# Time Shifting Property of Z-Transform

Signals and SystemsElectronics & ElectricalDigital Electronics

## Z-Transform

The Z-transform is a mathematical tool which is used to convert the difference equations in discrete time domain into the algebraic equations in z-domain. Mathematically, if $\mathrm{\mathit{x\left ( n \right )}}$ is a discrete time function, then its Z-transform is defined as,

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

## Time Shifting Property of Z-Transform

Statement – The time shifting property of Z-transform states that if the sequence $\mathrm{\mathit{x\left ( n \right )}}$ is shifted by n0 in time domain, then it results in the multiplication by $\mathrm{\mathit{z^{-n_{\mathrm{0}}}}}$ in the z-domain. Therefore, if

$$\mathrm{\mathit{x\left ( n \right )\overset{ZT}{\leftrightarrow}X\left ( z \right );\: \: \mathrm{ROC}\mathrm{\, =\, }\mathit{R} }}$$

With zero initial conditions.

Then, according to the time shifting property,

$$\mathrm{\mathit{x\left ( n-n_{\mathrm{0}} \right )\overset{ZT}{\leftrightarrow}z^{-n_{\mathrm{0}}}\, X\left ( z \right )}}$$

With ROC = R, except for the possible addition and deletion of 𝑧 = 0 or 𝑧 = ∞

## Proof

From the definition of the Z-transform, we have,

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

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

Substituting $\mathrm{\mathit{\left ( n-n_{\mathrm{0}}\right )=m }}$ in the above summation, then we have,

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

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

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

Also, it can be represented as,

$$\mathrm{\mathit{x\left ( n-n_{\mathrm{0}} \right )\overset{ZT}{\leftrightarrow}z^{-n_{\mathrm{0}}}X\left ( z \right )}}$$

Similarly, if signal is advanced in time, then according to the time shifting property, we get,

$$\mathrm{\mathit{x\left ( n\mathrm{\, +\, }n_{\mathrm{0}} \right )\overset{ZT}{\leftrightarrow}z^{n_{\mathrm{0}}}X\left ( z \right )}}$$

Also, if the initial conditions are not neglected, then

• The time shift property for time delay is,

$$\mathrm{\mathit{Z\left [ x\left ( n-n_{\mathrm{0}} \right ) \right ]\mathrm{\, =\, }z^{-n_{\mathrm{0}}}X\left ( z \right )\mathrm{\, +\, }z^{-n_{\mathrm{0}}}\sum_{p\mathrm{\, =\, }\mathrm{1}}^{n_{\mathrm{0}}}x\left ( -p \right )z^{p}}}$$

• The time shifting property for time advance is,

$$\mathrm{\mathit{Z\left [ x\left ( n\mathrm{\, +\, }n_{\mathrm{0}} \right ) \right ]\mathrm{\, =\, }z^{n_{\mathrm{0}}}X\left ( z \right )-z^{n_{\mathrm{0}}}\sum_{p\mathrm{\, =\, }\mathrm{0}}^{n_{\mathrm{0}}-\mathrm{1}}x\left ( p \right )z^{-p}}}$$

## Numerical Example (1)

Using the time shifting property of Z-transform, find the Z-transform of the sequence,

$$\mathrm{\mathit{x\left ( n \right )\mathrm{\, =\, }u\left ( n-\mathrm{3} \right ) }}$$

### Solution

The given sequence is,

$$\mathrm{\mathit{x\left ( n \right )\mathrm{\, =\, }u\left ( n-\mathrm{3} \right ) }}$$

Since the transform of a unit step sequence is given by,

$$\mathrm{\mathit{Z\left [ u\left ( n \right ) \right ]\mathrm{\, =\, }\frac{z}{z-\mathrm{1}};\: \: \mathrm{ROC}\to \left|z \right|>\mathrm{1}}}$$

Therefore, using the time shifting property of Z-transform $\mathrm{\mathit{\left [ \mathrm{i.e.,}\: x\left ( n-n_{\mathrm{0}} \right )\overset{ZT}{\leftrightarrow}z^{-n_{\mathrm{0}}}X\left ( z \right ) \right ]}}$, we get,

$$\mathrm{\mathit{Z\left [ u\left ( n-\mathrm{3} \right ) \right ]\mathrm{\, =\, }z^{-\mathrm{3}}Z\left [ u\left ( n \right ) \right ]\mathrm{\, =\, }z^{-\mathrm{3}}\left ( \frac{z}{z-\mathrm{1}} \right )}}$$

$$\mathrm{\mathit{\therefore Z\left [ u\left ( n-\mathrm{3} \right ) \right ]\mathrm{\, =\, }\frac{\mathrm{1}}{z^{\mathrm{2}}\left ( z-\mathrm{1} \right )};\; \; \mathrm{ROC}\to \left|z \right|>\mathrm{1}}}$$

## Numerical Example (2)

Using the time shifting property of Z-transform, find the Z-transform of the following sequence

$$\mathrm{\mathit{x\left ( n \right )\mathrm{\, =\, }\delta \left ( n\mathrm{\, +\, }\mathrm{5} \right )}}$$

### Solution

The given sequence is,

$$\mathrm{\mathit{x\left ( n \right )\mathrm{\, =\, }\delta \left ( n\mathrm{\, +\, }\mathrm{5} \right )}}$$

Since the Z-transform of the impulse sequence is given by,

$$\mathrm{\mathit{Z\left [ \delta \left ( n \right ) \right ]\mathrm{\, =\, }\mathrm{1}}}$$

Now, using the time shifting property of Z-transform $\mathrm{\mathit{\left [ \mathrm{i.e.,}\: x\left ( n\mathrm{\, +\, }n_{\mathrm{0}} \right )\overset{ZT}{\leftrightarrow}z^{n_{\mathrm{0}}}X\left ( z \right ) \right ]}}$, we get,

$$\mathrm{\mathit{Z\left [ \delta \left ( n\mathrm{\, +\, }\mathrm{5} \right ) \right ]\mathrm{\, =\, }z^{\mathrm{5}}\left ( \mathrm{1} \right )\mathrm{\, =\, }z^{\mathrm{5}}}}$$

Updated on 29-Jan-2022 08:14:46