π 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 Transforming Piecewise Functions
Until now, youβve learned how step functions act like ONβOFF switches, and youβve 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, weβre 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.
Thatβs it. Once each term is transformed, you can add them up thanks to linearity. Youβll be left with a fully transformed piecewise function living in the Laplace domain.
Checkpoint 270. πβ Strategy for Transforming a Piecewise Function.
(a) Whatβs Should you do First?
True.
- Thatβs the foundation of this method! Once a piecewise function is rewritten using step functions, you can apply the shift rule to each piece.
False.
- Thatβs 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
Letβs 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)\text{.}
\end{equation*}
Combine like switch terms:
\begin{equation*}
g(t) = 2t\ u_0(t) + (3 - 2t)\ u_1(t) - 3\ u_4(t)\text{.}
\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{1οΈβ£}}
+ \ub{\lap{(3 - 2t)\ u_1(t)}}_{\ds\text{2οΈβ£}}
- 3\ub{\lap{u_4(t)}}_{\ds\text{3οΈβ£}}\text{.}
\end{equation*}
1οΈβ£ Apply L\(_{10}\) with: \(\quad c=0\quad\) & \(\quad f(t) = t \quad\Rightarrow\quad f(t+0) = t\text{:}\)
\begin{align*}
\lap{t\cdot u_0(t)} \amp = e^{-0s}\lap{f(t+0)} = \lap{t} = \frac1{s^2}
\end{align*}
2οΈβ£ Apply L\(_{10}\) with: \(\quad c=1\quad\) & \(\quad f(t) = 3 - 2t \quad\Rightarrow\quad f(t+1) = 1-2t\text{:}\)
\begin{align*}
\lap{(3 - 2t)\cdot u_1(t)} \amp = e^{-s}\lap{f(t+1)} = e^{-s}\lap{1-2t} = e^{-s}\left(\frac1{s} - \frac{2}{s^2}\right)
\end{align*}
Writing these together under one sum completes the Laplace transform:
\begin{align*}
\lap{g(t)} \amp = \frac2{s^2} + \left(\frac1{s} - \frac{2}{s^2}\right)e^{-s} - \frac{3}{s}e^{-4s}
\end{align*}
Subsection Examples
These worked examples illustrate how the three step-function Laplace rules combine into one powerful method for handling any piecewise function you might encounter.
π Example 271. 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*}
\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 β \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*}
β\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*}
π Example 272. 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:
-
Term 1: Apply L10 with \(c=2\) and \(f(t)=t\text{.}\) Then \(f(t+2)=t+2\text{,}\) so:\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*}
-
Term 2: Apply L10 with \(c=5\) and \(f(t)=4-t\text{.}\) Then \(f(t+5)=4-(t+5)=-t-1\text{,}\) so:\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*}
Subsection π€ Wrap-Up
ποΈ \(\textbf{Key Takeaways...}\)
-
Rewrite piecewise functions in step form before doing anything else β this unifies the cases into one formula.
-
Combine like step terms to minimize how many Laplace transforms you have to compute.
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