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Section Chapter 11 Exercises
Reading Questions α―β
β Quick-Answer Questions
1. True-False.
(a) Key Strategy Summary.
When applying the Laplace transform to a piecewise function, the key strategy is to rewrite each piece using unit step functions and then apply the shift rule.
True.
This strategy allows you to use a single formula for the entire function and apply Laplace transforms systematically.
False.
This strategy allows you to use a single formula for the entire function and apply Laplace transforms systematically.
2. Multiple-Choice.
(a) Choosing the Right Switch.
Suppose a function
\(f(t)\) is ON for
\(t \ge 6\text{.}\) Which expression correctly switches it ON at that point?
This turns OFF at \(t = 6\text{,}\) which is the opposite of what we want.
Thatβs right, \(u_6(t)\) flips ON at \(t = 6\) and stays ON from there.
Close! This is also correct, but we prefer to write it as \(u_6(t)\) for clarity in this context.
This turns ON from \(t = 0\) to \(t = 6\text{,}\) then OFF. Not what we want here.
(b) Unit Step Behavior.
What is the only possible set of values that any unit step function
\(u_c(t)\) can take?
Unit step functions are OFF before \(t = c\text{,}\) so they can also be 0.
Correct! Unit step functions are always either 0 (OFF) or 1 (ON).
No, unit step functions are discrete switches, not continuous functions.
Even though the values are nonnegative, only 0 and 1 are ever used.
(c) Rewriting an ON Interval.
Which expression represents a function that is ON from
\(t = 1\) to
\(t = 4\) and OFF otherwise?
This expression turns ON at \(t = 1\) and back OFF at \(t = 4\text{.}\)
This stays ON after \(t = 1\text{,}\) but never switches OFF at \(t = 4\text{.}\)
This is ON before \(t = 4\text{,}\) but never turns ON at \(t = 1\text{.}\)
That would actually be negative during the interval from \(t = 1\) to \(t = 4\text{,}\) not what we want.
(d) Multiplying by a Step Function.
What is the effect of multiplying a function
\(f(t)\) by
\(u_6(t)\text{?}\)
It shifts
\(f(t)\) 6 units to the left.
Shifting the input would require \(f(t + 6)\text{,}\) not multiplication by \(u_6(t)\text{.}\)
It keeps
\(f(t)\) OFF before
\(t = 6\) and turns it ON at
\(t = 6\text{.}\)
Correct. \(u_6(t)\) acts as a switch that activates \(f(t)\) at \(t = 6\text{.}\)
It subtracts 6 from the output of
\(f(t)\text{.}\)
Step functions donβt affect the output directly, they control when the function is active.
It delays
\(f(t)\) by 6 units.
Delaying would require rewriting the input as \(f(t - 6)\text{,}\) not just multiplying by \(u_6(t)\text{.}\)
(e) Equivalent to Piecewise Form.
Which step-function expression is equivalent to the piecewise function
\begin{equation*}
f(t) = \left\{
\begin{array}{ll}
3, \amp 0 \le t \lt 2 \\
0, \amp \text{otherwise}
\end{array}
\right.
\end{equation*}
\(3 \cdot (u_0(t) - u_2(t))\)
This turns 3 ON at \(t = 0\) and OFF again at \(t = 2\text{.}\)
This turns ON at \(t = 2\text{,}\) not \(t = 0\text{.}\)
This would be ON before \(t = 2\text{,}\) but it wouldnβt stop at \(t = 0\text{,}\) itβs ON too early.
This is ON at \(t = 0\text{,}\) but it stays ON forever, it doesnβt turn OFF at \(t = 2\text{.}\)
(f) Reading a Step Expression.
Which interval does
\(5 \cdot (1 - u_6(t))\) activate over?
\(1 - u_6(t)\) is ON before \(t = 6\) and OFF afterward.
That would be true of \(u_6(t)\text{,}\) not its reversal.
This is a step function, not a spike, it applies to intervals, not points.
Again, thatβs when \(u_6(t)\) turns ON, not when its complement is active.
(g) Laplace of a Shifted Step.
What is
\(\lap{u_3(t)}\text{?}\)
\(\ds \dfrac{e^{-3s}}{s}\)
This is the standard formula for the Laplace transform of \(u_c(t)\text{.}\)
This is the transform of \(e^{-3t}\text{,}\) not a step function.
\(\ds \dfrac{1}{s} - e^{-3s}\)
This isnβt a correct transformation of a step function, it doesnβt match the form.
\(\ds \dfrac{s}{e^{3s}}\)
This inverts the formula incorrectly, look carefully at the units and exponents.
(h) Using the Shift Rule.
Suppose
\(\lap{f(t)} = F(s)\text{.}\) What is
\(\lap{f(t)\cdot u_4(t)}\text{?}\)
\(e^{-4s} \cdot \lap{f(t+4)}\)
This is the shift rule in action: delay the function, then shift its input left.
The exponent should be negative, delaying adds \(e^{-cs}\text{.}\)
This shifts the function right but doesnβt match the form used in the shift rule.
\(\lap{f(t)} \cdot u_4(t)\)
The Laplace transform applies to the whole product, you canβt apply it to one piece separately.
(i) Shifting Inside the Transform.
Which of the following is equal to
\(\lap{(t^2)\cdot u_2(t)}\text{?}\)
\(e^{-2s} \cdot \lap{(t + 2)^2}\)
This follows directly from the shift rule, replace \(t\) with \(t + 2\text{.}\)
\(e^{-2s} \cdot \lap{t^2}\)
We must shift the input, \(t^2\) becomes \((t + 2)^2\text{.}\)
\(\lap{t^2} \cdot u_2(t)\)
You canβt break apart the transform like this, it applies to the whole product.
\(e^{-2s} \cdot (t^2 + 4t + 4)\)
Thatβs the shifted polynomial, but we want the Laplace transform of that expression, not the polynomial itself.
(j) Interpreting a Full Step Form.
The function \(f(t)\) is written as:
\begin{equation*}
f(t) = 2t \cdot u_0(t) + (3 - 2t) \cdot u_1(t) - 3 \cdot u_4(t)
\end{equation*}
What can you infer about the original piecewise function?
It had three intervals:
\(0 \le t \lt 1\text{,}\) \(1 \le t \lt 4\text{,}\) and
\(t \ge 4\text{.}\)
Each change in step function corresponds to a new piece of the function.
It is active only for
\(t \ge 1\text{.}\)
No, \(u_0(t)\) activates the function at \(t = 0\text{.}\)
It is always equal to
\(2t\text{.}\)
Only the first term is \(2t\text{,}\) and it gets overridden by the later terms.
The function turns off completely at
\(t = 1\text{.}\)
It doesnβt turn OFF at \(t = 1\text{,}\) it switches to a new expression.
(k) Transform of a function that turns off.
Which of the following expressions is equivalent to the Laplace transform of \(t(1 - u_4(t))\text{?}\)
\(\lap{t} - \lap{t \cdot u_4(t)}\)
(l) Compute: step function activation.
Compute \(\lap{(t - 1) u_1(t)}\text{.}\)
\(\dfrac{1}{s^2} - \dfrac{e^{-s}}{s^2}\)
(m) Compute: step-off function.
Compute \(\lap{(1 - t)(1 - u_1(t))}\text{.}\)
\(\dfrac{1}{s} - \dfrac{1}{s^2} - \dfrac{e^{-s}}{s^2}\)
\(\dfrac{1}{s} - \dfrac{1}{s^2} + \dfrac{e^{-s}}{s^2}\)
\(\dfrac{e^{-s}}{s^2} - \dfrac{1}{s}\)
(n) Compute: windowed linear function.
What is \(\lap{t\, [u_1(t) - u_3(t)]}\text{?}\)
\(e^{-s} \left( \dfrac{1}{s^2} + \dfrac{1}{s} \right) - e^{-3s} \left( \dfrac{1}{s^2} + \dfrac{3}{s} \right)\)
\(\dfrac{1}{s^2} \cdot (e^{-s} - e^{-3s})\)
\(e^{-s} \cdot \dfrac{1}{s^2}\)
\(e^{-3s} \cdot \dfrac{1}{s^2}\)
(o) Piecewise to Laplace.
3. Other.
(a) Matching Shifts of \(y=x^2\) .
Consider the function
\(y=x^2\text{.}\) Match each of the shifted versions of
\(y=x^2\) with the correct description of how its graph is affected.
\(y=(x - 1)^2\) Horizontal shift by 1 (β)
Vertical shift by -1 (β)
\(y=(x + 1)^2\) Horizontal shift by -1 (β)
\(y=x^2 + 1\) Vertical shift by 1 (β)
\(y=(x - 1)^2 + 1\) Horizontal & vertical shift by 1 (β)
(b) Shifted Unit-Step Function.
(c) Multiplying a Function by \(u_c(t)\) .
Consider the parabola multiplied by two different shifted unit step functions:
\begin{equation*}
\left(\dfrac{1}{5}t^2 - 1\right) u_2(t), \qquad \left(\dfrac{1}{5}t^2 - 1\right) u_{-1}(t)\text{.}
\end{equation*}
Describe the ON/OFF behavior of the parabola in each case.
Exercises ποΈ Unit Step Functions
Exercise Group. Let
\(u_c(t)\) be the shifted unit step function and
\(f(t) = t^2-2 \text{.}\) Then compute the following values
1. \(u_0(2.7)\) 2. \(u_0(-5.32)\) 3. \(u_0(2)\cdot f(2)\) 4. \(u_0(-3)\cdot f(-3)\) 5. \(u_0(k)\cdot f(k)\text{,}\) where \(k \ge 0\) 6. \(u_0(k)\cdot f(k)\text{,}\) where \(k < 0\)
Sketch & Transform the Step Functions. Sketch the graph of the given function and determine its Laplacetransform.
7. \(p(t) = -2\ U(t-2)\) 8. \(w(t) = u(t - 3)\) 9. \(w(t) = u(t-1) - u(t-4)\) 10. \(y(t) = (t-1)^2u(t-3)\) 11. \(x(t) = te^{3t}u(t+\pi/4)\) 12. \(\lap{e^{3t}\left(1 - u_1(t)\right)}\)
Exercises ποΈ Piecewise Functions
Exercise Group. Rewrite the step function forms in piecewise form.
1. \(g(t) = (3 - t^2)\left(1 - u_2(t)\right)\) 2. \(h(t) = (t^2 - 4)\ u_3(t)\)
Exercise Group. Express each function below in terms of unit step functions and then compute its Laplace transform.
3. \(h(t) = \left\{
\begin{array}{ll}
0, & t \lt 2\\
t - 2, & t \ge 2
\end{array}
\right.\) 4. \(m(t) =
\left\{
\begin{array}{ll}
0, \amp t \lt 1 \\
t^2, \amp 1 \le t \lt 3 \\
0, \amp t \ge 3
\end{array}
\right.\) 5. \(f(t) = \left\{
\begin{array}{ll}
2e^{-\sfrac{t^2}{2}}, & t \lt 1\\
0, & t \ge 1
\end{array} \right.\) 6. \(\ds g(t) =
\left\{
\begin{array}{ll}
0, \amp t \le 1\\
2, \amp 1 \le t \le 2\\
1, \amp 2 \le t \le 3\\
3, \amp t \ge 3\\
\end{array}
\right.\)
Exercises Piecewise Inverse Transforms
Exercise Group. Find the inverse Laplace transform of each function.
1. \(Y(s) = e^{-3s} \cdot \dfrac{2}{s^2 + 9}\) 2. \(F(s) = e^{-2s} \cdot \dfrac{s}{s^2 + 4}\) 3. \(Y(s) = \dfrac{e^{-2s}}{s-1}\) 4. \(H(s) = \dfrac{e^{-2s} - 3e^{-4s}}{s+2}\) 5. \(M(s) = \dfrac{e^{-3s}}{s^2+9}\) 6. \(X(s) = \dfrac{e^{-s}(s-5)}{(s+1)(s+2)}\)
Exercises Solving Differential Equations
Solve each of the following initial-value problems using Laplace Transforms.
1. \(y'' + 4y = u_3(t), \quad y(0) = 0,\quad y'(0) = 0\) 2. \(z'' + 3z' + 2z = e^{-3t}U(t-2), \hspace{0.5cm}z(0) = 2,\hspace{0.5cm}z'(0) = -3\) 3. \(w'' + w = U(t-2) - U(t-4), \hspace{0.5cm}w(0) = 1,\hspace{0.5cm}w'(0) = 0\) 4. \(\begin{array}{l}
y'' + 9y = h(t) \\
y(0) = 0,\ y'(0) = 0
\end{array}
\quad\) where \(\quad
h(t) = \left\{
\begin{array}{ll}
1, \amp 2 \le t \lt 3 \\
0, \amp \text{otherwise}.
\end{array}
\right.\) 5. \(\begin{array}{l}
y'' + y = g(t) \\
y(0) = 0,\ y'(0) = 0
\end{array}
\quad\) where \(\quad g(t) =
\left\{
\begin{array}{ll}
2t, \amp 0 \le t \lt 1 \\
3, \amp 1 \le t \lt 4 \\
0, \amp t \ge 4
\end{array}
\right.\) 6. \(\begin{array}{l}
y'' + 2y' = g(t) \\
y(0) = 0,\ y'(0) = 0
\end{array}
\quad \) where \(\quad
g(t) = \left\{
\begin{array}{ll}
3, \amp 0 \le t \lt 1 \\
0, \amp 1 \le t \lt 4 \\
1, \amp t \ge 4 \\
\end{array}
\right.\)
Exercises Using the Definition of the Laplace Transform
1. \(d(t)\text{:}\)
\begin{equation*}
d(t) =
\left\{
\begin{array}{ll}
0, \amp t \lt 0\\
7, \amp 0 \le t \lt 3\\
0, \amp t \ge 3\\
\end{array}
\right.
\end{equation*}
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