I saw this question was being asked here a few times, but none got actual answer that helped me ( at other Transform of course ).
$$X\left(z\right)=\frac{1-3z^{-1}}{\left(1-0.2z^{-1}\right)\left(1+0.6z^{-1}\right)}$$ If I do by partial fractions, I will receive: $$X\left(z\right)=-\frac{7}{2}\cdot \frac{1}{1-0.2z^{-1}}+\frac{9}{2}\cdot \frac{1}{1+0.6z^{-1}}$$ Final answer by my way: $$x\left[n\right]=-\frac{7}{2}\cdot \left(0.2\right)^nu\left[n\right]\:-\frac{9}{2}\cdot \left(-0.6\right)^nu\left[-n-1\right]$$ With: $$ROC\::\:\left\{z\in \mathbb{C}|0.2<\left|z\right|<0.6\right\}$$ But if I will do by other way: $$X\left(z\right)=\frac{1}{\left(1-0.2z^{-1}\right)\left(1+0.6z^{-1}\right)}-\frac{1}{\left(1-0.2z^{-1}\right)\left(1+0.6z^{-1}\right)}\cdot 3z^{-1}$$ And define: $$G\left(z\right)=\frac{1}{\left(1-0.2z^{-1}\right)\left(1+0.6z^{-1}\right)}$$ With that, to partial fractions to each function, you will get two expressions for each part.

Final answer This way: $$x\left[n\right]=\frac{1}{4}\cdot \left(\frac{1}{5}\right)^nu\left[n\right]-\frac{3}{4}\cdot \left(-\frac{5}{3}\right)^nu\left[-n-1\right]-\frac{3}{4}\cdot \left(\frac{1}{5}\right)^{n-1}u\left[n-1\right]+\frac{9}{4}\cdot \left(\frac{3}{5}\right)^{n-1}u\left[-n\right]$$ I guess from here you understand what I mean, you will receive different output basically. The inverse will be different.

My question: Which is the "True" way, why is it the the "True" way, why the other way was not allowed? According to solutions, they did it the other way and thus I got with my partial fractions a bad solution.

Adding proof of my partial fractions:
$$y\left(x\right)=\frac{1-3x}{\left(1-0.2x\right)\left(1+0.6x\right)}\:=\:\frac{A}{1-0.2x}+\frac{B}{1+0.6x},$$ You will get: $$\:A=-\frac{7}{2},\:B=\frac{9}{2}$$ You can check at symbolab partial fraction calculator \ calculate easily, but its better to check at symbolab.
Just define $z^{-1}=x$ and go with it.

  • $\begingroup$ You appear to have at least a sign mistake in your partial fraction expansion. $\endgroup$
    – Andy Walls
    Jan 23 at 20:08
  • $\begingroup$ why? if you do partial fraction exepansion as when you call $z^{-1}=x$, you will receive it, I checked also at symbolab $\endgroup$ Jan 23 at 20:14
  • $\begingroup$ @AndyWalls Added proof at edit for my partial fractions. look $\endgroup$ Jan 23 at 20:17
  • $\begingroup$ My mistake. Your partial fraction expansion is correct. You haven't shown the inversion of the Z transforms back to the sample domain, so it is unclear what the difference is about which you are speaking. $\endgroup$
    – Andy Walls
    Jan 23 at 20:24
  • $\begingroup$ @AndyWalls Hmm, its kind of long to write it at latex, I am a hard writer at latex. I just know by the solutions, you will get by my Partial fraction expansion 2 expressions for $x[n]$ and if you do partial fractions in the other way, as I said, you will receive 4 expressions for $x[n]$. If I have to write the whole way for the answer, it will take me an hour, but actually, I can upload the final answer, I will EDIT $\endgroup$ Jan 23 at 20:30

1 Answer 1


First of all, I assume that the ROC is given, because the sequence $x[n]$ is only uniquely defined by $X(z)$ and the ROC.

Your first result is correct. Of course, both methods must lead to the same result if done correctly. Your second result is also correct, apart from two typos: the second and the fourth term should have the pole $-3/5$ in parenthesis (not $-5/3$ and $+3/5$).

Now it only remains to show that the two results are identical. This can be done in a few basic steps. Let's first look at the causal part of the second solution:

\begin{align*} x_1[n] &= \frac14\left(\frac15\right)^nu[n] - \frac34\left(\frac15\right)^{n-1}u[n-1] \\ &= \frac14\left(\frac15\right)^nu[n] - \frac34\cdot 5\left(\frac15\right)^{n}u[n] + \frac{15}{4}\delta[n] \\ &=-\frac72\left(\frac15\right)^{n}u[n] + \frac{15}{4}\delta[n] \end{align*}

In the same way you can rewrite the anti-causal part of the second solution, and you'll see that the term $\frac{15}{4}\delta[n]$ cancels out and the final result is identical to the first solution, which shouldn't come as a surprise.

  • $\begingroup$ Great answer, Thank you very much!! This post will help other people too who will have trouble in the future, as seen in other posts, none got really a good answer that fit ( atleast to me ). If I am good with it, probably others who will need it also!! $\endgroup$ Jan 24 at 16:48

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