Subsection 10.1.2 Mismatched Forms
When working with Laplace transforms, it’s common to encounter algebraic expressions that don’t perfectly match the standard forms we use for the inverse Laplace transform. In these cases, we need to manipulate the expression to make it fit one of the known forms.
For example, consider the expression:
\begin{equation*}
\frac{1}{s^2 + 25} \qquad \leftarrow \ \text{looks like} \quad \frac{b}{s^2 + b^2}.
\end{equation*}
While it’s not an exact match, we can observe from the denominator that \(b = 5.\)
However, the numerator doesn’t match yet. To correct this, we need a 5 in the numerator. We can’t just multiply the numerator by 5 without changing the value, so instead, we multiply the entire expression by \(\ds\frac{5}{5}\text{,}\) which is equivalent to multiplying by 1:
\begin{align*}
\frac{1}{s^2 + 25} =\amp\ \frac{1}{s^2 + 5^2} \\
=\amp\ \frac{5}{5} \cdot \frac{1}{s^2 + 5^2} \\
=\amp\ \frac{1}{5} \cdot \frac{5}{s^2 + 5^2} \quad \us{\text{match}}{\os{\text{perfect}}{\leftarrow}}\ \frac{b}{s^2 + b^2}
\end{align*}
Multiplying by \(\ds\frac{5}{5}\) doesn’t change the value of the expression, but it allows us to match the standard form.
One of the most common adjustments is introducing a missing constant in the numerator, as we did earlier. By multiplying both the numerator and denominator by this constant, we maintain the expression’s value while transforming it into a recognizable form. Let’s look at some examples:
Example 19.
Compute the inverse Laplace transforms for each.
| \(\ds \frac{12}{s}\) |
\(\hspace{27em}\) Solution. Solution
The \(s\) in the denominator tells us that we need \(L_1\text{.}\) Before we do, let’s factor out \(12\text{:}\)
\begin{equation*}
\ds \ilap{\frac{12}{s}} = 12 \cdot \ilap{\frac{1}{s}} = 12 \cdot 1 = 12\text{.}
\end{equation*}
|
| \(\ds \frac{5}{s^2 + 4}\) |
\(\hspace{27em}\) Solution. Solution
Since the denominator has the form, \(s^2 + \text{number}\text{,}\) and there is no \(s\) in the numerator, we should apply \(L_4\text{.}\) As before, it is helpful to first factor out the constant \(5\text{,}\)
\begin{equation*}
\ds \ilap{\frac{5}{s^2 + 4}} = 5\ \ilap{\frac{1}{s^2 + 4}} \quad \leftarrow \knowl{./knowl/xref/L4.html}{L_4} (b=2)\text{.}
\end{equation*}
According to \(L_4\text{,}\) we are missing \(2\) in the numerator. Let’s put it there by multiplying by \(2/2\text{,}\) like so
\begin{align*}
= 5\ \ilap{\frac{2}{2}\frac{1}{s^2 + 4}} =\amp\ 5\ \ilap{\frac{1}{2}\frac{2}{s^2 + 4}}\\
=\amp\ \frac{5}{2} \ilap{\frac{2}{s^2 + 4}}\\
=\amp\ \frac{5}{2} \sin(2t)
\end{align*}
|
| \(\ds \frac{17}{s^4}\) |
\(\hspace{27em}\) Solution. Solution
This denominator has the form \(s^{\text{power}}\text{,}\) which matches \(L_3\) with \(n = 3\text{.}\)
\begin{equation*}
\ds \ilap{\frac{17}{s^4}} = 17\ \ilap{\frac{1}{s^4}} \quad \leftarrow \knowl{./knowl/xref/L3.html}{L_3} (n=3)\text{.}
\end{equation*}
In this case, the numerator is missing a \(3!\text{.}\) We can introduce it by multiplying by \(3!/3!\text{,}\) like so
\begin{gather*}
= 17\ \ilap{\frac{3!}{3!}\frac{1}{s^4}} = \frac{17}{3!} \ilap{\frac{3!}{s^4}} = \frac{17}{6} t^3
\end{gather*}
|
| \(\ds \frac{7s}{s^2 + 25}\) |
\(\hspace{27em}\) Solution. Solution
\begin{equation*}
\knowl{./knowl/xref/L5.html}{L_5}, b=5
\end{equation*}
\begin{equation*}
\ilap{\frac{7s}{s^2 + 25}} = 7\ \ilap{\frac{s}{s^2 + 25}} = 7\ \cos(5t)
\end{equation*}
|
| \(\ds \frac{10}{(s - 3)^2 + 11}\) |
\(\hspace{27em}\) Solution. Solution
The form of this denominator is \(\ul{(s \pm \text{shift})^{2} + \text{number}}\) and has no \(s\) in the numerator. Therefore,
\begin{align*}
\ilap{\frac{10}{(s - 3)^2 + 11}} \amp\ \quad \knowl{./knowl/xref/L7.html}{\text{\(L_7\)}} (a=3,\ b=\sqrt{11})\\
=\amp\ 10\ \ilap{\frac{\sqrt{11}}{\sqrt{11}}\frac{1}{(s - 3)^2 + 11}}\\
=\amp\ \frac{10}{\sqrt{11}} \ilap{\frac{\sqrt{11}}{(s - 3)^2 + 11}}\\
=\amp\ \frac{10}{\sqrt{11}} e^{3t} \sin(\sqrt{11}t)
\end{align*}
|
| \(\ds \frac{2}{(s+7)^5}\) |
\(\hspace{27em}\) Solution. Solution
The denominator, \(\ul{(s \pm \text{shift})^{\text{power}}}\text{,}\) indicates an \(e^{at}\) and a \(t^n\) term. Therefore,
\begin{align*}
\ilap{\frac{2}{(s+7)^5}} \amp\ \quad \knowl{./knowl/xref/L6.html}{\text{\(L_6\)}} (a=-7, n=4)\\
=\amp\ 2\ \ilap{\frac{4!}{4!}\frac{1}{(s+7)^5}}\\
=\amp\ \frac{2}{4!} \ilap{\frac{4!}{(s+7)^5}}\\
=\amp\ \frac{1}{12} t^4 e^{-7t}
\end{align*}
|
Reading Questions Check-Point Questions
1. The most challenging part of the Laplace transform method is preparing the \(s\)-function for the backward Laplace transform.
The most challenging part of the Laplace transform method is preparing the \(s\)-function for the backward Laplace transform
True.
True. Preparing the s-function to match a known form in the backward Laplace transform table can be the most challenging step in the Laplace transform method.
False.
True. Preparing the s-function to match a known form in the backward Laplace transform table can be the most challenging step in the Laplace transform method.
2. \(\ds\quad \ilap{\frac{7}{s^2}} = \) .
3. Fill-in-the-blank. \(\ds\quad \ilap{\frac{5}{s^4}} = \frac{5}{6}\ilap{\frac{\fillinmath{X}}{s^4}} \).
Fill-in-the-blank. \(\ds\quad \ilap{\frac{5}{s^4}} = \frac{5}{6}\ilap{\frac{\fillinmath{X}}{s^4}} \)
- \(9\)
Incorrect.
- \(3\)
Incorrect.
- \(4!\)
Incorrect.
- \(3!\)
Correct!
4. Which of the following \(s\)-functions require adjustment to match one of the common inverse Laplace transforms?
Which of the following \(s\)-functions require adjustment to match one of the common inverse Laplace transforms?
- \(\ds\frac{s}{s^2 + 16}\)
Incorrect. This function already matches a known cosine form and does not require any missing constants.
- \(\ds\frac{8}{s^2 + 16}\)
Correct! This function would require a missing constant adjustment to match the sine form, with the numerator needing to be \(4\text{.}\)
- \(\ds\frac{s - 3}{(s - 3)^2 + 25}\)
Incorrect. This function already matches a known form and does not require any missing constants.
5. \(\ds\quad \ilap{\frac{10}{(s - 2)^2 + 25}} = \) .
\(\ds\quad \ilap{\frac{10}{(s - 2)^2 + 25}} = \)
- \(2e^{2t}\sin(5t)\)
Correct! Factoring out \(10\) and placing the missing constant gives the correct form: \(2e^{2t}\sin(5t)\text{.}\)
- \(5e^{2t}\sin(5t)\)
Incorrect. The correct answer requires factoring and rebalancing, giving \(2e^{2t}\sin(5t)\text{.}\)
- \(e^{2t}\cos(5t)\)
Incorrect. The sine form, not cosine, matches this function.
6. Fill-in-the-blank. \(\ds\quad \ilap{\frac{10}{s^2 + 25}} = \frac{\fillinmath{X}}{5}\ilap{\frac{5}{s^2 + 25}} \).
Fill-in-the-blank. \(\ds\quad \ilap{\frac{10}{s^2 + 25}} = \frac{\fillinmath{X}}{5}\ilap{\frac{5}{s^2 + 25}} \)
- \(10\)
Correct!
- \(5\)
Incorrect.
- \(2\)
Incorrect.
- \(1\)
Incorrect.
