# Laplace To Z Transform Calculator

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 $\mathrm\mathitx\left ( n \right )$ is a discrete time function, then its Z-transform is defined as,

## Laplace To Z Transform Calculator

In order to solve the difference equation, first it is converted into the algebraic equation by taking its Z-transform. Then, the solution of the equation is calculated in z-domain and finally, the time-domain solution of the equation is obtained by taking its inverse Z-transform.

What is now called the Z-transform (named in honor of Lotfi Zadeh) was known to, mathematician and astronomer, Pierre-Simon Laplace around 1785. With the introduction of digitally sampled-data, the transform was re-discovered by Hurewicz in 1947, and developed by Lotfi Zadeh and John Ragazzinie around 1952, as a way to solve linear, constant-coefficient difference equations.\(\)

Unlike the continuous-time case, causal difference equations can be iterated just like a computer program would do. All one needs to do, is to rewrite the difference equation so that the term \(y[n]\) is on the left and then iterating forward in time. This will give each value of the output sequence without ever obtaining a general expression for \(y[n]\). In this article, we however will look for a general analytical expression for \(y[n]\) using the Z-transform.

The Z-transform can be thought of as an operator that transforms a discontinuous sequence to a continuous algebraic function of complex variable \(z\). As we will see, one of the nice feature of this transform is that a convolution in time, transforms to a simple multiplication in the \(z\)-domain.

We solve the difference equations, by taking the Z-transform on both sides of the difference equation, and solve the resulting algebraic equation for output \(Y(z)\), and then do the inverse transform to obtain \(y[n]\).

Start with the Laplace transform of sampled signal \(x_s(t)\) from equation \(\eqrefeq:fstarnT\) $$ \def\lfz#1\overset\Large#1\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\, \def\laplace\lfz\mathscrL \beginalign x_s(t) \laplace X_s(s)\triangleq&\int_0^-^\infty e^-st\overbrace\sum_n=0^\inftyx_c(nT)\ \delta(t-nT)^x_s(t)\ \mathrmdt \nonumber \\ =& \int_0^-^\infty \sum_n=0^\inftye^-st\,x_c(nT)\ \delta(t-nT)\ \mathrmdt \labeleq:laplace0 \endalign $$

After substituting \(\eqrefeq:est\) in \(\eqrefeq:laplace0\), the terms \(\mathrme^-snT\) and \(x(nT)\) are independent of \(t\) and can be taken outside of the integration. $$ \beginalign X(s)&=\int_0^-^\infty \sum_n=0^\infty e^-s\colorbluenTx_c(\colorbluenT)\ \delta(t-nT)\ \mathrmdt \nonumber \\ &= \sum_n=0^\infty e^-snTx_c(nT)\underbrace\int_0^-^\infty \delta(t-nT)\ \mathrmdt_\text=1 according to equation (\refeq:combcond) \nonumber \\ &= \sum_n=0^\infty\ e^-snT\ x_c(nT) \labeleq:Fs \endalign $$

The unilateral Z-transform of the discrete function \(x[n]\) follows as $$ \def\lfz#1\overset\Large#1\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\, \def\ztransform\lfz\mathcalZ \shaded x[n] \ztransform X(z)=\sum_n=0^\infty z^-n\ x[n] \labeleq:ztransform $$ in this complex polynomial, \(z\) is any \(z\in\mathbbC\).

Note that we use the notation \(\def\lfz#1\overset\Large#1\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\, \def\ztransform\lfz\mathcalZ \ztransform\) as equivalent to the more common Z-transform notation \(\mathfrakZ\left\\,x[n]\,\right\\).

Apply the Fourier transform of a product to \(x_s(t)\) and call it \(X_s(\omega)\) $$ \def\lfz#1\overset\Large#1\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\, \def\fourier\lfz\mathcalF \beginalign x_s(t) = x_c(t)\,s(t) \fourier&\frac12\pi\Large(\colorpurpleX_c(\omega)*S(\omega)\Large)\triangleq X_s(\omega) \labeleq:fstar2 \endalign $$

Recall the Fourier transform of the Dirac comb \(s(t)\) $$ \def\lfz#1\overset\Large#1\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\, \def\fourier\lfz\mathcalF s(t)=\sum _n=-\infty ^\infty \delta(t-nT) \fourier \frac 2\pi T\sum _k=-\infty ^\infty \delta \left(\omega -\frac 2\pi kT\right)\triangleq S(\omega) \nonumber $$

Substituting the Fourier transform of the Dirac comb in \(\eqrefeq:fstar2\) where \(\frac2\piT=2\pi f_s=\omega_s\), the angular sample frequency $$ \def\lfz#1\overset\Large#1\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\, \def\fourier\lfz\mathcalF \newcommand\ccancel[2][black]\color#1\cancel\colorblack#2 \newcommand\ccancelto[3][black]\color#1\cancelto#2\colorblack#3 \beginalign x_s(t) \fourier X_s(\omega) &= \frac1\ccancel[red]2\pi\colorpurpleX_c(\omega) * \left(\frac\ccancel[red]2\piT\sum_k=-\infty^\infty\delta(\omega-k\omega_s)\right) \nonumber \\ &= \frac1T\sum_k=-\infty^\infty\colorpurpleX_c(\omega) * \delta(\omega-k\omega_s)\labeleq:XsjOmega0 \endalign $$

Equation \(\eqrefeq:XsjOmega\) implies that the Fourier transform of \(x_s(t)\) consists of periodically repeated copies of the Fourier transform of \(x_c(t)\). The copies of \(\colorpurpleX_c(\omega)\) are shifted by integer multiples of the sampling frequency and then superimposed as depicted below. Frequency-domain representation of sampling

For those familiar with the Laplace transform, we will map specific features between the \(s\) and \(z\)-domain: The origin \(s=0\) of the \(s\)-plane is mapped to \(z=e^0=1\) on the real axis in \(z\)-plane.

Each vertical line \(\sigma=\sigma_0\) in \(s\)-plane is mapped to a circle \(z=e^\sigma_0\) centered about the origin in \(z\)-plane. E.g. Leftmost vertical line \(\sigma\to-\infty\) is mapped as the origin, where \(z=\mathrme^-\infty=0\)

The imaginary axis \(\sigma=0\) is mapped as the unit circle, where \(z=\mathrme^0=1\)

Rightmost vertical line \(\sigma\to\infty\) is mapped as a circle with an infinite radius, where \(z=\mathrme^\infty=\infty\).

Each horizontal line \(j\omega=j\omega_0\) in \(s\)-plane is mapped to an angle from the origin in \(z\)-plane of angle \(\omega_0 T\) with respect to the positive horizontal direction.

Recall the definition of the Z-transform from equations \((\refeq:ztransform,\refeq:zalt,\refeq:zphi)\) $$ \def\lfz#1\overset\Large#1\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\, \def\ztransform\lfz\mathcalZ f[n]\ztransform\sum_n=0^\infty z^-n\ f[n]\\ \textwhere z\triangleqz\,\mathrme^j\varphi\text, z\triangleq \mathrme^\sigma T\text, \varphi\triangleq \omega T $$

The power series for the Z-transform is called a Laurent series. The Laurent series, represents an analytic function at every point inside the region of convergence. Therefore, the Z-transform and all its derivatives must be continuous function of \(z\) inside the region of convergence.

Laurent series converge in an annular (=ring shaped) region of the \(z\)-plane, bounded by poles. The set of values of \(z\) for which the Z-transform converges is called the region of convergence (ROC). This means that anytime we use the Z-transform, we need to keep the region of convergence in mind.

The goal of the subject is to gain necessary theoretical background in the field of integral calculus and integral transforms and practical skills in solving various examples and problems of fundamental integral calculus, and integral transforms.

I am currently taking an engineering course on Signals and Systems,where we spend a big deal of our time studying many types of transforms ( e.g. Laplace Transform,Fourier transform, Z-transform,Hilbert transform,etc...). - We study these transforms to solve a wide range of engineering problems which would be very hard and tedious to solve without utilizing a transform.For example, it is very easy to solve ordinary diffrential equations and convolution integrals using the laplace transform. - My professor pledged that if someone in class submitted a new transform with some helpful properties,he would put a full grade on the course for that student. - So,can somebody please refer me to some pages where i can learn about the mathematics behind transforms and their theory? - And under what branch of mathematics do they fall after all? - - Thanks in Advance - Hisham1987 13:07, 11 April 2007 (UTC)Reply[reply]