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LAPLACE TRANSFORM PPT PRESENTATION

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Published in: Mathematics
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In Mathematics, the Laplace transform is an integral transform named after its inventor Pierre-Simon Laplace. It takes a function of a real variable t (often time) to a function of a complex variable s (complex frequency). The Laplace transform is very similar to the Fourier transform.

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  1. Laplace Transform
  2. The French Newton Pierre-Simon Laplace Developed mathematics in astronomy, physics, and statistics Began work in calculus which led to the Laplace Transform Focused later on celestial mechanics One of the first scientists to suggest the existence of black holes
  3. Istory of the Transform Euler began looking at integrals as solutions to differential equations in the mid 1700's: re; a) — f eat X (r)eaz dc Lagrange took this a step further while working on probability density functions and looked at forms of the following equation: X (r)e-aæaz dc, Finally, in 1785, Laplace began using a transformation to solve equations of finite differences which eventually lead to the current transform Ys f e —s x +(x) clx,
  4. ni Ion The Laplace transform is a linear operator that switched a function f(t) to F(s). Specifically: e stf(t) dt. where: Go from time argument with real input to a complex angular frequency input which is complex.
  5. nctions There are two governing factors that determine whether Laplace transforms can be used: ' f(t) must be at least piecewise continuous for t 2 0 ' Meytwhere M and y are constants
  6. Inuity Since the general form of the Laplace transform is: it makes senseF(s) least piecewise e st f (t) dt. continuous for t 2 0. o If f(t) were very nasty, the integral would not be computable.
  7. e ness This criterion also follows directly from the general O definition: st f (t) dt. o If f(t) is not bounded by MeYt then the integral will not converge.
  8. aplace Transform Theorv dt lim — co 'General Theory 'Example 'Convergence 1 lim = lim —st e stldt lim -s t) 1 s e -st f (t) dt 0 t2 e
  9. TABLE 62.1 Elementary Laplace Transforms F(S) CLf(t)} ce Trans orms 'Some Laplace Transforms 'Wide variety of function can be transformed 1. 2. 4. 5. 6. 7. 8. 9. 10. 12. 13. 14. 17. 18. 19. 1 eat tn, n = positive integer 1 s 1 sn+l ' sp+l 'Inverse Transform 'Often requires partial fractions or other manipulation to find a form that is easy to apply the inverse sin at cos at Sinh at cosh at eat sin bt eat COS bt t n eat , n positive integer ectf(t) f(ct) — CIT (-tYf(t) s > lal s2 s2 a a 2' b (s — a)2 + b2 (s — a)2 + b2 (S — a)n+l s e-CSF(s) 1 FOG(s) e-cs ST(s) - sn-lf(0) - — f(n-l) (0)
  10. Laplace Transform for ODEs 'Equation with initial conditions 'Laplace transform is linear 'Apply derivative formula 'Rearrange 'Take the inverse d2y COY") + C(y) s2C(y) — sy(0) 1 1 s 1 s s(s2 + 1) S 2+1 s -1 — cos t Y
  11. Laplace- form in PDEs Laplace transform in two variables (always taken with respect to time variable, t): Inverse laplace of a 2 dimensional PDE: Can be used for any dimension PDE: The Transform reduces dimension by 1 : 'ODEs reduce to algebraic equations 'PDEs reduce to either an ODE (if original equation dimension 2) or another PDE (if original equation dimension >2) du —st — dt dt
  12. Consider the case where: ux+ut=t with u(x,o)=o and u(o,t)=t2 and Taking the Laplace of the initial equation leaves Ux+ U=1/s2 (note that the partials with respect to "x" do not disappear) with boundary condition Solving this as an ODE of variable x, U(x,s)=c(s)e-X+ 1/s2 Plugging in B.C., 2/s3=c(s) + 1/s2 so c(s)=2/s3- 1/s2 U(x,s)=(2/s3 - 1/S2) e-X+ 1/S2 Now, we can use the inverse Laplace Transform with respect to s to find - te-x + t
  13. Example Solutions
  14. slon qua ut = kuxx in (0,1) Initial Conditions: u(o,t) = u(l,t) = 1, u(x,o) = 1 + sin(ltx/l) Using af(t) + bg(t) -5 aF(s) + bG(s) df/dt -5 sF(s) - f(o) and and noting that the partials with respect to x commute with the transforms with respect to t, the Laplace transform U(x,s) satisfies sU(x,s) — u(x,0) = With eat -5 1/(s-a) and a=o, the boundary conditions become U(o,s) = U(l,s) = l/s. So we have an ODE in the variable x together with some boundary conditions. The solution is then: U(x,s) l/s + Therefore, when we invert the transform, using the Laplace table: u(x,t) = 1+ e -kit t/l sin(ax/l)
  15. e quatlon utt = C uxx in 0 < X < Initial Conditions: u(0,t) = f(t), u(x,0) = = 0 For x 00, we assume that u(x,t) -5 0. Because the initial conditions vanish, the Laplace transform satisfies s2U = c2U xx Solving this ODE, we get U(x,s) = + Where a(s) and b(s) are to be determined. From the assumed property of u, we expect that U(x,s) o as x 00. Therefore, b(s) = o. Hence, U(x,s) = F(s) e-sx/c. Now we use -Y e-bSF(s) To get u(x,t) = H(t — x/c)f(t — x/c).
  16. Real-Life App Ica Ions O Semiconductor mobility 0 Call completion in wireless networks O Vehicle vibrations on compressed rails O Behavior of magnetic and electric fields above the atmosphere
  17. x. Semicon uc or Mobl lty O Motivation semiconductors are commonly made with superlattices having layers of differing compositions need to determine properties of carriers in each layer o concentration of electrons and holes o mobility of electrons and holes conductivity tensor can be related to Laplace transform of electron and hole densities
  18. Nota non O R = ratio of induced electric field to the product of the current density and the applied magnetic field O p = electrical resistance o H = magnetic field O J = current density o E = applied electric field O n = concentration of electrons o u = mobility .v=pJt-RHJ Ev=RHJr+pJ
  19. Equation neg ampulation negc ne xx and xy — neg H
  20. uming a continuous distribution and that it follows: oxx(H) xy s+lp) + s-(p) de
  21. pplying the Lap ace rans orm HCG = x e-Y sin(xt)dt = — O o e—Yt o e—Yt o dt 0 ex(px+ dt, O dt o dt. O
  22. Source Johnson, William B. Transform method for semiconductor mobility, Journal of Applied Physics 99 (2006).

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