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Group 2 Bhadouria, Arjun Singh Glave, Theodore Dean Han, Zhe Chapter 5. Laplace Transform Chapter 19. Wave Equation

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Laplace Transform Chapter 5

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5. Laplace Transform 5.1. Introduction & Definition 5.2. Calculation of the Transform 5.3. Properties of the Transform 5.4. Application to the Solution of Differential Equations 5.4. Application to the Solution of Differential Equations 5.5. Discontinuous Forcing Functions; Heaviside Step Function 5.5. Discontinuous Forcing Functions; Heaviside Step Function 5.6. Impulsive Forcing Functions; Dirac Impulse Function 5.6. Impulsive Forcing Functions; Dirac Impulse Function 5.7. Additional Properties

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5.1. Introduction & Definition The Laplace transform is a widely used integral transform. Denoted, it is a linear operator of a function f(t) with a real argument t (t ≥ 0) that transforms it to a function F(s) with a complex argument s. The Laplace transform has the useful property that many relationships and operations over the originals f(t) correspond to simpler relationships and operations over the images F(s). The Laplace transform has many important applications throughout the sciences. The Laplace transform is used for solving differential and integral equations. In physics and engineering, it is used for analysis of linear time- invariant systems. In this analysis, the Laplace transform is often interpreted as a transformation from the time-domain, in which inputs and outputs are functions of time, to the frequency-domain, where the same inputs and outputs are functions of complex angular frequency, in radians per unit time. Given a simple mathematical or functional description of an input or output to a system, the Laplace transform provides an alternative functional description that often simplifies the process of analyzing the behavior of the system, or in synthesizing a new system based on a set of specifications.

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Basic idea

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Given a known function K(t,s), an integral transform of a function f is a relation of the form The Laplace Transform of f is defined as where the kernel function is K(s,t) = e -st, a=0, b= . F (s) is the symbol for the Laplace transform, L is the Laplace transform operator, and f(t) is some function of time, t.

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How to solve problems Time Domain Frequency Domain Solve algebraic equation Laplace transform Inverse Laplace transform Douglas Wilhelm Harder, University of Waterloo.

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5.2. Calculation of the Transform Since the Laplace Transform is defined by an improper integral, thus it must be checked whether the transform F(s) of a given function f(t) exists, that is whether the integral converges.

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EXAMPLE 2.1. Consider the following improper integral. We can evaluate this integral as follows: Note that if s = 0, then e st = 1. Thus the following two cases hold:

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EXAMPLE 2.2. Consider the following improper integral. We can evaluate this integral using integration by parts: Since this limit diverges, so does the original integral.

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Exponential order Suppose that f is a function for which the following hold: (1) f is piecewise continuous on [0, b] for all b > 0. (2) | f(t) | Ke ct when t T, for constants c, K, M, with K, M > 0. A function f that satisfies the conditions specified above is said to to have exponential order as t EXAMPLE 2.3. Is f(t)=exp(4t)cost of exponential order? Solution: Yes. Therefore, K=10, c=4, and T=10 for instance EXAMPLE 2.4. Is f(t)=ln(t) of exponential order? Solution: Yes. Therefore, K=100, c=1, and T=10 for instance

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Piecewise Continuous If an interval [a, b] can be partitioned by a finite number of points a = t 0 < t 1 < … < t n = b such that Then the function f is piecewise continuous Or we can say f is piecewise continuous on [a, b] if it is continuous there except for a finite number of jump discontinuities. Picture from Paul's Online Math Notes

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EXAMPLE 2.5. Piecewise continuous function Consider the following piecewise-defined function f. (a) From this definition of f, and from the graph of f below, we see that f is piecewise continuous on [0, 3]. (b) From this definition of f, and from the graph of f below, we see that f is NOT piecewise continuous on [0, 3].

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THEOREM 2.1 Existence of the Laplace Transform Let f(t) satisfy these conditions: (i)f(t) is piecewise continuous on for every A>0, (ii)f(t) is of exponential order as That is to say (1) f is piecewise continuous on [0, b] for all b > 0. (2) | f(t) | Ke ct when t T, for constants c, K, M, with K, M > 0. Then the Laplace Transform of f exists for s > c.

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Inverse Laplace transform operator By definition, the inverse Laplace transform operator, L -1, converts an s-domain function back to the corresponding time domain function: Importantly, both L and L -1 are linear operators. Thus,

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Examples of Calculation

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Laplace transform table

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5.3. Properties of the Transform THEOREM 3.0 The transform of an expression that is multiplied by a constant is the constant multiplied by the transform. That is: A various types of problems that can be treated with the Laplace transform include ordinary and partial differential equations as well as integral equations.

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Suppose u and v are functions whose Laplace transforms exist for s > a 1 and s > a 2, respectively. Then, for s greater than the maximum of a 1 and a 2, the Laplace transform of au(t) + bv(t) exists. That is, With Therefore THEOREM 3.1 Linearity of the Transform

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EXAMPLE 3.1. f (t) = 5e -2t - 3sin(4t) for t 0. by linearity of the Laplace transform, and using results of Laplace transform table, the Laplace transform F(s) of f is:

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For any U(s) and V(s) such that the inverse transforms For any constants a,b. THEOREM 3.2 Linearity of the Inverse Transform Basic idea : Consider a general expression, Expand a complex expression for F(s) into simpler terms, each of which appears in the Laplace Transform table. Then you can take the L -1 of both sides of the equation to obtain f(t).

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Perform a partial fraction expansion (PFE) where coefficients and have to be determined. EXAMPLE 3.2. To find : Multiply both sides by s + 1 and let s = -1 To find : Multiply both sides by s + 4 and let s = -4 Therefore,

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Let f(t) be continuous and f’(t) be piecewise continuous on 0≤t≤t 。 For every finite t 。, and let f(t) be of exponential order as t so that there are constants K, c, T such that | f(t) | Ke ct when t T. Then L{f’(t)} exists for all s>c. This is a very important transform because derivatives appear in the ODEs we wish to solve. Similarly, for higher order derivatives: where: THEOREM 3.3 Transform of the Derivative

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Deriving the Laplace transform of f (t ) often requires integration by parts. However, this process can sometimes be avoided if the transform of the derivative is known: For example, if f (t ) = t then f ‘ (t ) = 1 and f (0) = 0 so that, since: That is: Proof It has already been established that if: then: Now let so that: Therefore:

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For example, if: then: and Similarly: And so the pattern continues. EXAMPLE 3.3 Then substituting in: yields So:

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EXAMPLE 3.4 COMMENT: Difference in The values are only different if f(t) is not continuous at t=0 Example of discontinuous function: u(t) *Additional Section

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EXAMPLE 3.5 Try to solve the differential equation: take the Laplace transform of both sides of the differential equation to yield: Resulting in: The right-hand side can be separated into its partial fractions to give: From the table of transforms it is then seen that: Thus,

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THEOREM 3.4 Laplace Convolution Theorem If both exist for s>c, then As Laplace convolution of f and g.

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Proof: We have, therefore

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EXAMPLE 3.6 If f(t)=exp(t), g(t)=t, then EXAMPLE 3.7 Schiff. Joel L. The Laplace transform: theory and applications. P92

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5.4. Application to the Solution of Differential Equations Laplace transforms play a key role in important engineering concepts and techniques. Examples: Transfer functions Frequency response Control system design Stability analysis …

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Solution of ODEs by Laplace Transforms Procedure: 1.Take the L of both sides of the ODE. 2.Rearrange the resulting algebraic equation in the s domain to solve for the L of the output variable, e.g., F(s). 3.Perform a partial fraction expansion. 4.Use the L -1 to find f(t) from the expression for F(s).

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Equation with initial conditions Laplace transform is linear Apply derivative formula Rearrange Take the inverse EXAMPLE 4.1. Solve

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resulted in the expression EXAMPLE 4.2. Take L -1 of both sides :

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EXAMPLE 4.3. Taking Laplace transforms of both sides of this equation gives: K.A. Stroud. Engineering Mathematics. P1107

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EXAMPLE 4.4. A mass m is suspended from the end of a vertical spring of constant k (force required to produce unit stretch). An external force f(t) acts on the mass as well as a resistive force proportional to the instantaneous velocity. Assuming that x is the displacement of the mass at time t and that the mass starts from rest at x=0, (a)Set up a differential equation for the motion (b)Find x at any time t Solution: (a)The resistive force is given by –βdx/dt. The restoring force is given by –kx. Then by Newton’s law,

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(b) Taking the Laplace transform of (1), using we obtain So that on using (2) Where There are three cases to be considered. EXAMPLE 4.4.

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Case 1, R>0. In this case let We have Then we find from (3) Case 2, R=0. In this case thus EXAMPLE 4.4.

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Case 3. R<0, In this case let We have EXAMPLE 4.4. Schaum’s Outline of Theory and Problems of Advanced Mathematics for Engineers and Scientists. P115, Problem 4.46

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5.5. Discontinuous Forcing Functions; Heaviside Step Function The unit step function is widely used. It is defined as: Because the step function is a special case of a “constant”, it follows

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More generally, It is possible to express various discontinuous functions in terms of the unit step function. Therefore,or

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EXAMPLE 5.1 Schiff. Joel L. The Laplace transform: theory and applications. P92

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EXAMPLE 5.2 Determine L{g(t)} for Schiff. Joel L. The Laplace transform: theory and applications. P92

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5.6. Impulsive Forcing Functions; Dirac Impulse Function Pictorially, the unit impulse appears as follows: Mathematically: Picture from web.utk.edu

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We can prove that Where thus is known as Dirac delta function, or unit impulse function further

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The Laplace transform of a unit impulse: if we let f(t) = (t) and take the Laplace * Rectangular Pulse Function *Additional Section

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EXAMPLE 6.1 Some useful properties of Dirac Impulse Function

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5.7. Additional Properties THEOREM 7.1 S-plane (frequency) shift If exists for s>s 0, then for any real constant a, for s+a>s 0, or equivalently, Proof : EXAMPLE 7.1. EXAMPLE 7.2.

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THEOREM 7.2 Time Shift If exists for s>s 0, then for any real constant a>0, for s>s 0, or, equivalently, Proof : EXAMPLE 7.3.

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THEOREM 7.3 Multiplication by 1/s (Integrals) If exists for s>s 0, then for s>max{0,s 0 }, or, equivalently, Proof : Notice: EXAMPLE 7.4.

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COMMENT: Another way to prove Time Integration Property From Douglas Wilhelm Harder, University of Waterloo.

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THEOREM 7.4 Differentiation with Respect to s (Multiplication by t n ) If exists for s>s 0, then for s>s 0, or, equivalently, Proof : EXAMPLE 7.5.

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EXAMPLE 7.6. Firstly where substitute Douglas Wilhelm Harder, MMath

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EXAMPLE 7.6. substitute therefore

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THEOREM 7.5 Integration with Respect to s If there is a real number s 0 such that exists for s>s 0, and exists, then for s>s 0, or, equivalently, EXAMPLE 7.7. To evaluate

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THEOREM 7.6 Large s Behavior of F(s) Let f(t) be piecewise continuous on for each finite and of exponential order as, then (i) (ii) Proof : Since f(t) is of exponential order as then there exist real constants K and c, with K 0 such that | f(t) | Ke ct for all t T. Since f(t) is piecewise continuous on There must be a finite constant M such that | f(t) | M, on For all s>c:

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THEOREM 7.7 Initial-Value Theorem Let f be continuous and f’ be piecewise continuous on for each finite, and let f and f’ be of exponential order as then Proof : with the stated assumptions on f and f’, it follows that NOTE: The utility of this theorem lies in not having to take the inverse of F(s) in order to find out the initial condition in the time domain. This is particularly useful in circuits and systems. Since f’ satisfies the conditions of THEOREM 4.7, it follows that thus

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Given Find f(0) EXAMPLE 7.8.

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THEOREM 7.8 Final Value Theorem Let f be continuous and f’ be piecewise continuous on, and let f and f’ be of exponential order as then EXAMPLE 7.9. Suppose then NOTE: It can be used to find the steady-state value of a closed loop system (providing that a steady-state value exists. Again, the utility of this theorem lies in not having to take the inverse of F(s) in order to find out the final value of f(t) in the time domain. This is particularly useful in circuits and systems. *Additional Section

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Given:. EXAMPLE 7.10.

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THEOREM 7.9 Transform of Periodic Function If f is periodic with period T on and piecewise continuous one period, then EXAMPLE (a)cos(t) is repeated with period 2 (b)cos(t) is repeated with period From Douglas Wilhelm Harder, University of Waterloo.

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Consider f(t) below: (a) (b) EXAMPLE From Douglas Wilhelm Harder, University of Waterloo.

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THEOREM 7.10 Scaling in Time If exists for s>s 0, then for any real constant a>0, for s>s 0 Proof : EXAMPLE *Additional Section

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THEOREM 7.11 Time delay Time delays occur due to fluid flow, time required to do an analysis (e.g., gas chromatograph). The delayed signal can be represented as then EXAMPLE *Additional Section

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Common Transform Properties Table f(t)F(s)

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EXAMPLE 7.15 Solve the ODE, First, take L of both sides of the equation, Rearrange, Take L -1, From Table,

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References Schiff. Joel L. The Laplace transform: theory and applications. Murray. R. SPIEGEL. Schaum’s Outline of Theory and Problems of Advanced Mathematics for Engineers and Scientists. K.A. Stroud. Engineering Mathematics. Douglas Wilhelm Harder, Math. University of Waterloo ransforms.ppt ransforms.ppt d d

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