Subsection 9.3.2 Multiplication by \(e^{at}\)
The translation property, also known as the first shifting theorem, allows us to handle functions multiplied by an exponential term, \(e^{at}\text{.}\) This property is particularly useful for simplifying the Laplace transforms of products of exponential functions and other functions, such as sine, cosine, or polynomials.
Example 9.
\(\ \ \)\(\lap{e^{7t}\cos(3t)} \text{.}\)Solution. Solution
By the definition of the Laplace transform, we have:
\begin{align*}
\lap{e^{7t}\cos(3t)}
=\amp \int_0^{\infty} e^{-st} e^{7t}\cos(3t)\ dt\\
=\amp \int_0^{\infty} e^{-st+7t} \cos(3t)\ dt\\
=\amp \int_0^{\infty} e^{-(s-7)t} \cos(3t)\ dt\\
=\amp \ub{\int_0^{\infty} e^{-(s_0)t} \cos(3t)\ dt}_{\lap{\cos(3t)}}, \quad \text{where } s_0 = s-7\\
=\amp\ \frac{s_0}{s_0^2 + 9} \qquad (s_0 \gt 0)\\
=\amp\ \frac{s-7}{(s-7)^2 + 9} \qquad (s-7 \gt 0 \text{ or } s \gt 7)
\end{align*}
Thus, the Laplace transform of \(e^{7t}\cos(3t)\) is:
\begin{equation*}
\ds \lap{e^{7t}\cos(3t)} = \frac{s-7}{(s-7)^2 + 9}, \quad s \gt 7.
\end{equation*}
The translation property can be generalized for any function \(f(t)\) multiplied by \(e^{at}\text{.}\) The property is formally stated as:
\begin{equation*}
\lap{ e^{at} f(t) } = F(s-a),
\end{equation*}
where \(F(s)\) is the Laplace transform of \(f(t)\text{.}\)
Laplace Transform Property \(P_2\).
Let \(F(s) = \lap{f(t)}\text{.}\)
- \(P_2\)
- \(\ds \lap{ e^{at} f(t) } = F(s-a), \quad a \) is a constant.
By applying this property to the functions \(t^n\text{,}\) \(\cos(bt)\text{,}\) and \(\sin(bt)\text{,}\) we can derive additional common Laplace transforms:
Common Laplace Transforms \(L_6-L_8\).
- \({\LARGE \vphantom{\int}}L_6\)
- \(\lap{ t^n e^{at} } = \ds\frac{n!}{(s-a)^{n+1}}, \quad s >a \)
- \({\LARGE \vphantom{\int}}L_7\)
- \(\ds \lap{ e^{at}\sin(bt) } = \frac{b}{(s-a)^2+b^2}, \quad s >a \)
- \({\LARGE \vphantom{\int}}L_8\)
- \(\ds \lap{ e^{at}\cos(bt) } = \frac{s-a}{(s-a)^2+b^2}, \quad s >a\)
Reading Questions Check-Point Questions
1. \(\quad \lap{e^{5t} \sin(2t)} = \) .
\(\quad \lap{e^{5t} \sin(2t)} = \)
- \(\ds\frac{2}{(s-5)^2+4}\)
Correct! The translation property shifts the transform of \(\sin(2t)\) by 5 units, giving \(2/(s-5)^2+4\)
- \(\ds\frac{5}{(s-2)^2+4}\)
No, the translation property involves shifting the transform of \(\sin(2t)\) by 5, not by 2.
- \(\ds\frac{2}{(s+5)^2+4}\)
No, the correct shift should be \(s-5\text{,}\) not \(s+5\text{.}\)
- \(\ds\frac{5}{(s-2)^2+2}\)
No, the denominator should have \(2^4 b^2=4\text{,}\) not 2.
2. \(\quad \lap{e^{2t} t^3} = \) .
\(\quad \lap{e^{2t} t^3} = \)
- \(6/(s-2)^4\)
Correct! The translation property applied to \(t^3\) gives \(6/(s-2)^4\)
- \(6/(s+2)^4\)
No, the correct shift should be \(s-2\text{,}\) not \(s+2\text{.}\)
- \(3/(s-2)^3\)
No, the denominator should have \(3^4=6\text{,}\) not 3.
- \(3/(s+2)^3\)
No, the correct shift should be \(s-2\) and the power should be 4.
3. \(\quad \lap{e^{4t} \cos(5t)} = \) .
\(\quad \lap{e^{4t} \cos(5t)} = \)
- \(4/(s-4)^2+25\)
Correct! The Laplace transform of \(4 \cos 5 e^{4t} \cos(5t)\) is \(4/(s-4)^2+25\)
- \(4/(s+4)^2+25\)
No, the correct shift should be \(s-4\text{,}\) not \(s+4\text{.}\)
- \(5/(s-5)^2+16\)
No, the shift should be \(s-4\) and the denominator should have \(25=5^2\)
- \(4/(s-4)^2+16\)
No, the correct denominator should be \(4^2 25/(s-4)^2+25\text{,}\) not \(16^2\)
4. The translation property only works for exponential functions multiplied by sine and cosine functions.
The translation property only works for exponential functions multiplied by sine and cosine functions
True.
False. The translation property applies to any function \(f(t)\) multiplied by an exponential term \(e^{at}\)
False.
False. The translation property applies to any function \(f(t)\) multiplied by an exponential term \(e^{at}\)
5. If \(\ds\lap{f(t)}=\frac{1}{s(s+1)}\text{,}\) what is the Laplace transform of \(e^{2t}f(t)\text{?}\)
If \(\ds\lap{f(t)}=\frac{1}{s(s+1)}\text{,}\) what is the Laplace transform of \(e^{2t}f(t)\text{?}\)
- \(\ds\frac{1}{(s+2)(s+1)}\)
No, the correct shift should be \(s-2\text{,}\) not \(s+2\text{.}\)
- \(\ds\frac{1}{s(s+2)}\)
No, the correct shift should be \(s-2\text{,}\) not \(s\text{.}\)
- \(\ds\frac{1}{(s-2)(s+2)}\)
No, the correct shift should be \(s-2\text{,}\) not \(s+2\text{.}\)
- \(\ds\frac{1}{(s-2)(s+1)}\)
Correct! The translation property shifts the transform of \(f(t)\) by 2 units, giving \(1/(s-2)(s+1)\)
