- 1. SHREE SA’D VIDYA MANDAL INSTITUTE OF TECHNOLOGY DEPARTMENT OF CIVIL ENGINEERING
- 2. Damped force vibrating Model Laplace Transforms
- 3. Prepared by:- Name Arvindsai Nair Dhaval Chavda Saptak Patel Abhiraj Rathod Enrollment no. 130454106002 130454106001 140453106015 140453106014
- 4. 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
- 5. 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 .
- 6. 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))
- 8. provided of course that s > 0 (if s is real).Thus we have L(1) = (s > 0).
- 9. The Laplace Transform of δ(t – a) To obtain the Laplace transform of δ(t – a), we write and take the transform
- 10. 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)
- 11. Some Functions ƒ(t) and Their Laplace Transforms
- 12. Inverse of the Laplace Transform In 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,
- 13. s-Shifting: Replacing s by s – a in the Transform
- 14. EXAMPLE of s-Shifting: Damped Vibrations Q. To find the inverse of the transform :-
- 15. Solution:- Applying the inverse transform, using its linearity and completing the square, we obtain
- 16. • We now see that the inverse of the right side is the damped vibration (Fig. 1)
- 18. Example : Unrepeated Complex Factors. Damped Forced Vibrations Q.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
- 19. 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
- 20. 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
- 21. Fig.