 अमस्ते, मैशल्क मिस्टर भीराजदार भादा साहिप, अशिस्टन प्रपेसर, दिपार्ट्मेंट अप हुमनिती साईन्साईन्स, वाल्चन्द अप तेकनोलोगी सोलापुर. इन दिस वीडो लक्चार, लाप्लास त्रान्सपार्म अप पुराटिक फुंक्षेंस. टिस अप प्रपेसर लाप्लास प्रपार्म अप पुराटिक फुंक्षेंस अप पुराटिक लग्ची आप पुराटिक. A function f of t is said to be periodic if there exists a constant capital T where t greater than 0 such that f of t plus capital T is equal to f of t for all small t. For example, sine t and cos t are periodic functions of period 2 pi. Note that in general f of t plus n into capital T is equal to f of t for all small t implies that function f of t is periodic where n is either positive or negative integer. Now, let us see the statement of Laplace transform of periodic function. If f of t is a piecewise continuous periodic function with period capital T then Laplace transform of f of t equal to 1 upon 1 minus e raise to minus s into capital T into integration with limit 0 to capital T e raise to minus s t f of t dt. Now, using this formula we can find Laplace transform of periodic function. Let us consider examples. Example 1 find the Laplace transform of function f of t defined as f of t equal to 1 for 0 less than equal to t less than a and f of t equal to minus 1 for a less than t less than equal to 2 a where f of t is periodic function with period 2 a solution. Here already says that the function f of t is periodic with period 2 a therefore denote t equal to 2 a and we have to use formula of Laplace transform of periodic function f of t which is equal to 1 upon 1 minus e raise to minus s into capital T into integration with limit 0 to capital T e raise to minus s t f of t dt. In this substitute capital T equal to 2 a we get Laplace of f of t equal to 1 upon 1 minus e raise to minus 2 a s into integration with limit 0 to 2 a e raise to minus s t f of t dt. Now, to substitute the value of f of t we have to split this integration into 2 parts therefore Laplace of f of t equal to 1 upon 1 minus e raise to minus 2 a s in bracket integration with limit 0 to a e raise to minus s t f of t dt plus integration with limit a 2 to a e raise to minus s t f of t dt bracket close. Which is equal to 1 upon 1 minus e raise to minus 2 a s as it is and in bracket integration with limit 0 to a e raise to minus s t and value of f of t for this range is 1 q 1 in hypothesis into dt plus integration with limit a 2 to a e raise to minus s t and f of t is minus 1 for this range into dt bracket close. Which is equal to 1 upon 1 minus e raise to minus 2 a s as it is and in bracket integration of e raise to minus s t with respect to t is e raise to minus s t upon minus s with limit 0 to a. Now, when we take minus 1 is outside the integration we get minus sign and again integration of e raise to minus s t is e raise to minus s t upon minus s with limit a 2 to a bracket close. Which is equal to 1 upon 1 minus e raise to minus 2 a s in bracket when we put t equal to upper limit a we get minus e raise to minus a s upon s minus when t equal to lower limit 0 we get minus e raise to 0 upon s. And from second bracket when we take minus sign is outside minus minus plus and when t equal to upper limit 2 a we get e raise to minus 2 a s upon s minus when t equal to lower limit a we get e raise to minus a s upon s which is equal to keeping 1 upon 1 minus e raise to minus 2 a s as it is and in bracket. E raise to minus 2 a s upon s is single term but minus e raise to minus a s upon s repeated twice in this so that minus 2 times e raise to minus a s upon s and minus minus plus e raise to 0 is 1 so that 1 upon s bracket close. Now take LSM in this bracket so which is equal to 1 upon 1 minus e raise to minus 2 a s as it is and into enumerator e raise to minus 2 a s minus 2 times e raise to minus a s plus 1 upon s which is equal to keeping 1 upon 1 minus e raise to minus 2 a s as it is and e raise to minus 2 a s minus 2 into e raise to minus a s plus 1 is a square of 1 minus e raise to minus a s upon s as it is. Which is equal to keeping numerator 1 minus e raise to minus a s bracket square as it is upon s into. Now factors of 1 minus e raise to minus 2 a s are 1 minus e raise to minus a s into 1 plus e raise to minus a s. Now factor 1 minus e raise to minus a s gets cancel from numerator and denominator and we get laplace of f of t equal to 1 minus e raise to minus a s upon s in bracket 1 plus e raise to minus a s. This is the required answer to the given question. Now pause the video for a file and write the answer to the question. Find laplace transform of f of t which is equal to e raise to t for 0 less than t less than 2 pi if f of t plus 2 pi equal to f of t. Come back I hope you have written answer to this question. Here I will going to explain the solution. Question is find the laplace transform of f of t equal to e raise to t for 0 less than t less than 2 pi. Solution here if f of t plus 2 pi equal to f of t this statement implies that the given function f of t is periodic with period capital t equal to 2 pi. So that we start with formula of laplace transform of periodic function f of t which is equal to 1 upon 1 minus e raise to minus s into capital t into integration with limit 0 to capital t of e raise to minus s to f of t d t. In this substitute capital t is 2 pi and f of t is e raise to t which is equal to 1 upon 1 minus e raise to minus 2 pi s into integration with limit 0 to 2 pi e raise to minus s t into e raise to t d t which is equal to keeping 1 upon 1 minus e raise to minus 2 pi s as it is integration with limit 0 to 2 pi but e raise to minus s t into e raise to t can be simplified as e raise to minus in bracket s minus 1 into t d t which is equal to 1 upon 1 minus e raise to minus 2 pi s as it is and in bracket integration of e raise to minus of s minus 1 into t with respect to 2 t is e to power minus in bracket s minus 1 into t upon minus in bracket s minus 1 with limit t equal to 0 to 2 pi which is equal to 1 upon 1 minus e raise to minus 2 pi s as it is and in bracket when t equal to upper limit 2 pi we get e raise to minus 2 pi in bracket s minus 1 upon minus in bracket s minus 1 now minus minus plus when t equal to lower limit 0 we get e raise to 0 upon s minus 1 bracket close which is equal to now in numerator after simplifying using e raise to 0 is 1 we get 1 minus e raise to minus 2 pi in bracket s minus 1 upon in denominator s minus 1 into in another bracket 1 minus e raise to minus 2 pi s it is the required answer to the given question consider example 2 point the Laplace transform of function f of t defined as f of t equal to e into sin omega t for 0 less than t less than pi upon omega and f of t equal to 0 for pi upon omega less than t less than 2 pi upon omega if f of t plus 2 pi upon omega is equal to f of t solution again here from the statement f of t plus 2 pi upon omega is equal to f of t which implies that given function f of t is periodic function with period capital t equal to 2 pi upon omega so that again we start from formula of Laplace transform of periodic function f of t which is equal to 1 upon 1 minus e raise to minus s into capital t into integration with limit 0 to capital t of e raise to minus s t f of t dt which is equal to in this put capital t equal to 2 pi upon omega we get 1 upon 1 minus e to power minus 2 pi s upon omega into integration with limit 0 to 2 pi upon omega e raise to minus s t f of t dt so again to put the value of f of t we have to split the integration into sum up to integrals which is equal to keeping 1 upon 1 minus e to power minus 2 pi s upon omega as it is in bracket integration with limit 0 to pi upon omega e raise to minus s t and f of t sine omega t for this range into dt plus integration with limit pi upon omega 2 2 pi upon omega e raise to minus s t and f of t for this range is 0 into dt bracket close which is equal to keeping 1 upon 1 minus e raise to minus 2 pi s upon omega as it is and into integration with limit 0 to pi upon omega of e raise to minus s t into sine omega t dt which is equal to 1 upon 1 minus e raise to minus 2 pi s upon omega into bracket now value of this integration using standard formula of definite integration we write e raise to minus s t upon square of minus s plus omega square and in open bracket keeping minus s as it is sine omega t minus keeping omega into cos of omega t open bracket close with limit t equal to 0 to pi upon omega which is equal to keeping 1 upon 1 minus e raise to minus 2 pi s upon omega as it is and in bracket when t equal to upper limit we get e raise to minus s pi upon omega upon s square plus omega square in open bracket minus s as it is sine of omega into pi upon omega minus omega as it is into cos of omega into pi upon omega open bracket close minus 20 equal to lower limit 0 we get e raise to 0 upon s square plus omega square in open bracket minus s as it is sine 0 minus omega as it is cos of 0 open bracket close and then square bracket close which is equal to keeping 1 upon 1 minus e raise to minus 2 pi s upon omega as it is and into bracket e raise to minus pi s upon omega upon s square plus omega square and in open bracket omega omega cancel sine pi 0 therefore minus s into 0 is 0 minus omega as it is and cos of omega omega cancel cos pi is minus 1 open bracket close minus e raise to 0 is 1 1 upon s square plus omega square into bracket sine 0 0 therefore minus s into 0 is 0 minus cos 0 is 1 into omega that is omega open bracket close and square bracket close therefore laplace of f of t equal to from numerator we can take omega common and from denominator s square plus omega square common and 1 minus e raise to minus 2 pi s upon omega common so that we get omega upon s square plus omega square in bracket 1 minus e raise to minus 2 pi s upon omega is a constant coefficient and in bracket 1 plus e raise to minus s pi upon omega therefore laplace of f of t equal to numerator omega in bracket 1 plus e raise to minus s pi upon omega and in denominator s square plus omega square into the factors of 1 minus e raise to minus 2 pi s upon omega or 1 minus e raise to minus pi s upon omega into 1 plus e raise to minus pi s upon omega and 1 plus e raise to minus pi s upon omega factor gets cancelled so that laplace of f of t equal to omega upon s square plus omega square into 1 minus e raise to minus pi s upon omega so this is the required answer for the given question to prepare this video lecture I refer these two books as references thank you