Step function Laplace transform

Now, to find the current/voltage relationship in the time domain, we only need to substitute V(5) for the particular excitation of interest and carry out the inverse Laplace transform for the entire equation. However, first we need to know how the particular excitation we want to apply looks in the Laplace domain. Here is an example of how to do it for a voltage step. In the time domain it can be expressed as v(0 = Heaviside(t) Vi where the function Heaviside(/) takes a value 1 if i > 0 and 0 otherwise. Vi is the value of voltage applied during the step. The Laplace transform of this function, V s) =. [Heaviside(i) Vj] = VJs. To find the Laplace current of the above mentioned network to pulse excitation, we substitute this V(x) into Eq. (4) and perform inverse Laplace transformation of the resulting equation. We obtain, ... 

In the context of chemical kinetics, the eigenvalue technique and the method of Laplace transforms have similar capabilities, and a choice between them is largely dependent upon the amount of algebraic labor required to reach the final result. Carpenter discusses matrix operations that can reduce the manipulations required to proceed from the eigenvalues to the concentration-time functions. When dealing with complex reactions that include irreversible steps by the eigenvalue method, the system should be treated as an equilibrium system, and then the desired special case derived from the general result. For such problems the Laplace transform method is more efficient. 

We first define the unit step function (also called the Heaviside function in mathematics) and its Laplace transform 1... 

The Laplace transform of the unit step function (Fig. 2.3) is derived as follows ... 

With the result for the unit step, we can see the results of the Laplace transform of any step function f(t) = Au(t). 

The Laplace transform of a step function is essentially the same as that of a constant in (2-7). When you do the inverse transform of A/s, which function you choose depends on the context of the problem. Generally, a constant is appropriate under most circumstances. 

The rectangular pulse can be generated by subtracting a step function with dead time T from a step function. We can derive the Laplace transform using the formal definition... 

The final step should also has zero initial condition C (0) = 0, and we can take the Laplace transform to obtain the transfer functions if they are requested. As a habit, we can define x = V/Qin s and the transfer functions will be in the time constant form. 

Instead of spacing out in the Laplace-domain, we can (as we are taught) guess how the process behaves from the pole positions of the transfer function. But wouldn t it be nice if we could actually trace the time profile without having to do the reverse Laplace transform ourselves Especially the response with respect to step and impulse inputs Plots of time domain dynamic calculations are extremely instructive and a useful learning tool.1... 

For the design of the actively compensated RF pulse, experimental and numerical determination of the response function h(t) of the circuit is necessary. We should also keep in mind that modification to the circuit, such as probe timing, insertion or removal of RF filters, and so on, can alter h(t). In practice, it is convenient to measure the response y t) to a step excitation u(t) instead of that to the impulse excitation. By performing Laplace transformation to... 

Figure 22C summarizes the procedure for calculating the programming pulse shape v(t) from the target pulse shape bi(t) and the step response u(t). In addition to the measurement of y t), we need to perform a number of data operations such as multiplication, division, Laplace and inverse Laplace transformations. All of these functions can be performed on the software. 

Figure 23 Calculation of the shape of the actively compensated pulse can be carried out on the software. (A) shows the real (red line) and the imaginary (green line) component of an example of the target pulse shape t>,(f). Its leading and the trailing edges have a cosine shape with a transition time of 1.25 xs in 50 steps, and the width of the plateau is 5 ps. (B) Laplace transformation B(s) multiplied by the Laplace transformed step function U(s). (C) It was then divided by the Laplace transformation Y(s) of the measured step response y(t) of the proton channel of a 3.2-mm Varian T3 probe tuned at 400.244 MHz to obtain V(s). (D) Finally, inverse Laplace transformation was performed on V(s) to obtain the compensated pulse that results in the RF pulse with the target shape. Time resolution was 25 ns, and o = 20 was used for the Laplace and inverse Laplace transformations.

Next, bi(t) was Laplace transformed into B(s), and then multiplied by the Laplace transformation U(s) of the step function u(t). The result B(s)U(s) is displayed in Figure 23B. In this example, the step response y(t) was measured for the 1H channel of a Varian 3.2 mm T3 probe tuned at 400.244 MHz with a time resolution of 25 ns, and Laplace transformed into Y(s). By dividing B(s)U(s) by Y(s), the function plotted in Figure 23C was obtained, from which, by performing inverse Laplace transformation, the programming pulse shape v(t) was finally obtained, as shown in Figure 23D. The amplitude and the phase of the complex function v(t) give the intensity and the phase of the transient-compensated shaped pulse. 

Let us now apply the definition of the Laplace transformation to some important time functions steps, ramps, exponential, sines, etc. 

Note that the step function is just a constant (for time greater than zero). Laplace-transforming this function gives... 

Therefore the Laplace transformation of a step function (or a constant) of magnitude K is simply K(l/s). 

The 1/s is an operator or a transfer function showing what operation is performed on the input signal. This is a completely different idea than the simple Laplace transformation of a function. Remember, the Laplace transform of the unit step function was also equal to l/s. But this is the Laplace transformation of a function. The 1/s operator discussed above is a transfer function, not a function. 

J0i Use Laplace transforms to prove mathematically that a P controller produces steadystate ofiMt and that a PI controller does not. The disturbance is a step change in the load variable. The process openloop transfer functions, Gm and G[, are both liist-order lags with dUTerent gains but identical time constants. 

Inversion of the Laplace transformation gives the time function. If we have a transfer function the unit step function is C [G(,y s] and the impulse response is C" [G(,)]. 

We do not lose generality by considering such a unit step function. Because the differential equation is linear, by making superposition of the step function, the response from any surface contour can be treated. The Laplace transform of a step function is... 

Consider the transfer function given by equation 7.109. For a unit step change in the set point R, the Laplace transform of the controlled variable C is given by ... 

Furthermore, it is also not necessary to discuss different excitations in detail as long as we restrict ourselves to the linear response regime. There it holds that the response to any excitation allows the calculation of the response to other excitations via the convolution theorem of cybernetics.213 In the galvanostatic mode, e.g., we switch the current on from zero to /p (or switch it off from 7p to zero) and follow IKj) as a response to the current step. The response to a sinusoidal excitation then is determined through the complex impedance which is given by the Laplace transform of the response to the step function multiplied with jm (j = V-I,w = angular frequency). 

According to the superposition theorem of system theory for linear responses, this response to a step-function in the current can be employed to deduce the impedance behavior. As regards a qualitative discussion, one can adopt the above description by just replacing short/long times by high/small frequencies. Quantitatively the impedance is given by a Laplace transformation of Eq. (64) (or equivalently by applying Kirchhoff s laws to the equivalent circuit (Eq. (63))) with the result... 

APPLIED COMPLEX VARIABLES, John W. Dettman. Step-by-step coverage of fundamentals of analytic function theory—plus lucid exposition of 5 important applications Potential Theory Ordinary Differential Equations Fourier Transforms Laplace Transforms Asymptotic Expansions. 66 figures. Exercises at chapter ends. 512pp. 5)4 x 8)4. 64670-X Pa. 10.95... 

In the above relationships, u (t) is the step Heaviside function. The Laplace transform of the intravascular retention-time distribution is... 

The traditional way is to measure the impedance curve, Z(co), point-after-point, i.e., by measuring the response to each individual sinusoidal perturbation with a frequency, to. Recently, nonconventional approaches to measure the impedance function, Z(a>), have been developed based on the simultaneous imposition of a set of various sinusoidal harmonics, or noise, or a small-amplitude potential step etc, with subsequent Fourier- and Laplace transform data analysis. The self-consistency of the measured spectra is tested with the use of the Kramers-Kronig transformations [iii, iv] whose violation testifies in favor of a non-steady state character of the studied system (e.g., in corrosion). An alternative development is in the area of impedance spectroscopy for nonstationary systems in which the properties of the system change with time. 

The operation of Laplace transformation performed on the function y - f z) consists of two steps ... 

On the other hand, for a step input pressure p t) = pQH t), where H t) is the Heaviside step function, the Laplace transform of Eq, (16.66a), in combination with Eq. (16.68), gives the displacement in terms of the operational... 

In the transient (step-off) Kerr-effect response, it is also possible to obtain from Eqs. (273)—(275) for l = 2 the system of recurrence equations for the Laplace transforms of the corresponding relaxation functions < " (t) (m = 0,1,2) pertaining to that response, namely,... 

Suppose we suddenly increase the load current of a converter from 4 A to 5 A. This is a step load and is essentially a nonrepetitive stimulus. But by writing all the transfer functions in terms of s rather than just as a function of jco, we have created the framework for analyzing the response to such disturbances too. We will need to map the stimulus into the s-plane with the help of the Laplace transform, multiply it by the appropriate transfer function, and that will give us the response in the s-plane. We then apply the inverse Laplace transform and get the response with respect to time. This was the procedure symbolically indicated in Figure 7-3, and that is what we need to follow here too. However, we will not perform the detailed analysis for arbitrary load transients here, but simply provide the key equations required to do so. 

The last term in Eq. (A-3) is the Laplace transform of E(a) Aweq(a), where a plays the role of t in Eq. (A-4) and t plays the role of s in Eq. (A-4). Thus the problem is reduced to this experimental observations give us the Laplace transform of the desired distribution function [modified by multiplication by Aweq(A)], We want the distribution function. The required calculation is a numerical inverse Laplace transform. This is clearly feasible, and numerical techniques are discussed in the literature 56). It is not a simple matter to carry out, however, and the accuracy requirements on the data are likely to be stringent. No direct method of computing the distribution from frequency response is known, although the step response can be computed from the frequency response by standard techniques. In view of the foregoing discussion, it appears that in principle, at least, the distribution can be computed from the frequency response. 

The function with Laplace transform 1/s is a unit step function. The function with Laplace transform l/(s + a) is e a. ... 

T (t) given by eq. (8.4) is the solution to our initial differential equation (5.3). Indeed, taking the Laplace of eq. (8.4), it yields eq. (8.3). The procedure by which we find the time function when its Laplace transform is known is called the inverse Laplace transformation and is the most critical step while solving linear differential equations using Laplace transforms. To summarize the solution procedure described in the example above, we can identify the following steps ... 

Find the time function that has as its Laplace transform the right-hand side of the equation obtained in step 2. This function is the desired solution since it satisfies the differential equation and the initial conditions. 

---