1
$\begingroup$

I was going through some old exams and found this question:

Find the inverse $Z$-transform of $z^{-1/2}$.

I tried using the properties table, but I couldn't find a single useful property that would help me solve this question. I don't have the solution of the past exam, so I would really appreciate a solution.

$\endgroup$

2 Answers 2

3
$\begingroup$

The first thing to note and keep in mind is the $\mathcal{Z}^{-1} \{ z^{-1}\}$ which is $\delta(n-1)$.

Now this is not as straightforward as the case mentioned above, but it's not that difficult of a task either. We will make use of contour integration. The inverse $Z$-transform relation is given by:

\begin{equation} x[n] = \frac{1}{2 \pi j} \oint_C X(z) z^{n-1} dz \tag{1} \end{equation}

Plug in our $X(z) = z^{-1/2}$:

\begin{equation} x[n] = \frac{1}{2 \pi j} \oint_C z^{-1/2} z^{n-1} dz \tag{2} \end{equation}

At this stage we will apply a change of variables as follows:

\begin{align*} z &= e^{j\omega}\\ \frac{dz}{d\omega} &= je^{j\omega}= jz\\ dz &= je^{j\omega} \; d\omega \end{align*}

And for the contour, as you may know, the ROC of a Z-transform is considered based on the unit circle. As such, we can take the contour limits from either $0$ to $2\pi$ or from $-\pi$ to $\pi$; it doesn't matter ($\omega$ follows such a path on the circle when integrating counterclockwise). Using the latter limits and the above change of variable, we have:

\begin{equation} x[n] = \frac{1}{2 \pi j} \int_{-\pi}^{\pi} e^{-j\omega/2} e^{j\omega(n-1)} je^{j\omega}\; d\omega \tag{3} \end{equation}

The $j$'s cancel and so does $e^{j\omega}$ with $e^{-j\omega}$, leaving us with the following:

\begin{equation} x[n] = \frac{1}{2 \pi} \int_{-\pi}^{\pi} e^{-j\omega/2} e^{j\omega n} \; d\omega \tag{4} \end{equation}

Does it look familiar? This is nothing, but the inverse DTFT of $e^{-j\omega/2}$.

Let's solve the integral:

\begin{equation} \frac{1}{2 \pi} \int_{-\pi}^{\pi} e^{j(n-\frac12)\omega} \; d\omega \tag{5} \end{equation}

\begin{equation} \frac{1}{2 \pi} \left(\frac{e^{j(n-\frac12)\omega}}{j(n-\frac12)} \right)\bigg\rvert_{-\pi}^{\pi} \tag{6} \end{equation}

\begin{equation} \frac{e^{j(n-\frac12)\pi} - e^{-j(n-\frac12)\pi}}{2 \pi j(n-\frac12)} \tag{7} \end{equation}

Now we will make use of the identity:

\begin{align*} \sin(\theta) = \frac{e^{j\theta} - e^{-j\theta}}{2j} \end{align*}

In our case, this $\theta$ is just $\pi(n-\frac12)$ and we get:

\begin{equation} \frac{\sin(\pi(n-\frac12))}{\pi(n-\frac12)} \tag{8} \end{equation}

as our final answer!

I mentioned the $\mathcal{Z}^{-1} \{ z^{-1}\}$ just to motivate what I think the reason behind asking the question was. The inverse transform of a single delay is just a shifted impulse, but what if that delay was fractional?

$\endgroup$
7
  • $\begingroup$ Thanks a lot! I have learned contour integration in my complex analysis course but never knew it could be used in this context. May I ask how you got the relation in $(1)$? $\endgroup$
    – user64710
    May 16, 2023 at 17:44
  • $\begingroup$ Wonderful answer. Just just one question, is this change of variables always valid? Indeed, such a change simplified it ridiculously as we avoid the contour integral and deal with an ordinary integral instead. $\endgroup$ May 17, 2023 at 6:09
  • $\begingroup$ @RubemPacelli The change comes from the relation between the Z-Transform and DTFT. $\endgroup$ May 17, 2023 at 8:26
  • $\begingroup$ What is "C.O.V."? Calculus of variations? $\endgroup$ May 17, 2023 at 9:57
  • 1
    $\begingroup$ "[...] the ROC of a Z-transform is considered based on the unit circle." I'm not sure what you're trying to say here. That's also where the problem of this answer lies. There is no annulus $r<|z|<R$ for which the corresponding series converges. That's why we should say that only the IDTFT exists, but not the inverse Z-transform. $\endgroup$
    – Matt L.
    May 17, 2023 at 13:26
6
$\begingroup$

The correct answer to the question

What is the inverse $\mathcal{Z}$-transform of $F(z)=z^{-\frac12}?$

is

$F(z)=z^{-\frac12}$ is not a valid $\mathcal{Z}$-transform, hence its inverse transform doesn't exist.

For $F(z)$ to be a valid $\mathcal{Z}$-transform, its Laurent series about $z_0=0$ must exist. However, for the given function, $z_0=0$ is a branch point and there is no annulus $r_1<|z|<r_2$ for which $F(z)$ is analytic. Consequently, the inverse $\mathcal{Z}$-transform doesn't exist. In other words, there is no series of the form

$$\sum_{n=-\infty}^{\infty}a_nz^{-n}$$

which converges uniformly to the function $F(z)=z^{-\frac12}$ in some annulus $r_1<|z|<r_2$.

The result derived in Ahson Yousef's answer is the inverse discrete-time Fourier transform (IDTFT) of $F(e^{j\omega})=e^{-j\omega /2}$.

In this example the situation is similar to the inverse transformation of a rectangular function or of a Dirac delta impulse in the frequency domain, in the sense that in these cases only the IDTFT exists, but not the inverse $\mathcal{Z}$-transform. The corresponding series only converge on the unit circle $|z|=1$.

$\endgroup$
1
  • 3
    $\begingroup$ Fixed the missing +1 :-). $\endgroup$
    – Royi
    May 28, 2023 at 12:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.