17-26 INVERSE TRANSFORMS BY CONVOLUTION Showing details,...

17-26 INVERSE TRANSFORMS BY CONVOLUTION Showing details, find $f(t)$ if $\mathcal{L}\{f\}$ equals: 17. $\frac{5.5}{(s+1.5)(s-4)}$ 18. $\frac{1}{(s-a)^2}$ 19. $\frac{2\pi s}{(s^2 + \pi^2)^2}$ 20. $\frac{9}{s(s+3)}$ 21. $\frac{\omega}{s^2(s^2 + \omega^2)}$ 22. $\frac{e^{-ax}}{s(s-2)}$ 23. $\frac{40.5}{s(s^2 - 9)}$ 24. $\frac{240}{(s^3 + 1)(s^2 + 25)}$ 25. $\frac{18s}{(s^2 + 36)^2}$

Bcrypt_Sha256$$2B$12$We1Wwocamog01O5I.V2Tkouxdh4Ofnmgpwkor7Leaonfpu0Ubfpua Bcrypt_Sha256$$2B$12$We1Wwocamog01O5I.V2Tkokttmmj7Lscvwvlptp4Rlhbswcdg9.Wy and 87 other subject Calculus 3 educators are ready to help you.

Ask a new question

Transcript

00:01 In this problem, we are going to compute the inverse laplace transform of the following expressions using the convolution theorem.

00:10 We have 6 questions.

00:13 The first one is number 17.

00:16 We have the laplace transform given by 5 .5 divided by s plus 1 .5 times s minus 4.

00:28 Let us write this as 5 .5 times 1 over s plus 1 .5 times 1 over s minus 4.

00:40 The second factor is the laplace transform of exponential minus 1 .5t.

00:49 And the last factor is the laplace transform of exponential 4t.

00:54 So when we take the inverse laplace transform of both sides, we get this.

01:00 This is the convolution of exponential 4t.

01:04 And this convolution is defined as follows.

01:07 Integral from 0 to t, d tau exponential.

01:12 We just replace t by tau in either one of these functions.

01:19 And the other t by t minus tau.

01:22 And it doesn't matter.

01:23 So let's do it this way.

01:25 Exponential minus 1 .5 t minus tau times exponential 4 tau.

01:34 We have 5 .5 exponential minus 1 .5 t from 0 to t, d tau exponential 5 .5 tau.

01:47 So we have 5 .5 exponential minus 1 .5 t exponential 5 .5 tau over 5 .5 between 0 and t.

02:00 There is this cancellation.

02:03 We have exponential minus 1 .5.

02:07 So we have 5 .5 t times exponential 5 .5 t minus 1.

02:12 And if we expand this, we get exponential 4t minus exponential minus 1 .5 t...

Question

Please give Ace some feedback