La transformada de Laplace

La transformada de Laplace Sea f(t) una función definida para t ≥ 0, su transformada de Laplace se define como: donde s es una variable compleja Se dice que la transformada de Laplace de f(t) existe si la integral converge.

Pierre-Simon Laplace (1749 - 1827) "Podemos mirar el estado presente del universo como el efecto del pasado y la causa de su futuro. Se podría condensar un intelecto que en cualquier momento dado sabría todas las fuerzas que animan la naturaleza y las posiciones de los seres que la componen, si este intelecto fuera lo suficientemente vasto para someter los datos al análisis, podría condensar en una simple fórmula el movimiento de los grandes cuerpos del universo y del átomo más ligero; para tal intelecto nada podría ser incierto y el futuro así como el pasado estarían frente sus ojos." Pierre-Simon Laplace (1749 - 1827)

Notación: Observa que la transformada de Laplace es una integral impropia, uno de sus límites es infinito: Notación:

Condiciones suficientes de existencia de la TL Si f(t) es continua a trozos en [0, ∞) y Es decir, f(t) es de orden exponencial en el infinito: Entonces: L{f(t)} = F(s) existe s > a.

Unicidad de la TL Si f1(t) y f2(t) poseen la misma TL: L{f1(t) } = L{f2(t) }= F(s), Entonces el teorema de Lerch garantiza que

Calcula la transformada de f(t) = 1: Nota: Obviamente L{a} = a/s y L{0} = 0.

Calcula la transformada de f(t) = tn:

Calcula la transformada de f(t) = e-t:

Calcula la transformada de f(t) = Aeat:

Calcula la transformada de f(t) = sen(at): Ejercicio: calcula F(s) para f(t) = cos(at)

Calculemos la transformada de f(t) = eiat:

La función Heaviside o escalón unidad: 1 1 c c t

Función delta de Dirac área = 1 Sea la función parametrizada: Observemos que

Así la transformada de la función delta de Dirac es:

Step function and delta function There are two common functions which are used to represent very rapidly changing quantities. The first of these is the step function, u(t), defined by:

Step function and delta function If the step function ‘switches on’ at t = a it is defined by: u(t-a) t =a t

Step function and delta function The step function can be considered as the limiting case of a very steep “ramp” function: 2b

Step function and delta function The 2nd function is the delta function which is used to represent point loads and other large inputs applied over very small areas. It is defined by: But it also satisfies:

Step function and delta function The delta function can be considered as a limiting case as shown below: 1/2b 2b d(t) t

Step function and delta function The delta function for a point load at t=a is given by: It has the properties:

Laplace transform of the step function For the step function The Laplace transform is:

Laplace transform of the step function For the step function centred at t=a The Laplace transform is:

Laplace transform of the delta function For the delta function The Laplace transform is:

Laplace transform of the delta function For the delta function centred at t=a The Laplace transform is:

Funciones periódicas Supongamos que f (t) es una función periódica de periodo T. Entonces: donde F1(s) es la tranformada de Laplace de la función f(t) sobre el primer periodo y cero fuera. T

Demostración

Ejemplo: onda cuadrada

Tabla de transformadas de Laplace ( ) a s e n t at + - 1 ! 2 d

La TF es un caso particular de la TL Supongamos que  es complejo:  =  + i Antitransformando tendríamos:

todo  perteneciente a la región en rojo. Haciendo s = i( + i) Recordemos que  =  + i: Im() es analítica para Re () todo  perteneciente a la región en rojo. Haciendo s = i( + i) llegamos a la transformada de Laplace. -γ 

Transformada inversa de Laplace Al proceso inverso de encontrar f(t) a partir de F(s) se le conoce como transformada inversa de Laplace y se obtiene mediante: conocida también como integral de Bromwich o integral de Fourier-Mellin.

γ determina un contorno vertical en el plano complejo,tomado de Im(s) γ γ determina un contorno vertical en el plano complejo,tomado de tal manera que todas las singularidades de F(s) queden a su izquierda. Re(s) Con condiciones de existencia:

Por ejemplo, determinemos: Puesto que la función a invertir tiene un polo en s = -1, entonces basta con tomar γ > -1. Tomemos γ = 0 y el contorno de integración C de la figura. Im(s) R C1 γ=0 -1 Re(s) -R Haciendo R→∞ y utilizando teoría de residuos: 0 por la desigualdad ML cuando R→∞ con t≥0.

Sea F(s) una función analítica, salvo en un número finito de polos que se encuentran a la izquierda de cierta vertical Re(s) = γ. Y supongamos que existen m, R, k > 0 tq. para todo s del semiplano Re(s)  γ y |s| > R, tenemos que Entonces si t > 0: En particular, sea F(s) = N(s)/D(s), con N(s) y D(s) polinomios de grado n y d respectivamente, d > n; entonces podemos usar la igualdad anterior.

Ejemplo, determinar:

Propiedades 1. Linealidad: Si c1 y c2 son constantes, f1(x) y f2(x) son funciones cuyas transformadas de Laplace son F1(x) y F2(x), respectivamente; entonces: La transformada de Laplace es un operador lineal.

Demostración:

ò ò ò ò ( ) 2. Desplazamiento temporal F ( s ) = e f ( t ) dt X ( s ) ¥ ò F ( s ) = e - st f ( t ) dt ¥ X ( s ) = ò e - st f ( t - t ) u ( t - t ) dt ¥ = ò e - st f ( t - t ) dt ( ) l = t - t t ¥ = - st ò e e - s l f ( l ) d l = - e st F ( s )

Ejemplo: t 3

Shift in t Remember the definition of the Laplace transform: Question: What happens to f(t) for t<0 ? Answer: The Laplace transform assumes all functions are zero for t<0. Mostly we do not need to know this.

Shift in t Define a shifted function by:

Shift in t The shifted function can also be defined by: The Laplace transform of the shifted function is given by:

Shift in t Substitute t =t-a:

Example - Shift in t Calculate the Laplace transform of a square wave shown by the diagram below

Example - Shift in t Note that the 1st pulse can be constructed as superposition of two step functions:

Example - Shift in t Then, the Laplace transform of the first pulse is :

Example - Shift in t To obtain the Laplace transform of the 2nd pulse, we note that it is the 1st pulse shifted in time by 2a:

Example - Shift in t Thus the Laplace transform is given by: Similarly the Laplace transform of the 3rd pulse is (it is shifted by 4a):

Example - Shift in t Thus the Laplace transform of the whole square wave is given by:

Example - Shift in t In this case we can sum the series (it is a geometric series): Thus:

3. Desplazamiento en frecuencias Ejemplo:

4. Cambio de escala en tiempo

5. Derivada de la transformada de Laplace

6. Transformada de Laplace de las derivadas de una función La transformada de Laplace de la derivada de una función está dada por: donde f(0) es el valor de f(t) en t = 0. La transformada de Laplace de la segunda derivada de una función está dada por:

En forma similar: Demostración:

Supongamos que: Entonces:

Gracias a esta propiedad y a la linealidad de la TL podemos convertir una ec. diferencial como Resolver para y(t) en una ec. algebraica Resolver para Y(s)

Ec. Diferencial Transformada de Laplace Ec. Algebraica

Si resolvemos la ec. algebraica: y encontramos la transformada inversa de Laplace de la solución, Y(s), encontraremos la solución de la ec. diferencial.

Ec. Algebraica Inversa de la Transformada de Laplace Solución de la Ec. Diferencial

La transformada inversa de Laplace de: es

De modo que: es la solución de la ec. diferencial:

Para conseguirlo hemos aplicado: Primero, que la TL y su inversa son lineales: Y segundo, la TF de las derivadas de una función son: etc...

A este método se le conoce como cálculo de Heaviside. Por ejemplo: Y antitransformando obtendremos la solución.

Veamos un ejemplo concreto: Resolver la ec. diferencial

Ejemplo Resolver

Ejemplo: Resolver

7. Transformada de Laplace de la integral de una función Si existe la TL de f(t) cuando Re(s) > p ≥ 0, entonces: para Re(s) > p.

8. Transformada de Laplace de f(t)/t Ejemplo:

Useful theorems - Integration Integration. Given: Then

Example - Integration Use the integration result to find the inverse transform of: Put Then

9. TF de f(t)cos(at) y f(t)sen(at) Ejemplo:

10. Teorema del valor final Si existe, entonces: 11. Teorema del valor inicial El valor inicial f(0) de la función f(t) cuya transformada de Laplace es F(s), es:

como la convolución de y y se denota como 12. Integral de convolución Recordemos que la operación se conoce como la convolución de y y se denota como La transformada de Laplace de esta operación está dada por:

Si trabajamos con funciones que son cero para para t < 0, entonces la convolución queda: Así que para estas funciones podemos definirla convolución como:

Ejemplo: Verificar que funciona para f(t) = t y g(t) = e-2t con valores 0 para t < 0.

De hecho, podemos utilizar la convolución para encontrar transformadas inversas de Laplace:

Convolution theorem For any functions f(t) and g(t) with: Then convolution integral

Convolution theorem and inversion For any functions F(s) and G(s) with: Then Or Do not use unless really necessary!

Convolution theorem – example Use the convolution theorem to invert: Put Thus

Convolution theorem – example Now use Thus

Convolution theorem – example Expand out After much work (see notes)

Resolver la ec.integro-diferencial:

Antitransformando:

Desarrollo en fracciones parciales: Se utiliza para facilitar el cálculo de la transformada inversa, descomponiendo la función en componentes más sencillos. Raíces del denominador D(s) o polos de F(s): Caso I – Polos reales simples Caso II – Polos reales múltiples Caso III – Polos complejos conjugados Caso IV – Polos complejos conjugados múltiples

Caso I – Polos reales simples Ejemplo

método alternativo y resolver...

La transformada inversa de Laplace es:

Otro ejemplo Transformada inversa de Laplace:

Caso II – Polos reales múltiples Ejemplo Polos reales múltiples Polos reales simples

Transformada inversa de Laplace:

En general, para polos reales múltiples:

Caso III – Polos complejos conjugados conjugados complejos ejemplo Transformada inversa de Laplace:

ejemplo Transformada inversa de Laplace: donde

Caso IV – factores complejos conjugados múltiples Se trata de repetir los métodos usados en los casos II y III, teniendo en cuenta que trabajamos con complejos.

Partial Fractions – General Case In the solution of ODEs by the Laplace Transform method, expressions of the form often occur. Here P(s) and Q(s) are both polynomials. These are easy to invert if they are written in partial fraction form: Here

Partial Fractions – Complex Roots If a1 and a2 are a complex conjugate pair, then we can avoid complex numbers by combining these factors into a quadratic expression. Say a1 = -a+ib, a2= -a-ib, then: Then

Partial Fractions – Repeated Roots If a1 = a2 then the partial fraction form is: If a1 = a2 = a3 then the partial fraction form is: Etc, etc

Method 1 - Cover-up Rule Multiply by a factor (s - ai) Put s = ai Basic idea: Multiply by a factor (s - ai) Put s = ai Evaluate ai

Cover-up Rule Find the partial fraction form for: The partial fraction form is:

Cover-up Rule To calculate A, multipy by s and put s=0:

Cover-up Rule To calculate B, multipy by s+1 and put s=-1: Thus the partial fraction form is:

Cover-up Rule The above procedure can be carried out by “covering up”. For A use: Now put s = 0, ignoring the covered up items:

Cover-up Rule Similarly for B, “covering up” gives: Now put s = -1, ignoring the covered up items:

Complex Cover-up Rule Find the coefficients A, B, C in: Find A using the standard cover-up idea:

Complex Cover-up Rule Put s+1 = 0, that is s = -1: The coefficients B and C are more difficult to calculate. A modified version of the cover-up rule involves using complex factors

Complex Cover-up Rule Multiply the whole equation by (s+i):

Complex Cover-up Rule Now put (s+i)=0, that is s = -i: After tidying up:

Complex Cover-up Rule Equate real and imaginary parts: Final partial fraction:

Method 2 - Substitution of values Basic idea Put s = bi for i=1,2,…,n. Here bi are convenient, easy to work with, numbers Obtain n equations in the n unknown ai Solve for ai

Substitution of values Find the coefficients A, B, C in: Use cover-up for B : Put s=0:

Substitution of values Use cover-up for C: Put s=-1:

Substitution of values We cannot use cover-up for A. So far we have: One good way to calculate A is to substitute a convenient value for s. Say s =1:

Substitution of values Substitute back:

Example - substitution of values Substitute values to work out A and B in: Put s=1: Put s=2:

Example - substitution of values Subtract two equations Substitute back

Method 3 - Equate coefficients Basic idea: Multiply whole equation by P(s) Equate coefficients of each power of s Solve resulting equations for ai

Example - Equate coefficients Calculate A and B in: Multiply by s(s+1): Equate coefficients:

Example - Equate coefficients Calculate A and B in: Multiply by (s+1)(s2 +1): Equate coefficients:

Preferred method for partial fractions Calculate as many coefficients as possible using the simple cover-up rule Calculate the remaining coefficients by: Substituting values for s Using the complex cover-up method Equating coefficients

More on differential equations ODE problem Apply Laplace transform

More on differential equations Rearrange This term comes from the initial conditions. To invert, convert into partial fraction form, then use tables and useful rules This term comes from the right hand side of the ODE. To invert, convert into partial fraction form (if possible) then use tables. Otherwise use partial fractions on 1/P(s) , invert, and then apply the convolution theorem

Transfer function (optional extra) If we forget about the initial transient in Then

Transfer function (optional extra) Thus the transfer function Can be written as: If f(t)=d(t) then F(s)=1 and Thus the transfer function is the Laplace transform of the response of the system to an impulse (delta function)

Example (Laplace transform solution of an ODE) Solve the following problem using the Laplace transform method: Step 1. Define your transform

Example (Laplace transform solution of an ODE) Step 2. Transform the ODE:

Example (Laplace transform solution of an ODE) Use formulae from the tables Tidy up

Example (Laplace transform solution of an ODE) Step 3. Use the initial conditions and solve for Y(s):

Example (Laplace transform solution of an ODE) Step 4. Find the partial fraction forms. First Cover-up for A: Put s = -1:

Example (Laplace transform solution of an ODE) Cover-up for B: Put s = -2: Substitute back:

Example (Laplace transform solution of an ODE) Now work out the partial fraction form for: Use cover-up for C Put s = -1:

Example (Laplace transform solution of an ODE) Use cover-up for D Put s = -2: Result so far:

Example (Laplace transform solution of an ODE) To find B, put s = 0 :

Example (Laplace transform solution of an ODE) To find A, put s = 1 :

Example (Laplace transform solution of an ODE) Substitute back: Combine both partial fractions:

Example (Laplace transform solution of an ODE) Tidy up: Step 5. Invert using tables. Look at each term separately:

Example (Laplace transform solution of an ODE) Combine to invert Y(s):

Inversion of typical terms in partial fractions (I) Inversion of From tables: (II) Inversion of

Inversion of typical terms in partial fractions (III) Inversion of

Inversion of typical terms in partial fractions From tables:

Even more on Laplace transforms and ODE’s The Laplace transform method gives the result: Here:

Even more on Laplace transforms and ODE’s In the special case when f(t) is a polynomial, exponential, sine or cosine or a sum of such terms, then F(s) can be written in the partial fraction form: Thus the partial fraction form for F(s)/P(s) will be:

Even more on Laplace transforms and ODE’s Thus the part of y(t) arising from the inverse of F(s)/P(s) will contain the following parts: The same kinds of functions as in the rhs f(t). (This part corresponds to the particular integral of Module 3) A sum of several exponentials of the form below (This part corresponds to the complementary function of Module 3)

Even more on Laplace transforms and ODE’s The real part of all the ai determine whether or not the system is stable: If any Re{ai} are positive the solution will grow exponentially with time. That means the system is unstable. If all Re{ai} are negative the solution will decay exponentially with time. That means the system is stable.

Even more on Laplace transforms and ODE’s If all Re{ai} are zero the solution will oscillate with time. That means the system is stable – except for the case of resonance. The case of resonance occurs when one of the ai is the same as one of the bi. In this case terms like that below will occur: This leads to terms in the solution of the form below which correspond to weak instability.

Final example (summary of all methods)   Solve the following problem using the integrating factor method, the guessing method and the Laplace transform method: with

Final example (summary of all methods) (a) Integrating factor method (Module 1). Compare with the standard form: In our case g(t)=1, thus the integrating factor is given by:

Final example (summary of all methods) Multiply the ODE by the integrating factor: Use the product rule in reverse: Integrate:

Final example (summary of all methods) Use integration by parts on the rhs: Divide by the integrating factor: Use the initial condition y(0)=1: Substitute back:

Final example (summary of all methods) (b) Guessing method (see Module 3). Step 1. Find the complementary function. Solve: Try y=Celt. This gives the characteristic equation: Thus:

Final example (summary of all methods) Step 2. Find the particular integral. Try Substitute in the ODE: Equate coefficients:

Final example (summary of all methods) Step 3. Combine the particular integral and the complementary function. The general solution is: Step 4. Use the initial condition y(0)=1: Substitute back:

Final example (summary of all methods) (c) Laplace Tranform method (see Module 5) Step1. Define your transform: Step 2. Transform the ODE:

Final example (summary of all methods) Tidy up: Step 3. Use the initial condition y(0)=1 and solve for Y(s):

Final example (summary of all methods) Step 4. Put in partial fraction form. Find A,B,C for: Use cover-up to find B:

Final example (summary of all methods) Use cover-up to find C: At this stage we have: Put s = 1 :

Final example (summary of all methods) Substitute back: Combine all terms in Y(s):

Final example (summary of all methods) Step 5. Invert using tables: