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Section 13.7 Exercises
Exercises 💡 Conceptual Quiz
📝: Abbreviations.
1.
(a) Unit Step Behavior.
What is the only possible set of values that any unit step function
\(u_c(t)\) can take?
1 only
Unit step functions are OFF before \(t = c\text{,}\) so they can also be 0.
0 and 1
Correct! Unit step functions are always either 0 (OFF) or 1 (ON).
any real number
No, unit step functions are discrete switches, not continuous functions.
any nonnegative number
Even though the values are nonnegative, only 0 and 1 are ever used.
(b) Rewriting an ON Interval.
Which expression represents a function that is ON from
\(t = 1\) to
\(t = 4\) and OFF otherwise?
\(\quad u_1(t) - u_4(t)\)
This expression turns ON at \(t = 1\) and back OFF at \(t = 4\text{.}\)
\(\quad u_1(t)\)
This stays ON after \(t = 1\text{,}\) but never switches OFF at \(t = 4\text{.}\)
\(\quad 1 - u_4(t)\)
This is ON before \(t = 4\text{,}\) but never turns ON at \(t = 1\text{.}\)
\(\quad u_4(t) - u_1(t)\)
That would actually be negative during the interval from \(t = 1\) to \(t = 4\text{,}\) not what we want.
(c) 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 don’t 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 \(u_6(t)\text{.}\)
(d) 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*}
\(\quad 3 \cdot (u_0(t) - u_2(t))\)
This turns 3 ON at \(t = 0\) and OFF again at \(t = 2\text{.}\)
\(\quad 3 \cdot u_2(t)\)
This turns ON at \(t = 2\text{,}\) not \(t = 0\text{.}\)
\(\quad 3 \cdot (1 - u_2(t))\)
This would be ON before \(t = 2\text{,}\) but it wouldn’t stop at \(t = 0\text{,}\) it’s ON too early.
\(\quad 3 \cdot u_0(t)\)
This is ON at \(t = 0\text{,}\) but it stays ON forever, it doesn’t turn OFF at \(t = 2\text{.}\)
(e) Reading a Step Expression.
What interval is the function
\(5 \cdot (1 - u_6(t))\) activated on?
\(\quad t \lt 6\)
\(1 - u_6(t)\) is ON before \(t = 6\) and OFF afterward.
\(\quad t \gt 6\)
That would be true of \(u_6(t)\text{,}\) not its reversal.
\(\quad t = 6\) only
This is a step function, not a spike, it applies to intervals, not points.
\(\quad t \ge 6\)
Again, that’s when \(u_6(t)\) turns ON, not when its complement is active.
(f) Using the Shift Rule.
Suppose
\(\lap{f(t)} = F(s)\text{.}\) What is
\(\lap{f(t)\cdot u_4(t)}\text{?}\)
\(\quad e^{-4s} \cdot \lap{f(t+4)}\)
This is the shift rule in action: delay the function, then shift its input left.
\(\quad e^{4s} \cdot F(s)\)
The exponent should be negative, delaying adds \(e^{-cs}\text{.}\)
\(\quad \lap{f(t - 4)}\)
This shifts the function right but doesn’t match the form used in the shift rule.
\(\quad \lap{f(t)} \cdot u_4(t)\)
The Laplace transform applies to the whole product, you can’t apply it to one piece separately.
(g) Shifting Inside the Transform.
Which of the following is equal to
\(\lap{(t^2)\cdot u_2(t)}\text{?}\)
\(\quad e^{-2s} \cdot \lap{(t + 2)^2}\)
This follows directly from the shift rule, replace \(t\) with \(t + 2\text{.}\)
\(\quad e^{-2s} \cdot \lap{t^2}\)
We must shift the input, \(t^2\) becomes \((t + 2)^2\text{.}\)
\(\quad \lap{t^2} \cdot u_2(t)\)
You can’t break apart the transform like this, it applies to the whole product.
\(\quad e^{-2s} \cdot (t^2 + 4t + 4)\)
That’s the shifted polynomial, but we want the Laplace transform of that expression, not the polynomial itself.
(h) 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 the 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 doesn’t turn OFF at \(t = 1\text{,}\) it switches to a new expression.
(i) Transform of a function that turns off.
Which of the following expressions is equivalent to the Laplace transform of \(t(1 - u_4(t))\text{?}\)
(j) 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 🏋️♂️ Practice Drills
1.
(a) Laplace of a Shifted Step.
What is
\(\lap{u_3(t)}\text{?}\)
\(\quad \dfrac{e^{-3s}}{s}\)
This is the standard formula for the Laplace transform of \(u_c(t)\text{.}\)
\(\quad \dfrac{1}{s + 3}\)
This is the transform of \(e^{-3t}\text{,}\) not a step function.
\(\quad \dfrac{1}{s} - e^{-3s}\)
This isn’t a correct transformation of a step function, it doesn’t match the form.
\(\quad \dfrac{s}{e^{3s}}\)
This inverts the formula incorrectly, look carefully at the units and exponents.
(b) Select the Transform.
\(\lap{(t - 1) u_1(t)} = \)
(c) Select the Transform.
\(\lap{(1 - t)(1 - u_1(t))} = \)
\(\dfrac{1}{s} - \dfrac{1}{s^2} - \dfrac{e^{-s}}{s^2}\)
-
\(\dfrac{1}{s} - \dfrac{1}{s^2} + \dfrac{e^{-s}}{s^2}\)
-
\(\dfrac{e^{-s}}{s^2} - \dfrac{1}{s}\)
-
\(\dfrac{e^{-s}}{s^2}\)
-
(d) Select the Transform.
\(\lap{t\, [u_1(t) - u_3(t)]} = \)
\(e^{-s} \left( \dfrac{1}{s^2} + \dfrac{1}{s} \right) - e^{-3s} \left( \dfrac{1}{s^2} + \dfrac{3}{s} \right)\)
-
\(\dfrac{1}{s^2} \cdot (e^{-s} - e^{-3s})\)
-
\(e^{-s} \cdot \dfrac{1}{s^2}\)
-
\(e^{-3s} \cdot \dfrac{1}{s^2}\)
-
(e) Piecewise to Laplace.
A function is defined as:
\begin{equation*}
f(t) = \left\{
\begin{array}{ll}
0, \amp t \lt 2 \\
4, \amp 2 \le t \lt 5 \\
t - 5, \amp t \ge 5
\end{array}
\right.
\end{equation*}
What is \(f(t)\) in terms of step functions?
\(\quad 4\cdot(u_2(t) - u_5(t)) + (t - 5)\cdot u_5(t)\)
- Perfect. This captures \(4\) ON from \(t = 2\) to \(t = 5\) and \(t - 5\) turning ON at \(t = 5\text{.}\)
\(\quad 4\cdot (1 - u_2(t)) + (t - 5)\cdot u_5(t)\)
Incorrect. This function corresponds to
\begin{equation*}
\left\{
\begin{array}{ll}
4, \amp t \lt 2 \\
0, \amp 2 \le t \lt 5 \\
t - 5, \amp t \ge 5
\end{array}
\right.
\end{equation*}
\(\quad 4\cdot u_2(t) + (t - 5)\cdot u_5(t)\)
Incorrect. This function corresponds to
\begin{equation*}
\left\{
\begin{array}{ll}
0, \amp t \lt 2 \\
4, \amp 2 \le t \lt 5 \\
4 + t - 5, \amp t \ge 5
\end{array}
\right.
\end{equation*}
\(\quad 4\cdot(u_0(t) - u_2(t)) + (t - 5)\cdot u_2(t)\)
Incorrect. This function corresponds to
\begin{equation*}
\left\{
\begin{array}{ll}
0, \amp t \lt 0 \\
4, \amp 0 \le t \lt 2 \\
t - 5, \amp t \ge 2
\end{array}
\right.
\end{equation*}
(f) Shifted Unit-Step Function.
Evaluate the Expression.
Let
\(u_c(t)\) be the shifted unit step function and
\(f(t) = t^2-2 \text{.}\) Then compute the following values
2.
3.
4.
5.
6.
\(u_0(k)\cdot f(k)\text{,}\) where
\(k \ge 0\)
7.
\(u_0(k)\cdot f(k)\text{,}\) where
\(k < 0\)
Sketch & Transform the Step Functions.
Sketch the graph of the given function and determine its Laplace transform.
8.
9.
10.
\(w(t) = u(t-1) - u(t-4)\)
11.
12.
\(x(t) = te^{3t}u(t+\pi/4)\)
13.
\(\lap{e^{3t}\left(1 - u_1(t)\right)}\)
Exercise Group.
Rewrite the step function forms in piecewise form.
14.
\(g(t) = (3 - t^2)\left(1 - u_2(t)\right)\)
15.
\(h(t) = (t^2 - 4)\ u_3(t)\)
Exercise Group.
Express each function below as a sum of unit step functions, then compute its Laplace transform.
16.
\(h(t) = \left\{
\begin{array}{ll}
0, & t \lt 2\\
t - 2, & t \ge 2
\end{array}
\right.\)17.
\(m(t) =
\left\{
\begin{array}{ll}
0, \amp t \lt 1 \\
t^2, \amp 1 \le t \lt 3 \\
0, \amp t \ge 3
\end{array}
\right.\)18.
\(f(t) = \left\{
\begin{array}{ll}
2e^{-\sfrac{t^2}{2}}, & t \lt 1\\
0, & t \ge 1
\end{array} \right.\)19.
\(\ds g(t) =
\left\{
\begin{array}{ll}
0, \amp t \le 1\\
2, \amp 1 \le t \le 2\\
1, \amp 2 \le t \le 3\\
3, \amp t \ge 3\\
\end{array}
\right.\)
Exercises ✍🏻 Problems
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.
\(X(s) = \dfrac{e^{-s}(s-5)}{(s+1)(s+2)}\)
Solving Differential Equations.
Solve each of the following initial-value problems using Laplace Transforms.
7.
\(y'' + 4y = u_3(t),\ \ y(0) = 0,\ \ y'(0) = 0\)
8.
\(z'' + 3z' + 2z = e^{-3t}U(t-2),\ \ z(0) = 2,\ \ z'(0) = -3\)
9.
\(w'' + w = U(t-2) - U(t-4),\ \ w(0) = 1,\ \ w'(0) = 0\)
10.
\(\begin{array}{l}
y'' + 9y = h(t) \\
y(0) = 0,\ y'(0) = 0
\end{array}
\quad\) where \(\quad
h(t) = \left\{
\begin{array}{ll}
1, \amp 2 \le t \lt 3 \\
0, \amp \text{otherwise}.
\end{array}
\right.\)11.
\(\begin{array}{l}
y'' + y = g(t) \\
y(0) = 0,\ y'(0) = 0
\end{array}
\quad\) where \(\quad g(t) =
\left\{
\begin{array}{ll}
2t, \amp 0 \le t \lt 1 \\
3, \amp 1 \le t \lt 4 \\
0, \amp t \ge 4
\end{array}
\right.\)12.
\(\begin{array}{l}
y'' + 2y' = g(t) \\
y(0) = 0,\ y'(0) = 0
\end{array}
\quad \) where \(\quad
g(t) = \left\{
\begin{array}{ll}
3, \amp 0 \le t \lt 1 \\
0, \amp 1 \le t \lt 4 \\
1, \amp t \ge 4 \\
\end{array}
\right.\)
13.
Compute the Laplace transform of \(d(t)\) using the definition:
\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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