# How do I examine if the signal can be z-transformed?

It's given signal

$$x[n]=\sin(\frac{2 \pi}{N} m n) u[n]$$

where $$u[n]$$ is the unit step function. Can I calculate Z-Transform?

$$\mathcal{Z}$$ transform exists when $$\sum_{n=-\infty}^{\infty} x[n]z^{-n} < \infty$$

Now I'm confused I would say it doesn't exist because it has an infinite number of elements and its sum is infinite but then I check table for transform and transform from sigma function exist? It just jumps into infinity. Can someone tell me where I am making mistakes..

If you just want to check whether the $$\mathcal{Z}$$-transform of a sequence exists, it is not necessary to actually compute its $$\mathcal{Z}$$-transform. Fat32's answer showed you how to do that for your example.

The $$\mathcal{Z}$$-transform of a sequence $$x[n]$$ exists if for a set of points in the complex plane

$$\left|\sum_{n=-\infty}^{\infty}x[n]z^{-n}\right|<\infty\tag{1}$$

is satisfied. Note that in the formula in your question you forgot the magnitude, which is essential.

For your example it is straightforward to show that $$(1)$$ is satisfied for $$|z|>1$$:

\begin{align}\left|\sum_{n=-\infty}^{\infty}x[n]z^{-n}\right|&=\left|\sum_{n=0}^{\infty}\sin(\omega_0n)z^{-n}\right|\\&\le \sum_{n=0}^{\infty}\big|\sin(\omega_0n)\big|\;|z|^{-n}\\&\le\sum_{n=0}^{\infty}|z|^{-n}=\frac{1}{1-|z|^{-1}},\qquad |z|>1\tag{2}\end{align}

where I've used $$|\sin(\omega_0n)|\le 1$$.

Eq. $$(2)$$ shows that the $$\mathcal{Z}$$-transform of the given sequence exists for $$|z|>1$$. The region $$|z|>1$$ in the complex plane is called region of convergence (ROC).

You can use the basic properties of Z-transform to find the result:

Given, $$x[n] = \sin(\frac{2 \pi}{N} m n) u[n] = \sin(\omega_m n) u[n]$$

Decompose the sine wave into complex exponentials:

$$x[n] = \frac{1}{2 j} \left( e^{j \omega_m n} - e^{-j \omega_m n} \right) u[n] \tag{1}$$

And observe the following :

• 1 : $$U(z) = \mathcal{Z} \{ u[n] \} = \frac{1}{1-z^{-1}} ~~~~,~~~ |z| > 1$$

• 2 :

if $$x[n] \longleftrightarrow X(z)$$ then $$e^{j \omega_m n} ~ x[n] \longleftrightarrow X( z ~e^{ -j \omega_m} ) ~~~,~~~\text{ ROC the same}$$

Then apply these two observations to Eq (1) :

\begin{align} X(z) &= \mathcal{Z} \{ x[n] \} = \mathcal{Z} \{ \frac{1}{2 j} \left( e^{j \omega_m} - e^{-j \omega_m} \right) u[n] \} \\ &= \frac{1}{2 j} \left( \mathcal{Z} \{ e^{j \omega_m} u[n] \} - \mathcal{Z} \{ e^{-j \omega_m} u[n] \} \right) \\ &= \frac{1}{2 j} \left( U( z e^{-j \omega_m} ) - U( z e^{j \omega_m} ) \right) \\ &= \frac{1}{2 j} \left( \frac{1}{1-e^{j \omega_m} z^{-1}} - \frac{1}{1-e^{-j \omega_m} z^{-1}} \right) \\ \end{align}