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RolyPoly

  • 3 years ago

Convolution \[sin\omega t * cos \omega t\]

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  1. RolyPoly
    • 3 years ago
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    \[sin\omega t * cos \omega t\]\[=\int_0^tsin\omega u \ cos \omega (t-u)du\]\[=-\frac{1}{\omega}\int_0^t\ cos \omega (t-u)d(cos\omega u)\]\[=\frac{1}{2\omega}[ cos ^2\omega (t-u)]_0^t\] Doesn't seem right

  2. experimentX
    • 3 years ago
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    \[ =\int_0^t \sin\omega u \; \cos \omega (t-u)du \\ =\int_0^t \sin\omega u \; (\cos \omega t \cos \omega u - \sin \omega t \sin \omega u)du \\ = \int_0^t \sin\omega u \; \cos \omega t \;\cos \omega u \;du - \int_0^t \sin \omega u \;\sin \omega t \; \sin \omega u \;du \]

  3. experimentX
    • 3 years ago
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    \[ =\int_0^t \sin\omega u \; \cos \omega (t-u)du \\ =\int_0^t \sin\omega u \; (\cos \omega t \cos \omega u - \sin \omega t \sin \omega u)du \\ = \cos \omega t \; \int_0^t \sin\omega u \;\cos \omega u \;du - \sin \omega t \;\int_0^t \sin \omega u \; \sin \omega u \;du \]

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