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# Answer to Question #111960 in Electrical Engineering for Dn

Question #111960
find continued fraction expansion of the following impedance function.
z(s)=(s^5 + 4 s^3 + 5 s)/(3 s^4 + 6 s^2)
1
2020-04-24T13:38:55-0400

First, let us divide "x^5 + 4 x^3 + 5 x" by "3x^4 + 6 x^2" as usual numbers. We need to multiply "3x^4" by "\\frac{x}{3}", to cancel "x^5". Subtracting "\\frac{x}{3} \\cdot (3 x^4 + 6 x^2)", obtain remainder "2 x^3 + 5 x", so no further reduction is possible.

Hence, "x^5 + 4 x^3 + 5 x = \\frac{x}{3} \\cdot ( 3 x^4 + 6 x^2) + (2 x^3 + 5 x)", from where "\\frac{x^5 + 4 x^3 + 5 x}{3 x^4 + 6 x^2} = \\frac{x}{3} + \\frac{2 x^3 + 5 x}{3 x^4 + 6 x^2}".

The remainder "\\frac{2 x^3 + 5 x}{3 x^4 + 6 x^2} = \\frac{x(5+2x^2)}{3x^2(x^2+2)} = \\frac{1}{3}\\frac{5 + 2 x^2}{x(x^2+2)}" has degree of polynomial in the numerator smaller, than degree of the denominator. Hence, we can decompose it into partial fractions: "\\frac{5 + 2 x^2}{x(x^2+2)} = \\frac{A}{x} + \\frac{B x + C}{x^2+2} = \\frac{A x^2 + 2 A + B x^2 + C x}{x(x^2+2)}", from where equating the coefficients next to powers of "x" in the left and right hand sides of previous equation, obtain linear system"A+B = 2; C = 0; 2 A = 5", from where "A = \\frac{5}{2}, B = -\\frac{1}{2}, C = 0".

Therefore, "\\frac{2 x^3 + 5 x}{3 x^4 + 6 x^2} = \\frac{1}{3}\\left[ \\frac{5}{2 x} - \\frac{1}{2(x^2+2)} \\right]", and finally "\\frac{x^5 + 4 x^3 + 5 x}{3 x^4 + 6 x^2} = \\frac{x}{3} + \\frac{5}{6 x} - \\frac{1}{6(x^2+2)}"

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