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Lesson 1: Definition and Basic Transforms

Estimated time: 45-55 minutes

Learning Objectives

By the end of this lesson, you will be able to:

The Laplace Transform Definition

Laplace Transform: For a function f(t) defined for t ≥ 0, the Laplace transform is L{f(t)} = F(s) = ∫0 e-st f(t) dt, provided the integral converges. The variable s is typically a real number (or complex) large enough for convergence.

The Laplace transform converts a function of time t into a function of the frequency variable s. It turns differentiation into multiplication by s, which is why it simplifies differential equations.

Computing Basic 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.

Result: L{eat} = 1/(s-a).

Example 3: L{t}

L{t} = ∫0 t e-st dt. Using 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.

More generally: L{tn} = n!/sn+1.

Example 4: L{sin(at)} and L{cos(at)}

Using integration by parts twice (or Euler's formula):

L{sin(at)} = a/(s² + a²), s > 0.

L{cos(at)} = s/(s² + a²), s > 0.

Linearity

Linearity Property: L{αf(t) + βg(t)} = αF(s) + βG(s). The Laplace transform of a linear combination is the same linear combination of the transforms.

Example 5: Using Linearity

Find L{3t² - 5e2t + 7sin(4t)}.

Step 1: L{t²} = 2!/s³ = 2/s³.

Step 2: L{e2t} = 1/(s-2).

Step 3: L{sin 4t} = 4/(s²+16).

Answer: L{3t² - 5e2t + 7sin 4t} = 6/s³ - 5/(s-2) + 28/(s²+16).

Table of Basic Laplace Transforms

f(t)F(s) = L{f(t)}Condition
11/ss > 0
tnn!/sn+1s > 0
eat1/(s-a)s > a
sin(at)a/(s²+a²)s > 0
cos(at)s/(s²+a²)s > 0
eatsin(bt)b/[(s-a)²+b²]s > a
eatcos(bt)(s-a)/[(s-a)²+b²]s > a
tneatn!/(s-a)n+1s > a

Existence of the Transform

Existence Theorem: If f(t) is piecewise continuous on [0, ∞) and of exponential order (|f(t)| ≤ Mect for some constants M, c), then L{f(t)} exists for s > c.

All polynomials, exponentials, sines, cosines, and their combinations satisfy these conditions. Functions like e do not (they grow too fast).

Check Your Understanding

1. Compute L{5 + 2t - e-3t}.

Answer: 5/s + 2/s² - 1/(s+3).

2. What is L{t³}?

Answer: 3!/s4 = 6/s4.

3. Find L{e2tcos(3t)}.

Answer: (s-2)/[(s-2)²+9].

4. Does L{e} exist? Why or why not?

Answer: No. e is not of exponential order because it grows faster than any Mect.

Key Takeaways

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In Lesson 2, learn to go backward: inverse Laplace transforms.

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