# Z transform of finite signals

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 :

Taking z- transform using time-delay property and keeping in mind that delta[n] has z-transform = 1; I get: 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: which on making the denominators of the 2 fraction equal and simplifying becomes : 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?

• 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. – abhilash Jun 12 at 17:27
• @abhilash: The ROC is $|z|>0$. This is true for any causal sequence of finite length. – Matt L. Jun 12 at 19:01
• 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 – abhilash Jun 13 at 8:09

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 $$|z|>0$$.

• Thank you, Matt. This is very helpful. – Adarsh Jun 13 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.