1. (a) Calculate the Laplace Transform (F(s)) for each of...

1. (a) Calculate the Laplace Transform (F(s)) for each of the following functions: (i) (f(t) = 2 cos^2(t) - 2 sin^2(t)) (ii) (f(t) = 3 sin(t) cos(t)) (iii) (f(t) = t^2 e^{3t}) [6 marks] (b) Use the First Shift Theorem to find the inverse Laplace transform of (frac{s - 3}{s^2 - 6s + 20}) [4 marks] (c) Use the method of partial fractions to find the inverse Laplace transform of (frac{5}{(s - 1)^2 (1 + s^2)}) [5 marks] (d) Use the Second Shift Theorem to calculate the Laplace Transform (F(s)) for (f(t) = frac{e^{3t - 6} + e^{-3t + 6}}{2} H(t - 2)) where (H(t)) is the standard Heaviside function. [5 marks]

Transcript

00:01 Hello, let's look at the question.

00:02 We have to calculate the laplace transform for fs for each of the following function.

00:06 For the cos particle c, ft is equal to 2 cos square t minus 2 sin square t.

00:16 Let's see how we can solve this.

00:18 For this we will take 2 sin square a is equal to 1 minus cos 2a and 2 cos square a is equal to 1 minus sin 2a.

00:31 So we will take 1 plus cos 2t minus cos 2t.

00:43 So our ft will be 2 cos 2t.

00:49 So our lft will be 2l cos 2t which will be 2s pi s square plus 2 square as our l cos at is s pi s square plus a square.

01:07 So our answer will be 2s pi s square plus 4.

01:13 For next part we have ft is equal to sin t cos t.

01:23 As we know for 3 by 2, 2 sin cos t is equal to 3 by 2 sin 2t.

01:39 Then for lft we will have 3 by 2 l sin 2t by 3 by 2 2 by s square plus 4.

01:53 As we know our sin at laplace is a by s square plus a square.

02:03 Now for the next part we have to given ft is equal to t square e to the power 3t.

02:13 We will see this as the laplace will be n factorial by s to the power n plus 1 l e ct ft minus f s minus c.

02:29 So e to the power ct t square will be 2 factorial by s minus c whole cube as l t square e to the power ct will be 2 factorial by s minus c whole cube.

02:50 This is the answer for this.

02:53 Now let's look at part b.

02:55 We have to use the shift theorem to find the inverse laplace transform of s minus 3 by s square minus 6s plus 20.

03:02 Let's see how we can do this.

03:07 We have s minus 3 by s minus 3 whole square minus 9 plus 20.

03:13 Let's transform into this.

03:14 Then we will have s minus 3 by s minus 3 whole square plus 11.

03:19 We can write l inverse s minus 3 by s minus 3 whole square plus 11.

03:29 We can write it as e to the power 3t cos root 11t.

03:36 Our l cos at is s by s square plus a square.

03:43 So for let's look for part c where we have been given to use the method of partial fraction to find the inverse laplace transform of 5 divided by s minus 1 whole square 1 plus s square.

04:12 Let's see how we can do this.

04:18 We find 5 by s minus 1 whole square 1 plus s square is equal to a s plus b by s minus 1 square plus c s plus d by s square plus 1...

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