Damped force vibrating Model of Laplace Transforms

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))

7. Example:
Then,

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.