📜 Laplace Transform of a Piecewise Function.
To find \(\lap{g(t)}\) for a piecewise function:
-
Combine and simplify terms so you have the fewest possible \(u_c(t)\)-switches.
Section 13.5 Transforming Piecewise Functions
Until now, you’ve learned how step functions act like ON–OFF switches, and you’ve built the three key Laplace rules (L\(_{9}\text{,}\) L\(_{10}\text{,}\) and L\(_{11}\)) that describe how those switches behave under the Laplace transform. With these tools, we’re ready for the real payoff: transforming entire piecewise functions.
Subsection General Strategy
The game plan for handling a piecewise function in the Laplace domain is always the same three steps.
That’s it. Once each term is transformed, you can add them up thanks to linearity. You’ll be left with a fully transformed piecewise function living in the Laplace domain.
Checkpoint 263. 📖❓ Strategy for Transforming a Piecewise Function.
(a) What’s Should you do First?
True.
- That’s the foundation of this method! Once a piecewise function is rewritten using step functions, you can apply the shift rule to each piece.
False.
- That’s the foundation of this method! Once a piecewise function is rewritten using step functions, you can apply the shift rule to each piece.
To take the Laplace transform of a piecewise function, you should first rewrite it in terms of unit step functions.
(b) Matching Activation Intervals to Switches.
Subsection Piecewise Functions in the Laplace Domain
Let’s apply this strategy to the forcing function of our model problem, \(g(t)\text{:}\)
| Piece | Active Interval | Step Function Switch |
\begin{equation*}
2t
\end{equation*}
|
\begin{equation*}
0 \le t \lt 1
\end{equation*}
|
\begin{equation*}
u_0(t) - u_1(t)
\end{equation*}
|
\begin{equation*}
3
\end{equation*}
|
\begin{equation*}
1 \le t \lt 4
\end{equation*}
|
\begin{equation*}
u_1(t) - u_4(t)
\end{equation*}
|
\begin{equation*}
0
\end{equation*}
|
\begin{equation*}
t \ge 4
\end{equation*}
|
\begin{equation*}
u_4(t)
\end{equation*}
|
Multiply each piece by its switch, then sum:
\begin{equation*}
g(t) = 2t\Big(u_0(t) - u_1(t)\Big) + 3\Big(u_1(t) - u_4(t)\Big) + 0\ u_4(t).
\end{equation*}
Combine like switch terms:
\begin{equation*}
g(t) = 2t\ u_0(t) + (3 - 2t)\ u_1(t) - 3\ u_4(t).
\end{equation*}
Now, we are ready to find the Laplace transform of \(g\text{.}\) Taking the Laplace transform of both sides and using linearity, we get \(3\) transforms:
\begin{equation*}
\lap{g(t)}
= 2\ub{\lap{t\ u_0(t)}}_{\ds\text{1️⃣}}
+ \ub{\lap{(3 - 2t)\ u_1(t)}}_{\ds\text{2️⃣}}
- 3\ub{\lap{u_4(t)}}_{\ds\text{3️⃣}}\text{.}
\end{equation*}
1️⃣ Apply L\(_{10}\) with: \(\quad c=0\quad\) & \(\quad f(t) = t \quad\Rightarrow\quad f(t+0) = t\text{:}\)
\begin{equation*}
\lap{t\cdot u_0(t)} = e^{-0s}\lap{f(t+0)} = \lap{t} = \frac1{s^2}
\end{equation*}
2️⃣ Apply L\(_{10}\) with: \(\quad c=1\quad\) & \(\quad f(t) = 3 - 2t \quad\Rightarrow\quad f(t+1) = 1-2t\text{:}\)
\begin{equation*}
\lap{(3 - 2t)\cdot u_1(t)} = e^{-s}\lap{f(t+1)} = e^{-s}\lap{1-2t} = e^{-s}\left(\frac1{s} - \frac{2}{s^2}\right)
\end{equation*}
Writing these together under one sum completes the Laplace transform:
\begin{equation*}
\lap{g(t)} = \frac2{s^2} + \left(\frac1{s} - \frac{2}{s^2}\right)e^{-s} - \frac{3}{s}e^{-4s}
\end{equation*}
Subsection Examples
These worked examples illustrate how the step-function rules combine to handle any piecewise function you might encounter.
🌌 Example 264. Transforming a 2-Part Piecewise Function.
Find the Laplace transform of
\begin{equation*}
g(t) = \left\{
\begin{array}{ll}
e^{-t}, \amp t \lt 1 \\
3, \amp t \ge 1
\end{array}
\right.
\end{equation*}
Solution. \(\rightarrow \lap{g(t)}\)
First, determine the step function switches
\begin{equation*}
g(t) =
\left\{
\begin{array}{llll}
e^{-t}, & t \lt 1 & \leftarrow & 1-u_{1}(t)\\
3, & t \ge 1 & \leftarrow & u_{1}(t).
\end{array}
\right.
\end{equation*}
Now, contruct \(g(t)\) and combine the \(u_1(t)\) terms:
\begin{align*}
g(t) \amp = e^{-t} \cdot \left(1 - u_{1}(t)\right) + 3 \cdot u_{1}(t) \\
\amp = e^{-t} + \left(-e^{-t} + 3\right) \cdot u_{1}(t)
\end{align*}
Apply the Laplace transform to both terms:
\begin{align*}
\lap{g(t)}
\amp = \us{\large \knowl{./knowl/xref/lt-L2.html}{\text{L\(_{2}\)}}}{\ul{\lap{e^{-t}}}}
+ \us{\large ⭐ \knowl{./knowl/xref/lt-L10.html}{\text{L\(_{10}\)}}}{\ul{\lap{\left(3 - e^{-t}\right) \cdot u_{1}(t)}}}\\
\amp = \frac{1}{s+1} + e^{-s}\lap{3 - e^{-1}e^{-t}}\\
\amp = \frac{1}{s+1} + e^{-s}\left(\lap{3} - e^{-1}\lap{e^{-t}}\right)\\
\amp = \frac{1}{s+1} + e^{-s}\left(\frac3{s} - e^{-1}\frac{1}{s+1}\right)
\end{align*}
\begin{align*}
⭐\quad c \amp = 1\\
f(t) \amp = 3 - e^{-t}\\
f(t+1) \amp = 3 - e^{-(t+1)}\\
\amp = 3 - e^{-1}e^{-t}
\end{align*}
Simplifying completes the transform:
\begin{equation*}
\lap{g(t)}
= \frac{3e^{-s}}{s} + \frac{1 - e^{-s-1}}{s + 1}\text{.}
\end{equation*}
🌌 Example 265. Transforming a 3-Part Piecewise Function.
Find the Laplace transform of
\begin{equation*}
h(t) = \left\{
\begin{array}{ll}
0, \amp t \lt 2 \\
t, \amp 2 \le t \lt 5 \\
4, \amp t \ge 5
\end{array}
\right.
\end{equation*}
Solution. \(\rightarrow \lap{h(t)}\)
Write each piece using step functions:
\begin{align*}
h(t) \amp = t \cdot (u_{2}(t) - u_{5}(t)) + 4 \cdot u_{5}(t)
\end{align*}
Combine like terms:
\begin{align*}
h(t) \amp = t \, u_{2}(t) - t \, u_{5}(t) + 4 \, u_{5}(t) \\
\amp = t \, u_{2}(t) + (4 - t) u_{5}(t)
\end{align*}
Transform each term:
-
\begin{equation*} \lap{t \, u_{2}(t)} = e^{-2s}\lap{t+2} = e^{-2s}\left( \frac{1}{s^2} + \frac{2}{s} \right)\text{.} \end{equation*}
-
\begin{equation*} \lap{(4-t) u_{5}(t)} = e^{-5s}\lap{-t-1} = e^{-5s}\left(-\frac{1}{s^2} - \frac{1}{s}\right)\text{.} \end{equation*}
Combine the pieces:
\begin{equation*}
\lap{h(t)} = e^{-2s}\left( \frac{1}{s^2} + \frac{2}{s} \right) + e^{-5s}\left( -\frac{1}{s^2} - \frac{1}{s} \right)\text{.}
\end{equation*}
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