Arvindmaths 150316024240 Conversion Gate01
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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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