Interactive Lecture
Transfer Functions, Block Diagrams, and Stability Analysis
by Dr Arnaldo Delli Carri
A control system manages, commands, directs, or regulates the behaviour of other devices using control loops. Everything from an aircraft autopilot to the temperature controller in your home runs on one.
Feedback lets a system correct itself: it continuously measures the actual output, compares it with the desired output (reference), and adjusts the input to minimise the error. The transfer function describes mathematically how the system transforms inputs to outputs.
Open-Loop Systems: These systems have no feedback mechanism. The output is not measured or compared to the input. Examples include a toaster timer or a washing machine on a fixed cycle.
Closed-Loop (Feedback) Systems: These systems continuously monitor the output and adjust the input based on the error. Examples include cruise control in cars, thermostats, and autopilot systems.
A transfer function is a mathematical representation of the relationship between the input and output of a linear time-invariant (LTI) system in the Laplace domain.
Transfer Function Definition:
\[G(s) = \frac{Y(s)}{U(s)}\]
Where:
General Form:
\[G(s) = \frac{b_m s^m + b_{m-1} s^{m-1} + \cdots + b_1 s + b_0}{a_n s^n + a_{n-1} s^{n-1} + \cdots + a_1 s + a_0}\]
Or in factored form:
\[G(s) = K\frac{(s - z_1)(s - z_2)\cdots(s - z_m)}{(s - p_1)(s - p_2)\cdots(s - p_n)}\]
Where \(z_i\) are zeros and \(p_i\) are poles
A first-order system has a transfer function with a single pole:
\[G(s) = \frac{K}{\tau s + 1}\]
Where:
Step Response:
\[y(t) = K(1 - e^{-t/\tau})\]
The system reaches 63.2% of final value at \(t = \tau\)
Second-order systems are the most common in control engineering:
\[G(s) = \frac{\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}\]
Where:
Damping Classification:
Block diagrams provide a visual representation of control systems. Key reduction rules:
Series Connection: \(G_{total} = G_1 \cdot G_2\)
Parallel Connection: \(G_{total} = G_1 + G_2\)
Feedback Loop: \(G_{total} = \frac{G}{1 \pm GH}\)
Where \(+\) is used for negative feedback and \(-\) for positive feedback
For a standard feedback control system with forward path gain G(s) and feedback path H(s):
\[T(s) = \frac{Y(s)}{R(s)} = \frac{G(s)}{1 + G(s)H(s)}\]
Where:
Control systems are evaluated based on time-domain specifications:
Explore how system parameters affect dynamic behaviour. Watch the physical system respond, see poles move on the s-plane, track the step response with annotations, and observe signal flow through the block diagram — all synchronised in real time.
\[G(s) = \frac{1.0}{1.0\,s + 1}\]
| Concept | Equation | Description |
|---|---|---|
| Transfer Function | \(G(s) = \frac{Y(s)}{U(s)}\) | Output to input ratio in Laplace domain |
| First-Order System | \(G(s) = \frac{K}{\tau s + 1}\) | Single pole system |
| Second-Order System | \(G(s) = \frac{\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}\) | Standard second-order form |
| Closed-Loop TF | \(T(s) = \frac{G(s)}{1 + G(s)H(s)}\) | Feedback system transfer function |
| Series Connection | \(G_{total} = G_1 \cdot G_2\) | Cascaded blocks multiply |
| Parallel Connection | \(G_{total} = G_1 + G_2\) | Parallel blocks add |
| Time Constant | \(\tau = -\frac{1}{\text{pole}}\) | First-order system response speed |
| Settling Time (2%) | \(t_s = \frac{4}{\zeta\omega_n}\) | Time to reach ±2% of final value |
| Peak Time | \(t_p = \frac{\pi}{\omega_n\sqrt{1-\zeta^2}}\) | Time to first peak (underdamped) |
| Percent Overshoot | \(PO = e^{-\frac{\pi\zeta}{\sqrt{1-\zeta^2}}} \times 100\%\) | Maximum overshoot percentage |
A first-order temperature control system has a transfer function \(G(s) = \frac{5}{2s + 1}\). If a unit step input is applied, calculate:
Show Solution
Solution:
Given: \(G(s) = \frac{5}{2s + 1}\)
Step 1: Rewrite in standard form:
\[G(s) = \frac{5/2}{s + 1/2} = \frac{2.5}{s + 0.5}\]
Or: \(G(s) = \frac{5}{2s + 1} = \frac{5}{2(s + 0.5)} = \frac{2.5}{s + 0.5}\)
Standard form: \(G(s) = \frac{K}{\tau s + 1}\)
(a) Steady-state gain: K = 5
(b) Time constant: τ = 2 seconds
(c) Time to reach 95%:
For a first-order system: \(y(t) = K(1 - e^{-t/\tau})\)
At 95%: \(0.95 = 1 - e^{-t/\tau}\)
\[e^{-t/\tau} = 0.05\]
\[t = -\tau \ln(0.05) = 2 \times 2.996 = 5.99\,\text{s}\]
(d) Final temperature: \(y(\infty) = K \times \text{input} = 5 \times 1 = 5\,°\text{C}\)
A second-order control system has the transfer function:
\[G(s) = \frac{16}{s^2 + 4s + 16}\]
Determine:
Show Solution
Solution:
Given: \(G(s) = \frac{16}{s^2 + 4s + 16}\)
Standard form: \(G(s) = \frac{\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}\)
(a) Natural frequency:
\[\omega_n^2 = 16 \implies \omega_n = 4\,\text{rad/s}\]
(b) Damping ratio:
\[2\zeta\omega_n = 4 \implies \zeta = \frac{4}{2 \times 4} = 0.5\]
(c) Type of damping:
Since ζ = 0.5 < 1, the system is underdamped
(d) Percent overshoot:
\[PO = e^{-\frac{\pi\zeta}{\sqrt{1-\zeta^2}}} \times 100\%\]
\[PO = e^{-\frac{\pi \times 0.5}{\sqrt{1-0.5^2}}} \times 100\% = e^{-\frac{\pi \times 0.5}{\sqrt{0.75}}} \times 100\%\]
\[PO = e^{-\frac{1.571}{0.866}} \times 100\% = e^{-1.814} \times 100\% = 16.3\%\]
(e) Settling time (2%):
\[t_s = \frac{4}{\zeta\omega_n} = \frac{4}{0.5 \times 4} = \frac{4}{2} = 2\,\text{s}\]
A unity feedback control system has a forward path transfer function:
\[G(s) = \frac{10}{s(s + 2)}\]
Calculate:
Show Solution
Solution:
Given: \(G(s) = \frac{10}{s(s + 2)}\), H(s) = 1 (unity feedback)
(a) Closed-loop transfer function:
\[T(s) = \frac{G(s)}{1 + G(s)H(s)} = \frac{G(s)}{1 + G(s)}\]
\[T(s) = \frac{\frac{10}{s(s+2)}}{1 + \frac{10}{s(s+2)}} = \frac{10}{s(s+2) + 10} = \frac{10}{s^2 + 2s + 10}\]
(b) Characteristic equation:
\[1 + G(s)H(s) = 0 \implies s^2 + 2s + 10 = 0\]
(c) System poles:
Using quadratic formula:
\[s = \frac{-2 \pm \sqrt{4 - 40}}{2} = \frac{-2 \pm \sqrt{-36}}{2} = \frac{-2 \pm j6}{2} = -1 \pm j3\]
Poles: \(s_1 = -1 + j3\), \(s_2 = -1 - j3\) (complex conjugate pair)
(d) Steady-state error for unit step:
For a Type 1 system (one integrator), the steady-state error to a step input is:
\[e_{ss} = \lim_{s \to 0} \frac{sR(s)}{1 + G(s)} = \lim_{s \to 0} \frac{s \cdot \frac{1}{s}}{1 + \frac{10}{s(s+2)}}\]
For step input with unity feedback Type 1 system: \(e_{ss} = 0\)