// Numbas version: finer_feedback_settings {"name": "simple linear factors partial fractions and delta function", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"rulesets": {}, "extensions": [], "tags": [], "ungrouped_variables": ["a", "b", "c", "P", "R", "k", "k2"], "functions": {}, "name": "simple linear factors partial fractions and delta function", "advice": "

\\(X(s)=\\frac{\\var{R}s+\\var{P}}{(s+\\var{a})(s+\\var{b})(s+\\var{c})}+\\frac{\\var{k}e^{-\\var{k2}s}}{(s+\\var{a})(s+\\var{b})}\\)

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Take the first fraction and mutiply across by the denominator \\((s+\\var{a})(s+\\var{b})(s+\\var{c})\\) to get

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\\(\\var{R}s+\\var{P}=A(s+\\var{b})(s+\\var{c})+B(s+\\var{a})(s+\\var{c})+C(s+\\var{a})(s+\\var{b})\\)

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let s = \\(-\\var{a}\\)

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\\(\\simplify{{R}*{-{a}}+{P}}=\\simplify{(-{a}+{b})*(-{a}+{c})}A\\)

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\\(A=\\simplify{{{R}*{-{a}}+{P}}/((-{a}+{b})*(-{a}+{c}))}\\)

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let s = \\(-\\var{b}\\)

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\\(\\simplify{{R}*{-{b}}+{P}}=\\simplify{(-{b}+{a})*(-{b}+{c})}B\\)

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\\(B=\\simplify{{{R}*{-{b}}+{P}}/((-{b}+{a})*(-{b}+{c}))}\\)

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let s = \\(-\\var{c}\\)

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\\(\\simplify{{R}*{-{c}}+{P}}=\\simplify{(-{c}+{a})*(-{c}+{b})}C\\)

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\\(C=\\simplify{{{R}*{-{c}}+{P}}/((-{c}+{b})*(-{c}+{a}))}\\)

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Take the second fraction, ignore the exponential function and mutiply across by the denominator \\((s+\\var{a})(s+\\var{b})\\) to get

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\\(\\var{k}=A(s+\\var{b})+B(s+\\var{a})\\)

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let s = \\(-\\var{a}\\)

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\\(\\var{k}=A(\\simplify{-{a}+{b}})+B(0)\\)

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\\(A=\\simplify{{k}/({b}-{a})}\\)

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let s = \\(-\\var{b}\\)

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\\(\\var{k}=A(0)+B(\\simplify{{a}-{b}})\\)

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\\(B=\\simplify{{k}/({a}-{b})}\\)

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Determine the partial fraction breakdown of the following expression:

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     \\(X(s)=\\frac{\\var{R}s+\\var{P}}{(s+\\var{a})(s+\\var{b})(s+\\var{c})}+\\frac{\\var{k}e^{-\\var{k2}s}}{(s+\\var{a})(s+\\var{b})}\\)

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.

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Express your answer as a sum of five fractions:

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               \\(X(s) =\\) [[0]]

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