values that make the numerator zero.
• Stability: the system is stable if and only if all poles have negative real part (lie in the left half-plane).
• First-order system: $G(s) = \dfrac{K}{\tau s+1}$, with a pole at $s=-1/\tau$. Final value for a unit step: $G(0)=K$. After one time constant $\tau$, the response is at 63% of the final value.
• Second-order standard form: $G(s) = \dfrac{\omega_n^2}{s^2+2\zeta\omega_ns+\omega_n^2}$, with natural frequency $\omega_n$ and damping ratio $\zeta$.
• Overshoot (for $0<\zeta<1$): $M_p = e^{-\pi\zeta/\sqrt{1-\zeta^2}}$. Peak time $t_p = \pi/\omega_d$ and settling time (2%) $t_s \approx 4/(\zeta\omega_n)$, where $\omega_d=\omega_n\sqrt{1-\zeta^2}$ is the damped (oscillation) frequency.
• Dead time $T$ adds an extra factor $e^{-sT}$ to the transfer function: a pure delay before the output reacts.

How to solve the problems

1. Identify whether the system is first- or second-order, and read off $K$, $\tau$ (or $\omega_n$, $\zeta$) directly from the form of $G(s)$.
2. Find the poles by setting the denominator to zero, and check the sign of the real part for stability.
3. Find the final value for a unit step using $G(0)$ (set $s=0$).
4. Use the standard formulas for overshoot, peak time and settling time for second-order systems instead of solving the differential equation.
5. Check the answer against the physics: a system without an integrator and with positive $K$ should go to a finite, positive final value.

Example

A system has the transfer function $G(s) = \dfrac{36}{s^2+3.6s+36}$. Find the damping ratio, the final value for a unit step, and the overshoot.

1. Compare with the standard form: $\omega_n^2=36$, so $\omega_n=6$ rad/s.
2. $2\zeta\omega_n=3.6$ gives $\zeta = 3.6/(2\cdot6) = 0.3$.
3. Final value: $G(0)=36/36=1$.
4. Overshoot: $M_p=e^{-\pi\cdot0.3/\sqrt{1-0.3^2}}\approx e^{-0.988}\approx0.372$, i.e. about 37%.

Answer: $\zeta=0.3$, final value $=1$, overshoot $\approx37$%.

Common mistakes

• Thinking that zeros determine stability – it is the poles that determine stability, not the zeros.
• Forgetting to set $s=0$ correctly when finding the final value (only valid for stable systems without an integrator).
• Confusing $\omega_n$ with the damped frequency $\omega_d=\omega_n\sqrt{1-\zeta^2}$ – they are not the same when $\zeta>0$.
• Thinking dead time $e^{-sT}$ is the same as a pole – it changes the phase, not the number of poles in a rational approximation.

Concepts in this part

Practise laplace and transfer functions in the app →

2. PID control

What is it about?

A PID controller is the most widely used control algorithm in industry: it compares a measured value with a reference and computes an actuator signal from the error. The three terms – proportional (P), integral (I) and derivative (D) – each have their own job, and good control is about balancing them against each other: fast response, little overshoot, zero steady-state error and robustness to noise.

This unit is about understanding what each term does, how they combine in a feedback loop, and how to find reasonable starting values for the gains.

Concepts and formulas

• Ideal PID controller: $C(s) = K_p + \dfrac{K_i}{s} + K_ds$, where $K_i=K_p/T_i$ and $K_d=K_pT_d$ ($T_i$ = integral time, $T_d$ = derivative time).
• The P term gives a fast response, but leaves a steady-state error for most processes.
• The I term integrates the error over time and removes steady-state error, but can make the system slower and more oscillatory.
• The D term reacts to how fast the error is changing and damps overshoot, but amplifies measurement noise and is therefore often filtered.
• Closed loop with forward gain $G$ and feedback $H$: $\dfrac{Y}{R} = \dfrac{G}{1+GH}$ (negative feedback).
• Steady-state error for a step: $e_\infty = \dfrac{1}{1+K_pG(0)}$ for pure P control; with an I term, $e_\infty=0$ for a step.
• Steady-state error for a ramp (type-1 system, PI control of $G(s)=K/(\tau s+1)$): the velocity constant $K_v = K_pK/T_i$, and $e_\infty = 1/K_v$.
• Ziegler–Nichols (ultimate gain method): find the ultimate gain $K_u$ and period $T_u$ (pure P control until sustained oscillation). PID table: $K_p=0.6K_u$, $T_i=T_u/2$, $T_d=T_u/8$.
• Integrator windup: the I term keeps growing while the actuator is saturated, causing large overshoot once the error reverses.

How to solve the problems

1. Identify which term (P, I or D) the question is about, and what kind of error or effect it produces.
2. For steady-state error: check whether the loop has an I term (type 1) – the error is then zero for a step, but can be finite for a ramp.
3. For closed-loop expressions: set up $G/(1+GH)$ and simplify before finding poles or the final value.
4. For Ziegler–Nichols: plug $K_u$ and $T_u$ directly into the table formulas – do not mix up $K_p$, $T_i$ and $T_d$.
5. Check whether the answer makes sense: increasing $K_p$ should give a faster but more unsettled response; an I term should remove the steady-state error for a step.

Example

A process $G(s) = \dfrac{2}{s+1}$ is controlled by a PI controller with $K_p=3$ and $T_i=0.5$ s. The reference is a ramp with slope 1 (unit/s). Find the steady-state error.

1. Velocity constant: $K_v = K_pK/T_i = 3\cdot2/0.5 = 12$.
2. Steady-state error: $e_\infty = 1/K_v = 1/12 \approx 0.0833$.

Answer: $e_\infty \approx 0.083$ (units).

Common mistakes

• Thinking an I term removes steady-state error for *all* types of references – it removes the error for a step, but not necessarily for a ramp.
• Increasing $K_p$ a lot to get rid of steady-state error instead of adding an I term – that causes more overshoot and can make the system unstable.
• Confusing $T_i$ (integral time, a large $T_i$ means weak I action) with $K_i$ (where a large $K_i$ means strong I action) – they are each other's inverse, scaled by $K_p$.
• Forgetting anti-windup in a controller with actuator saturation, so the I term "locks up" and causes large overshoot.

Concepts in this part

Practise pID control in the app →

3. Frequency analysis

What is it about?

The Bode plot shows how a system responds to sinusoidal input signals at different frequencies: how much the amplitude is boosted or attenuated (in decibels), and how much the phase is shifted. This frequency response is used to judge how stable a closed-loop control system will be – without having to find the poles explicitly – and to understand how far you can raise the gain before the system starts to oscillate uncontrollably.

Phase margin and gain margin are the most important "safety margins" a control engineer reads off a Bode plot before putting a controller into service.

Concepts and formulas

• Gain in decibels: $|G|_{dB} = 20\log_{10}|G|$. Inverse: $|G| = 10^{|G|_{dB}/20}$.
• A single pole ($1/(\tau s+1)$) gives an asymptotic $-20$ dB/decade above the corner frequency $\omega_c = 1/\tau$; a single zero gives $+20$ dB/decade.
• The phase contribution from a single pole goes gradually from $0^\circ$ to $-90^\circ$ around the corner frequency; from a zero, from $0^\circ$ to $+90^\circ$.
• A pure dead time $e^{-sT}$ does not change the magnitude, but gives a phase lag $\varphi = -\omega T$ (radians) that grows linearly with frequency.
• Crossover frequency $\omega_c$: where $|L(j\omega)| = 1$ (0 dB), with $L$ the loop gain $= GH$.
• Phase margin: $PM = 180^\circ + \angle L(j\omega_c)$ at the crossover frequency. A common target is 45–60°.
• Gain margin: how many dB $|L|$ can be increased before the system becomes unstable, read where the phase is $-180^\circ$.
• Bandwidth: the frequency where the closed-loop magnitude has dropped 3 dB from its low-frequency value.

How to solve the problems

1. Find the corner frequency of each pole/zero ($\omega=1/\tau$) and reason your way to the asymptotes.
2. Add the magnitude and phase contributions from each term (poles, zeros, any dead time) to get the total magnitude and phase.
3. Find the crossover frequency $\omega_c$ where $|L|=1$, and read the phase there to compute the phase margin.
4. For dead time: compute the phase contribution as $-\omega T$ in radians, and convert to degrees by multiplying by $180/\pi$.
5. Check that the answer is reasonable: a stable loop should have a positive phase margin and a positive gain margin.

Example

A loop has $L(s) = \dfrac{4}{s}e^{-sT}$ with dead time $T = 0.05$ s. Find the phase margin.

1. Crossover frequency: $|L(j\omega)| = 4/\omega = 1$ gives $\omega_c = 4$ rad/s.
2. Phase from the integrator: $-90^\circ$.
3. Phase from the dead time: $\varphi_T = -\omega_cT\cdot(180/\pi) = -4\cdot0.05\cdot(180/\pi) \approx -11.46^\circ$.
4. Total phase: $-90^\circ - 11.46^\circ = -101.46^\circ$.
5. Phase margin: $PM = 180^\circ - 101.46^\circ \approx 78.5^\circ$.

Answer: $PM \approx 78.5^\circ$ (a good margin).

Common mistakes

• Thinking dead time changes the magnitude – it only changes the phase, never $|G|$.
• Reading the gain margin at the crossover frequency instead of where the phase is $-180^\circ$ (and vice versa for the phase margin).
• Forgetting to convert the dead-time phase contribution from radians to degrees.
• Thinking a higher bandwidth is always better – it gives a faster response, but also more sensitivity to noise.

Concepts in this part

Practise frequency analysis in the app →

4. Step response and first-order systems

What is it about?

Many processes react in the same way when you change the input abruptly: a kettle that is switched on, a tank that fills, a motor that gets voltage. The output rises quickly at first and then levels off towards a new value. This is a first-order system, and it is fully described by two numbers: the gain $K$ and the time constant $\tau$.

Concepts and formulas

• Transfer function:

• Step response: when the input jumps by $A$ at $t = 0$, the output becomes

• Final value: $y(\infty) = KA$. The gain $K$ is the change in output divided by the change in input once everything has settled.
• The time constant $\tau$ is the time it takes to reach 63.2 % of the change. After $3\tau$ you are at 95 %, after $4\tau$ at 98 % and after $5\tau$ at 99.3 %.
• Rise time (from 10 % to 90 %): $t_r = \tau\ln 9 \approx 2.2\,\tau$.
• RC circuit: $\tau = RC$. A thermal process: $\tau$ = thermal capacitance times thermal resistance.
• Time to reach a fraction $p$ of the final value: $t = -\tau\ln(1 - p)$.

How to solve the problems

1. Read $K$ and $\tau$ from $G(s)$. Make sure the denominator has the form $\tau s + 1$ (divide by the constant term if needed).
2. Final value: $KA$. Value at time $t$: $KA(1 - e^{-t/\tau})$.
3. Time to a given fraction: $t = -\tau\ln(1-p)$.

Example

$G(s) = \dfrac{5}{2s + 1}$, and the input jumps from 0 to 3.

1. $K = 5$ and $\tau = 2$ s.
2. The final value is $5\cdot 3 = 15$.
3. After 2 s the output is $15\,(1 - e^{-1}) \approx 9.48$, i.e. 63.2 % of the way.
4. Settling time to 98 %: $4\tau = 8$ s.

Common mistakes

• Reading the wrong time constant when the denominator is not normalized: $\dfrac{10}{4s + 2} = \dfrac{5}{2s + 1}$ gives $\tau = 2$, not 4.
• Forgetting to multiply by the step size $A$.
• Mixing up 63 % (one time constant) and 98 % (four time constants).

Concepts in this part

Practise step response and first-order systems in the app →