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opdracht2 - 0 5cos 2π50t Fourriercoefficient van A0 = = =...

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Opdracht 2 Fourrier Analyse Vraag 1: Vraag 2 en 3: zie labview bestand Vraag 4: Enkelfasig gelijkgericht signaal is de som van de dubbelfasig gelijkgericht signaal en het oorspronkelijke sinusvormig signaal, beide met de halve amplitude. Dit heb ik geprobeerd te maken in labview maar het is niet gelukt om de dubbelfasig geljkgericht signaal te maken. Daarom heb ik moeten improviseren met het afkappen van de oorspronkelijke sinus. Maar ik kan wel de waarde uitrekenen van de eerste 10 harmonische en de dc-term. Het oorspronkelijke signaal met de halve amplitude is
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Unformatted text preview: . ( ) 0 5cos 2π50t . Fourriercoefficient van A0 = = * . = . A0 2A π 2 0 5π 0 32 Volt Fourriercoefficient van An = *- = - .- An 4Aπ 14 n2 1 0 644 n2 1 = - . A1 0 21 = - . A2 0 042 = - . A3 0 018 = - . A4 0 010 = - . A5 0 0064 = - . A6 0 0045 = - . A7 0 0033 = - . A8 0 0025 = - . A9 0 0020 =- . A10 0 0016 Dit komt neer dat de reeks: ( )= . + .- .- .- .-V t 0 32 0 5cos2π50t 0 21cos2π100t 0 042cos2π200t 0 018cos2π300t .- .- .- .-0 010cos2π400t 0 0064cos2π500t 0 0045cos2π600t 0 0033cos2π700t .- .- . 0 0025cos2π800t 0 0020cos2π900t 0 0016cos2π1000t...
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