Control SystemsSection 1

Mathematical Modelling of Systems

Converting physical systems into sets of equations to analyze their dynamic response using frequency-domain tools like Fourier and Laplace transforms.

1. Introduction to Automatic Control

We use automatic control to reach a desired response for a given system. The design and implementation of control systems generally follow three fundamental steps:

Mathematical Modelling: Representing the physical system using differential equations.
Dynamic Analysis: Analyzing how the system responds to different inputs over time.
Controller Design: Creating a controller (e.g., PID, Compensators) to correct the system's behavior and meet performance specifications.

Types of Control Systems

Open-Loop Control System

The output of the system does not affect the control action. The output is affected significantly by any external disturbances.

Example: A standard toaster or an immersion water heater.

Input → Controller → Actuator → Output

Closed-Loop Control System

The output is measured and fed back to the controller to compare it with the reference value. The controller takes action based on the calculated error.

Example: Speed control of a motor.

Input → Controller → Motor → Output
↑ ——— Feedback ——— ↓

2. Mathematical Modelling (Time vs. S-Domain)

To analyze a system, we must translate its physical dynamics into mathematical equations. In the Time Domain, systems are described by non-separable differential equations. For instance, consider a system with input $x(t)$ and output $y(t)$:

\ddot{y}(t) + 5\dot{y}(t) + 10y(t) = 2\dot{x}(t) + 3x(t)

In this time-domain form, it is extremely difficult to mathematically separate the output $y(t)$ as an explicit function of the input $x(t)$. To solve this structural issue, engineers utilize the S-Domain (Laplace Domain).

By transforming the differential equation into the S-domain, time derivatives simply become algebraic multiplications by $s$. The equation transforms into a separable algebraic equation:

$$s^2Y(s) + 5sY(s) + 10Y(s) = 2sX(s) + 3X(s)$$

$$Y(s)[s^2 + 5s + 10] = X(s)[2s + 3]$$

This allows us to explicitly define the Transfer Function—the direct ratio of output to input $H(s) = \frac{Y(s)}{X(s)}$ :

H(s) = \frac{2s + 3}{s^2 + 5s + 10}

3. Fourier Series & Fourier Transform

Fourier Series

To deeply understand the S-Domain, we must first look at the Frequency Domain. The Fourier Series decomposes any periodic signal into a DC component plus a sum of sinusoidal (sine and cosine) functions of different fundamental and harmonic frequencies.

It is used to analyze and represent periodic signals in terms of their harmonic components. Mathematically, a periodic function $f(t)$ with period $T$ can be expressed as:

f(t) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos \left( \frac{2\pi nt}{T} \right) + b_n \sin \left( \frac{2\pi nt}{T} \right) \right)

Where the coefficients are calculated as follows:

a_0 = \frac{1}{T} \int_{0}^{T} f(t) dt

a_n = \frac{2}{T} \int_{0}^{T} f(t) \cos \left( \frac{2\pi nt}{T} \right) dt

b_n = \frac{2}{T} \int_{0}^{T} f(t) \sin \left( \frac{2\pi nt}{T} \right) dt

Here, $\frac{a_0}{2}$ represents the DC component (average value) of the signal, while the $a_n$ and $b_n$ terms represent the amplitudes of the harmonic frequencies.

Application: This is highly applicable in power systems for analyzing harmonic distortion caused by non-linear loads. By identifying the dominant harmonics and their magnitudes using the Fourier series, engineers can design specific filters to eliminate them.

Fourier Transform

The Fourier Transform extends this concept to non-periodic signals (signals that do not repeat over time). It transforms a signal from the time domain $f(t)$ to the frequency domain $F(\omega)$ using the following integral:

F(\omega) = \int_{-\infty}^{\infty} f(t) e^{-j\omega t} dt

Application: In audio processing, a recording might contain high-frequency noise. By applying a Fourier Transform, we visualize the signal in the frequency spectrum, apply a low-pass filter to mathematically kill the high-frequency noise, and then use an Inverse Fourier Transform to recover the clean audio in the time domain.

4. The Laplace Transform (S-Domain)

The Fourier transform is an excellent tool for frequency domain analysis, but it has a significant limitation: it mainly handles purely oscillatory (steady-state) signals. What if a system has an exponential decaying response (a transient response), which is extremely common in physical systems?

Extending Fourier to Laplace

To handle these transient responses, we must extend the Fourier Transform by adding a damping (decaying) factor. Let's start with the standard Fourier transform integral:

X(\omega) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} dt

We modify our time-domain signal $x(t)$ by multiplying it by an exponential decaying term $e^{-\sigma t}$, where $\sigma$ is a real number representing the damping factor. The integral becomes:

\int_{-\infty}^{\infty} \left[ x(t) e^{-\sigma t} \right] e^{-j\omega t} dt = \int_{-\infty}^{\infty} x(t) e^{-(\sigma + j\omega)t} dt

We then define a new complex variable, $s$, to replace the $(\sigma + j\omega)$ term:

s = \sigma + j\omega

This transforms our time-domain signal into a complex frequency domain (the S-Domain). The variable $s$ dictates the exact nature of the system's response:

Only $\sigma$ ($s = \sigma$): The system has a purely exponential (decaying or growing) response.
Only $j\omega$ ($s = j\omega$): The system has a purely oscillatory response (sine/cosine waves). Notice that setting $\sigma = 0$ perfectly reduces the Laplace transform back to the Fourier transform.
Both $\sigma + j\omega$: The system exhibits a decaying oscillatory response (a damped sine wave), which is the standard transient response of practical electrical systems.

Unilateral vs. Bilateral Laplace Transform

Bilateral Laplace Transform

Defined over the entire time axis from $-\infty$ to $\infty$. It handles both causal and non-causal signals.

\mathcal{L}[f(t)] = \int_{-\infty}^{\infty} f(t) e^{-st} dt

Unilateral Laplace Transform

Considers only $t \ge 0$. This is the standard in Control Systems because physical electrical systems are causal (they respond only after an input is applied at $t=0$).

\mathcal{L}[f(t)] = \int_{0}^{\infty} f(t) e^{-st} dt

5. Linear Time-Invariant (LTI) Systems & Transfer Functions

In control theory, we focus on Linear Time-Invariant (LTI) systems because their mathematical properties allow us to predict their behavior reliably. An LTI system satisfies two fundamental properties:

Linearity: A linear system possesses the properties of superposition and homogeneity.
- Superposition (Additivity): If the system's response to input $x_1(t)$ is $y_1(t)$ and to $x_2(t)$ is $y_2(t)$, then the combined input $x_1(t) + x_2(t)$ will strictly yield $y_1(t) + y_2(t)$.
- Scaling (Homogeneity): If input $x(t)$ yields $y(t)$, then scaling the input $A \cdot x(t)$ will scale the output to $A \cdot y(t)$.
Time-Invariance: The behavior and characteristics of the system are fixed over time. A delay in the input produces an identical delay in the output without altering the signal shape.

Real-world Note: No physical system is perfectly linear. However, we approximate non-linear systems as linear around a specific operating point to simplify the design of controllers. This approximation is typically valid for small deviations.

Transfer Function and System Response

The Transfer Function (T.F) is defined as the ratio of the output signal to the input signal in the S-domain, assuming zero initial conditions.

Time Domain

In the time domain, the output $y(t)$ is the mathematical convolution ($*$) of the input $x(t)$ and the system's impulse response $h(t)$. Convolution is computationally difficult.

$$y(t) = h(t) * x(t)$$

Frequency Domain (S-Domain)

By applying the Laplace transform, the complex convolution becomes a simple algebraic multiplication.

Y(s) = H(s) \cdot X(s) \implies H(s) = \frac{Y(s)}{X(s)}

Deriving the Laplace Transform Table

How do we convert standard time-domain inputs (like a step function or a sine wave) into the S-domain? We substitute the time-domain function $f(t)$ into the fundamental unilateral Laplace integral:

F(s) = \int_{0}^{\infty} f(t) \cdot e^{-st} dt

Proof: Laplace Transform of a Unit Step Input

A unit step input $u(t)$ has a constant value of $1$ for $t \ge 0$. Let's plug $f(t) = 1$ into the integral:

\mathcal{L}[1] = \int_{0}^{\infty} 1 \cdot e^{-st} dt

Evaluating the integral of the exponential:

\mathcal{L}[1] = \left[ -\frac{1}{s} e^{-st} \right]_{0}^{\infty}

Substitute the limits ($\infty$ and $0$). Note that $e^{-\infty} = 0$ and $e^{0} = 1$:

\mathcal{L}[1] = \left( -\frac{1}{s} \cdot 0 \right) - \left( -\frac{1}{s} \cdot 1 \right) = \frac{1}{s}

This proves why a constant input of 1 becomes $\frac{1}{s}$ in the frequency domain!

By applying this exact integration process to various functions (ramps, exponentials, sines), engineers have built a standard lookup table to speed up system analysis:

Time Domain $f(t)$	S-Domain $F(s)$
Unit Step $u(t)$ or Constant $1$	$\frac{1}{s}$
Ramp $t$	$\frac{1}{s^2}$
Exponential $e^{-at}$	$\frac{1}{s+a}$
Sine Wave $\sin(\omega t)$	$\frac{\omega}{s^2 + \omega^2}$
Cosine Wave $\cos(\omega t)$	$\frac{s}{s^2 + \omega^2}$
Derivative $\frac{d}{dt}f(t)$	$sF(s) - f(0)$
Integral $\int f(t)dt$	$\frac{F(s)}{s}$

6. Electrical Systems in S-Domain

Transforming passive electrical components (Resistors, Inductors, Capacitors) into the S-domain simplifies circuit analysis. By taking the Laplace transform of their time-domain differential equations (assuming zero initial conditions), we can define their complex impedance $X(s)$ or $Z(s) = \frac{V(s)}{I(s)}$.

Resistor

Time Domain

v(t) = R \cdot i(t)

Apply Laplace

V(s) = R \cdot I(s)

Impedance

$$Z_R = R$$

Inductor

Time Domain

v(t) = L \frac{di(t)}{dt}

Apply Laplace

V(s) = L \cdot s \cdot I(s)

V(s) = I(s) \cdot X_L

Impedance

$$X_L = sL$$

Capacitor

Time Domain

i(t) = C \frac{dv(t)}{dt}

Apply Laplace

I(s) = C \cdot s \cdot V(s)

V(s) = \frac{I(s)}{Cs} = I(s) \cdot X_c

Impedance

X_c = \frac{1}{sC}

Steady-State AC Analysis Note: In electrical steady-state AC circuits, the transient response has died out, meaning the damping factor $\sigma = 0$. By substituting $s = j\omega$, the impedances directly map to the familiar AC reactances: $X_L = j\omega L$ and $X_C = \frac{1}{j\omega C}$.

7. Solved Examples

Example 1: Differential Equation to Transfer Function

Find the Transfer Function $H(s) = Y(s)/X(s)$ for the following differential equation (assuming zero initial conditions):

5\dot{y} + \ddot{y} + 10y = 2\dot{x} + 3x

Solution Steps:

1. Apply the Laplace transform to both sides, converting derivatives to S-multipliers:

$$5sY(s) + s^2Y(s) + 10Y(s) = 2sX(s) + 3X(s)$$

2. Factor out $Y(s)$ on the left and $X(s)$ on the right:

$$Y(s)[s^2 + 5s + 10] = X(s)[2s + 3]$$

3. Isolate the ratio $\frac{Y(s)}{X(s)}$ to find the Transfer Function:

\frac{Y(s)}{X(s)} = \frac{2s + 3}{s^2 + 5s + 10}

Example 2: RLC Circuit Transfer Function

Consider an electrical circuit with an input voltage $V_i$, a series resistor $R$, and an output voltage $V_o$ taken across a parallel combination of an Inductor $L$ and Capacitor $C$. Find the transfer function $\frac{V_o(s)}{V_i(s)}$.

Solution Steps:

1. First, find the equivalent impedance of the parallel LC combination ($Z_{eq}$):

Z_{eq} = sL \parallel \frac{1}{sC} = \frac{sL \cdot \frac{1}{sC}}{sL + \frac{1}{sC}} = \frac{\frac{L}{C}}{\frac{s^2LC + 1}{sC}} = \frac{sL}{s^2LC + 1}

2. Apply the standard voltage divider rule to find $V_o$:

V_o(s) = V_i(s) \left( \frac{Z_{eq}}{R + Z_{eq}} \right)

3. Substitute $Z_{eq}$ and multiply numerator and denominator by $(s^2LC + 1)$ to simplify:

\frac{V_o(s)}{V_i(s)} = \frac{\frac{sL}{s^2LC + 1}}{R + \frac{sL}{s^2LC + 1}} = \frac{sL}{R(s^2LC + 1) + sL}

Example 3: Active Filter (Inverting Op-Amp)

Find the transfer function of an inverting amplifier where the input impedance is $R_1$ and the feedback impedance is $R_2$ in parallel with a capacitor $C_1$.

Solution Steps:

1. For an inverting amplifier, the standard transfer function is given by $\frac{V_o}{V_i} = -\frac{Z_{feedback}}{Z_{input}}$.

2. Define the impedances: Let $Z_{input} = R_1$ and calculate $Z_{feedback}$ as $R_2 \parallel C_1$:

Z_{feedback} = \frac{R_2 \cdot \frac{1}{sC_1}}{R_2 + \frac{1}{sC_1}} = \frac{R_2}{1 + sR_2C_1}

3. Substitute into the Op-Amp formula to calculate the final transfer function:

\frac{V_o(s)}{V_i(s)} = -\frac{Z_{feedback}}{Z_{input}} = -\frac{\frac{R_2}{1 + sR_2C_1}}{R_1} = -\frac{R_2}{R_1(1 + sR_2C_1)}