Relation between Laplace Transform and Z-Transform

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 x(n) is a discrete-time signal or sequence, then its bilateral or two-sided Z-transform is defined as −

$$\mathrm{Z[x(n)] \:=\: X(z) \:=\: \sum_{n=-\infty}^{\infty}\: x(n) z^{-n} \quad \dotso\: (1)}$$

Where, z is a complex variable.

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

$$\mathrm{Z[x(n)] \:=\: X(z) \:=\: \sum_{n=0}^{\infty}\: x(n) z^{-n} \quad \dotso\:(2)}$$

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 x(t) is a continuous-time function, then its Laplace transform is defined as −

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

Equation (1) gives the bilateral Laplace transform of the function x(t). But for the causal signals, the unilateral Laplace transform is applied, which is defined as −

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

Relation between Laplace Transform and Z-Transform

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

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

Taking the Laplace transform on both sides, we get,

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

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

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

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

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

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

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

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

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