// Numbas version: exam_results_page_options {"name": "Harmonic component of the complex form of a Fourier series", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Harmonic component of the complex form of a Fourier series", "statement": "

Given the function:

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\$$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] \\,\\,\\,\\,f(t+\\simplify{2*{L}})=f(t)\$$

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Determine an expression for the complex exponential Fourier coefficients \$$c_{k}\$$ and hence calculate the value for the amplitude of the \$$\\var{k}\$$th harmonic component.

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Amplitude = [[0]]

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Calculate a particular harmonic component of the complex form of a Fourier series expansion.

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\$$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] \\,\\,\\,\\,f(t+\\simplify{2*{L}})=f(t)\$$

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\$$c_{k}=\\frac{1}{2L}\\int_{-L}^{-L}f(t)e^{-\\frac{jk\\pi}{L}t}\$$

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In this example \$$2L=\\simplify{2*{L}}\$$

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\$$c_{k}=\\frac{1}{\\simplify{2*{L}}}\\left(\\int_{-\\var{L}}^{\\simplify{-{L}/2}}\\var{a}e^{-\\frac{j{k}\\pi}{\\var{L}}t}+\\int_{-{\\simplify{{L}/2}}}^{\\simplify{{L}/2}}\\var{b}e^{-\\frac{j{k}\\pi}{\\var{L}}t}dt+\\int_{\\simplify{{L}/2}}^{\\var{L}}\\var{c}e^{-\\frac{j{k}\\pi}{\\var{L}}t}\\right)\$$

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\$$c_{k}=\\frac{1}{\\simplify{2*{L}}}\\left(-\\frac{\\var{a}*\\var{L}}{j{k}\\pi}e^{-\\frac{j{k}\\pi}{\\var{L}}t}|_{-\\var{L}}^{-\\simplify{{L}/2}}-\\frac{\\var{b}*\\var{L}}{j{k}\\pi}e^{-\\frac{{k}\\pi}{\\var{L}}t}|_{-\\simplify{{L}/2}}^{\\simplify{{L}/2}}-\\frac{\\var{c}*\\var{L}}{j{k}\\pi}e^{-\\frac{j{k}\\pi}{\\var{L}}t}|_{\\simplify{{L}/2}}^{\\var{L}}\\right)\$$

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\$$c_{k}=-\\frac{1}{2{k}\\pi j}\\left(\\var{a}e^{-\\frac{j{k}\\pi}{\\var{L}}t}|_{-\\var{L}}^{-\\simplify{{L}/2}}+\\var{b}e^{-\\frac{j{k}\\pi}{\\var{L}}t}|_{-\\simplify{{L}/2}}^{\\simplify{{L}/2}}+\\var{c}e^{-\\frac{j{k}\\pi}{\\var{L}}t}|_{\\simplify{{L}/2}}^{\\var{L}}\\right)\$$

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\$$c_{k}=\\frac{j}{2{k}\\pi}\\left(\\var{a}e^{\\frac{j{k}\\pi}{{2}}}-\\var{a}e^{j{k}\\pi}+\\var{b}e^{-\\frac{j{k}\\pi}{2}}-\\var{b}e^{\\frac{j{k}\\pi}{2}}+\\var{c}e^{-j{k}\\pi}-\\var{c}e^{-\\frac{j{k}\\pi}{2}}\\right)\$$

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\$$c_{k}=\\frac{j}{2{k}\\pi}\\left(\\simplify{{a}-{b}}e^{\\frac{j{k}\\pi}{{2}}}-\\var{a}e^{j{k}\\pi}+\\simplify{{b}-{c}}e^{-\\frac{j{k}\\pi}{2}}+\\var{c}e^{-j{k}\\pi}\\right)\$$

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Recall \$$e^{j\\theta}=cos(\\theta)+jsin(\\theta)\$$    and     \$$e^{-j\\theta}=cos(\\theta)-jsin(\\theta)\$$

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\$$c_{k}=\\frac{j}{2{k}\\pi}\\left(\\simplify{{a}-{b}}cos(\\frac{{k}\\pi}{2})+\\simplify{{a}-{b}}jsin(\\frac{{k}\\pi}{2})-\\var{a}cos({k}\\pi)-\\var{a}jsin({k}\\pi)+\\simplify{{b}-{c}}cos(\\frac{{k}\\pi}{2})-\\simplify{{b}-{c}}jsin(\\frac{{k}\\pi}{2})+\\var{c}cos({k}\\pi)-\\var{c}jsin({k}\\pi)\\right)\$$

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Recall \$$sin({k}\\pi)=0\$$ for all whole number values of \$$k\$$

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\$$c_{k}=\\frac{j}{2{k}\\pi}\\left(\\simplify{{c}-{a}}cos({k}\\pi)+\\simplify{{a}-{c}}cos(\\frac{{k}\\pi}{{2}})+\\simplify{{c}-2*{b}+{a}}jsin(\\frac{{k}\\pi}{2}))\\right)\$$

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\$$c_{k}=\\frac{1}{2{k}\\pi}\\left(\\simplify{-{c}+2*{b}-{a}}sin(\\frac{{k}\\pi}{2})+j(\\simplify{{c}-{a}}cos({k}\\pi)+\\simplify{{a}-{c}}cos(\\frac{{k}\\pi}{{2}}))\\right)\$$

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\$$c_{k}=\\frac{1}{2{k}\\pi}\\left(\\simplify{(-{c}+2*{b}-{a})}sin(\\frac{{k}\\pi}{2})+(\\simplify{({c}-{a})}cos({k}\\pi)+\\simplify{({a}-{c})}cos(\\frac{{k}\\pi}{2}))j\\right)\$$

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Therefore

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\$$c_{\\var{k}}=\\frac{1}{\\simplify{2*{k}pi}}\\left(\\simplify{-{c}+2*{b}-{a}}sin(\\frac{\\var{k}\\pi}{2})+j(\\simplify{{c}-{a}}cos(\\var{k}\\pi)+\\simplify{{a}-{c}}cos(\\frac{\\var{k}\\pi}{{2}}))\\right)\$$

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\$$c_{\\var{k}}=\\frac{1}{\\simplify{2*{k}pi}}\\left(\\simplify{(-{c}+2*{b}-{a})sin({k}*pi/2)}+(\\simplify{(({c}-{a})cos({k}*pi)+({a}-{c})cos({k}*pi/2))})j\\right)\$$

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The \$$\\var{k}\$$th harmonic component equals \$$2|c_{\\var{k}}|\$$

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\$$2*|c_{\\var{k}}|=\\frac{1}{\\simplify{{k}pi}}\\sqrt{\\simplify{((-{c}+2*{b}-{a})sin({k}*pi/2))^2+(({c}-{a})cos({k}*pi)+({a}-{c})cos({k}*pi/2))^2}}\$$

", "type": "question", "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}]}]}], "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}]}