I was trying to solve the Z-transform for u[n] - u[n-N], where u[n] means discrete unit step function, and N is some finite integer. I solved this using 2 methods.

Method 1 :

u[n] - u[n] = 1 +

Taking z- transform using time-delay property and keeping in mind that delta[n] has z-transform = 1; I get:

enter image description here

which suggests that ROC is |Z| > 0

Method 2:

I know that z transform of $u[n] = \frac{z}{z-1}$ with ROC |Z| > 1

using this and the time-delay property on both u[n] and u[n-N] I say that:

enter image description here

which on making the denominators of the 2 fraction equal and simplifying becomes : enter image description here

Which suggests ROC |z| > 1 .

The result in method-1 makes sense since the signal is a finite duration signal and taking z=0 would essentially mean a divide by zero situation while calculating the z transform.

But Method-2 is something that results from simply applying the properties of z-transform on some pre-known result for a special signal.

Why are the results different then?

  • $\begingroup$ First method , sum of N terms, a=1, r=$z^{-1}$ , provided |r|<1. So there also ROC is |z|>1 not 0 as you assumed. $\endgroup$
    – abhilash
    Jun 12, 2020 at 17:27
  • 1
    $\begingroup$ @abhilash: The ROC is $|z|>0$. This is true for any causal sequence of finite length. $\endgroup$
    – Matt L.
    Jun 12, 2020 at 19:01
  • $\begingroup$ Yep Matt, my mistake, |r|<1 for infinite series, also i forgot (z-1) is a factor of $(z^N -1)$ no pole at Z=1 $\endgroup$
    – abhilash
    Jun 13, 2020 at 8:09

2 Answers 2


Note that

$$1+z^{-1}+\ldots + z^{-(N-1)}=\sum_{n=0}^{N-1}z^{-n}=\frac{1-z^{-N}}{1-z^{-1}}\tag{1}$$

where I've used the formula for a finite geometric series.

So both your results are identical and correct.

The ROC is $|z|>0$, which is the case for all causal sequences of finite length. Note that in the expression on the right-hand side of $(1)$ there is a pole-zero cancellation at $z=1$, so in fact there is no pole at $z=1$, hence the ROC $|z|>0$.

  • $\begingroup$ Thank you, Matt. This is very helpful. $\endgroup$
    – Adarsh
    Jun 13, 2020 at 8:39

Maybe I am wrong but here is how I look at it: The two methods you wrote give the same result.

Let's choose $z=2 + j0$ for example and length $n=5$

If we sum $Z\{x[n]\} = Z\{ u[n]-u[n-5] \} = 1+2^{-1}+2^{-2}+2^{-3}+2^{-4}=1.9375$

Same thing goes for your method 2:

$\dfrac{1}{1-z^{-1}}-z^{-n}\dfrac{1}{1-z^{-1}} $ would be $\dfrac{1}{1-2^{-1}}-\dfrac{2^{-5}}{1-2^{-1}}=1.9375$

You get the same result for both. They just happen to look different but they mean the same.

  • $\begingroup$ Thank you for your answer. Really appreciate it. $\endgroup$
    – Adarsh
    Jun 13, 2020 at 8:40

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