| by Kenneth Chase | 8 comments

Change of Scale Property – Laplace Transform – Engineering Mathematics 3


Hello friends so today in this video we are gonna derive the formula for change of scale property so what is change of scale property so change of scale property is given by if Laplace of F of T or function of T is equal to Phi of s that is function of s then the change of scale property says that Laplace of function of AT is equal to 1 by a function of s by a now to derive this property we’ll start with the definition of Laplace transform now what is the definition of Laplace transform so here and say by the definition of Laplace transform we have Laplace of function of T is equal to integration 0 to infinity e raised to minus s T F of T DT I’ll say this as equation number 1 now as I want to find out Laplace of function of a T I will say that this my F of T is f of a t so on the right hand side also I will put it as function of a T so here therefore Laplace of function of a T is equal to integration 0 to infinity e raise to minus st f of a T DT so this is by the definition of Laplace transform now let’s evaluate this integration and let’s prove that it is equal to 1 by Phi of s by a now for that what we’ll do is here I use the integration by substitution method so here I will put a T as you so if I put a T equal to you then here I get T as u upon a and if I will find the differentiation with respect to u then here we’ll get DT by D u is equal to 1 by constant and derivative of U with respect to u is 1 so therefore this step will become DT equal to 1 by D u so this is the substitution that we got for DT in terms of U now let us find out the limits so when it is integration with respect to T we have limits from 0 to infinity now we want to convert it into u so when T is 0 we’ll find out what is the value of U so when T is 0 here we’ll substitute that T 0 so a into 0 is 0 so therefore value of U is also 0 when T is equal to 0 now when T is infinity then here we’ll get infinity into a that is infinity it means U is also infinity so it means the limits are unchanged or limits remains same for you so now it’s substitute this lower and upper limits the substitution for T and the substitution for DT in the integration so we will give therefore Laplace of function of a T is equal to integration 0 to infinity now here we have erased 2 minus s into T but the value of T is U upon a so here we’ll get minus s you buy a here I’ll say he raised to minus s EU by ay next 80 is you so your will get F of U and DT is 1 by D u so f of U into 1 by D u now I can take this one by outside since it is a constant and here we’ll get integration 0 to infinity now integration is with respect to u so this s and a are constant so here I will say he raised to minus s upon a into u into f of u D u now 1 by as it is and to get the substitution of this integration let us again observe the definition of Laplace transform so according to definition here we have e raised to minus s T F of T DT and the Laplace of F of T is given as function of s so here I will say that this value is equal to function of s so I can say it here or here I will say that it is equal to function of s it means whatever is the coefficient of T you will get function of that variable so here we have s as a coefficient of T hence we are getting the function of s in the answer so similarly it will observe the current term then here the coefficient of U is s by a it means here we are going to get the function of s by a so I got this from the definition of Laplace transform so here if you’ll observe then we got the Laplace of function of a T as 1 upon a Phi of s by a which is the change of scale property thank you

8 Comments

Avinash Bajgude

Aug 8, 2016, 12:59 pm Reply

Easy to understand!!

Hoyt Volker

Feb 2, 2017, 6:54 pm Reply

Continue, ekeeda. Nice.

蔡宇晴

May 5, 2017, 4:16 pm Reply

bravo!

ME STRUCTURAL ENGINEERING

Oct 10, 2017, 1:44 pm Reply

Super

Phanindra Reddy

Apr 4, 2018, 12:18 pm Reply

What is region of convergence

SUBASH SINGH PATEL SONEBHDRA

May 5, 2018, 9:35 pm Reply

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Varun Mangrulkar

Oct 10, 2018, 3:13 pm Reply

Thank you so much sir, well explained 💗

Ekeeda

May 5, 2019, 1:48 pm Reply

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