 Namaste. Myself, Dr. Baswaraz Amiradhar, Assistant Professor, Department of Communities and Sciences, Walsh and Institute of Technology, Sholabon. In this studio, I explain the universal aplot transform by the convalescent theorem. Learning outcomes. At the end of this session, the student will be able to find the universal aplot transform using convalescent theorem. Come to an example. Using the convalescent theorem, find the universal aplot transform of s plus 1 upon s squared plus 2s plus 5 whole squared. Solution. Let s plus 1 upon s squared plus 2s plus 5 whole squared can split as the product of the two functions. That is 1 upon s squared plus 2s plus 5 into s plus 1 upon s squared plus 2s plus 5. Let f of s is equal to 1 upon s squared plus 2s plus 5. And g of s is equal to s plus 1 upon s squared plus 2s plus 5. Now, find the f of t and g of t. That is f of t is equal to L inverse of f of s, which is equal to L inverse of 1 upon s squared plus 2s plus 5. f of t is equal to L inverse of 1 upon s plus 1 whole squared plus 4 because by making the perfect square and that is a pop t is equal to e to the power of minus t into L inverse of 1 upon s cot plus 4 by first shifting property that is equal to e to the power of minus t into sin 2 t by 2 and g of t is equal to L inverse of g of s which is equal to L inverse of s plus 1 upon s cot plus 2 s plus 5 then by making the perfect square the s cot plus 2 s plus 5 can be written as s plus 1 whole square plus 4 that is L inverse of s plus 1 upon s plus 1 whole square plus 4. Therefore, the g of t is equal to e to the power of minus t into L inverse of s upon s cot plus 4 because by first shifting property therefore, the g of t is equal to e to the power of minus t into cos 2 t by the convolution theorem we know that L inverse of f of s into g of s is equal to the integration u is equal to 0 to t f of u into g of t minus u into d o. Therefore, the L inverse of s plus 1 upon s cot plus 2 s plus 5 whole square is equal to integration u is equal to 0 to t and then replacing the f of t t by u and in g of t t minus u it becomes the e to the power of minus u into sin 2 u up by 2 into e to the power minus t minus u into cos of 2 into t minus u into d o. Therefore, the L inverse of s plus 1 upon s cot plus 2 s plus 5 whole square is equal to 1 by 2 integration u is equal to 0 to t e to the power of minus t into sin 2 u into cos 2 t minus 2 u into d o because by using the law of indices and the basis are same the add is their powers where that is e to the power of minus u and minus of minus plus u will get cancelled that is the e to the power of minus t is left now. Now, while you have to integrating with respect to u the t is it is a constant now take it is e to the power of minus t is outside the integral sin that is e to the power of minus t by 2 integration u is equal to 0 to t 1 by 2 into sin 2 u plus 2 t minus 2 u plus sin 2 u minus 2 t plus 2 u into d o by using the basic telemetric formula that is 2 sin a into cos b is equal to sin a plus b plus sin a minus b. Therefore, that is equal to e to the power of minus t by 4 into integration u is equal to 0 to t in the plus 2 u minus 2 u get cancelled that is sin 2 t and 2 u plus 2 u that is 4 u. Now, on integrating with respect to u e to the power of minus by 4 into sin 2 t integration of d u is u bit limit 0 to t plus the integration of sin is minus cos 4 u minus 2 t by 4 bit limit is 0 to t and substituting the upper minus lower limit that is l inverse of s plus 1 upon s cot plus 2 s plus 5 whose cot is equal to e to the power of minus t by 4 into t sin 2 t plus 1 by 4 into minus cos 2 t plus cos 2 t that is plus cos 2 t minus cos 2 t get cancelled and that is equal to e to the power of minus t into t into sin 2 t by 4. First I will do and add the answer for the question what is the l inverse of s plus 2 upon s cot plus 4 s plus 13 solution by making the perfect square the denominator can be written as l inverse of s plus 2 upon s cot plus 4 s plus 13 is equal to l inverse of s plus 2 upon s plus 2 whose cot plus 9 then by first shifting property l inverse of s plus 2 upon s cot plus 4 s plus 13 is equal to e to the power of minus 2 t into l inverse of s upon s cot plus 9 that is equal to e to the power of minus 2 t into the l inverse of s upon s cot plus 9 becomes the cos 3 t. Now come to another example find l inverse of s upon s to the power of 4 minus 16 by conolation theorem solution let s upon s to the power of 4 minus 16 which can be written as the product of the two function s upon s cot plus 4 into 1 upon s cot minus 4 let f of s is equal to s upon s cot plus 4 and g of s is equal to 1 upon s cot minus 4 to find the f of t and g of t that is f of t is equal to l inverse of f of s which is equal to cos 2 t and g of t is equal to l inverse of g of s which is equal to sin h to t by 2. Then by the conolation theorem we know that l inverse of f of s into g of s is equal to integration u is equal to 0 to t f of u into g of t minus u into d u. Therefore, the l inverse of s upon s to the power 4 minus 16 is equal to integration u is equal to 0 to t by replacing in f of t t by u and in g of t t by t minus u and it becomes the cos 2 u into sin h 2 into t minus u by 2 into d u. Now, is equal to 1 by 2 which is the conolation take out sin 1 by 2 integration u is equal to 0 to t cos 2 u into e to the power 2 into t minus u minus e to the power of minus 2 into t minus u by 2 into d u because the by using the definition of the hyperbolic function the sin h theta is equal to e to the power theta minus e to the power of minus theta by 2. Now, here the theta is equal to 2 into t minus u non that 1 by 2 is the conolation taking outside which is equal to 1 by 2 1 by 4 integration u is equal to 0 to t e to the power 2 into t minus u into cos 2 u into d u minus integration u is equal to 0 to t e to the power of minus 2 into t minus u into cos 2 u into d u which is equal to 1 by 4. While I have to integrate with respect to u, theta has a constant which is taking outside the integral sin that become e to the power 2 t integration u is equal to 0 to t e to the power of minus 2 u into cos 2 u into d u minus e to the power minus 2 t integration u is equal to 0 to t e to the power 2 u into cos 2 u into d u. But we know that integration e to the power a x into cos b x into d x is equal to e to the power a x upon a square plus b square into a cos b x plus b sin b x that is l inverse of s upon s to the power 4 minus 16 is equal to 1 by 4 into e to the power 2 t into e to the power of minus 2 u upon minus 2 square plus 2 square into minus 2 into cos 2 u plus 2 into sin 2 u between the limit 0 to t minus e to the power of minus 2 t into e to the power 2 u upon 2 square plus 2 square into 2 cos 2 u plus 2 sin 2 u between 2 sin 2 u bit limit 0 to t. Now, substituting the upper minus lower limit it becomes the 1 by 32 into e to the power 2 t into e to the power of minus 2 t into minus 2 into cos 2 t plus 2 sin 2 t minus of bracket minus 2 plus 0 close the bracket minus e to the power of minus 2 t into e to the power 2 t into 2 cos 2 t plus 2 sin 2 t minus of 2 plus 0. Now, the multiply the entire term by e to the power 2 t and the this second term by e to the power of minus 2 t and is simplifying we get that is equal to 1 by 32 into minus 2 into cos 2 t plus 2 into sin 2 t plus 2 into e to the power 2 t minus 2 cos 2 t minus 2 sin 2 t plus 2 into e to the power of minus 2 t minus 2 sin 2 t plus 2 sin 2 t it will get cancel and that the remaining terms are becomes is equal to 1 by 32 into minus 2 cos 2 t plus 2 into e to the power of 2 t minus 2 cos 2 t plus 2 into e to the power minus 2 t is equal to 1 by 32 into minus 2 cos 2 t minus 2 cos 2 t that is a minus 4 cos 2 t that is a minus 4 into cos 2 t plus 2 into e to the power 2 t plus e to the power of minus 2 t, but we know that cos h theta is equal to e to the power theta plus e to the power of minus theta by 2 implies the 2 cos h theta is equal to e to the power theta plus e to the power of minus theta. Therefore, the L inverse of s upon s to the power 4 minus 16 is equal to 1 by 32 into minus 4 into cos 2 t plus 2 into the e to the power 2 t plus e to the power of minus 2 t become the 2 into cos h 2 theta that is 2 into 2 that is 4 the cos h 2 theta which is equal to 4 is the common 4 by 32 into cos h 2 t minus cos 2 t. Forans are that is the L inverse of s upon s to the power 4 minus 16 is equal to 1 by 8 into cos h 2 t minus cos 2 t reference thank you.