Converting Calculus to Algebra
The Laplace transform is like a translator between two languages. In the time domain, we speak calculus (derivatives, integrals). In the s-domain, we speak algebra (polynomials, fractions). This translation makes complex engineering problems tractable. Instead of solving differential equations (hard), we solve algebraic equations (easy), then translate back.
Definition: The Laplace Transform of a function \(f(t)\) defined for \(t \geq 0\) is:
where \(s\) is a complex variable (typically \(s = \sigma + i\omega\)) and the integral converges for \(\text{Re}(s) > \sigma_0\) for some convergence abscissa \(\sigma_0\).
Let's derive the Laplace transforms of some fundamental functions by evaluating the integral directly:
Solution:
Result: \(\mathcal{L}\{1\} = \dfrac{1}{s}\) for \(s > 0\)
Solution:
Result: \(\mathcal{L}\{e^{at}\} = \dfrac{1}{s-a}\) for \(s > a\)
Solution: Using integration by parts repeatedly (or the Gamma function):
Result: \(\mathcal{L}\{t^n\} = \dfrac{n!}{s^{n+1}}\) for \(s > 0\)
In particular: \(\mathcal{L}\{t\} = \dfrac{1}{s^2}\), \(\mathcal{L}\{t^2\} = \dfrac{2}{s^3}\), \(\mathcal{L}\{t^3\} = \dfrac{6}{s^4}\)
Solution: Using the identity \(\sin(bt) = \dfrac{e^{ibt} - e^{-ibt}}{2i}\):
Result: \(\mathcal{L}\{\sin(bt)\} = \dfrac{b}{s^2 + b^2}\) for \(s > 0\)
Linearity: The Laplace transform is a linear operator. If \(\mathcal{L}\{f(t)\} = F(s)\) and \(\mathcal{L}\{g(t)\} = G(s)\), then:
for any constants \(c_1, c_2\).
First Shifting Theorem: If \(\mathcal{L}\{f(t)\} = F(s)\), then:
provided the transform of \(f(t)\) exists for \(s > c\); the transform of \(e^{at}f(t)\) exists for \(s > c + a\).
Key idea: Multiplying by \(e^{at}\) in the time domain corresponds to shifting the transform by \(a\) in the s-domain. This is why damped oscillations like \(e^{at}\sin(bt)\) are easy to handle — just shift the \(\sin(bt)\) transform by \(a\)!
Example: Since \(\sin(bt) \to \dfrac{b}{s^2+b^2}\), we have \(e^{at}\sin(bt) \to \dfrac{b}{(s-a)^2+b^2}\).
The Laplace transform exists when \(f(t)\) is piecewise continuous on \([0, \infty)\) and of exponential order. This means there exist constants \(M > 0\), \(c\), and \(T\) such that:
Most functions encountered in engineering satisfy these conditions. The Laplace transform typically exists for \(\text{Re}(s) > c\).
The following table contains the most commonly used Laplace transform pairs. Rather than deriving each from scratch, these can be looked up and combined using linearity and shifting properties.
| \(f(t)\) | \(F(s) = \mathcal{L}\{f(t)\}\) | Condition |
|---|---|---|
| 1 | \(\dfrac{1}{s}\) | \(s > 0\) |
| \(t\) | \(\dfrac{1}{s^2}\) | \(s > 0\) |
| \(t^n\) | \(\dfrac{n!}{s^{n+1}}\) | \(s > 0\) |
| \(e^{at}\) | \(\dfrac{1}{s-a}\) | \(s > a\) |
| \(\sin(bt)\) | \(\dfrac{b}{s^2 + b^2}\) | \(s > 0\) |
| \(\cos(bt)\) | \(\dfrac{s}{s^2 + b^2}\) | \(s > 0\) |
| \(e^{at}\sin(bt)\) | \(\dfrac{b}{(s-a)^2 + b^2}\) | \(s > a\) |
| \(e^{at}\cos(bt)\) | \(\dfrac{s-a}{(s-a)^2 + b^2}\) | \(s > a\) |
| \(t^n e^{at}\) | \(\dfrac{n!}{(s-a)^{n+1}}\) | \(s > a\) |
| \(\sinh(bt)\) | \(\dfrac{b}{s^2 - b^2}\) | \(s > |b|\) |
| \(\cosh(bt)\) | \(\dfrac{s}{s^2 - b^2}\) | \(s > |b|\) |
| \(t\sin(bt)\) | \(\dfrac{2bs}{(s^2 + b^2)^2}\) | \(s > 0\) |
| \(t\cos(bt)\) | \(\dfrac{s^2 - b^2}{(s^2 + b^2)^2}\) | \(s > 0\) |
| \(u(t - \tau)\) | \(\dfrac{e^{-\tau s}}{s}\) | \(s > 0\) |
| \(\delta(t)\) | \(1\) | All \(s\) |
| \(f'(t)\) | \(sF(s) - f(0)\) | |
| \(f''(t)\) | \(s^2F(s) - sf(0) - f'(0)\) |
Here are eight detailed examples demonstrating the use of the definition, linearity, and shifting properties.
Solution: Use linearity to split the transform of the sum:
From the table:
Answer: \(F(s) = \dfrac{10}{s^3} - \dfrac{3}{s-2} + \dfrac{12}{s^2+9}\) for \(s > 2\)
Solution: Use the First Shifting Theorem. First, find the transform of \(\cos(2t)\):
Now apply the shifting theorem with \(a = 3\): replace \(s\) with \((s-3)\):
Expand the denominator:
Answer: \(F(s) = \dfrac{s-3}{s^2 - 6s + 13}\) for \(s > 3\)
Solution: Use the table formula for \(t^n e^{at}\) with \(n = 3\) and \(a = -2\):
Answer: \(F(s) = \dfrac{6}{(s+2)^4}\) for \(s > -2\)
Solution: Expand the square and use linearity:
From the table:
Combine over a common denominator \(s^3\):
Answer: \(F(s) = \dfrac{s^2 + 2s + 2}{s^3}\) for \(s > 0\)
Solution: First, recall the transform of \(\sin(4t)\):
Apply the First Shifting Theorem with \(a = -1\): replace \(s\) with \((s - (-1)) = s + 1\):
Simplify the denominator:
Answer: \(F(s) = \dfrac{4}{s^2 + 2s + 17}\) for \(s > -1\)
Solution: Use linearity and the hyperbolic transform formulas:
From the table:
Factor the denominator if needed: \(s^2 - 9 = (s-3)(s+3)\)
Answer: \(F(s) = \dfrac{s + 6}{s^2 - 9} = \dfrac{s+6}{(s-3)(s+3)}\) for \(s > 3\)
Solution: First, simplify using the trig identity \(\sin(t)\cos(t) = \frac{1}{2}\sin(2t)\):
Now apply linearity:
Answer: \(F(s) = \dfrac{5}{s} + \dfrac{3}{s^2} - \dfrac{2}{s+4} + \dfrac{1}{s^2+4}\) for \(s > 0\)
Solution: First, apply linearity inside the exponential:
Now use the table formulas for shifted trig functions. With \(a = 2\), \(b = 1\):
Expand the denominator: \((s-2)^2 + 1 = s^2 - 4s + 4 + 1 = s^2 - 4s + 5\)
Answer: \(F(s) = \dfrac{3s - 11}{s^2 - 4s + 5}\) for \(s > 2\)
These problems are similar in style and difficulty to past exam questions. Click each problem to reveal the step-by-step solution.
Step 1: Identify the form. We have a polynomial multiplied by an exponential: $\mathcal{L}\{t^n e^{at}\}$.
Step 2: Use the first shifting theorem (multiplication by exponential):
Step 3: Identify parameters: $n = 3$ and $a = -2$.
Step 4: Substitute:
Answer: $\boxed{\dfrac{6}{(s+2)^4}}$ for $s > -2$
Step 1: Identify the form: exponential times cosine, $\mathcal{L}\{e^{at}\cos(bt)\}$.
Step 2: Use the first shifting theorem for cosine:
Step 3: Identify parameters: $a = 3$ and $b = 4$.
Step 4: Substitute:
Step 5: Expand denominator (optional for verification):
Answer: $\boxed{\dfrac{s-3}{(s-3)^2+16}}$ or $\boxed{\dfrac{s-3}{s^2-6s+25}}$ for $s > 3$
Find $\mathcal{L}\{f(t)\}$ where $f(t) = \begin{cases} 2t & t < 3 \\ t^2 - 3 & t \geq 3 \end{cases}$
Step 1: Rewrite the piecewise function using the unit step function $u_3(t)$:
Step 2: For the step-shifted term, express $t^2 - 2t - 3$ in terms of $(t-3)$:
Step 3: Rewrite $f(t)$:
Step 4: Apply the Laplace transform:
Step 5: Combine:
Answer: $\boxed{\dfrac{2}{s^2} + e^{-3s}\left(\dfrac{2}{s^3} + \dfrac{4}{s^2}\right)}$ for $s > 0$
Select a function and adjust parameters to see how the time-domain function maps to the s-domain transform. Watch how scaling and shifting properties change the transform.
Left plot: The function \(f(t)\) in the time domain (0 to 10 seconds).
Right plot: The magnitude of the Laplace transform \(|F(s)|\) in the s-domain (s from 0.1 to 10).
Observation: Notice how changing the parameter shifts peaks and changes the shape. The exponential decay in the time domain creates peaks in the s-domain.
Answer the following 6 questions. Feedback appears after each choice.
Find \(\mathcal{L}\{3t^2 + 2e^{5t}\}\)
Find \(\mathcal{L}\{e^{-3t}\cos(4t)\}\)
Which condition ensures the Laplace transform of \(f(t)\) exists?
Find \(\mathcal{L}\{\sin(2t)\cos(2t)\}\) (Hint: use \(\sin A \cos A = \frac{1}{2}\sin(2A)\))
If \(\mathcal{L}\{f(t)\} = F(s)\), what is \(\mathcal{L}\{e^{-2t}f(t)\}\)?
Find \(\mathcal{L}\{t^4\}\)