Laplace Transforms, Dirac Delta, and Periodic Functions

A mass m = 1 is attached to a spring with constant k = 4; there is no damping. The mass is released from rest with y(0) = 3. At the instant t = 2π the mass is struck with a hammer, providing an impulse 8δ(t – 2π). Determine the equation of motion of the mass.

Example 1.

4 8 2 0 3 0 0IVP: y'' y (t ) , y( ) ,y'( )

Taking Laplace transforms we get 2 23 4 8 ss Y(s) s Y(s) e

Solving for Y(s) we get

2

2 2

3 8

4 4

ss eY(s)

s s

Taking inverse transforms we get

2

2

3 2 4 2 2

3 2 4 2

y(t) cos( t) u (t)sin( (t ))

cos( t) u (t)sin( t)

In terms of piecewise functions we have

3 2 0 2

3 2 4 2 2

cos( t), [ , )y(t)

cos( t) sin( t), [ , )

At t = 2π there is a discontinuity in the velocity (the derivative) since amplitude increases from 3 to 5 (5 = sqrt(32 + 42)).

Let’s look at the graph

0 2 4 6 8 10 12 14 16 18 20-5

-4

-3

-2

-1

0

1

2

3

4

53.0*cos(2.0*t) + 4.0*sin(2.0*t)*heaviside(t - 6.2832)

Sharp Point

The derivative of the solution is

y’(t) = 8*cos(2*t)*u2*pi (t ) - 6*sin(2*t) + 4*sin(2*t)*δ(t - 2*pi)

The “jump” due to the Dirac delta term.

Significant jump in the velocity.

Example 2.

Consider a mass spring system with m = k = 1 with y(0) = y’(0) = 0. At each of the time instants t = 0, π, 2π, 3π, . . . . , nπ, . . . the mass is struck with a hammer with unit impulse. Determine the equation of the resulting motion.

2 0 0 0 0IVP: y'' y (t) (t ) (t ) ... (t n ) ... , y( ) ,y'( )

0

0 0 0 0

n

IVP : y'' y (t n ) , y( ) ,y'( )

Taking Laplace transforms we get 2

0

n s

n

s Y(s) Y(s) e

Solving for Y(s) we get2

01

n s

n

eY(s)

s

0

n

n

y(t) u (t) sin(t n )

Compute the inverse transform term by term to get

Using some trig we have then we have1 nsin(t n ) ( ) sin(t)

0

0 2 0

0 3

0 4 0

[ , ), sin(t)

[ , ), sin(t) sin(t)

y(t) [ , ), sin(t) sin(t) sin(t) sin(t)

[ , ), sin(t) sin(t) sin(t) sin(t)

ETC.

So the graph is if n is even

0 if n i odd

sin(t),y(t)

, s

0 2 4 6 8 10 12 14 16 18 20-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1Half wave rectification of sin(t).

Periodic Functions and Laplace transforms.

Periodic forcing functions in practical mechanical or electrical systems often are more complicated than pure sines or cosines. The nonconstant function f(t) defined for t ≥ 0 is said to be periodic if there is a number p > 0 such that

f(t + p) = f(t) (*) for all t ≥ 0. The least positive value of p (if any) for which Eq. (*) holds is called the period of f. Such a function is shown

The following theorem simplifies the computation of the Laplace transform of a periodic function.

Let f(t) be periodic with period p and piecewise continuous for t ≥ 0. Then the Laplace transform F(s) = L{f(t)} exists for s > 0 and is given by

The proof uses the definition of a Laplace Transforms, properties of periodic functions, and geometric series. We omit the proof here.

The principal advantage of this Theorem is that it enables us to find the Laplace transform of a periodic function without the necessity of an explicit evaluation of an improper integral.

Several periodic functions that occur in applications are shown next with their Laplace Transforms.

Example

Consider a mass-spring-dashpot system with m = 1, c = 4, and k = 20 in appropriate units. Suppose that the system is initially at rest at equilibrium x(0) = 0, x’(0) = 0 and that the mass is acted on beginning at time t = 0 by the external force f(t) whose graph is shown below which is a square wave with amplitude 20 and period 2π. Find the solution for the position of the mass at time t.

Using the previous theorem we have that the Laplace transform of the forcing function is

and the Laplace transform of the solution to the IVP is

After some “work” we can get that the solution of the IVP for the interval nπ< t < (n + 1)π consistent of a transient part and a steady state part:

where τ = t –nπ , for t in the interval nπ < t < (n+1)π

The figure shows the graph of xsp(t).

τ = t –nπ, for t in the interval nπ < t < (n+1)π;

Source: Edwards & Penney Differential Equations