How to draw a finite-duration sequence

Question

On page $359$ of the 2nd edition of the book "Signals & Systems", by Oppenheim and Willsky, the authors wrote the following words:

"Consider a general sequence $x[n]$ that is of finite duration. That is, for some integers $N1$ and $N2$, $x[n] = 0$ outside the range $-N1\leq n\leq N2$. A signal of this type is illustrated in $\text{Figure}\; 5.1\:(\text{a})$"

My three questions are:

1. Regarding $x[n]$ in the figure, do the leading and trailing zero-valued samples exist?

2. If the leading and trailing zero-valued samples exist, then what is the duration of (how many samples are in) the finite-duration $x[n]$ sequence?

3. If the leading and trailing zero-valued samples do not exist, then why are they shown in $\text{Figure}\; 5.1\:(\text{a})?$

Answer 1

The plot as shown makes sense as nothing has limited the time axis to be going from $-\infty$ to +$\infty$. So we are showing the finite duration sequence as it would appear on such a time axis.

We can limit the time axis to a finite number of samples as used in the Discrete Fourier Transform (DFT) or the an unlimited time axis as used in the Discrete Time Fourier Transform (DTFT). Neither approach would change the number of samples in a finite duration signal.

So the plot and text is implying the time axis extends from $-\infty$ to +$\infty$. The number of samples in the finite duration sequence would be the non-zero samples.

My confusion when first answering this question was the case of the DFT and how adding zeros is significant to the result, but really this is just approximating the DTFT and not changing the duration of the sequence itself.

Answer 2

1. if you want the zero padding to exist, it exists. if you don't want those samples to exist as zero, you must define what they are. the most common alternative is that outside the interval $-N_1 \ge n \ge N_2$ is to periodically extend the the sequence.
2. $N_1 + N_2 +1$
3. they're shown as existing because whomever made that illustration intends them to exist and for them to equal zero.

if you define: $N \triangleq N_1 + N_2 +1$ and impose periodicity

$$ x[n+N] = x[n] \qquad \qquad \forall n \in \mathbb{Z} $$

then the DFT is a mapping of one periodic function (with a finite period) to another periodic function having the same period $N$.

$$ X[k] = \sum\limits_{n=n_0}^{n_0+N} x[n] e^{-j 2 \pi nk/N} \qquad \qquad n_0 \in \mathbb{Z} $$

you are allowed to set $n_0 = -N_1$ and the upper limit becomes $N_2$.

If, instead, you define it as

$$ x[n] = 0 \qquad \qquad \text{for} \ n<-N_1 \ \text{or} \ n>N_2 $$

then the DTFT is

$$\begin{align} X \big( e^{j\omega} \big) &\triangleq \sum\limits_{n=-\infty}^{\infty} x[n] e^{-j \omega n} \\ &= \sum\limits_{n=n_0}^{n_0+N} x[n] e^{-j \omega n} \qquad \qquad n_0 = -N_1 \\ \end{align} $$

Now it turns out that the two are equal when $ \omega = 2 \pi k / N $.