I am solving a problem to find the zeros and poles. Subsequently, it is requires to determine the impulse response. Below is the system function:


I am able to compute the poles by quadratic factor: $z=\tfrac{-b \pm \sqrt{b^2-4ac}}{2a}$ $$=\tfrac{-(-4) \pm \sqrt{4^2-4(20)(1)}}{2(20)}$$ $$=\tfrac{1}{10} \pm \tfrac{1}{5}j$$ $$= 0.1 \pm 0.2j $$

The poles are $p_{1,2}= 0.1 \pm 0.2j $

However, I am stuck in the Inverse Z transform in order to determine the impulse. Below is my step:

$$H(z)=\tfrac{z}{(z-0.1-0.2j)(z-0.1+0.2j)} \cdot \tfrac{z^{-2}}{z^{-2}}$$ $$ =\tfrac{z^{-1}}{(z^{-1}-0.1z^{-2}-0.2jz^{-2})(z^{-1}-0.1z^{-2}+0.2jz^{-2})} $$ $$ =\tfrac{z^{-1}}{z^{-1}(1-0.1z^{-1}-0.2jz^{-1})z^{-1}(1-0.1z^{-1}+0.2jz^{-1})} $$ $$ =\tfrac{1}{z^{-1}(1-0.1z^{-1}-0.2jz^{-1})(1-0.1z^{-1}+0.2jz^{-1})} $$ $$ =\tfrac{1}{z^{-1}(1-z^{-1}(0.1+0.2j))(1-z^{-1}(0.1-0.2j))} $$

Before I move to partial fraction expansion, may I know what can I do to the common factor $z^{-1}$ in the denominator? I think it will be a long operation if I substitute pole value in order to find value $A$ and $B$ in the partial fraction expansion.

  • $\begingroup$ I wouldn't bother with the negative exponents of $z$ until you're ready to realize this thing and turn your transfer function (whether it's broken into partial fractions or not) into a difference equation, which from that, you can write code to implement the digital filter. $\endgroup$ Mar 21 at 19:26
  • $\begingroup$ And I would continue to use "$p_1$" and "$p_2$" instead of the actual values until later. $\endgroup$ Mar 21 at 19:27

1 Answer 1


You need to write your transfer function $H(z)$ as a sum of partial fractions.This is called partial fraction decomposition.

Then since the Z transform is linear the impulse response $h(t)$ will be the sum of the reverse Z transform $Z^{-1}$ of the fractions.


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