# Signals and Systems – Relation between Laplace Transform and Z-Transform

Signals and SystemsElectronics & ElectricalDigital Electronics

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

$$\mathrm{\mathit{Z}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n}\right)}\right]}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\sum_{\mathit{n=-\infty}}^{\infty}\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathit{z^{-\mathit{n}}}\:\:\:\:\:\:...(1)}$$

Where, z is a complex variable.

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

$$\mathrm{\mathit{Z}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{n}\right)}\right]}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{z}\right)}\:\mathrm{=}\:\sum_{\mathit{n=\mathrm{0}}}^{\infty}\mathit{x}\mathrm{\left(\mathit{n}\right)}\mathit{z^{-\mathit{n}}}\:\:\:\:\:\:...(2)}$$

## Laplace Transform

The Laplace transform is a mathematical tool which is used to convert the differential equation in time domain into the algebraic equations in the frequency domain or s-domain.

Mathematically, if $\mathit{x}\mathrm{\left(\mathit{t}\right)}$ is a continuous-time function, then its Laplace transform is defined as −

$$\mathrm{\mathit{L}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{t}\right)}\right]}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{s}\right)}\:\mathrm{=}\:\int_{-\infty }^{\infty }\mathit{x}\mathrm{\left(\mathit{t}\right)}\mathit{e^{-\mathit{st}}\:\mathit{dt}}\:\:\:\:\:\:...(3)}$$

Equation (1) gives the bilateral Laplace transform of the function $\mathit{x}\mathrm{\left(\mathit{t}\right)}$. But for the causal signals, the unilateral Laplace transform is applied, which is defined as −

$$\mathrm{\mathit{L}\mathrm{\left[\mathit{x}\mathrm{\left(\mathit{t}\right)}\right]}\:\mathrm{=}\:\mathit{X}\mathrm{\left(\mathit{s}\right)}\:\mathrm{=}\:\int_{\mathrm{0} }^{\mathrm{\infty} }\mathit{x}\mathrm{\left(\mathit{t}\right)}\mathit{e^{-\mathit{st}}\:\mathit{dt}}\:\:\:\:\:\:...(4)}$$

## Relation between Laplace Transform and Z-Transform

Let $\mathit{x}\mathrm{\left(\mathit{t}\right)}$ is a continuous-time signal. The discrete-time version of this signal is $\mathit{x}^{*}\mathrm{\left(\mathit{t}\right)}$ and the signal $\mathit{x}^{*}\mathrm{\left(\mathit{t}\right)}$ is obtained by sampling $\mathit{x}\mathrm{\left(\mathit{t}\right)}$ with a sampling period of T seconds, in other words, the sequence $\mathit{x}^{*}\mathrm{\left(\mathit{t}\right)}$ is obtained by multiplying the signal $\mathit{x}\mathrm{\left(\mathit{t}\right)}$ with a sequence of impulses which are T seconds apart, i.e.,

$$\mathrm{\mathit{x}^{*}\mathrm{\left(\mathit{t}\right)}\:\mathrm{=}\sum_{\mathit{n=\mathrm{0}}}^{\infty }\mathit{x}\mathrm{\left(\mathit{nT}\right)}\mathit{\delta \mathrm{\left(\mathit{t-nT}\right)}}}$$

Taking the Laplace transform on both sides, we get,

$$\mathrm{\mathit{L}\mathrm{\left[ \mathit{x}^{*}\mathrm{\left(\mathit{t}\right)}\right]}\:\mathrm{=}\:\mathit{X}^{*}\mathrm{\left(\mathit{s}\right)}\:\mathrm{=}\:\mathit{L}\mathrm{\left[\sum_{\mathit{n=\mathrm{0}}}^{\infty }\mathit{x}\mathrm{\left(\mathit{nT}\right)}\mathit{\delta \mathrm{\left(\mathit{t-nT}\right)}}\right]}}$$

$$\mathrm{\Rightarrow \mathit{X}^{*}\mathrm{\left(\mathit{s}\right)}\:\mathrm{=}\:\sum_{\mathit{n=\mathrm{0}}}^{\infty }\mathit{x}\mathrm{\left(\mathit{nT}\right)}\mathit{L}\:\mathrm{\left[\mathit{\delta \mathrm{\left ( \mathit{t-nT}\right)}} \right ]}}$$

$$\mathrm{\Rightarrow \mathit{L}\mathrm{\left[ \mathit{x}^{*}\mathrm{\left(\mathit{t}\right)}\right]}\:\mathrm{=}\:\mathit{X}^{*}\mathrm{\left(\mathit{s}\right)}\:\mathrm{=}\:\sum_{\mathit{n=\mathrm{0}}}^{\infty}\mathit{x\mathrm{\left (\mathit{nT}\right)}}\mathit{e^{-nsT}}\:\:\:\:\:\:...(\mathrm{5})}$$

Now, the Z-transform of the sequence x(nT) is given by,

$$\mathrm{\mathit{Z}\mathrm{\left[ \mathit{x}\mathrm{\left(\mathit{nT}\right)}\right]}\:\mathrm{=\mathit{Z}\mathrm{\left[\mathit{x}^{*}\mathrm{\left(\mathit{t}\right)} \right ]}\:\mathrm{=}\:\sum_{\mathit{n=\mathrm{0}}}^{\infty}\mathit{x\mathrm{\left (\mathit{nT}\right)}}\mathit{z^{-n}}}\:\:\:\:\:\:...(\mathrm{6})}$$

From eqns. (5)&(6), we have,

$$\mathrm{ \mathit{L}\mathrm{\left[ \mathit{x}^{*}\mathrm{\left(\mathit{t}\right)}\right]}\:\mathrm{=}\:\mathrm{\left[\:\sum_{\mathit{n=\mathrm{0}}}^{\infty}\mathit{x\mathrm{\left (\mathit{nT}\right)}}\mathit{z^{-n}} \right ]}_{\mathit{z=e^{sT}}}}$$

Therefore, the relation between the Laplace transform and Z-transform is given by,

$$\mathrm{\mathit{L}\mathrm{\left[ \mathit{x}^{*}\mathrm{\left(\mathit{t}\right)}\right]}\:\mathrm{=}\:\mathit{Z\mathrm{\left[\mathit{x}^{*}\mathrm{\left(\mathit{t}\right)}\right]}}_{\mathit{z=e^{sT}}}}$$

Updated on 07-Jan-2022 07:01:24