# How can I solve such an inverse Z-transform?

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.

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:

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

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?

• 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 $(1)$?
– user64710
May 16 at 17:44
• 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. May 17 at 6:09
• @RubemPacelli The change comes from the relation between the Z-Transform and DTFT. May 17 at 8:26
• What is "C.O.V."? Calculus of variations? May 17 at 9:57
• "[...] 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. May 17 at 13:26

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 $$r_1<|z| for which $$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|.

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 $$|z|=1$$.

• Fixed the missing +1 :-).
– Royi
May 28 at 12:23