Difference Between Laplace Transform And Z-Transform

If you are Searching for the difference between Laplace and z transform then here are the Relationship and Difference Between The Z-Transform And The Laplace Transform.

Z-transform

The unilateral or one-sided Z-transform is simply the Laplace transform of an ideally sampled signal with the substitution of

z =def esT $$z\stackrel{\mathrm{d}\mathrm{e}\mathrm{f}}{=}{e}^{sT}$$

where T=1/fs $T=1/f_s$is the sampling period (in units of time e.g., seconds) and fs $f_s$is

 the sampling rate (in samples per second or hertz)

Let

ΔT(t) =def ∑n=0∞δ(t−nT)$${\mathrm{\Delta }}_T(t)\stackrel{\mathrm{d}\mathrm{e}\mathrm{f}}{=}\sum _{n=0}^{\mathrm{\infty }}\delta (t-nT)$$

be a sampling impulse train (also called a Dirac comb) and

xq(t)=def x(t)ΔT(t)=x(t)∑n=0∞δ(t−nT)=∑n=0∞x(nT)δ(t−nT)=∑n=0∞x[n]δ(t−nT)$$\begin{array}{rl}{x}_q(t)& \stackrel{\mathrm{d}\mathrm{e}\mathrm{f}}{=}x(t){\mathrm{\Delta }}_T(t)=x(t)\sum _{n=0}^{\mathrm{\infty }}\delta (t-nT)\\ & =\sum _{n=0}^{\mathrm{\infty }}x(nT)\delta (t-nT)=\sum _{n=0}^{\mathrm{\infty }}x[n]\delta (t-nT)\end{array}$$

be the continuous-time representation of the sampled x(t) \

x[n] =def x(nT) $$x[n]\stackrel{\mathrm{d}\mathrm{e}\mathrm{f}}{=}x(nT)$$

are the discrete samples of x(t) The Laplace transform of the sampled signal x_q(t) \ is

Xq(s)=∫∞0−xq(t)e−stdt=∫∞0−∑n=0∞x[n]δ(t−nT)e−stdt=∑n=0∞x[n]∫∞0−δ(t−nT)e−stdt=∑n=0∞x[n]e−nsT.$$\begin{array}{rl}{X}_q(s)& ={\int }_{0^{-}}^{\mathrm{\infty }}{x}_q(t)e^{-st}dt\\ & ={\int }_{0^{-}}^{\mathrm{\infty }}\sum _{n=0}^{\mathrm{\infty }}x[n]\delta (t-nT)e^{-st}dt\\ & =\sum _{n=0}^{\mathrm{\infty }}x[n]{\int }_{0^{-}}^{\mathrm{\infty }}\delta (t-nT)e^{-st}dt\\ & =\sum _{n=0}^{\mathrm{\infty }}x[n]e^{-nsT}.\end{array}$$

This is precisely the definition of the unilateral Z-transform of the discrete function x[n] $x[n]$.

X(z)=∑n=0∞x[n]z−n$$X(z)=\sum _{n=0}^{\mathrm{\infty }}x[n]z^{-n}$$

with the substitution of z←esT $z\leftarrow e^{sT}$.

Comparing the last two equations, we find the relationship between the unilateral Z-transform

 and the Laplace transform of the sampled signal:

Xq(s)=X(z)∣∣z=esT$$X_q(s)=X(z){|}_{z=e^{sT}}$$

The similarity between the Z and Laplace transforms is expanded upon in the theory of

 time scale calculus.

Read Also What Is Z-Transform and Inverse Z-Transform In Easy Way

Laplace Transform

The main drawback of fourier transform (i.e. continuous F.T.) is that it can be defined only for stable systems. Where as, Laplace Transform can be defined for both stable and unstable systems.

Following are the Laplace transform and inverse Laplace transform equations.

Following table mentions Laplace transform of various functions.

To convert Laplace transform to Fourier tranform, replace s with j*w, where w is the radial frequency. in units of radians per second (rad/s).

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