Zeitreihenanalyse - Aufgabe 6

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Exercise

Show that g(t) = \mu + R \sin(\omega t + \phi)\, is periodic with period p = \frac{2 \pi}{\omega}\,


\begin{align}
g(t + k p) 
& = g \left(t + k \frac{2 \pi}{\omega} \right) \\
& = \mu + R \sin \left(\omega \left(t + k \frac{2 \pi}{\omega} \right) + \phi \right) \\
& = \mu + R \sin(\omega t + 2 \pi k + \phi) \\
& = \mu + R \sin(\omega t + \phi) \\
& = g(t)
\end{align}
\,

Exercise

Determine the value of \omega \, such that g(t) = \mu + R \sin(\omega t + \phi)\, is periodic with p=12\,


\begin{align}
g(t + 12 k) 
& = g \left(t + 12 k \right) \\
& = \mu + R \sin \left(\omega (t + 12 k) + \phi \right) \\
& = \mu + R \sin(\omega t + 12 \pi k \omega + \phi) \\
\end{align}
\,

Wähle \omega = \frac{2 \pi}{12}\,


\begin{align}
g(t + 12 k)
& = \mu + R \sin \left(\frac{2 \pi}{12} t + 12 \pi k \frac{2 \pi}{12} + \phi \right) \\
& = \mu + R \sin \left(\frac{2 \pi}{12} t + 2 \pi k + \phi \right) \\
& = \mu + R \sin \left(\frac{2 \pi}{12} t + \phi \right) \\
& = g(t)
\end{align}
\,

Exercise