Training Control Systems Transfer Functions & Block Diagrams
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Transfer Functions & Block Diagrams

30 min Control Systems

Transfer Functions & Block Diagrams

A control system has a plant (what you want to govern), a sensor (what you measure), and a controller (the decision maker). The cleanest way to analyse such systems is in the Laplace domain, where differential equations become algebraic. The transfer function $G(s) = Y(s)/U(s)$ is the ratio of the Laplace transform of the output to that of the input, with zero initial conditions.

Complex systems built from multiple subsystems are summarized by block diagrams. Three reduction rules handle almost everything: series blocks multiply, parallel blocks add, and a feedback loop with forward gain $G$ and feedback $H$ has closed-loop transfer $G/(1+GH)$.

In this lesson we take differential equations of simple physical systems, transform to the $s$-domain, and reduce block diagrams to closed-loop transfer functions.

Transfer Function

Given $a_n y^{(n)} + \cdots + a_0 y = b_m u^{(m)} + \cdots + b_0 u$ with zero initial conditions:

$$G(s) = \frac{Y(s)}{U(s)} = \frac{b_m s^m + \cdots + b_0}{a_n s^n + \cdots + a_0}$$

Feedback Formula

For unity-feedback: $T(s) = \dfrac{G(s)}{1 + G(s)}$. For non-unity feedback $H$: $T(s) = \dfrac{G(s)}{1 + G(s)H(s)}$.

Example 1 — Mass-spring transfer function

The equation $m\ddot{y} + c\dot{y} + ky = f(t)$ has input $f$ and output $y$. Find $G(s)$.

Taking Laplace (zero IC): $(ms^2 + cs + k)Y(s) = F(s)$, so $G(s) = 1/(ms^2 + cs + k)$.

Example 2 — Closed-loop for unity feedback with $G = 10/(s+2)$

Find $T(s)$.

$T(s) = \dfrac{10/(s+2)}{1 + 10/(s+2)} = \dfrac{10}{s+12}$. DC gain $T(0) = 10/12 = 0.833$.

Example 3 — Series + Parallel reduction

Two blocks $G_1 = 2/(s+1)$ in series with $G_2 = 5/s$, placed in parallel with $G_3 = 3$. Find the equivalent transfer function.

Series first: $G_{12} = G_1 G_2 = 10/[s(s+1)]$.

Parallel with $G_3$: $G_{eq} = G_{12} + G_3 = 10/[s(s+1)] + 3 = [3s(s+1) + 10]/[s(s+1)] = (3s^2 + 3s + 10)/[s(s+1)]$.

Interactive Demo: First-Order Closed-Loop
T(s) =10/(s+12)
DC gain T(0) =0.833
τ (closed-loop) =0.083s

Practice Problems

1. Laplace-transform $\dot y + 3 y = u$ with zero IC.
2. For $G(s) = 5/(s^2+4)$, what are the poles?
3. Reduce two blocks in series: $G_1 = 3/s$, $G_2 = 2/(s+4)$.
4. For unity feedback with $G = K/(s+1)$, find the closed-loop pole as a function of $K$.
5. What is the DC gain of $G(s) = 8/(s^2 + 4s + 16)$?
6. Why is the Laplace transform convenient for linear ODEs?
Show Answer Key

1. $(s+3)Y = U$, so $G = 1/(s+3)$.

2. Poles at $s = \pm 2j$ (pure imaginary).

3. $G_1 G_2 = 6/[s(s+4)]$.

4. $T = K/(s+1+K)$ — pole at $s = -(1+K)$.

5. $G(0) = 8/16 = 0.5$.

6. It turns differential equations into algebraic equations and handles initial conditions automatically.

Step Response of First & Second Order