# Signals and Systems β Z-Transform of Sine and Cosine Signals

## Z-Transform

The Z-transform (ZT) is a mathematical tool which is used to convert the difference equations in time domain into the algebraic equations in z-domain.

Mathematically, if $\mathrm{\mathit{x\left ( n \right )}}$ is a discrete-time signal or sequence, then its bilateral or two-sided Z-transform is defined as −

$$\mathrm{\mathit{Z\left [ x\left ( n \right ) \right ]\mathrm{\, =\,}X\left ( z \right )\mathrm{\, =\,}\sum_{n\mathrm{\, =\,}-\infty }^{\infty }x\left ( n \right )z^{-n}}}$$

Where, z is a complex variable.

Also, the unilateral or one-sided z-transform is defined as −

$$\mathrm{\mathit{Z\left [ x\left ( n \right ) \right ]\mathrm{\, =\,}X\left ( z \right )\mathrm{\, =\,}\sum_{n\mathrm{\, =\,}\mathrm{0} }^{\infty }x\left ( n \right )z^{-n}}}$$

## Z-Transform of Causal Sine Sequence

The causal sine sequence is defined as,

$$\mathrm{\mathit{x\left ( n \right )\mathrm{\, =\,}\mathrm{sin}\,\omega n\: u\left ( n \right )\mathrm{\, =\,}\begin{Bmatrix} \mathrm{sin}\: \omega n &\mathrm{for}\: n\geq \mathrm{0} \ \mathrm{0} &\mathrm{for}\: n< \mathrm{0} \ \end{Bmatrix}}}$$

Therefore, the Z-transform of the sine sequence is obtained as follows −

$$\mathrm{\mathit{Z\left [ x\left ( n \right ) \right ]\mathrm{\, =\,}X\left ( z \right )\mathrm{\, =\,}Z\left [ \mathrm{sin}\: \omega n\: u\left ( n \right ) \right ]}}$$

$$\mathrm{\mathit{\Rightarrow X\left ( z \right )\mathrm{\, =\,}\sum_{n\mathrm{\, =\,}\mathrm{0}}^{\infty }\mathrm{sin}\left ( \omega n\:\right )z^{-n}}}$$

$$\mathrm{\mathit{\because \mathrm{sin}\, \omega n\mathrm{\, =\,}\frac{e^{j\, \omega n}-e^{-j\, \omega n}}{\mathrm{2}j}}}$$

$$\mathrm{\mathit{\therefore X\left ( z \right )\mathrm{\, =\,}\sum_{n\mathrm{\, =\,}\mathrm{0}}^{\infty }\left ( \frac{e^{j\, \omega n}-e^{-j\, \omega n}}{\mathrm{2}j} \right )z^{-n}}}$$

$$\mathrm{\mathit{\Rightarrow X\left ( z \right )\mathrm{\, =\,}\frac{\mathrm{1}}{\mathrm{2}j}\sum_{n\mathrm{\, =\,}\mathrm{0}}^{\infty }\left ( e^{j\, \omega n}-e^{-j\, \omega n} \right )z^{-n}}}$$

$$\mathrm{\mathit{\mathrm{\, =\,}\frac{\mathrm{1}}{\mathrm{2}j}\left [ \sum_{n\mathrm{\, =\,}\mathrm{0}}^{\infty }\left ( e^{j\, \omega\, }z^{-\mathrm{1}} \right )^{n}-\sum_{n\mathrm{\, =\,}\mathrm{0}}^{\infty }\left ( e^{-j\, \omega\, }z^{-\mathrm{1}} \right )^{n} \right ]}}$$

$$\mathrm{\mathit{\Rightarrow X\left ( z \right )\mathrm{\, =\,}\frac{\mathrm{1}}{\mathrm{2}j}\left [ \frac{\mathrm{1}}{\left ( \mathrm{1}-e^{j\, \omega\, }z^{-\mathrm{1}} \right )}-\frac{\mathrm{1}}{\left ( \mathrm{1}-e^{-j\, \omega\, }z^{-\mathrm{1}} \right )} \right ]}}$$

$$\mathrm{\mathit{\Rightarrow X\left ( z \right )\mathrm{\, =\,}\frac{\mathrm{1}}{\mathrm{2}j}\left [ \frac{z}{\left ( z-e^{j\, \omega\, } \right )}-\frac{z}{\left ( z-e^{-j\, \omega\, } \right )} \right ]\mathrm{\, =\,}\frac{\mathrm{1}}{\mathrm{2}j}\left [ \frac{z\left ( z-e^{-j\, \omega\, } \right )-z\left ( z-e^{j\, \omega\, } \right )}{\left ( z-e^{j\, \omega\, } \right )\left ( z-e^{-j\, \omega\, } \right )} \right ]}}$$

$$\mathrm{\mathit{\Rightarrow X\left ( z \right )\mathrm{\, =\,}\frac{\mathrm{1}}{\mathrm{2}j}\left [ \frac{z\left ( e^{j\, \omega}-e^{-j\, \omega\, } \right )}{z^{\mathrm{2}}-z\left (e^{j\, \omega}\mathrm{\, +\,}e^{-j\, \omega} \right ) \mathrm{\, +\,}\mathrm{1}} \right ]}}$$

$$\mathrm{\mathit{\because \left ( \frac{e^{j\, \omega}-e^{-j\, \omega}}{\mathrm{2}} \right )\mathrm{\, =\,}\mathrm{sin}\: \omega \; \; \mathrm{and}\; \; \left ( \frac{e^{j\, \omega}\mathrm{\, +\,}e^{-j\, \omega}}{\mathrm{2}} \right )\mathrm{\, =\,}\mathrm{cos}\: \omega }}$$

$$\mathrm{\mathit{\therefore X\left ( z \right )\mathrm{\, =\,}\frac{z\, \mathrm{sin}\, \omega }{z^{\mathrm{2}}-\mathrm{2}\, z\, \mathrm{cos}\, \omega \mathrm{\, +\,}\mathrm{1}}}}$$

This series converges for |π§| > 1. Therefore, the ROC of the Z-transform of the causal sine sequence is |π§| > 1. Hence, the Z-transform of a sine function along with its ROC is represented as,

$$\mathrm{\mathit{\mathrm{sin}\, \omega n\: u\left ( n \right )\overset{ZT}{\leftrightarrow}\left ( \frac{z\, \mathrm{sin}\, \omega }{z^{\mathrm{2}}-\mathrm{2}\, z\, \mathrm{cos}\, \omega \mathrm{\, +\,}\mathrm{1}} \right );\; \; }ROC\rightarrow \left|\mathit{z}\right|>1}$$

## Z-Transform of Causal Cosine Sequence

The causal cosine sequence is defined as −

$$\mathrm{\mathit{x\left ( n \right )\mathrm{\, =\,}\mathrm{cos}\, \omega n\: u\left ( n \right )\mathrm{\, =\,}\begin{Bmatrix} \mathrm{cos}\: \omega n &\mathrm{for}\: n\geq \mathrm{0} \ \mathrm{0} &\mathrm{for}\: n< \mathrm{0} \ \end{Bmatrix}}}$$

Thus, the Z-transform of the cosine sequence is obtained as follows −

$$\mathrm{\mathit{Z\left [ x\left ( n \right ) \right ]\mathrm{\, =\,}X\left ( z \right )\mathrm{\, =\,}Z\left [ \mathrm{cosin}\: \omega n\: u\left ( n \right ) \right ]}}$$

$$\mathrm{\mathit{\Rightarrow X\left ( z \right )\mathrm{\, =\,}\sum_{n\mathrm{\, =\,}\mathrm{0}}^{\infty }\mathrm{cos}\left ( \omega n\:\right )z^{-n}}}$$

$$\mathrm{\mathit{\because \mathrm{cos}\, \omega n\mathrm{\, =\,}\frac{e^{j\, \omega n}\mathrm{\, +\,}e^{-j\, \omega n}}{\mathrm{2}}}}$$

$$\mathrm{\mathit{\therefore X\left ( z \right )\mathrm{\, =\,}\sum_{n\mathrm{\, =\,}\mathrm{0}}^{\infty }\left ( \frac{e^{j\, \omega n}\mathrm{\, +\,}e^{-j\, \omega n}}{\mathrm{2}} \right )z^{-n}}}$$

$$\mathrm{\mathit{\Rightarrow X\left ( z \right )\mathrm{\, =\,}\frac{\mathrm{1}}{\mathrm{2}}\sum_{n\mathrm{\, =\,}\mathrm{0}}^{\infty }\left ( e^{j\, \omega n}\mathrm{\, +\,}e^{-j\, \omega n} \right )z^{-n}\mathrm{\, =\,}\frac{\mathrm{1}}{\mathrm{2}}\left [ \sum_{n\mathrm{\, =\,}\mathrm{0}}^{\infty }\left ( e^{j\, \omega\, }z^{-\mathrm{1}} \right )^{n}\mathrm{\, +\,}\sum_{n\mathrm{\, =\,}\mathrm{0}}^{\infty }\left ( e^{-j\, \omega\, }z^{-\mathrm{1}} \right )^{n} \right ] }}$$

$$\mathrm{\mathit{\Rightarrow X\left ( z \right )\mathrm{\, =\,}\frac{\mathrm{1}}{\mathrm{2}}\left [ \frac{\mathrm{1}}{\left ( \mathrm{1}-e^{j\, \omega\, }z^{-\mathrm{1}} \right )}\mathrm{\, +\,}\frac{\mathrm{1}}{\left ( \mathrm{1}-e^{-j\, \omega\, }z^{-\mathrm{1}} \right )} \right ]}}$$

$$\mathrm{\mathit{\Rightarrow X\left ( z \right )\mathrm{\, =\,}\frac{\mathrm{1}}{\mathrm{2}}\left [ \frac{z}{\left ( z-e^{j\, \omega\, } \right )}\mathrm{\, +\,}\frac{z}{\left ( z-e^{-j\, \omega\, } \right )} \right ]\mathrm{\, =\,}\frac{\mathrm{1}}{\mathrm{2}}\left [ \frac{z\left ( z-e^{-j\, \omega\, } \right )\mathrm{\, +\,}z\left ( z-e^{j\, \omega\, } \right )}{\left ( z-e^{j\, \omega\, } \right )\left ( z-e^{-j\, \omega\, } \right )} \right ]}}$$

$$\mathrm{\mathit{\Rightarrow X\left ( z \right )\mathrm{\, =\,}\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{z\left [ \mathrm{2}z-\left ( e^{j\, \omega}-e^{-j\, \omega\, } \right ) \right ]}{z^{\mathrm{2}}-z\left (e^{j\, \omega}\mathrm{\, +\,}e^{-j\, \omega} \right ) \mathrm{\, +\,}\mathrm{1}} \right\}}}$$

$$\mathrm{\mathit{\therefore X\left ( z \right )\mathrm{\, =\,}\frac{z\, \left ( z-\mathrm{cos}\, \omega \right ) }{z^{\mathrm{2}}-\mathrm{2}\, z\, \mathrm{cos}\, \omega \mathrm{\, +\,}\mathrm{1}}}}$$

This series also converges for |π§| > 1. Therefore, the ROC of the Z-transform of the causal cosine sequence is |π§| > 1. Hence, the Z-transform of a cosine function along with its ROC is represented as,

$$\mathrm{\mathit{\mathrm{cos}\, \omega n\: u\left ( n \right )\overset{ZT}{\leftrightarrow}\left [ \frac{z\, \left ( z-\mathrm{cos}\, \omega \right ) }{z^{\mathrm{2}}-\mathrm{2}\, z\, \mathrm{cos}\, \omega \mathrm{\, +\,}\mathrm{1}} \right ];\; \; }ROC\rightarrow \left|\mathit{z}\right|>1}$$