18.2 Laplace transform and its applications

Laplace Transform Fundamentals

The Laplace transform converts time-domain signals into a complex frequency domain, turning differential equations into algebraic ones that are much easier to solve. For an intro EE course, this is one of the most important tools you'll encounter for analyzing circuits and LTI systems.

Definition and Properties

The Laplace transform takes a time-domain signal $f(t)$ and produces a complex frequency-domain representation $F(s)$:

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

Here, $s = \sigma + j\omega$ is a complex variable. The real part $\sigma$ controls exponential growth or decay, and the imaginary part $\omega$ controls oscillation frequency.

The inverse Laplace transform recovers the original signal: $f(t) = \mathcal{L}^{-1}\{F(s)\}$. In practice, you'll almost always find inverse transforms using partial fraction expansion and a table of common transform pairs rather than doing the complex integral directly.

The region of convergence (ROC) specifies which values of $s$ make the integral converge. The ROC matters because different time-domain signals can produce the same algebraic expression for $F(s)$ but with different ROCs.

• For a causal, stable system, the ROC is a right half-plane that extends to the right of the rightmost pole
• Example: If a causal system has poles at $s = -1$ and $s = -2$, the ROC is $\text{Re}(s) > -1$

S-Plane Representation

The s-plane is the complex plane where you plot the poles and zeros of $F(s)$. The horizontal axis is the real part of $s$ (damping), and the vertical axis is the imaginary part (oscillation frequency).

• Poles are values of $s$ where $F(s) \to \infty$. Each pole corresponds to a natural mode of the system (e.g., a decaying exponential or a sinusoidal oscillation).
• Zeros are values of $s$ where $F(s) = 0$. Zeros shape how the system responds at different frequencies, creating cancellations or notches.

A pole's position tells you a lot: how far left it sits determines how fast that mode decays, and how far up or down from the real axis determines its oscillation frequency. A pole at $s = -3 + j4$, for instance, corresponds to a mode that decays with time constant $1/3$ seconds and oscillates at $4$ rad/s.

System Characterization

Transfer Function

The transfer function $H(s)$ captures the entire input-output behavior of an LTI system in one expression:

$H(s) = \frac{Y(s)}{X(s)}$

where $Y(s)$ is the Laplace transform of the output and $X(s)$ is the Laplace transform of the input. This only works when initial conditions are zero (or have been accounted for separately).

The pole and zero locations of $H(s)$ determine everything about the system's dynamics:

• Poles in the left half-plane (LHP): stable modes that decay over time
• Poles in the right half-plane (RHP): unstable modes that grow over time
• Poles on the imaginary axis: marginally stable modes that neither grow nor decay (pure oscillation)

Example: A simple low-pass filter with $H(s) = \frac{1}{s+1}$ has a single pole at $s = -1$. That pole is in the LHP, so the system is stable. The cutoff frequency is $1$ rad/s, meaning signals below that frequency pass through with little attenuation.

Initial and Final Value Theorems

These two theorems let you find the starting and ending values of a signal without computing the full inverse transform.

Initial Value Theorem:

$\lim_{t \to 0^+} f(t) = \lim_{s \to \infty} sF(s)$

This gives you the value of the signal right at $t = 0^+$.

Final Value Theorem:

$\lim_{t \to \infty} f(t) = \lim_{s \to 0} sF(s)$

This gives the steady-state value, but only if all poles of $sF(s)$ are in the LHP. If any pole is on the imaginary axis or in the RHP, the final value theorem does not apply (the signal doesn't settle to a constant).

Example: Suppose you apply a unit step $U(s) = \frac{1}{s}$ to a system with transfer function $H(s)$. The steady-state output is:

$\lim_{t \to \infty} y(t) = \lim_{s \to 0} s \cdot H(s) \cdot \frac{1}{s} = \lim_{s \to 0} H(s)$

So you just evaluate $H(0)$ to find where the output settles.

Applications of Laplace Transform

Solving Differential Equations

The Laplace transform turns calculus into algebra. Here's the process:

1. Take the Laplace transform of both sides of the differential equation. Derivatives become powers of $s$, and initial conditions appear as extra terms.
2. Solve algebraically for the output $Y(s)$.
3. Apply the inverse Laplace transform (usually via partial fractions and a table) to get $y(t)$.

Worked example: Solve $\frac{d^2y}{dt^2} + 3\frac{dy}{dt} + 2y = u(t)$ with $y(0) = 0$ and $y'(0) = 0$.

Step 1: Take the Laplace transform. With zero initial conditions, $\frac{d^2y}{dt^2}$ becomes $s^2Y(s)$, $\frac{dy}{dt}$ becomes $sY(s)$, and the unit step $u(t)$ becomes $\frac{1}{s}$:

$s^2Y(s) + 3sY(s) + 2Y(s) = \frac{1}{s}$

Step 2: Factor and solve for $Y(s)$:

$Y(s) = \frac{1}{s(s^2 + 3s + 2)} = \frac{1}{s(s+1)(s+2)}$

Step 3: Use partial fraction expansion:

$Y(s) = \frac{1/2}{s} - \frac{1}{s+1} + \frac{1/2}{s+2}$

Step 4: Inverse transform each term using a table:

$y(t) = \frac{1}{2} - e^{-t} + \frac{1}{2}e^{-2t}, \quad t \geq 0$

Note: The original guide listed the answer as involving $\cos(t)$ and $\sin(t)$, but the characteristic roots are $s = -1$ and $s = -2$ (both real), so the solution contains only real exponentials.

System Analysis using Laplace Transform

Once you have $H(s)$, you can extract several types of information without going back to the time domain:

• Stability analysis: Check pole locations. All poles in the LHP means the system is stable.
• Frequency response: Substitute $s = j\omega$ into $H(s)$ to get $H(j\omega)$, which tells you the gain and phase shift at each frequency $\omega$.
• Transient response: Use the inverse Laplace transform to see how the system responds to specific inputs (step, impulse, ramp, etc.).

The Laplace framework also supports control system design. By adding feedback or compensation, you can move pole and zero locations to achieve desired stability margins, settling times, or frequency characteristics.

Example: Consider a unity feedback system with open-loop transfer function $G(s) = \frac{K}{s(s+2)}$.

The closed-loop transfer function is:

$H(s) = \frac{G(s)}{1 + G(s)} = \frac{K}{s^2 + 2s + K}$

For stability, both coefficients in the denominator must be positive (by the Routh criterion for a second-order system), which requires $K > 0$. The poles are at:

$s = \frac{-2 \pm \sqrt{4 - 4K}}{2} = -1 \pm \sqrt{1 - K}$

When $K > 1$, the poles become complex conjugates with negative real parts, producing an underdamped but stable response.

For the steady-state error to a unit step input, use the final value theorem on the error signal $E(s) = \frac{1}{1 + G(s)} \cdot \frac{1}{s}$:

$e_{ss} = \lim_{s \to 0} \frac{s}{s} \cdot \frac{1}{1 + G(s)} = \lim_{s \to 0} \frac{1}{1 + \frac{K}{s(s+2)}} = 0$

Because $G(s)$ has a pole at the origin (a "type 1" system), the steady-state error to a step input is actually zero. The gain $K$ affects the transient behavior (overshoot, settling time) but not the final value for a step input.