I've been working trying to analyze a causal stable system. Hopefully a numeric example such as the one I am working on, and the problems I face during it could be useful to others.

I have this system function:

$$H(z) = \frac{1-\frac{3}{10}z^{-1}+\frac{1}{3}z^{-2}}{1-\frac{3}{5}z^{-1}+\frac{38}{75}z^{-2}+\frac{2}{5}z^{-3}}$$

So I originally thought that since the numerator has the power $z^{-2}$ and the denominator has the power $z^{-3}$ I could use partial fractions, but now I believe that I am wrong to think that. I believe this case is the numerator has a higher degree than the denominator?

I multiplied the fraction through with $\frac{z^3}{z^3}$


Well now the numerator has a degree that is the same as the denominator. I think I need to use long division.

I get,

$$H(z) = 1+ \frac{\frac{9}{10}z^2-\frac{13}{75}z-\frac{2}{5}}{z^3-\frac{3}{5}z^2+\frac{38}{75}z+\frac{2}{5}}$$

Factoring the denominator,

$$H(z) = 1+ \frac{\frac{9}{10}z^2-\frac{13}{75}z-\frac{2}{5}}{(z-(\frac{2}{5}+j.711)(z-(\frac{2}{5}-j.711)(z+\frac{1}{5})}$$

After much later I find the coefficients,

$$H(z) = 1+ \frac{1.12+j.273}{z-(\frac{2}{5}+j.711)}+\frac{1.12-j.273}{z-(\frac{2}{5}-j.711)}-\frac{.3805}{z+\frac{1}{5}}$$

Writing it in terms of delays,

$$H(z) = 1+ \frac{(1.12+j.273)z^{-1}}{1-(\frac{2}{5}+j.711)z^{-1}}+\frac{(1.12-j.273)z^{-1}}{1-(\frac{2}{5}-j.711)z^{-1}}-\frac{.3805z^{-1}}{1+\frac{1}{5}z^{-1}}$$

Now taking the inverse Z-Transform

$$h[n]= \delta[n]+ (1.12+j.273)(\tfrac{2}{5}+j.711)^{(n-1)}u[n-1]+(1.12-j.273)(\tfrac{2}{5}-j.711)^{(n-1)}u[n-1]-.3805(\tfrac{-1}{5})^{(n-1)}u[n-1]$$

That was pretty difficult and time consuming for me too. I ended up find some sort of solution but I am not sure if it is correct. Is there an easier way to solve more difficult IIR systems such as these with complex roots? What do the complex coefficients mean? Is there a standard to simplifying these even more? Further more, what would the input output equation of this type be?

UPDATE: Using part of the example kindly provided below, noting that complex conjugate pairs can be simplified further to obtain an impulse function I was able to come up with the following: (Focusing on the complex conjugate part in particular)

$$H(z) = \frac{1.12+j.273}{z-(\frac{2}{5}+j.711)}+\frac{1.12-j.273}{z-(\frac{2}{5}-j.711)}$$

Which simplifies to,

$$H(z)= \frac{(2.24)-(.9381e^{-j.811}+.9381e^{j.811})z^{-1}}{1-.8z^{-1}+.665z^{-2}}z^{-1} $$

I am going to ignore the extra factor of the delay tacked on at the end. I found a Fourier Transform Pair that particularly looks like this! Not sure if this is what should be done but it looks hopeful!

$$r^ncos(w_0)u[n]<---DTFT--->\frac{1-r \cos(w_0)z^{-1}}{1-2r\cos(w_0)z^{-1}+r^2z^{-2}}$$

rearranging some things to make it fit better with the DTFT pair,

$$H(z)= 2.24\frac{1-(.8375)\cos(.811)z^{-1}}{1-2(.58)\cos(.811)z^{-1}+(.815)^2z^{-2}} $$

Well one thing that is alarming is that my r values (according to the DTFT pair) are slightly off.

The three different values:

$r=.58$ is worrisome.



The other two r values seem like it could be due to averaging errors. But $r=.58$ seems a little too off.

What I think I would do is take the average $r$, though I don't know if this is the right approach, and use that to find the new impulse function for this complex conjugate pair.

  • $\begingroup$ Darklink, without a definition of Region of Convergence (ROC) you cannot uniquely compute the impulse response associated with $H(z)$. There will be a distinct impulse response for each possible valid ROC. In your case you have 2 distinct poles (one conjugate pair) and this yields at most 3 different ROCs and hence 3 different impulse response. So please specify your ROC. A helper would be to call it causal and stable which reduces the possibilities to 1. Unique in that case... $\endgroup$
    – Fat32
    Oct 12, 2018 at 22:37
  • $\begingroup$ Ah yes, I am assuming causal and stable. I will update my question. $\endgroup$
    – DarkLink
    Oct 12, 2018 at 22:56

1 Answer 1


It turns out that those terms with complex conjugate scalers and exponentials can be combined. Rather than do it to your example, I would rather do it in the Z-domain. Let

$$\begin{align} H(z) &= \frac{A_1}{1-p_1\,z^{-1}} + \frac{A_2}{1-p_2\,z^{-1}} \\ \\ &= \frac{A_1(1-p_2\,z^{-1})+A_2(1-p_1\,z^{-1})}{(1-p_1\,z^{-1})(1-p_2\,z^{-1})} \\ \\ &= \frac{(A_1-A_1 p_2\,z^{-1})+(A_2-A_2 p_1\,z^{-1})}{1-(p_1+p_2)\,z^{-1} + p_1 p_2\, z^{-2}} \\ \\ &= \frac{(A_1+A_2) - (A_1 p_2 + A_2 p_1)z^{-1}}{1-(p_1+p_2)\,z^{-1} + p_1 p_2\, z^{-2}} \\ \\ \end{align}$$

Now it turns out that there are two ways for the impulse response, $h[n] = \mathcal{Z}^{-1} \{H(z)\}$, to be total real (with no imaginary component). One way is for both poles $p_1,p_2$ to be real and that will also require the two coefficients $A_1,A_2$ to also be real.

$$\begin{align} h[n] &= \mathcal{Z}^{-1}\{H(z)\} \\ \\ &= \mathcal{Z}^{-1}\left\{ \frac{A_1}{1-p_1\,z^{-1}} \right\} + \mathcal{Z}^{-1}\left\{ \frac{A_2}{1-p_2\,z^{-1}} \right\} \\ \\ &= A_1 p_1^n u[n] + A_2 p_2^n u[n] \\ \end{align}$$

The other way for $h[n]$ to be real is for both poles $p_1,p_2$ to be complex-conjugate of each other and that will also require the two coefficients $A_1,A_2$ to also be complex-conjugate.

$$\begin{align} h[n] &= \mathcal{Z}^{-1}\{H(z)\} \\ \\ &= \mathcal{Z}^{-1}\left\{ \frac{(A_1+A_2) - (A_1 p_2 + A_2 p_1)z^{-1}}{1-(p_1+p_2)\,z^{-1} + p_1 p_2\, z^{-2}} \right\} \\ \end{align}$$

(i need to return to this later.)

  • $\begingroup$ Looking forward to the completion of your answer.... $\endgroup$ Oct 13, 2018 at 14:55
  • $\begingroup$ oh dear, thanks for reminding me @DilipSarwate. i'm working now, so this has to happen later this evening. $\endgroup$ Oct 15, 2018 at 18:18

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