SHREE SA’D VIDYA MANDAL INSTITUTE OF TECHNOLOGY

DEPARTMENT OF CIVIL ENGINEERING

Damped force vibrating Model

Laplace Transforms

Prepared by:-Name Arvindsai Nair

Dhaval Chavda

Saptak Patel

Abhiraj Rathod

Enrollment no.130454106002

130454106001

140453106015

140453106014

The Laplace TransformThe Laplace Transform•Suppose that f is a real- or complex-valued function of the (time)variable t > 0 and s is a real or complex parameter. •We define the Laplace transform of f as

The Laplace TransformThe Laplace Transform•Whenever the limit exists (as a finite number). When it does, the integral is said to converge.•If the limit does not exist, the integral is said to diverge and there is no Laplace transform defined for f .

The Laplace TransformThe Laplace Transform•The notation L ( f ) will also be used to denote the Laplace transform of f.•The symbol L is the Laplace transformation, which acts on functions f =f (t) and generates a new function, F(s)=L(f(t))

Example:Then,

provided of course that s > 0 (if s is real). Thus we have L(1) = (s > 0).

The Laplace Transform of δ(t – a)To obtain the Laplace transform of δ(t –

a), we write

and take the transform

The Laplace Transform of δ(t – a) To take the limit as k → 0, use l’Hôpital’s

rule

This suggests defining the transform of δ(t – a) by this limit, that is,

(5)

Some Functions ƒ(t) and Their LaplaceTransforms

Inverse of the Laplace TransformIn order to apply the Laplace

transform to physical problems, it is necessary to invoke the inverse transform.

If L(f (t))=F(s), then the inverse Laplace transform

is denoted by,

s-Shifting: Replacing s by s – a in the Transform

EXAMPLE of s-Shifting: Damped Vibrations

Q. To find the inverse of the transform :-

Solution:-Applying the inverse transform, using its linearity and completing the square, we obtain

• We now see that the inverse of the right side is the damped vibration (Fig. 1)

Example : Unrepeated Complex Factors. Damped Forced VibrationsQ.Solve the initial value problem for a

damped mass–spring system, y + 2y + 2y = r(t), r(t) = 10 sin 2t if 0 < t < π and 0 if t > π; y(0) = 1, y(0)

= –5.Solution. From Table 6.1, (1), (2) in Sec. 6.2,

and the second shifting theorem in Sec. 6.3, we obtain the subsidiary equation

We collect the Y-terms, (s2 + 2s + 2)Y, take –s + 5 – 2 = –s + 3 to the right, and solve,

(6)

For the last fraction we get from Table 6.1 and the first shifting theorem

(7)continued

In the first fraction in (6) we have unrepeated complex roots, hence a partial fraction representation

Multiplication by the common denominator gives

20 = (As + B)(s2 + 2s + 2) + (Ms + N)(s2 + 4).

We determine A, B, M, N. Equating the coefficients of each power of s on both sides gives the four equations

Fig.