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math121a-f23:october_16_monday

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October 13 (Friday)

  • What is Laplace transform?
  • What's the difference between that and Fourier transform?
  • When to use it?

Definition

Given a function f(t)f(t) on the positive real line t>0t>0, we can define the following function of pp: F(p)=t=0f(t)eptdt. F(p) = \int_{t=0}^\infty f(t) e^{-pt} dt. Again, we require the function f(t)f(t) to have moderate growth at tt \to \infty for the integral to be well-defined.

Examples

  • f(t)=1f(t) = 1, F(p)=1/p,F(p) = 1/p, valid for Re(p)>0Re(p) > 0
  • f(t)=eatf(t) = e^{a t}, F(p)=1/(pa),F(p) = 1/(p-a), valid for Re(pa)>0Re(p-a) > 0.
  • f(t)=cos(at)f(t) = \cos(at), F(p)=(1/2)[1/(pia)+1/(p+ia)]=p/(p2+a2).F(p) = (1/2)[1/(p-ia) + 1/(p+ia)] = p/(p^2 + a^2). valid for Re(p)>0Re(p)>0 if aa is real.

properties

Suppose we know the Laplace transform of f(t)f(t), let's denote F=LT(f)F = LT(f) (note we just write the name of the function ff, not including its input variables t). What can we say about LT(f)LT(f')?

We can do integration by part LT(f)=0ept(d/dt)f(t)dt=t=0t=eptdf=t=0t=d[eptf]d[ept]f=eptf(t)0+0peptf(t)dt=f(0)+pF(p). LT(f') = \int_0^\infty e^{-pt} (d/dt) f(t) dt = \int_{t=0}^{t=\infty} e^{-pt} df = \int_{t=0}^{t=\infty} d[e^{-pt} f] - d[e^{-pt}] f= e^{-pt} f(t)|_0^\infty + \int_0^\infty p e^{-pt} f(t) dt = -f(0) + pF(p).

math121a-f23/october_16_monday.1697213091.txt.gz · Last modified: 2023/10/13 09:04 by pzhou