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", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Finn løsningen x(t) til følgende likninger. Du har lov til å bruke en tabell. Skriv inn bare løsningen uten x(t)=... .

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See 'show steps'.

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$x'+(\\var{c})x=(\\var{d}) e^{\\var{b}t}$ gitt at $x(0)=\\var{a}$.

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$L\\{t\\}=\\frac{1}{s^2}$

$L\\{e^{at}\\}=\\frac{1}{s-a}$

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$x''+(\\var{cb})x'+(\\var{db})x = e^{-t}$ gitt at $x(0)=0$ og $x'(0)=0$.

Laplacetransformer diff-likningen og skriv et utrykk for $X(s)$ som en brøk med 1 i telleren. Skriv bare brøken uten $X(s)$. Du skal delbrøkoppspalte svaret i neste oppgavedel.

$L\\{\\sin(at)\\}=\\frac{a}{s^2+a^2}$

$L\\{e^{at}\\}=\\frac{1}{s-a}$

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$x''+(\\var{cb})x'+(\\var{db})x = e^{-t}$ gitt at $x(0)=0$ og $x'(0)=0$.

Delbrøkoppspalt brøken fra deloppgave b).

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$L\\{\\sin(at)\\}=\\frac{a}{s^2+a^2}$

$L\\{e^{at}\\}=\\frac{1}{s-a}$

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$x''+(\\var{cb})x'+(\\var{db})x = e^{-t}$ gitt at $x(0)=0$ og $x'(0)=0$.

Ta $L^{-1}$ av det delbrøkoppspaltete utrykket og skriv løsningen av difflikningen uten $x(t)=$.

$L\\{\\sin(at)\\}=\\frac{a}{s^2+a^2}$

$L\\{e^{at}\\}=\\frac{1}{s-a}$

"}], "answer": "{ac}*e^(-t)+{bc}*e^({a}t)+{cc}*e^({b}t)", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": "0.001", "failureRate": 1, "vsetRangePoints": "5", "vsetRange": ["0", "1"], "checkVariableNames": false, "maxlength": {"length": "0", "partialCredit": "0", "message": ""}, "minlength": {"length": "0", "partialCredit": "0", "message": ""}, "valuegenerators": [{"name": "t", "value": ""}]}]}]}, {"name": "Fourier", "pickingStrategy": "all-ordered", "pickQuestions": "1", "questions": [{"name": "Norsk Trigonometric form of a Fourier series Evaluate ak & bk", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}, {"name": "Mathias Sandulescu", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3503/"}], "tags": [], "metadata": {"description": "

Calculating particular harmonic components of a Fourier series expansion.

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En funsksjon $f(t)$ er gitt ved:

\$f(t)=\\left[ \\begin{array}{cc}\\,\\,\\var{a}&\\,\\,-\\var{L}<t<-\\simplify{{L}/2}\\\\\\,\\,\\var{b}&\\,\\,-\\simplify{{L}/2}<t<\\simplify{{L}/2}\\\\\\,\\,\\var{c}&\\,\\,\\simplify{{L}/2}<t<\\var{L}\\end{array}\\right] \\,\\,\\,\\, \\text{der} f(t+\\simplify{2*{L}})=f(t)\$

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We derived the formulae for a trigonometric Fourier series of a periodic function having period \$2L\$

\$2L=\\simplify{2*{L}}\\implies L=\\var{L}\$

We can either evaluate \$a_0=\\frac{1}{L}\\int_{-L}^{L}f(t)dt\$ or use the shortcut:

\$\\frac{a_0}{2}=\$ the average value of the wave over one complete cycle \$=\\frac{Area}{Base}\$

\$\\frac{a_0}{2}=\\frac{\\var{a}*\\simplify{{L}/2}+\\var{b}*\\var{L}+\\var{c}*\\simplify{{L}/2}}{\\simplify{2*{L}}}=\\simplify{({a}+2*{b}+{c})/4}\$

The formula for the Fourier coefficient \$a_k\$ is given by: \$a_k=\\frac{1}{L}\\int_{-L}^{L}f(t)cos\\left(\\frac{{k}\\pi}{L}t\\right)dt\$

\$a_k=\\frac{1}{\\var{L}}\\left(\\int_{-\\var{L}}^{-\\frac{\\var{L}}{2}}\\var{a}cos\\left(\\frac{{k}\\pi}{\\var{L}}t\\right)dt+\\int_{-\\frac{\\var{L}}{2}}^{\\frac{\\var{L}}{2}}\\var{b}cos\\left(\\frac{{k}\\pi}{\\var{L}}t\\right)dt+\\int_{\\frac{\\var{L}}{2}}^{\\var{L}}\\var{c}cos\\left(\\frac{{k}\\pi}{\\var{L}}t\\right)dt\\right)\$

\$a_k=\\frac{1}{\\var{L}}\\left(\\frac{\\var{a}*\\var{L}}{{k}\\pi}sin\\left(\\frac{{k}\\pi}{\\var{L}}t\\right)|_{-\\var{L}}^{-\\frac{\\var{L}}{2}}+\\frac{\\var{b}*\\var{L}}{{k}\\pi}sin\\left(\\frac{{k}\\pi}{\\var{L}}t\\right)|_{-\\frac{\\var{L}}{2}}^{\\frac{\\var{L}}{2}}+\\frac{\\var{c}*\\var{L}}{{k}\\pi}sin\\left(\\frac{{k}\\pi}{\\var{L}}t\\right)|_{\\frac{\\var{L}}{2}}^{\\var{L}}\\right)\$

\$a_k=\\frac{\\var{a}}{{k}\\pi}sin(-\\frac{{k}\\pi}{2})-\\frac{\\var{a}}{{k}\\pi}sin(-{k}\\pi)+\\frac{\\var{b}}{{k}\\pi}sin(\\frac{{k}\\pi}{2})-\\frac{\\var{b}}{{k}\\pi}sin(-\\frac{{k}\\pi}{2})+\\frac{\\var{c}}{{k}\\pi}sin({k}\\pi)-\\frac{\\var{c}}{{k}\\pi}sin(\\frac{{k}\\pi}{2})\$

\$a_k=-\\frac{\\var{a}}{{k}\\pi}sin(\\frac{{k}\\pi}{2})+\\frac{\\var{b}}{{k}\\pi}sin(\\frac{{k}\\pi}{2})+\\frac{\\var{b}}{{k}\\pi}sin(\\frac{{k}\\pi}{2})-\\frac{\\var{c}}{{k}\\pi}sin(\\frac{{k}\\pi}{2})\$

\$a_k=\\frac{\\simplify{-{a}+2{b}-{c}}}{{k}\\pi}sin(\\frac{{k}\\pi}{2})\$

\$a_\\var{k}=\\simplify{(-{a}+2{b}-{c})/({k}*pi)sin({k}*pi/2)}\$

The formula for the Fourier coefficient \$b_k\$ is given by: \$b_k=\\frac{1}{L}\\int_{-L}^{L}f(t)sin\\left(\\frac{{k}\\pi}{L}t\\right)dt\$

\$b_k=\\frac{1}{\\var{L}}\\left(\\int_{-\\var{L}}^{-\\frac{\\var{L}}{2}}\\var{a}sin\\left(\\frac{{k}\\pi}{\\var{L}}t\\right)dt+\\int_{-\\frac{\\var{L}}{2}}^{\\frac{\\var{L}}{2}}\\var{b}sin\\left(\\frac{{k}\\pi}{\\var{L}}t\\right)dt+\\int_{\\frac{\\var{L}}{2}}^{\\var{L}}\\var{c}sin\\left(\\frac{{k}\\pi}{\\var{L}}t\\right)dt\\right)\$

\$b_k=\\frac{1}{\\var{L}}\\left(-\\frac{\\var{a}*\\var{L}}{{k}\\pi}cos\\left(\\frac{{k}\\pi}{\\var{L}}t\\right)|_{-\\var{L}}^{-\\frac{\\var{L}}{2}}-\\frac{\\var{b}*\\var{L}}{{k}\\pi}cos\\left(\\frac{{k}\\pi}{\\var{L}}t\\right)|_{-\\frac{\\var{L}}{2}}^{\\frac{\\var{L}}{2}}-\\frac{\\var{c}*\\var{L}}{{k}\\pi}cos\\left(\\frac{{k}\\pi}{\\var{L}}t\\right)|_{\\frac{\\var{L}}{2}}^{\\var{L}}\\right)\$

\$b_k=-\\frac{\\var{a}}{{k}\\pi}cos(-\\frac{{k}\\pi}{2})+\\frac{\\var{a}}{{k}\\pi}cos(-{k}\\pi)-\\frac{\\var{b}}{{k}\\pi}cos(\\frac{{k}\\pi}{2})+\\frac{\\var{b}}{{k}\\pi}cos(-\\frac{{k}\\pi}{2})-\\frac{\\var{c}}{{k}\\pi}cos({k}\\pi)+\\frac{\\var{c}}{{k}\\pi}cos(\\frac{{k}\\pi}{2})\$

\$cos(-\\theta)=cos(\\theta)\$

\$b_k=-\\frac{\\var{a}}{{k}\\pi}cos(\\frac{{k}\\pi}{2})+\\frac{\\var{a}}{{k}\\pi}cos({k}\\pi)-\\frac{\\var{b}}{{k}\\pi}cos(\\frac{{k}\\pi}{2})+\\frac{\\var{b}}{{k}\\pi}cos(\\frac{{k}\\pi}{2})-\\frac{\\var{c}}{{k}\\pi}cos({k}\\pi)+\\frac{\\var{c}}{{k}\\pi}cos(\\frac{{k}\\pi}{2})\$

\$b_k=\\frac{\\var{c}-\\var{a}}{{k}\\pi}cos(\\frac{{k}\\pi}{2})+\\frac{\\var{a}-\\var{c}}{{k}\\pi}cos({k}\\pi)\$

\$b_\\var{k}=\\simplify{({c}-{a})/({k}*pi)cos({k}*pi/2)}+\\simplify{(-{c}+{a})/({k}*pi)cos({k}*pi)}\$

Recall the amplitude of the \$\\var{k}\$th harmonic component is given by \$\\sqrt{(a_\\var{k})^2+(b_\\var{k})^2}\$

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Finn verdien på det første leddet i til Fourier serien \$\\frac{a_{0}}{2}\$. Angi svaret ditt med tre desimaler.

\$\\frac{a_{0}}{2}\$ = [[0]]

Finn et utrykk for Fourier koefissienten \$ a_{k}\$ og så regn ut \$a_{\\var{k}}\$. Angi svaret ditt med tre desimaler.

\$ a_{\\var{k}}\$ = [[0]]

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Finn et utrykk for Fourier koefissienten \$ b_{k}\$ og så regn ut \$b_{\\var{k}}\$. Angi svaret ditt med tre desimaler.

\$ b_{\\var{k}}\$ = [[0]]

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Regn ut amplituden til den \$\\var{k}\$te harmoniske komponenten. Angi svaret ditt med tre desimaler.

Amplituden = [[0]]

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