Lesson 1: Definition and Basic Transforms
Estimated time: 45-55 minutes
Learning Objectives
By the end of this lesson, you will be able to:
- State the definition of the Laplace transform as an improper integral
- Compute Laplace transforms of basic functions from the definition
- Use linearity to combine transforms
- Apply the first shifting theorem (s-shifting)
- Build and use a table of common Laplace transforms
What Is the Laplace Transform?
The Laplace transform converts a function of time f(t) into a function of a complex variable F(s). It turns derivatives into algebraic operations, making differential equations much easier to solve.
Laplace Transform: L{f(t)} = F(s) = ∫0∞ e-st f(t) dt, defined for values of s where the integral converges.
The key idea: differentiation in the t-domain becomes multiplication by s in the s-domain. This converts a differential equation into an algebraic equation.
Computing Transforms from the Definition
Example 1: L{1}
L{1} = ∫0∞ e-st dt = [-e-st/s]0∞ = 0 - (-1/s) = 1/s, for s > 0.
Example 2: L{eat}
L{eat} = ∫0∞ e-st eat dt = ∫0∞ e-(s-a)t dt = 1/(s-a), for s > a.
Example 3: L{t}
L{t} = ∫0∞ t e-st dt. Integration by parts: u = t, dv = e-stdt.
= [-te-st/s]0∞ + (1/s)∫0∞ e-st dt = 0 + (1/s)(1/s) = 1/s², for s > 0.
Example 4: L{sin(bt)}
L{sin(bt)} = ∫0∞ e-st sin(bt) dt. Using integration by parts twice (or the formula):
L{sin(bt)} = b/(s² + b²), for s > 0.
Table of Common Transforms
| f(t) | F(s) = L{f(t)} | Condition |
|---|---|---|
| 1 | 1/s | s > 0 |
| tn | n!/sn+1 | s > 0 |
| eat | 1/(s-a) | s > a |
| sin(bt) | b/(s²+b²) | s > 0 |
| cos(bt) | s/(s²+b²) | s > 0 |
| eatsin(bt) | b/((s-a)²+b²) | s > a |
| eatcos(bt) | (s-a)/((s-a)²+b²) | s > a |
| tneat | n!/(s-a)n+1 | s > a |
Linearity
Linearity Property: L{af(t) + bg(t)} = aL{f(t)} + bL{g(t)} = aF(s) + bG(s) for any constants a, b.
Example 5: Using Linearity
Find L{3t² - 5e2t + 4sin(3t)}.
= 3 L{t²} - 5 L{e2t} + 4 L{sin(3t)}
= 3(2!/s³) - 5/(s-2) + 4(3/(s²+9))
= 6/s³ - 5/(s-2) + 12/(s²+9).
First Shifting Theorem (s-Shifting)
First Shifting Theorem: If L{f(t)} = F(s), then L{eatf(t)} = F(s-a). Multiplying by eat in the t-domain shifts s to s-a in the s-domain.
Example 6: Applying s-Shifting
Find L{e3tcos(2t)}.
L{cos(2t)} = s/(s²+4). Shifting s to s-3:
L{e3tcos(2t)} = (s-3)/((s-3)²+4).
Transforms of Derivatives
Derivative Property: L{f'(t)} = sF(s) - f(0). L{f''(t)} = s²F(s) - sf(0) - f'(0). Each derivative multiplies by s and subtracts initial conditions.
This is the property that makes Laplace transforms so powerful for solving IVPs: derivatives become algebraic expressions involving s and the initial values.
Check Your Understanding
1. Compute L{5e-2t} from the table.
2. Find L{t³}.
3. Use the first shifting theorem to find L{e-tsin(4t)}.
4. What does the derivative property say about L{y''}?
Key Takeaways
- L{f(t)} = ∫0∞ e-st f(t) dt converts functions from the t-domain to the s-domain.
- The transform is linear: constants factor out, sums split.
- The s-shifting theorem: L{eatf(t)} = F(s-a).
- Derivatives become algebra: L{f'} = sF(s) - f(0).
- Memorize the basic transform table -- it is the foundation for everything that follows.
Ready for More?
Next Lesson
In Lesson 2, learn to go backwards: inverse Laplace transforms using partial fractions.
Start Lesson 2