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n

• L(t ) = • L(eat) =

n!

• L(floor(t/a)) =

s1+n 1

• L(sqw(t/a)) =

s−a

• L(cos bt) = • L(sin bt) =

e−as

s s2 + b2 b

s2 + b2 e−as • L(H(t − a)) = s −as • L(δ(t − a)) = e

s(1 − e−as) 1 s

• L(a trw(t/a)) = • L(tα) =

tanh(as/2) 1

s2

Γ(1 + α)

• L(t−1/2) =

s1+α r π s

tanh(as/2)

Proof of L(tn ) = n!/s1+n Slide 1 of 3 The first step is to evaluate L(f (t)) for f (t) = t0 [n = 0 case]. The function t0 is written as 1, but Laplace theory conventions require f (t) = 0 for t < 0, therefore f (t) is technically the unit step function.

L(1) =

R∞ 0

(1)e−stdt t=∞

= −(1/s)e−st|t=0 = 1/s

Laplace integral of f (t) = 1. Evaluate the integral. Assumed s > 0 to evaluate limt→∞ e−st .

Proof of L(tn ) = n!/s1+n Slide 2 of 3 The value of L(f (t)) for f (t) = t can be obtained by s-differentiation of the relation L(1) = 1/s, as follows. Technically, f (t) = 0 for t < 0, then f (t) is called the ramp function. d L(1) ds

R∞ d = ds 0 (1)e−stdt R ∞ d −st = 0 ds (e ) dt R∞ = 0 (−t)e−stdt

Laplace integral for f (t) = 1.

= −L(t)

Definition of L(t).

Used

d ds

Rb a

F dt =

R b dF

Calculus rule (eu )0 = u e .

Then d L(t) = − ds L(1)

dt.

a ds 0 u

Rewrite last display.

d = − ds (1/s)

Use L(1) = 1/s.

= 1/s2

Differentiate.

Proof of L(tn ) = n!/s1+n Slide 3 of 3 This idea can be repeated to give

L(t2) = −

d

L(t)

ds = L(t2) 2 = 3. s

d The pattern is L(tn ) = − ds L(tn−1), which implies the formula

L(tn) = The proof is complete.

n! s1+n

.

Proof of L(eat ) =

1

s−a The result follows from L(1) = 1/s, as follows. R∞ L(eat) = 0 eate−stdt R ∞ −(s−a)t dt = 0 e R ∞ −St = 0 e dt = 1/S = 1/(s − a)

Direct Laplace transform. Use eA eB = eA+B . Substitute S = s − a. Apply L(1) = 1/s. Back-substitute S = s − a.

Proof of L(cos bt) =

s s2 + b2

and L(sin bt) =

Slide 1 of 2 Use will be made of Euler’s formula

b s2 + b2

eiθ = cos θ + i sin θ, usually √ first introduced in trigonometry. In this formula, θ is a real number in radians and i = −1 is the complex unit.

eibte−st = (cos bt)e−st + i(sin bt)e−st R∞ 0

R∞ e−ibte−stdt = 0 R(cos bt)e−stdt ∞ + i 0 (sin bt)e−stdt

1 s − ib

=

R∞

(cos bt)e−stdt R∞ + i 0 (sin bt)e−stdt 0

Substitute θ = bt into Euler’s formula and multiply by e−st . Integrate t = 0 to t = ∞. Then use properties of integrals. Evaluate the left hand side using L(eat ) = 1/(s − a), a = ib.

Proof of L(cos bt) = Slide 2 of 2

1 s − ib s + ib s2 + b2 s s2 + b2 b s2 + b2

s s2 + b2

and L(sin bt) =

= L(cos bt) + iL(sin bt)

b s2 + b2 Direct Laplace transform definition.

= L(cos bt) + iL(sin bt)

Use complex rule 1/z = z/|z|2, z = A √ + iB , z = A − iB , |z| = A2 + B 2.

= L(cos bt)

Extract the real part.

= L(sin bt)

Extract the imaginary part.

Proof of L(H(t − a)) = e−as /s

L(H(t − a)) =

R∞

=

R∞

0

a

H(t − a)e−stdt (1)e−stdt

R∞

(1)e−s(x+a)dx R∞ (1)e−sxdx =e 0 =

0 −as

= e−as(1/s)

Direct Laplace transform. Assume a ≥ 0. Because H(t − a) = 0 for 0 ≤ t < a. Change variables t = x + a. Constant e−as moves outside integral. Apply L(1) = 1/s.

Proof of L(δ(t − a)) = e−as Slide 1 of 3 The definition of the delta function is a formal one, in which every occurrence of symbol δ(t − a)dt under an integrand is replaced by dH(t − a). The differential symbol dH(t − a) is taken in the sense of the Riemann-Stieltjes integral. This integral is defined in Rudin’s Real analysis for monotonic integrators α(x) as the limit

Z

b

f (x)dα(x) = lim a

N →∞

N X

f (xn)(α(xn) − α(xn−1))

n=1

where x0 = a, xN = b and x0 < x1 < · · · < xN forms a partition of [a, b] whose mesh approaches zero as N → ∞. The steps in computing the Laplace integral of the delta function appear below. Admittedly, the proof requires advanced calculus skills and a certain level of mathematical maturity. The reward is a fuller understanding of the Dirac symbol δ(x).

Proof of L(δ(t − a)) = e−as Slide 2 of 3

L(δ(t − a)) =

R∞

=

R∞

0

0

e−stδ(t − a)dt e−stdH(t − a)

= limM →∞ = e−sa

RM 0

e−stdH(t − a)

Laplace integral, a > 0 assumed. Replace δ(t − a)dt by dH(t − a). Definition of improper integral. Explained below.

Proof of L(δ(t − a)) = e−as Slide 3 of 3 To explain the last step, apply the definition of the Riemann-Stieltjes integral:

Z

M −st

e 0

dH(t − a) = lim

N →∞

N −1 X

e−stn (H(tn − a) − H(tn−1 − a))

n=0

where 0 = t0 < t1 < · · · < tN = M is a partition of [0, M ] whose mesh max1≤n≤N (tn − tn−1) approaches zero as N → ∞. Given a partition, if tn−1 < a ≤ tn, then H(tn − a) − H(tn−1 − a) = 1, otherwise this factor is zero. Therefore, the sum reduces to a single term e−stn . This term approaches e−sa as N → ∞, because tn must approach a.

Proof of L(floor(t/a)) =

e−as s(1 − e−as)

Slide 1 of 3 The library function floor present in computer languages C and Fortran is defined by floor(x) = greatest whole integer ≤ x, e.g., floor(5.2) = 5 and floor(−1.9) = −2. The computation of the Laplace integral of floor(t) requires ideas from infinite series, as follows.

F (s) =

R∞ 0

floor(t)e−st dt

=

P∞ R n+1

=

P∞ n

=

n=0

n

n=0

s

1−e

−s

s

(n)e−stdt

Laplace integral definition. On n ≤ t floor(t) = n.

<

n + 1,

(e−ns − e−ns−s)

Evaluate each integral.

P∞

Common factor removed.

n=0

ne−sn

Proof of L(floor(t/a)) =

e−as s(1 − e−as)

Slide 2 of 3

= = = = =

x(1 − x) P∞ n=0

nxn−1

s x(1 − x) d P∞ s dx x(1 − x) d

1

dx 1 − x

s x s(1 − x) e−s s(1 −

n=0

e−s)

xn

Define x = e−s . Term-by-term differentiation. Geometric series sum. Compute the derivative, simplify. Substitute x = e−s .

Proof of L(floor(t/a)) =

e−as s(1 − e−as)

Slide 3 of 3 To evaluate the Laplace integral of floor(t/a), a change of variables is made.

L(floor(t/a)) =

R∞ 0

floor(t/a)e−st dt

R∞ = a 0 floor(r)e−asr dr = aF (as) =

e−as s(1 − e−as)

Laplace integral definition. Change variables t = ar . Apply the formula for F (s). Simplify.

Proof of L(sqw(t/a)) =

1 s

tanh(as/2)

Slide 1 of 3 The square wave defined by sqw(x) = (−1)floor(x) is periodic of period 2 and R2 piecewise-defined. Let P = 0 sqw(t)e−st dt.

P = =

R1 R01 0 1

sqw(t)e−st dt +

e−stdt −

1

1

sqw(t)e−st dt

e−stdt 1

(1 − e ) + (e−2s − e−s) s s 1 = (1 − e−s)2 s =

−s

R2

R2

Apply

Rb a

=

Rc a

+

Rb c

.

Use sqw(x) = 1 on 0 ≤ x < 1 and sqw(x) = −1 on 1 ≤ x < 2. Evaluate each integral. Collect terms.

Proof of L(sqw(t/a)) =

1 s

tanh(as/2)

Slide 2 of 3 – Compute L(sqw(t))

R2 L(sqw(t)) = = = = = =

0

sqw(t)e−st dt

Periodic function formula.

1 − e−2s 1

(1 − e−s)2

s 1 1 − e−s s 1 + e−s

1 1−

.

s cosh(s/2) 1 s

Use the computation of P above.

1 − e−2s = (1 − e−s)(1 + e−s).

e−s/2

+ 1 sinh(s/2)

.

Factor

1 es/2 − e−s/2 s es/2

e−2s

.

tanh(s/2).

.

Multiply the fraction by es/2 /es/2 . Use sinh u = (eu − e−u )/2, cosh u = (eu + e−u)/2. Use tanh u = sinh u/ cosh u.

Proof of L(sqw(t/a)) =

1 s

tanh(as/2)

Slide 3 of 3 To complete the computation of L(sqw(t/a)), a change of variables is made:

L(sqw(t/a)) = = =

R∞ R0∞ 0 a

sqw(t/a)e−st dt

Direct transform.

sqw(r)e−asr (a)dr

Change variables r t/a.

tanh(as/2)

See L(sqw(t)) above.

as 1 = tanh(as/2) s

=

1

tanh(as/2) Rt The triangular wave is defined by trw(t) = 0 sqw(x)dx. Proof of L(a trw(t/a)) =

L(a trw(t/a)) = =

=

s2

f (0) + L(f 0(t)) s 1 s

L(sqw(t/a))

1 s2

tanh(as/2)

Let f (t) = a trw(t/a). Use

L(f 0(t)) = sL(f (t)) − f (0). Use f (0) = 0, then use R t/a (a 0 sqw(x)dx)0 = sqw(t/a). Table entry for sqw.

Proof of L(tα ) =

L(tα) = = = =

Γ(1 + α) s1+α

R∞

tαe−stdt 0 R∞ (u/s)αe−udu/s 0 1 R∞ s1+α 1

0

uαe−udu

Γ(1 + α). 1+α s

Definition of Laplace integral. Change variables u = st, du = sdt. Because s=constant for u-integration. Because Γ(x) ≡

R∞ 0

ux−1e−udu.

Gamma Function The generalized factorial function Γ(x) is defined for x > 0 and it agrees with the classical factorial n! = (1)(2) · · · (n) in case x = n + 1 is an integer. In literature, α! means Γ(1 + α). For more details about the Gamma function, see Abramowitz and Stegun or maple documentation. r Proof of L(t

−1/2

L(t−1/2) =

)=

π s

Γ(1 + (−1/2)) √

π = √ s

s1−1/2

Apply the previous formula. Use Γ(1/2) =

√

π.

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