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Laplace Transform

Definition

F(s)=L{f(t)}=0estf(t) dt

The Laplace Transforms of Basic Functions

f(t) F(s)
1 1/s
tn n!sn+1
exp(at) 1sa
tneat n!(sa)n+1
sin(kt) ks2+k2
cos(kt) ss2+k2
sinh(kt) ks2k2
cosh(kt) ss2k2

Constraints

  • constraint 1 For a function f(t), constant c, M>0, and T>0 s.t.

    |f(t)|M ect,  t>T

    f(t) 不能跑得比 exp(ct)

  • constraint 2

    f(t) should be piecewise continuous on [0,)

      t[0,), f(t) 為 discontinuous 的點有限

note : these constraints are sufficient conditions


五大定理

Five Theorem

Transforms of Derivatives

L{f(t)}=0est f(t) dt=[est f(t)]0+s 0est f(t) dt=sL{f(t)}f(0)L{f(t)}=s2 F(s)sf(0)f(0)L{f(t)}=s3F(s)s2f(0)sf(0)f(0)L{f(n)(t)}=snF(s)sn1f(0)sn2f(0)sf(n2)f(n1)(0)

First Translation Theorem

translation for s

L{eat f(t)}=F(sa)

proof :

L{eat f(t)}=0est eat f(t) dt=0e(sa)t f(t) dt=F(sa)

Second Translation Theorem

translation for t

L{f(ta)u(ta)}=eas F(s)

or

L{g(t) u(ta)}=eas L{g(t+a)}

in which u(t) is an unit step function.

proof :

L{f(ta) u(ta)}=0est f(ta) u(ta) dt=aest f(ta) dt(let t1=ta)=0es(t1+a) f(t1) dt1=eas F(s)

Derivatives of Transforms

L{tn f(t)}=(1)n dndsnF(s)

Convolution

Definition :

f(t)g(t)=f(τ) g(tτ) dτ

When f(t)=0 for t<0 and g(t)=0 for t<0

We have the same definition as below.

f(t)g(t)=0tf(τ) g(tτ) dτ

Convolution Theorem

L{f(t)g(t)}=F(s) G(s)

Integration

from theorem above

L{0tf(τ) dτ}=L{f(t)1}=F(s)s

Transform of a Periodic Function

When

f(t+T)=f(t)

then

L{f(t)}=11esT0Test f(t) dt
11esT

來自無窮等比級數

Dirac Delta Function

δ(tt0)={for t=t00for tt0

Integration

δ(tt0) dt=1

Sifting

pqf(t) δ(tt0) dt=f(t0)
t0[p,q]

Laplace Transform

L{δ(tt0)}=et0s
when t0>0

proof :

 δ(tt0)=u(tt0) L{δ(tt0)}=sL{u(tt0)}u(0)

due to second translation theorem, we have

L{δ(tt0)}=et0ss s=et0s