Laplace Transform of Sine/Proof 4
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Theorem
Let $\sin$ denote the real sine function.
Let $\laptrans f$ denote the Laplace transform of a real function $f$.
Then:
- $\laptrans {\sin at} = \dfrac a {s^2 + a^2}$
where $a \in \R_{>0}$ is constant, and $\map \Re s > 0$.
Proof
By definition of the Laplace transform:
- $\ds \laptrans {\sin a t} = \int_0^{\to +\infty} e^{-s t} \sin a t \rd t$
From Integration by Parts:
- $\ds \int f g' \rd t = f g - \int f' g \rd t$
Here:
\(\ds f\) | \(=\) | \(\ds \sin a t\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds f'\) | \(=\) | \(\ds a \cos a t\) | Derivative of $\sin a x$ | ||||||||||
\(\ds g'\) | \(=\) | \(\ds e^{-s t}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds g\) | \(=\) | \(\ds -\frac 1 s e^{-s t}\) | Primitive of $e^{a x}$ |
So:
\(\text {(1)}: \quad\) | \(\ds \int e^{-s t} \sin a t \rd t\) | \(=\) | \(\ds -\frac 1 s e^{-s t} \sin a t + \frac a s \int e^{-s t} \cos a t \rd t\) |
Consider:
- $\ds \int e^{-s t} \cos a t \rd t$
Again, using Integration by Parts:
- $\ds \int h j\,' \rd t = h j - \int h' j \rd t$
Here:
\(\ds h\) | \(=\) | \(\ds \cos a t\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds h'\) | \(=\) | \(\ds -a \sin a t\) | Derivative of Cosine Function | ||||||||||
\(\ds j\,'\) | \(=\) | \(\ds e^{-s t}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds j\) | \(=\) | \(\ds -\frac 1 s e^{-s t}\) | Primitive of Exponential Function |
So:
\(\ds \int e^{-s t} \cos a t \rd t\) | \(=\) | \(\ds -\frac 1 s e^{-st} \cos a t - \frac a s \int e^{-s t} \sin a t \rd t\) |
Substituting this into $(1)$:
\(\ds \int e^{-s t} \sin a t \rd t\) | \(=\) | \(\ds -\frac 1 s e^{-s t} \sin a t + \frac a s \paren {-\frac 1 s e^{-s t} \cos a t - \frac a s \int e^{-s t} \sin a t \rd t}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\frac 1 s e^{-s t} \sin a t - \frac a {s^2} e^{-s t} \cos a t - \frac {a^2} {s^2} \int e^{-s t} \sin a t \rd t\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {1 + \frac {a^2} {s^2} } \int e^{-s t} \sin a t \rd t\) | \(=\) | \(\ds -e^{-s t} \paren {\frac 1 s \sin a t + \frac a {s^2} \cos a t}\) |
Evaluating at $t = 0$ and $t \to +\infty$:
\(\ds \paren {1 + \frac {a^2} {s^2} } \laptrans {\sin a t}\) | \(=\) | \(\ds \intlimits {-e^{-s t} \paren {\frac 1 s \sin a t + \frac a {s^2} \cos a t} } {t \mathop = 0} {t \mathop \to +\infty}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0 - \paren {-1 \paren {\frac 1 s \times 0 + \frac a {s^2} \times 1} }\) | Boundedness of Real Sine and Cosine, Complex Exponential Tends to Zero | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac a {s^2}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \laptrans {\sin a t}\) | \(=\) | \(\ds \frac a {s^2} \paren {1 + \frac {a^2} {s^2} }^{-1}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac a {s^2} \paren {\frac {s^2} {a^2 + s^2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac a {s^2 + a^2}\) |
$\blacksquare$
Sources
- For a video presentation of the contents of this page, visit the Khan Academy.