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..


2 Answers 2


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


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).

  • $\begingroup$ regarding first equation, is it absolute summable -or- absolute value of z-transform? If it is absolute summable then the sigma symbol should be outside. $\endgroup$
    – abhilash
    Commented Jun 12, 2020 at 8:28
  • $\begingroup$ @abhilash: It is intended as written. $\endgroup$
    – Matt L.
    Commented Jun 12, 2020 at 8:35

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) :

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


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