Laplace Transform: Give a piecewise function f(t) = 2, 0...

Laplace Transform: Give a piecewise function f(t) = { 2, 0 <= t <= 1 { t, 1 <= t <= 4 { 0, t >= 4. Find F(s) using definition of Laplace Transform. Answer: F(s) = -e^-s / s + 2 / s - 4e^-4s / s - e^-4s / s^2 + e^-s / s^2 Linear Property of Laplace Transform: Find the Laplace transform of the function given, f(t) = 2t^4 - e^-4t. Answer: F(s) = 48 / s^5 - 1 / (s + 4) First shift Property of Laplace Transform: Find the Laplace transform of the function given, f(t) = e^-4t sin(3t). Answer: F(s) = 3 / (s^2 + 8s + 25) Laplace Transform of Multiplying by t^n functions: Find the Laplace transform of the function given, f(t) = t^2 cos t. Answer: F(s) = (-6s + 2s^3) / (s^2 + 1)^3

Transcript

00:01 A piecewise continuous function is given as f f of t is equal to 2 when 0 less than or equal to t less than or equal to 1 t when 1 less than or equal to 4 0 when t is greater than or equal to 4 the main objective is to find the laplace transformation l of f of t the formula for the laplace transformation l of f of t is given as integration over 0 to infinity f of d into e raised to minus s d d here since the function is a piecewise credit function is split the given integral into 3 according to the given function so integration over 0 to 1 2 e raise 2 minus s d g t plus integration over 1 to 4 t e -raise to minus s d t plus integration of 4 to infinity 0 into e -raise to minus s t d t we gave this function as equation number one now we will find the given obtained integral separate that is 0 to 1 2 erase to minus s d d t is equal to 2 into integration over 0 to 1 e raised to minus st d t which is equal to 2 into erase to minus st divided by minus s and also now applying the integrals which is equal to 2 into sorry minus 2 by s into e -r -r -raise to minus s b minus 2 y s a a raise to minus s minus e -raise to 0 minus 2 by s e -wrease to minus s minus 1 so we get minus 2 e -raise to minus s by s by s plus 2 by s we name this equation as equation number 2.

02:44 Now we find the value of integral over 1 2 4 t e raised 2 minus st dt which is equal to now we will use the formula integration of udv is equal to udv is equal to ub minus integration bd.

03:11 This equation is called the integration by parts of equation.

03:17 So for that first we substitute u for t and u for dt.

03:26 D v4e for e raised to minus s d d t then d becomes e raised to minus s d by minus s.

03:40 Now substituting these values we get uv, u .s .t, v is e raised to minus s t by minus s.

03:53 And applying the limits 1 to 4 minus integral over 1 to 4 v.

04:00 D .u.

04:00 V is e raised to minus s d by minus s d u .s.

04:10 D .t.

04:10 Which is equal to.

04:14 Now applying the limits we get minus 4 e raised to minus 4s divided by s plus e raised to minus s by s minus e raised to minus s d divided by minus s into minus s squared...

Question

Please give Ace some feedback