This question concerns the relation between the Z transform and the Laplace S transform. It is so that the z-Transform is the discrete time Laplace transform.

Please follow this proof:

F(S) = Integral f(t) exp -St dt from 0 to infinity.

Let us descretize t such t= nTs with Ts is the sampling time and and is the sampling instant. Ts=1/fs with fs is the sampling frequency.

Substituting the descretizing condition in the Laplace transform ,then we have:

F(S) = SUM f(nTs) exp - n S Ts

Now one can define F(z) as z= exp STs,

F(z) = SUM f(nTs) z^-n

The last equation is the z transform of the discrete time function f(nTs)

In order to map the z-plane one has to locate the jw axis of the S -plane on the Z-plane where sigma=0.

It follows z at the jw axcess = expjwTs,

So as a complex quantity the magnitude of z =1 and its angle =wTs= 2pif/fs

which is called normalized radian frequency.

So the line corresponds to jw axis in the S plane is a circle whose amplitude is eqaul to one and it is called the unity circle.

Best wishes

The general form is for Laplace. The assumptions that r=1 and sigma=0 are for Fourier, which is a special case of Laplace and gives us the frequency response of the system.

It is not true that we "always" set σ=0.

I suppose that you know WHY we have introduced the complex frequency s=σ + jw.

However, in case we want to

* measure/simulate the frequency respose of a system, or

* show (for example, in a BODE diagram) how the magnitude and phase changes with frequency ,

it is necessary to set σ=0 (s=jw) because this corresponds with real world conditions. You are not able to produce a test signal s=σ + jw that contains a real part σ .

The complex variables 'z' and 's' refer respectively to Zeta Transform (for discrete signals) and Laplace Transform (for continuous signals).

The Discrete Fourier Transform is a particular case of Zeta Transform, and can be obtained when z=e^jw. It means that the DFT is the zeta transform evaluated on the unit circle of the complex plane.

In the same way, the Fourier Transform is a particular case of Laplace Transform, and can be calculated at s=jw.

Dear Raghuram:

1. Z transform is discrete counter part of lapalace transform and it is also a generalization of Discrete time fourier transform (DTFT)

Actually Z=r*e^jw, and if you represent Z transform in the form of DTFT then, it is represented as X(z)= F{ x[n]*r^-n}

so if you take r=1, then X(z)= F{ x[n]} and we can analysed a system's stability by making it equivalent to DTFT. r=1 means a circle of radius one in Z plane and that's why we always check poles inside the unit circle for stability of signal.

DTFT exist for stable signals while Z transform may exists for unbounded /unstable signals (because of the presence of extra term r^-n) and by making this extra term as one we can check stability of signals

2. (a) From the perspective of DSP:

Similarly, Laplace transform is generalization of continuous time fourier transform and it can be represented as X(s)= F{ x(t)*e^-σ*t} and we make this extra term as one by σ =0 and check stability of signal/system. and that's why we check poles on left side of line σ =0 i.e. imaginary axis in s plane

(b) from network point of view:

If you want to analysed both transient and steady state behavior of a circuit then we can't take S=jw , because if you make σ =0 then you ignore transient behavior of circuit and only steady state behavior of circuit can be known

i.e. if you replace directly s=jw in a circuit then only steady state behavior of circuit can be known and to know the transient behavior of circuit , σ should be non zero

The reason behind this is " magnitude of e^-jw is always one and it already assume that circuit response attend steady state while magnitude of e^-( σ+jw) is not always one and magnitude depends upon the term e^-σ i.e. this σ consider the damping of circuit's response means transient of response.

I Hope You understand this.

The method to map variable s to z (and viceversa) is the bilinear transformation

http://en.wikipedia.org/wiki/Bilinear_transform

It is a basic question in Electric Circuit Analysis. When dealling with sinusoids, we rely on Euler's Identity definiton of e^jw=cosw+jsinw for sinisoids. And we add V*e^(jw*t)=Vm*e^j*(w*t+phi) for new definition as sinor representation. Now, we can define phasor notation as V=Vm*e^(j*phi) in specific frequency w for AC circuit analysis. It is all about frequency analysis fo circuit in sinusoids. Then, all periodic signals can be decomposed into finite terms in various frequency called harmonics. We can solved the circuit response for given periodic signals by superpositon of finite AC reponses.

Thank you for all contributor about this useful e discussion , I hope i found answer for my inquiry which is .

What is the correct way to perform Z transform on Laplaced functions f(S)

1) multiply f(S) by Z^-n under summation,

2) or just replace e^st by Z?

I have transformed function f(S) = Z tanh (a/2*S) ,,, z and a are constant .I want to perform Z transform for this function ,

I came across this discussion only today. So my answer is perhaps too late.

I feel that there are quite a lot of misunderstandings, not only in the questions, but also in the answers. Maybe, some comments on how these mathematical methods have been developed will be helpful.

It started with Fourier's attempt to solve some differential equations (in connection with problems of heat and the distribution of energy in a solid body). Fourier's idea was to approach solutions by sinusoidal functions, since these (in contrast to polynomials and quite a lot of other functions) do not increase to infinity, when the argument tends to infinity. This was important, since he wanted to let tend time-argument t to infinity.

Fourier used a superposition of sin- and cosine-functions. For simplicity, they all had a common period, i.e., periods were T, T/2, T/3, etc. The mathematical advantage of this choice is that these functions can be reformulated in a way that they form an orthonormal system of functions. By doing so, Fourier created what we nowadays call a Fourier series approach of functions in an interval of length T. The sin-function with fundamental period T is sin( t *2*pi/T) . That means: one period (or full oscillation) per time T. The frequency of oscillation is thus 1 oscillation per T. Therefore 1/T -- also expressed as f -- is frequency. However, for conciseness, it is easier to use the abbreviation W = 2*pi/T. (This is angular frequency and not frequency!). The sin-part of the fundamental oscillation is sin(W*t) thus.

For best approach of a function, we must use sin and cos-functions with certain amplitudes, from which our known formulas stem: sum(A_i cos(i*W*t) + B_i sin(i*W*t)), where index i runs from 0 to a finite natural number I.

Letting tend I towards infinity generates a mathematical expression that under certain conditions converges. These conditions are that the functions to be approached have finite energy per period (mathematically speaking: they must be elements of a suitable function space ).

It is easier to use complex exponentials rather than sin and cos functions. Therefore, reformulation results in sum(Ci*exp(j*i*W*t)), where index i runs from -infinity to +infinity.

Now, this is the origin of the exponential and of discrete angular frequency W.

The next step is to widen the period of the function to be approached from T to infinity. Under suitable conditions for the functions to be approached, the infinite sum (series) will then turn into an integral, the well-known Fourier integral. Suitable condition for the function to be approached means: the infinite integral of the squared absolute value of the function must exist and have a final value!

This is now the point where problems occur. Many functions of interest do not belong to that class of functions. However, in electrical engineering, there are many problems, where the exact description of an action is only interesting from points in time on, from which on the process is to be observed. Call that point in time t = 0.

Next, we are often only interested in the action for a certain interval of time only. Therefore, if we multiply the true solution of our problem by H(t) * exp(sigma*t), with H being Heaviside's jump-function, sigma being a positive, but sufficiently small real number, then we obtain a function that is 0 for negative times (the past) and that deviates from the true function only a little for a certain time. We approach our solution therefore by multiplication with this factor function and hope that now the Fourier integral converges. It does in many more cases than before. The result is the integral of the Laplace-transform by setting j*w+sigma equal to p or sometimes to s). Laplace-transform is thus a special case of Fourier transform, and not vice versa.

And what is the relation to z-transform?

In the Fourier approach , we approache a given function within a complete interval. When deriving discrete Fourier transform or -- in a variant similar to Laplace transform -- z-transform, we do this approach at certain equidistant points only. Therefore, we end up wit series here, rather than with integrals. Thats it!

Well... s = jW is the imaginary axis of the s-plan and z = e^jw is the unit circle of the z-plan, and it´s along these lines that the frequency response of continuous time and discrete time systems is measured.

The Laplace transform is a generalisation of the continuous-time Fourier transform. Certain signals do not have a Fourier transform (the integral diverges), such as the step function. Therefore, a convergence term e^(-σt) is multiplied with the signal before taking the Fourier transform. The idea is that for certain values of σ, the decaying exponential convergence term should allow the resulting integral to converge. Multiplying e^(-σt) with e^{-jωt} gives us e^-{σ+jω)t, where we now have a complex variable of two components (real and imaginary). Making the substitution of s=σ+jω , we have the Laplace transform. Note that when σ = 0, the Laplace transform reduces down to the Fourier transform. This can be visually seen in the complex plane as the imaginary axis (jω-axis). All complex values on this axis are imaginary, hence s=jω. All values of s (and therefore, σ) where the Laplace transform converges is called the region of convergence (ROC), and if this ROC encompasses the imaginary axis, this means the Fourier transform also exists (i.e. converges).

The Fourier transform of a discrete-time signal is called the discrete-time Fourier transform (DTFT). We can loosely arrive at the formula by assuming our discrete-time signal x(n) is modelled as the multiplication of the continuous-time signal x(t) with a train of impulses. Taking the Fourier transform of this product, rearranging, and using the sifting property of delta functions, we can easily arrive at the DTFT formula, which is an infinite summation of samples multiplied by e^(-jωn). Now again, based on similar logic to the rationale for the Laplace transform, certain signals do not have a converging DTFT, so a convergence term is multiplied r^(-n) to ensure the infinite sum converges to a finite value. Multiplying this with the exponential, we get [re^(jω)]^-n, which gives us a complex value in polar form (c.f. rectangular complex form of Laplace variable s). We make the substitution of z=re^(jω) and arrive at the familiar formula for the z-transform.

Now, for the special case when r=1 (which corresponds to the unit circle), the z-transform is equivalent to the DTFT and z=e^(jω), where ω is now the polar angle (c.f. with ω being the rectangular imaginary component in the Laplace case). This neatly explains why the DTFT has a periodicity of 2π radians, since when you trace around a unit circle, you reach 0 again after a revolution of 2π.

When you sample a continuous signal, you model this process by multiplying a train of impulses by the analogue signal. You then take the Laplace transform (you could take the Fourier transform). From that you end up with an exp(-snT) term and a summation. (where n is an integer n=0,1,2,3... and T the sampling interval). You then define a different transform, the Z transform from this by defining z=exp(sT). This is true for s=σ + jw. So for a complex s you get a corresponding complex z. So for a condition with s on the jw axis the real part of s is zero and z=exp(jwT). So when we work out frequency response of z transforms we use this expression (or the frequency normalised version when theta=wT and z=exp(jtheta) ). Otherwise we keep the real part there when for instance working out transient parts of say a step response.

Does it help you understand if one simply points out:

- In the case of the s-plane things are continuous (and perhaps limited extent) in time and the transforms have infinite extent. So the axes in the s-plane have infinite extent.

- In the case of the z-plane things are discrete and the transforms are periodic. "Frequency" is completely determined in a single period. So in the z-plane, one of the corresponding "frequency axis" is a unit circle.

Factoid: One can consider the periodic frequency values to be, as usual, over a single period only. But, for convenience of visualization or follow-on computations, one can consider 2 or more periods. This might be handy for example when doing temporal interpolation / sample rate increase when there will be computations or specifications given in the frequency domain.

so to compress some of the answers, such as Francisco Restivo's and Steven So's (and maybe others) the Laplace transform over the imaginary axis becomes the Continuous Fourier Transform (C-FT), and the Z-transform over the unit circle becomes the Discrete Time Fourier Transform (DTFT). This is the meaning of the substitutions that were stated in the original question. Whenever we are interested in the C-FT or the DTFT but we start from the Laplace or the Z- transforms, we just perform the stated substitutions.

When you assume r=1 and σ=0 you get to the particular case of Fourier transforms rather than the general case of Laplace transforms as Moshe Nazarathy summarised. The importance of the Fourier transform is that it gives us the frequency response of the system.

Do not forget: if you start from Laplace-transform, you implicitely suppose that the function to be transformed from time-domain to (complex) frequency domain is 0 for negative time arguments! In many cases, this does not matter. However, there are cases, where this may lead to mathematical problems, or where you may miss important terms (set-up terms).

In order not to get a wrong impression: Laplace-transform is a powerful mathematical tool. I like it very much. However, as with every all-days tool, we should always use the appropriate tool. Sometimes, Laplace transform is more appropriate, sometimes it is Fourier transform. (And for discrete applications or numerics, this applies for z-transform and DTFT).

What I am a bit surprised (and disappointed) is the following:

Nearly two month ago there was a question from RG member Raghuram Tr .

As a result, we had about 15 answers - and it would be interesting for all of us to know if we could satisfy the questioner. But he never gave any response.

Did he read all the answers at all?

Following from what Michael Hoffmann wrote, what the Laplace transform adds is the e^(σt) term, which guarantees that the whole thing is bound amplitudewise, i.e. it converges, so sometimes the FT does not exist but the Laplace Transform does :-)

To Mr. Wangenheim

At least i wrote these answers carefully , i get fruitful information , i hope all contributions do the same thing for all question .

This question concerns the relation between the Z transform and the Laplace S transform. It is so that the z-Transform is the discrete time Laplace transform.

Please follow this proof:

F(S) = Integral f(t) exp -St dt from 0 to infinity.

Let us descretize t such t= nTs with Ts is the sampling time and and is the sampling instant. Ts=1/fs with fs is the sampling frequency.

Substituting the descretizing condition in the Laplace transform ,then we have:

F(S) = SUM f(nTs) exp - n S Ts

Now one can define F(z) as z= exp STs,

F(z) = SUM f(nTs) z^-n

The last equation is the z transform of the discrete time function f(nTs)

In order to map the z-plane one has to locate the jw axis of the S -plane on the Z-plane where sigma=0.

It follows z at the jw axcess = expjwTs,

So as a complex quantity the magnitude of z =1 and its angle =wTs= 2pif/fs

which is called normalized radian frequency.

So the line corresponds to jw axis in the S plane is a circle whose amplitude is eqaul to one and it is called the unity circle.

Best wishes

TO GO FROM ANALOG

TO DIGITAL

WE MAP THE jw AXIS

TO THE UNIT CIRCLE .

THIS IS DONE

BY SAMPLING

THE ANALOG SIGNAL .