( displaystyle f(t)) | ( displaystyle F(s)) | |
---|---|---|

1 | ( displaystyle 1over s) | ( displaystyle (s > 0)) |

( displaystyle t^n) | ( displaystyle n!over s^{n+1}) | ( displaystyle (s > 0)) |

(( displaystyle n = mbox{ integer } > 0)) | ||

( displaystyle t^p,; p > -1) | ( displaystyle Gamma (p+1) over s^{(p+1)}) | ( displaystyle (s>0)) |

( displaystyle e^{at}) | ( displaystyle 1 over s-a) | ( displaystyle (s > a)) |

( displaystyle t^ne^{at}) | ( displaystyle n! over (s-a)^{n+1}) | ( displaystyle (s > 0)) |

(( displaystyle n= ext{ integer } > 0)) | ||

( displaystyle cos omega t) | ( displaystyle frac{s}{s^{2}+omega ^{2}}) | ( displaystyle (s > 0)) |

( displaystyle sin omega t) | ( displaystyle omega over s^2+omega^2) | ( displaystyle (s > 0)) |

( displaystyle e^{lambda t} cos omega t) | ( displaystyle s - lambda over (s-lambda)^2+omega^2) | ( displaystyle (s > lambda)) |

( displaystyle e^{lambda t} sin omega t) | ( displaystyle omega over (s-lambda)^2+omega^2) | ( displaystyle (s > lambda)) |

( displaystyle cosh bt) | ( displaystyle s over s^2-b^2) | ( displaystyle (s > |b|)) |

( displaystyle sinh bt) | ( displaystyle b over s^2-b^2) | ( displaystyle (s > |b|)) |

( displaystyle t cos omega t) | ( displaystyle s^2-omega^2 over (s^2+omega^2)^2) | ( displaystyle (s>0)) |

( displaystyle t sin omega t) | ( displaystyle 2omega s over (s^2+omega^2)^2) | ( displaystyle (s>0)) |

( displaystyle sin omega t -omega tcos omega t) | ( displaystyle 2omega^3over (s^2+omega^2)^2) | ( displaystyle (s>0)) |

( displaystyle omega t - sin omega t) | ( displaystyle omega^3 over s^2(s^2+omega^2)^2) | ( displaystyle (s>0)) |

( displaystyle frac{1}{t}sinomega t) | ( displaystyle arctan left({omega over s} ight)) | ( displaystyle (s>0)) |

( displaystyle e^{at}f(t)) | ( displaystyle F(s-a)) | |

( displaystyle t^kf(t)) | ( displaystyle (-1)^{k}F^{(k)}(s)) | |

( displaystyle f(omega t)) | ( displaystyle frac{1}{omega}Fleft(frac{s}{omega } ight), quad omega >0) | |

( displaystyle u(t- au)) | ( displaystyle e^{- au s} over s) | ( displaystyle (s>0)) |

( displaystyle u(t- au)f(t- au), ( au > 0)) | ( displaystyle e^{- au s}F(s)) | |

( displaystyle displaystyle {int^t_o f( au)g(t- au), d au}) | ( displaystyle F(s) cdot G(s)) | |

( displaystyle delta(t-a)) | ( displaystyle e^{-as}) | ( displaystyle (s>0)) |

## What is Laplace Transform? Formula, Properties, Conditions and Applications

Mathematics plays a decisive role to understand the behavior and working of **electrical** **and** **electronic systems**. Polynomials, Algebra, Probability, Integrations, and Differentiations etc…forms a significant part of the tools used to solve the systems. With the increasing complexity of systems, very sophisticated methods are required. Differential equations are prominently used for defining control systems. These equations are simple to solve. But complexity arises while solving higher order differential equations. To solve such complex higher order differential equations, the mathematical method that proved to be effective is **Laplace Transform**. As this transform is widely employed, it is useful to know what they really meant for and how do they work.

## 8.8: A Brief Table of Laplace Transforms - Mathematics

1 Department of Mathematics, Faculty of Basic Education, PAAET, Shaamyia, Kuwait

2 M. S. Software Engg, SITE, V. I. T. University, Vellore, India

Email: * [email protected], [email protected]

Copyright © 2013 Fethi Bin Muhammad Belgacem, Rathinavel Silambarasan. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Received February 22, 2013 revised April 28, 2013 accepted May 5, 2013

**Keywords:** Laplace Transform Natural Transform Sumudu Transform

In this paper, the Laplace transform definition is implemented without resorting to Adomian decomposition nor Homotopy perturbation methods. We show that the said transform can be simply calculated by differentiation of the original function. Various analytic consequent results are given. The simplicity and efficacy of the method are illustrated through many examples with shown Maple graphs, and transform tables are provided. Finally, a new infinite series representation related to Laplace transforms of trigonometric functions is proposed.

Integral transforms methods have been used to a great advantage in solving differential equations. Limitations of Fourier series technique, were overcome by the extensive coverage of the Fourier transform to functions, which need not be periodic [1]. The complex variable in the Fourier transform is substituted by a single variable to obtain the well known Laplace transform [1-8], a favorite tool in solving initial value problems (IVPs). The integral equation defined by Léonard Euler was first named as Laplace by Spitzer in 1878. However the very first Laplace transform applications were established by Bateman in 1910 to solve Rutherford’s radioactive decay, and Bernstein in 1920 with theta functions. For a real function with variable the Laplace transform, designated by the operator, , giving rise to a function in, , in the right half complex plane, is defined by,

While we completely focus on the Laplace transform, in this paper, many of the ideas herein stem from recent work on the Sumudu transform, and studies and observations connecting the Laplace transform with the Sumudu transform through the Laplace-Sumudu Duality (LSD) for and the Bilateral Laplace Sumudu Duality (BLSD) for [9-16]. Indeed, considering the -multiplied two-sided Laplace transform,

by making the parameter change s with in the equation above we get the two-sided Sumudu transform,

Here, the constants and may be finite (or) infinite, and are based on the exponential boundedness nature required on in the domain set

While Analyses about the properties of the Sumudu transform, transform tables, and many of its physical applications can be found in [9-12,14,16], investigations, applications, and transform tables stemming from the Natural transform can be found in [17-21]. This new integral transform combines both (one sided) Sumudu and (one sided) Laplace transforms by,

Obviously, taking, , in the Natural transform, leads to the Laplace transform, and taking, , results in one sided Sumudu transform. We note that while the Natural can be bilateral like both Sumudu and Bilateral Laplace, when the variable is chosen positive in the definition, both and the variable, , must be positive as well, as as this ought to correct a related one sided Natural Transform defintion misprint appearing in our papers [17,18].

The gist and essence of this work is solving the Laplace integral equation once by differention, and by integration by parts. Divided into two major sections, this paper in Section 2 explains the various multiple shift properties connected with the Laplace transform by just differentiating the original function. The new infinite series representation of trigonometric functions related with the Laplace transform is proved in Section 3. In consequence of our formulations and derivations, three tables are provided at the end of the Section 3 ended with concluding remarks and directions for some future work. The tables are respectively covering derivatives periods for the function E (in Table 1 ), 21 trignometric series expansions entries (in Table 2 ), and 16 main Laplace transform properties, as generated by Proposition 3 (in Table 3 ). Examples 1, 2, and 3 in the body of the text of Section 2, as well as Example 6 and Entry 17 of transform Table 2 in Section 3, are respectively afforded Maple graphs (see Figures 1-5), showing both the time function invoked in the corresponding example, and its resulting Laplace trasform.

2. Laplace Transforms by Function Differentiation

As stated earlier in the introduction, an ultimate goal of ours, among others, is calculating the Laplace transform of by simple differentiation rather than usual integration. We show that we can do this, without resorting to the Adomian nor homotopy methods, ADM, and HPM, as was done in [22,23]. Along with the Laplace series definition below, some elementary properties are proved.

Definition The Laplace transform (henceforth designated as F(s)) of the exponential order and sectionwise continuous function, is defined by,

(1)

Remark We observe that, from the traditional Laplace transform, taking, so that

and

, so that

. Now using

and in Bernoulli’s integration by parts,

and noting for n ≥ 0 Equation (1) follows.

Can’t one choose u = e − st and for solving the Laplace integral equation by parts? The detailed answer with analysis is given in Section 3. For simplicitywe use hereafter.

Multiple Shifts and Periodicity Results

Theorem 1 The Laplace transform of derivative of, with respect to is defined by,

Proof. The LHS of above equation is

, substituting Equation

(1) for, the proof is completed.

Table 1 . Period function calculations.

Table 2 . New infinite series representation of trigonometric functions.

Table 3 . Laplace transform properties with respect to the Proposition 3.

Theorem 2 The Laplace transform of antiderivative of f(t), in the domain with respect to, is given by,

Proof. Applying Equation (1) in and performing the usual computations, yields the RHS of the equation above and proves out theorem.

Theorem 3 For, the Laplace transform of the function, is given by,

(a) (b)

Figure 1 . Graph of example 1. (a) (b).

(a) (b)

Figure 2 . Graph of example 2 with a = 1, b = 2. (a) (b).

Proof. From the theory of Laplace transform

, when

(2)

When in Equation (2),

(a) (b)

Figure 3 . Graph of example 3. (a) (b).

(a) (b)

Figure 4 . Graph of example 6 with a = 5. (a) (b).

(3)

When in Equation (2),

(4)

Finally for the non-negative integer, after simplifi-

(a) (b)

Figure 5 . Graph of entry 17 of Table 2 and its Laplace transform. (a) (b) .

which yields the result of Theorem 3.

Theorem 4 The Laplace transform of the function

, for, is,

Proof. Substituting Equation (1) for in

and after the usual computations, Theorem 4 follows.

Example 1 As an application of Theorem 3, the Laplace transform of, where

, denotes the first kind order one Bessel function, is calculated as follows, (graph shown in Figure 1 ),

Now substituting the above derivatives in Equation (3), and after applying both the limits, and

for,

The multiple-shift theorems that follow are useful in treating differential and integral equations with polynomial coefficients.

Example 2 The Laplace transform of is calculated by taking,

( Figure 2 ), When in Theorem 4,

Theorem 5 Let when the derivative of the function, with respect to is shifted by, then the Laplace transform is given by,

(5)

Proof. The proof is simple, we have

where

is given by Theorem 1.

Theorem 6 For non-negative integers i and, when the derivative of the function, with respect to is shifted with, then the Laplace transform is given by,

(6)

Proof. The LHS of above equation is

and are given by Theorem 1, and after proper calculations, the proof is calculated.

Theorem 7 For, the Laplace transform of the antiderivative of the function with respect to in the interval shifted with, is given by,

Proof. Applying Theorem 2 in LHS yields the RHS of the Equation above.

Theorem 8 The Laplace transform of the antiderivative of the function with respect to in the interval shifted with is given by,

Proof. Computing the summation in the RHS of the Equation in Theorem 2 with respect to in the domain, times, yields the proof.

We now establish the following resultsTheorem 9 For, the Laplace transform of the derivative of with respect to is given by,

(7)

Proof. Substituting Theorem 3 in

.

Theorem 10 The Laplace transform of the

derivative with respect to of for non-negative integers, and is given by,

(8)

Proof. Substituting Theorem 4 in

.

Example 3 Consider the function, , then taking yields the expected derivatives,

and,. Next, for in Theorem 11,

Theorem 11 For non-negative integers and, the Laplace transform of the antiderivative with respect to in the domain of, is given by,

Proof. From the property of Laplace transform, the LHS of above equation is in which Theorem 3 is substituted and simplified.

Theorem 12 The Laplace transform of the antiderivative with respect to in the domain of

, where, , is given by,

Proof. The proof is straightforward where we multiplied to Theorem 4.

Example 4 Consider the function, , which Laplace transform we can find by taking, yielding, and,

Now, since from the theorem above, for we have,

From Theorem 5 through Theorem 12, there is no restriction on positive integers and, which means both can be same (or) different and either of the integer can less than (or) greater than to one another.

The Theorem 5 and the Theorem 9 varies only in the coefficients, that is the order of the derivative, the same holds for Theorem 6 and Theorem 10, again the Theorem 7 and Theorem 11 varies only in the coefficients, that is the order of the anti-derivative, similarly for Theorem 8 and 12. Hence we have the following propositions, respectively.

Proposition 1 If the function and its derivative with respect to go to zero as, then,

(9)

(10)

(11)

(12)

The following initial and final value, convolution, and function periodicity related theorems can be easily verified through conventional Laplace transform theory.

Theorem 13 Let the function, be Laplace transformable, then,

(13)

(14)

Theorem 14 The Laplace transform of the convolution of two functions, and, , is given by,

Theorem 15 The Laplace transform of the periodic function with period so that , is given by,

Proof. Writing Equation (1) as,

Now substituting in the second infinite series of the above equation so that the limits changes to and by having and after rearranging and evaluating completes the proof.

Example 5 The full sine-wave rectifier is given by the function, with the period.

Using Theorem 15, the Laplace transform of the full sine-wave rectifier is calculated by using the entries of column 5 of Table 1 ,

3. Laplace Transforms by Integration by Parts

The Laplace transform of is calculated by substituting in the Laplace integral transform, now by taking and evaluating by parts gives

. On the other hand, to calculate the Laplace transform of, we take and and after evaluation leads. Here we can also take u =

and again it gives the same Laplace transform. Hence, in this section, we solve the Laplace integral equation by taking, , and, and integrating by parts. Below, the sub-scripts in say represents the order of integration in the variable .

Subject to some constraints we then generally haveProposition 3 The Laplace transform of a Taylor’s seriezable trigonometric function, , is given by,

Proof. Now, so that Next, leads to

Substituting and in the Bernoulli’s formula of continuous integration by parts and observing is positive for all gives Proposition 3.

Example 6 The Laplace transform of with non-negative integer, a, is calculated by simply integrating the function. Now, for,

and,. Furthermore, in view of Proposition 3, when applying the upper and lower limits in the antiderivatives above, we get.

and, whence we get, (function and its Laplace transform in Figure 4 )

We agree that constants and polynomials cannot be Laplace transformed with the Proposition 3, since the continuous integration of constant and polynomials with respect to does not converge anywhere when and.

3.1. New Infinite Series Representation for Trig Functions

In the Proposition 3 the limitations of to be Taylor’s seriezable trigonometric function is acceptable only on a theoritical point of view, from the evaluation of Laplace transform of trigonometric functions vice-versa of Definition of Section 2 and Proposition 3. On the other hand, we show under what condition the Proposition 3 exists? Definitely the answer would be by finding the inverse Laplace transform of Proposition 3.

For simplicity’s sake, re-writing the Proposition 3, is akin to evaluating the limits and representing, , in Proposition 3. The Laplace transform of Taylor’s seriezable trigonometric function is simply defined by,

The inverse Laplace transform of Proposition 3 would be same as inverse Laplace transform of the above equation, and hence it is enough to find the inverse Laplace transform of.

For a start up, when in the inverse Laplace transform of would be which is Dirac delta function [2] since,.

Again when in the inverse Laplace transform of s is given by the first derivative of Dirac delta function with respect to,. In particular, readers are invited to consider connected relation to Dirac delta function but in the Sumudu transform context (see Equations (2.19), (2.20), (4.18), and (4.20) in [9]). In general, the inverse Laplace transform of is given by

, since. In all cases upto the n-th derivative the initial value theorem is undefined for and leads of course to the study of generalized functions (see [2], and references therein for more details).

We prove the inverse Laplace transform of singular functions that satisfy the Tauberian (initial value) theorem in the following proposition where the trigonometric functions are represented in new infinite series, where coefficients are calculated by integrating the function, [24].

Proposition 4 The necessary condition for the existence of Proposition 3 (and hence the above equation) is that, the Taylor seriezable trigonometric function can be expressed as,

Proof. Taking inverse Laplace transform of

(15)

In [14] the Bilateral Laplace Sumudu Duality (BLSD) was established. the inverse Laplace transform of is given by (see Equation (5.10) in [14]),

(16)

Thus, here the

is the first kind Bessel’s function of order zero. And this particular function will play the major role in the exponential kerneled integral transforms (see Equations (30) through (35) in [20]). And the Laplace transform is taken with respect to, since v and are independent, the permissibility of interchange of order of integration is considered in favour. Though the function gives no meaning (as becomes zero when evaluated) but as per the Laplace (integral) transform point of view this is worth (see Theorem 5.1. Equation (5.8) in [14]). By having,

(17)

The Laplace transform of the first derivative of

with respect to is and with the help of Equations (16) and (17),

(18)

Therefore the inverse Laplace transform of is

. In general, from the Laplace transform of the derivative of function with respect to,

(19)

(20)

Finally from Equations (19) and (20),

(21)

Since the second part of right hand side of Equation (21) is zero,

(22)

Substituting Equation (22) in Equation (15) for completes the proof of Proposition 4.

Thus, the function from the Equation (20) satisfies the initial value theorem (unlike Dirac delta function) which is zero. To concretize ideas, we give the following example.

Example 7 Consider the function, then

, , ,

,

, and,

. Applying Proposition 4, as t tends to zero, all the and

and since is the common factor,

, the function, , can be written in the new infinite series as,

From Equation (22), it is wothy to note that the Laplace transform of the integral of derivative of with respect to t, in the domain with respect to v, is simply the s power the order of the derivative of.

Along with that of the function, , in light of this new infinite series Proposition 4, Table 2 gives all new infinite series expansions of basic trigonometric functions. The extra factor in the infinite series of entries 5, 6, 9, 10, 14, 16, 20 and 21 are common for all while integrating. Furthermore, the following expression is easily derivable from the Bessel’s function,

Therefore, the Laplace transform of can be calculated through,

Entry 17 of Table 2 has the following Laplace transform (shown in Figure 5 ),

3.2. LaplaceTransform Properties in View of Proposition 3

The Laplace transform of multiple shifts functions can readily be derived with the help of Proposition 3. Since the derivation of the various properties are straightforward and similar to the theorems of Section 2.1, we give directly the Laplace transform of shifted functions, based on Proposition 3 in Table 3 where and i are non-negative integers.

Example 8 The Laplace transform of the function, , is obtained by simply integratingthus, , ,

,.

When, , in entry 3 of Table 3 ,

Therefore, when, and,

and yielding,

Example 9 The Laplace transform of is calculated by integrating cost. Now, , ,

and,

. Now, for, , and,

and,.

Finally, with in the entry 4 of Table 3 ,

Example 10 The Laplace transform of,

is calculated. For, , after taking limits, we get, Hence, applying the formula with, , in entry 11 of Table 3 ,

We consequently then have,

It is important to note that with respect to the entries 5 and 9 (and entries 6 and 10) of Table 3 , the proposition 1 Equation (9) (and Equation (10)) holds true. Similarlly with respect to the entries 7 and 11 (and entries 8 and 12) of Table 3 , the proposition 2 Equation (11) (and Equation (12)) remains the same.

3.3. Concluding Remarks and Future Work

As far as the Section 2 is concerned, when the function is Laplace transformed by differentiation, then the inverse Laplace is automatically an integration process. Having worked with various examples, our proposed methods lead to exact solutions. A remaining open query is that of defining the inverse for the Laplace transform by using similar tools and processes as in Proposition 3. But in view of the concept of Section 2 above, Laplace and inverse Laplace transform are the respective reciprocal processes of differentiation and integration of the function.

If so, then with the Proposition 3, the inverse Laplace transform will be the process of differentiating. For example consider the function, its Laplace transform by the Proposition 3 is given by

which gives. Hence for finding the original function, when equating the coefficients of identical powers of with Proposition 3, we get As the sub-scripts denote the order of integration. Now by differentiating times, one should get the infinite series of the function, cost as entry 4 of Table 2 .

As part of some future works in this regard, we aspire to pursing working schemes of this paper, and establishing more comprehensive tables as was done for the Sumudu transform in [10], and for the Natural transform in [17]. With this said, it is perhaps research worthy, in the near future, to put all considered aspects in the framework and applications of the theory of reproducing kernels [25].

The Authors wish to acknowledge and thank the AM Editorial Board, as well as SCRIP Grants Committee, for helping defray a major part (75%) of the processing charges of this article. We also thank the referees and AM staff for constructive comments that helped improve the structural flow of the paper.

## 8.8: A Brief Table of Laplace Transforms - Mathematics

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## 8.8: A Brief Table of Laplace Transforms - Mathematics

Here are a set of assignment problems for the Laplace Transforms chapter of the Differential Equations notes. Please note that these problems do not have any solutions available. These are intended mostly for instructors who might want a set of problems to assign for turning in. Having solutions available (or even just final answers) would defeat the purpose the problems.

If you are looking for some practice problems (with solutions available) please check out the Practice Problems. There you will find a set of problems that should give you quite a bit practice.

Here is a list of all the sections for which assignment problems have been written as well as a brief description of the material covered in the notes for that particular section.

The Definition – In this section we give the definition of the Laplace transform. We will also compute a couple Laplace transforms using the definition.

Laplace Transforms – In this section we introduce the way we usually compute Laplace transforms that avoids needing to use the definition. We discuss the table of Laplace transforms used in this material and work a variety of examples illustrating the use of the table of Laplace transforms.

Inverse Laplace Transforms – In this section we ask the opposite question from the previous section. In other words, given a Laplace transform, what function did we originally have? We again work a variety of examples illustrating how to use the table of Laplace transforms to do this as well as some of the manipulation of the given Laplace transform that is needed in order to use the table.

Step Functions – In this section we introduce the step or Heaviside function. We illustrate how to write a piecewise function in terms of Heaviside functions. We also work a variety of examples showing how to take Laplace transforms and inverse Laplace transforms that involve Heaviside functions. We also derive the formulas for taking the Laplace transform of functions which involve Heaviside functions.

Solving IVPs' with Laplace Transforms - In this section we will examine how to use Laplace transforms to solve IVP’s. The examples in this section are restricted to differential equations that could be solved without using Laplace transform. The advantage of starting out with this type of differential equation is that the work tends to be not as involved and we can always check our answers if we wish to.

Nonconstant Coefficient IVP’s – In this section we will give a brief overview of using Laplace transforms to solve some nonconstant coefficient IVP’s. We do not work a great many examples in this section. We only work a couple to illustrate how the process works with Laplace transforms.

IVP’s with Step Functions – This is the section where the reason for using Laplace transforms really becomes apparent. We will use Laplace transforms to solve IVP’s that contain Heaviside (or step) functions. Without Laplace transforms solving these would involve quite a bit of work. While we do not work one of these examples without Laplace transforms we do show what would be involved if we did try to solve on of the examples without using Laplace transforms.

Dirac Delta Function – In this section we introduce the Dirac Delta function and derive the Laplace transform of the Dirac Delta function. We work a couple of examples of solving differential equations involving Dirac Delta functions and unlike problems with Heaviside functions our only real option for this kind of differential equation is to use Laplace transforms. We also give a nice relationship between Heaviside and Dirac Delta functions.

Convolution Integral – In this section we giver a brief introduction to the convolution integral and how it can be used to take inverse Laplace transforms. We also illustrate its use in solving a differential equation in which the forcing function (*i.e.* the term without an y’s in it) is not known.

Table of Laplace Transforms – This section is the table of Laplace Transforms that we’ll be using in the material. We give as wide a variety of Laplace transforms as possible including some that aren’t often given in tables of Laplace transforms.

## Table of Laplace and Z Transforms

All time domain functions are implicitly=0 for t<0 (i.e. they are multiplied by unit step).

u(t) is more commonly used to represent the step function, but u(t) is also used to represent other things. We choose gamma ( &gamma (t)) to avoid confusion (and because in the Laplace domain ( &Gamma (s)) it looks a little like a step input).

atan is the arctangent (tan -1 ) function. The atan function can give incorrect results (typically the function is written so that the result is always in quadrants I or IV). To ensure accuracy, use a function that corrects for this. In most programming languages the function is atan2. Also be careful about using degrees and radians as appropriate.

The Laplace transform of a function f ( t )

### Linearity Edit

and is, therefore, regarded as a linear operator.

### Time shifting Edit

### Frequency shifting Edit

The unilateral Laplace transform takes as input a function whose time domain is the non-negative reals, which is why all of the time domain functions in the table below are multiples of the Heaviside step function, *u*(*t*) .

The entries of the table that involve a time delay *τ* are required to be causal (meaning that *τ* > 0 ). A causal system is a system where the impulse response *h*(*t*) is zero for all time t prior to *t* = 0 . In general, the region of convergence for causal systems is not the same as that of anticausal systems.

## Laplace and z transforms

### 15.11 Exercises

Using the definition of the Laplace transform

Find Laplace transforms of the following using Table 15.1 : (a)

Find inverse Laplace transforms of the following using tables: (a)

Find Laplace transforms of the following using the properties of the transform: (a)

Find inverse Laplace transforms of the following using the properties of the transform: (a)

Find inverse Laplace transforms using partial fractions: (a)

In each case solve the given differential equation using Laplace transforms: (a)

A capacitor of capacitance *C* in an RC circuit, as in Figure 15.7 , is charged so that initially its potential is *V*_{0}. At *t* = 0, it begins to discharge. Its charge *q* is then described by the differential equation

Figure 15.7 . An RC circuit for Exercise 15.8 .

Using Laplace transforms, find the charge on the capacitor at time *t* after the switch was closed.

Find the response of a system with zero initial conditions to an input of *f* (*t*) = 2e −3*t* , given that the impulse response of the system is *h*(*t*) *=* 1/2(e −*t* – e −2*t* ).

The impulse response of a system is given by *h*(*t*) = 3e −4 *t* . Find the system's step response, that is, the response of the system to an input of the step function, *u*(*t*).

Use the result that *Y*(*s*) = *H*(*s*)*F*(*s*) where *Y* is the Laplace transform of the system output, *F*(*s*) the Laplace transform of the input and *H*(*s*) the system transfer function to show that the step response can be found by

Given that the step response of a system is

find the system's transfer function and

find its response to an input of e −*t* .

A system has a known impulse response of *h*(*t*) = e −*t* sin(2*t*). Find the input function *f* (*t*) that would produce an output of

A system has transfer function

Find its steady state response to a single frequency input of *e* j5*t*

Find the steady-state response to an input of cos(5*t*) and sin(5*t*).

Using the definition of the *z* transform

Find *z* transforms of the following using Table 15.2 : (a)

Find inverse *z* transforms of the following using Table 15.2 : (a)

z 2 + z ( sin ( 1 ) − cos ( 1 ) ) z 2 − 2 z cos ( 1 ) + 1

Find *z* transforms of the following using the properties of the transform: (a)

Find inverse *z* transforms of the following using the properties of the transform: (a)

Find inverse *z* transforms using partial fractions (a)

In each case solve the given difference equation using *z* transforms, *n* ≥ 0 (a)

y n + y n − 1 = n , where y − 1 = 0

Find the response of a system with zero initial conditions, to an input of *f _{n}* =

_{2}(0.3)

*n*, given that the impulse response of the system is

*h*= (0.1)

_{n}*n*+ (−0.5)

*n*.

The impulse response of a discrete system is given by *h _{n}*=(0.8)

*n*. Find the system's step response, that is, the response of the system to an input of the step function,

*u*

_{n}.Use the result that *y*(*z*) =*H(z)F(z*), where *Y*is the *z* transform of the system output, *F (z*) is the *z* transform of the input and *H (z*) is the system transfer function to show that the step response can be found by

Given that the step response of a discrete system is

find the system's transfer functions and

find its response to an input of 6(0.5) *n* .

A system has a known impulse response of *h _{n}* = (0.5)

*n*. Find the input function

*f*that would produce an output of 2(0.5)

_{n}*n*+ 2

_{n}−2u

_{n}given zero initial conditions.

A system has transfer function *H (z*) = *z*/ (10*z* −3) (i)

find its steady state response to a single frequency input of e j5i

find the steady state response to an input of cos(5*n*) and sin(5*n*).

## 2 Answers 2

Let $widetilde~~= 0 $ The general solution is simply: $ widetilde~~~~x> -frac~~~~$ where $A$ and $B$ are two arbitrary constant that need two condition to be determined. The first condition is a physical one: the solution must be limited in time and so $A=0$. The second condition is that: $phi(0, t) = e^<-t>, t > 0 .$ Now the Laplace transform of $phi(0,t)$ is simply $widetilde~~~~$. Finally we have that the Laplace transform of our $phi(x,s)$ is: $ widetilde~~~~x>> ~~~~- frac~~~~$~~

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