Arvindmaths 150316024240 Conversion Gate01

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SHREE SA’D VIDYA MANDAL INSTITUTE OF TECHNOLOGY DEPARTMENT OF CIVIL ENGINEERING

description

aap of laplace transformation in civil engg.

Transcript of Arvindmaths 150316024240 Conversion Gate01

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SHREE SA’D VIDYA MANDAL INSTITUTE OF TECHNOLOGY

DEPARTMENT OF CIVIL ENGINEERING

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Damped force vibrating Model

Laplace Transforms

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Prepared by:-Name Arvindsai Nair

Dhaval Chavda

Saptak Patel

Abhiraj Rathod

Enrollment no.130454106002

130454106001

140453106015

140453106014

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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

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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 .

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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))

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Example:Then,

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

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

a), we write

and take the transform

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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)

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Some Functions ƒ(t) and Their LaplaceTransforms

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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,

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

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EXAMPLE of s-Shifting: Damped Vibrations

Q. To find the inverse of the transform :-

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

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

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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

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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

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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

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Fig.

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