How to find the partial fractions of the given irreducible equation?

Question

$$H(z) = \frac{(z^2 + \frac 13 z)(z^2 -\frac 78 z)}{(z^2 - 2z +2)(z^2 -\frac 34z + \frac 18)}.$$ How can i find the partial fractions of the given function above? I know the equation in the denominator is irreducible. Can anyone solve this equation?

Answer 1

Hint...the partial fractions will take the form $$1+\frac{Az+B}{z^2-2z+2}+\frac{C}{z-\frac 12}+\frac{D}{z-\frac 14}$$

Answer 2

Let's write your rational function as $$ {{\left( {z^{\,2} + 1/3\,z} \right)\left( {z^{\,2} - 7/8\,z} \right)} \over {\left( {z^{\,2} - 2\,z + 2} \right)\left( {z^{\,2} - 3/4\,z + 1/8\,} \right)}} = z^{\,2} {{\left( {z + 1/3} \right)\left( {z - 7/8} \right)} \over {\left( {z^{\,2} - 2\,z + 2} \right)\left( {z^{\,2} - 3/4\,z + 1/8\,} \right)}} $$

The term following $z^2$ is the ratio of a polynomial of $2$nd degree with a polynomial of $4$th degree.

Since you said that you are not going to work in the complex field, remaining in the real one, you can perform the partial fraction decomposition keeping the non-decomposable factors at the denominator with $2$nd degree but then putting a $1$st degree polynomial at the numerator of each summand.

That is you can put $$ {{\left( {z + 1/3} \right)\left( {z - 7/8} \right)} \over {\left( {z^{\,2} - 2\,z + 2} \right)\left( {z^{\,2} - 3/4\,z + 1/8\,} \right)}} = {{\left( {a\,z + b} \right)} \over {\left( {z^{\,2} - 2\,z + 2} \right)}} + {{\left( {c\,z + d} \right)} \over {\left( {z^{\,2} - 3/4\,z + 1/8\,} \right)}} $$

Now you can solve for the four unknowns $a,b,c,d$ in various ways.
The most common one is to perform the sum in the RHS, and then equate the coefficients at the numerator.
Another possible way is to assign to $z$ four "easy" values and solve the $4 \times 4$ linear system.
Either way you reach to $$ a = {1 \over {15}}\quad b = {{17} \over {15}}\quad c = - {1 \over {15}}\quad b = - {{13} \over {60}} $$