f(z)= z/(z+1)(z+1)^2
The form of the partial fraction decomposition is;
"\\displaystyle\n\\frac{z}{\\left(z + 1\\right)^{3}}=\\frac{A}{z + 1}+\\frac{B}{\\left(z + 1\\right)^{2}}+\\frac{C}{\\left(z + 1\\right)^{3}}=\\frac{z}{\\left(z + 1\\right)^{3}}=\\frac{\\left(z + 1\\right)^{2} A + \\left(z + 1\\right) B + C}{\\left(z + 1\\right)^{3}}\\\\\n\\Rightarrow z=\\left(z + 1\\right)^{2} A + \\left(z + 1\\right) B + C\\\\"
Expanding the right hand side yields;
"\\displaystyle\nz=z^{2} A + 2 z A + z B + A + B + C\\\\\n\\text{Collecting like terms yields;}\\\\\nz=z^{2} A + z \\left(2 A + B\\right) + A + B + C\\\\\n\\text{Comparing yields the following system of linear equations:}\\\\\n\\qquad\\qquad\\qquad\\qquad\\qquad\\qquad\\qquad\\begin{cases} A = 0\\\\2 A + B = 1\\\\A + B + C = 0 \\end{cases}\\\\\n\\text{Solving the above equations yields: }A=0,\\ B=1,\\ C=-1\\\\\n\\therefore\\frac{z}{\\left(z + 1\\right)^{3}}=\\frac{0}{z + 1}+\\frac{1}{\\left(z + 1\\right)^{2}}+\\frac{-1}{\\left(z + 1\\right)^{3}}=\\frac{z}{\\left(z + 1\\right)^{3}}=\\frac{1}{\\left(z + 1\\right)^{2}}-\\frac{1}{\\left(z + 1\\right)^{3}}"
Comments
Leave a comment