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# Engineering Analysis -Third Class.ppsx

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### Engineering Analysis -Third Class.ppsx

1. 1. MINISTRY OF SCIENTIFIC EDUCATION AND HIGHER RESEARCHES NORTHERN TECHNICAL UNIVERSITY ENGINEERING TECHNICAL COLLEGE / MOSUL DEPARTMENT OF COMPUTER TECHNOLOGY 1 ENGINEERING ANALYSIS LECTURE DEPARTMENT OF COMPUTER TECHNOLOGY –THIRD CLASS 2018 -2019 ARJUWAN MOHAMMED ABDULJAWAD ALJAWADI LECTURER
2. 2. 2 - ENGINEERING ANALYSIS One of important transforms used in linear- system analysis. It is named in honor of the great French mathematician, Pierre Simon De Laplace (1749-1827).
3. 3. 3 - Purpose of Laplace Transform • To convert from one type of operation to another operations of different types in more simple form. • A well-known technique for solving differential equations.
4. 4. 4
5. 5. - THE TIME DOMAIN SIGNAL IS CONTINUOUS , EXTENDS TO: 1. POSITIVE AND NEGATIVE INFINITY. 2. PERIODIC OR APERIODIC SIGNAL 5 - Laplace Transform Definition: The Laplace transform F(s) of a time function F (t) is given by the integral:
6. 6. 6 This definition is called the bilateral, or two-sided, Laplace transform—hence, the subscript b. Notice that the bilateral Laplace transform integral becomes the Fourier transform integral if is replaced by (𝑗𝑤 ) . The Laplace transform variable is complex 𝑠 = 𝛼 + 𝑗𝑤, we can rewrite (1) as: 𝐹 𝑠 = −∞ ∞ 𝑓(𝑡)𝑒−(𝜎+𝑗𝜔)𝑡 𝑑𝑡 = −∞ ∞ 𝑓(𝑡)𝑒−𝜎𝑡 𝑒−𝑗𝑤 𝑑𝑡
7. 7. 7 𝑓 𝑡 𝑖𝑠 𝑧𝑒𝑟𝑜 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑡 < 0 Thus the first integral in above equation is zero. The resulting transform, called the unilateral, or single-sided Laplace transform, is given by:
8. 8. 8 THE LAPLACE-TRANSFORM VARIABLE S IS COMPLEX, AND WE DENOTE ITS REAL PART AS ∝ AND ITS IMAGINARY PART AS 𝑗𝑤 THAT IS: 𝑠 = ∝ +𝑗𝑤 The S-plane
9. 9. 9 Some Elementary Functions F(t) and their Laplace Transform F(t) F(s) U(t) 1 s t 1/s2 tn n!/ sn+1 e−at 1 s + a eat 1 s − a sin wt w/s2+w2 cos at s/s2+a2 sin hat w/s2- w2 cos hat s/s2- w2
10. 10. 10 - Laplace Transform of some important functions 1. Laplace Transform of a unit –step function 𝑓(𝑡) = 1.
11. 11. 11 2. Laplace Transform of 𝑓(𝑡) = 𝑒𝑎𝑡
12. 12. 12 3. Laplace Transform of 𝑓(𝑡) = 𝑡𝑛 Where n= (1,2,3,4,…………………………….) f s = 0 ∞ e−st tn .dt udv = uv − v . du U=tn , du = (𝑛 𝑡𝑛−1) , dv= e−st , v=- e−st s f s = − e−st.tn/s |- 0 ∞ 𝑛 𝑒−𝑠𝑡 𝑠 .tn-1.dt
13. 13. 13 The first term limits will be form (0 to ∞) by substituting it yields zero while the second term by substitution we get: f s = 0 ∞ −ne−st s n tn-1.dt u=ntn-1 , du=n(n-1)tn-2 , dv= e−st s , v= - e−st/s2 f s = ne−st tn-1 /s2 | - 0 ∞ −n(n − 1)tn-2/s2 .e−st . dt f s = 0 ∞ n(n − 1)e−st tn-2 /s2 .dt Then after n times integration we have : f s = 0 ∞ n! sn tn−ne−st. dt f s = 0 ∞ n! sn e−st. dt = n! sn+1 e−st | = n! sn+1
14. 14. NOTE: IN THIS SAME METHOD YOU CAN FIND LAPLACE TRANSFORM OF COS 𝑤𝑡 14 𝐿𝑎𝑝𝑙𝑎𝑐𝑒 [𝑠𝑖𝑛 𝑤𝑡] = 0 ∞ 𝑠𝑖𝑛𝑤𝑡𝑒−𝑠𝑡 . 𝑑𝑡 𝑒𝑗𝑤𝑡 = 𝑐𝑜𝑠 𝑤𝑡 + 𝑗𝑠𝑖𝑛 𝑤𝑡 𝑠𝑖𝑛 𝑤𝑡 𝑖𝑠 𝑡ℎ𝑒 𝐼𝑚𝑎𝑔𝑖𝑛𝑎𝑟𝑦 𝑝𝑎𝑟𝑡 (𝐼𝑚) 𝑜𝑓 𝑒𝑗𝑤𝑡 𝑠𝑖𝑛 𝑤𝑡 = 𝐼𝑚 (𝑒𝑗𝑤𝑡 ) 𝐿𝑎𝑝𝑙𝑎𝑐𝑒 [𝑠𝑖𝑛 𝑤𝑡] = 𝐼𝑚 0 ∞ 𝑒𝑗𝑤𝑡𝑒−𝑠𝑡. 𝑑𝑡 = 𝐼𝑚 0 ∞ 𝑒−(𝑠−𝑗𝑤). 𝑑𝑡 = 𝐼𝑚 [− 𝑒− 𝑠−𝑗𝑤 𝑡 (𝑠 − 𝑗𝑤) ] = 𝐼𝑚 [0 − (− 𝑒0 𝑠 − 𝑗𝑤) ) = 𝐼𝑚 1 𝑠 − 𝑗𝑤 = 𝐼𝑚 [ 1 𝑎 − 𝑗𝑤 ∗ 𝑠 + 𝑗𝑤 𝑠 + 𝑗𝑤 ] = 𝐼𝑚 (𝑠 + 𝑗𝑤) / (𝑠2 + 𝑤2) = 𝑤 / (𝑠2 + 𝑤2) 4. Laplace Transform of f t = 𝑠𝑖𝑛 𝑤𝑡
15. 15. 5. LAPLACE TRANSFORM OF F T = 𝐶𝑜𝑠ℎ(𝑎𝑡) 15 𝐶𝑜𝑠ℎ(𝑎𝑡) = 𝑒𝑎𝑡 + 𝑒−𝑎𝑡 2 ℒ cosh 𝑎𝑡 = 1 2 1 𝑠 − 𝑎 + 1 𝑠 + 𝑎 = 1 2 ∗ [ 𝑠 + 𝑎 + 𝑠 − 𝑎 (𝑠 − 𝑎)(𝑠 + 𝑎) ] = 1 2 ∗ [ 2𝑠 𝑠2 + 𝑎𝑠 − 𝑎𝑠 − 𝑎2 ] = 1 2 [ 2𝑠 𝑠2 − 𝑎2 ] = 𝑠 𝑠2 − 𝑎2
16. 16. 16 - Laplace Transform Properties: 1. Multiplication by Constant: ℒ 𝑘. 𝐹 𝑡 = 0 ∞ 𝑘. 𝐹 𝑡 𝑒−𝑠𝑡 . 𝑑𝑡 = k. 0 ∞ 𝐹(𝑡)𝑒−𝑠𝑡 . 𝑑𝑡 =𝑘. 𝐹(𝑠) Example: ℒ3. 𝑒2𝑡 = 3. 1 𝑠 − 2 = 3 𝑠 − 2
17. 17. 17 2. Linearity: If 𝐹 𝑡 = 𝐹1 𝑡 + 𝐹2 𝑡 ℒ 𝐹 𝑡 = 0 ∞ 𝐹(𝑡)𝑒−𝑠𝑡 .dt = 0 ∞ 𝐹1 𝑡 + 𝐹2 𝑡 𝑒−𝑠𝑡 . 𝑑𝑡 = 0 ∞ 𝐹1 𝑡 𝑒−𝑠𝑡. 𝑑𝑡 + 0 ∞ 𝐹2 𝑡 𝑒−𝑠𝑡. 𝑑𝑡 = ℒ 𝐹1 𝑡 + ℒ[𝐹2 𝑡 ] = 𝐹1 𝑠 + 𝐹2(𝑠) If 𝐹 𝑡 = 𝑎1𝐹1 𝑡 + 𝑎2𝐹2 𝑡 Where a1 and a2 are constants Then ℒ[𝑎1𝐹1 𝑡 + 𝑎2𝐹2 𝑡 ] = 𝑎1ℒ 𝐹1 𝑡 + 𝑎2ℒ 𝐹2 𝑡
18. 18. 18 Examples: 1. ℒ[2𝑠𝑖𝑛3𝑡 + 𝑐𝑜𝑠3𝑡] = 2 ℒ 𝑠𝑖𝑛3𝑡 + ℒ[𝑐𝑜𝑠3𝑡] = 2. 3 𝑠2+9 + 𝑠 𝑠2+9 = 𝑠+6 𝑠2+9 2. ℒ[ 4 𝑒5𝑡 + 6𝑡3 − 3𝑠𝑖𝑛4𝑡 + 2𝑐𝑜𝑠2𝑡] = 4 𝑠 − 5 + 6.3! 𝑠4 − 12 𝑠2 + 16 + 2𝑠 𝑠2 + 4 = 4 𝑠 − 5 + 36 𝑠4 − 12 𝑠2 + 16 + 2𝑠 𝑠2 + 4 - Home Work: - ℒ[3𝑐𝑜𝑠6𝑡 − 5𝑠𝑖𝑛6] - ℒ[3𝑡10 − 8𝑒−3𝑡 + 5𝑐𝑜𝑠3𝑡 + 4𝑠𝑖𝑛2𝑡] - ℒ[3𝑐𝑜𝑠5𝑡 − 4𝑠𝑖𝑛ℎ5𝑡]
19. 19. 19 3. Multiplication by exponential: 𝑓1 𝑡 = 𝑓 𝑡 𝑒−𝑎𝑡 ℒ 𝑓 𝑡 𝑒−𝑎𝑡 = 0 ∞ 𝑓 𝑡 𝑒−𝑎𝑡 𝑒−𝑠𝑡 0 ∞ 𝑓 𝑡 𝑒− 𝑠+𝑎 𝑡 . 𝑑𝑡 Then 𝑓(𝑠 + 𝑎) The transform ℒ 𝑓 𝑡 𝑒−𝑎𝑡 is thus the same as ℒ 𝑓 𝑡 with everywhere in the result replaced by (𝑠 + 𝑎).
20. 20. 20 - Example: ℒ 𝑒−𝑎𝑡 𝑠𝑖𝑛𝑤𝑡 = 𝑤 (𝑠+𝑎)2+ 𝑤2 ℒ 𝑡2 𝑒−4𝑡 = 2 (𝑠 + 4)3 - Example: 1. ℒ 𝑡2𝑒3𝑡 = 2 (𝑠−3)2 2. ℒ 𝑒−2𝑡 𝑠𝑖𝑛4𝑡 = 4 (𝑠+2)2+16 = 4 𝑠2+2𝑠+20 3. ℒ 𝑒−4𝑡 𝑐𝑜𝑠ℎ5𝑡 = (𝑠−4) (𝑠−4)2−25 = (𝑠−4) 𝑠2−4𝑠+16−25 = 𝑠−4 𝑠2−4𝑠−9 - Home Work: Find ℒ 𝑒−2𝑡 (3𝑐𝑜𝑠6𝑡 − 5𝑠𝑖𝑛6𝑡 - Home Work: Find ℒ 𝑒−4𝑡 𝑐𝑜𝑠ℎ5𝑡 using cosh = 1 2 ( 𝑒𝑎𝑡 + 𝑒−𝑎𝑡 )
21. 21. 21 4. Multiplication by t (frequency derivative) If ℒ 𝑓(𝑡) = 𝑓(𝑠) Then ℒ 𝑡. 𝑓 𝑡 = − 𝑑 𝑑𝑠 𝑓(𝑠) In general it can be ℒ 𝑡𝑛 𝑓 𝑡 = (−1)𝑛 𝑑𝑛 𝑑𝑠𝑛 𝑓(𝑠) - Example: ℒ 𝑡. 𝑠𝑖𝑛2𝑡 ℒ 𝑠𝑖𝑛2𝑡 = 2 𝑠2 + 4 ℒ 𝑡 𝑠𝑖𝑛2𝑡 = − 𝑑 𝑑𝑠 . 2 𝑠2 + 4 = 𝑠2+4 ∗0−2∗2𝑠 (𝑠2+4)2 = 4𝑠 (𝑠2+4)2
22. 22. 22 - Example: ℒ 𝑡 3𝑠𝑖𝑛2𝑡 − 2𝑐𝑜𝑠2𝑡 = 3𝑡𝑠𝑖𝑛2𝑡 − 2𝑡𝑐𝑜𝑠𝑡 ℒ2𝑠𝑖𝑛2𝑡 = 6 𝑠2 + 4 ℒ𝑐𝑜𝑠2𝑡 = 2𝑠 𝑠2 + 4 = 3𝑡𝑠𝑖𝑛2𝑡 − 2𝑡 𝑐𝑜𝑠2𝑡 ℒ 3𝑡 𝑠𝑖𝑛2𝑡 = 𝑠2 + 4 ∗ 0 − 6 ∗ 2𝑠 (𝑠2 + 4)2 = −12𝑠 (𝑠2 + 4)2 ℒ 2𝑡 𝑐𝑜𝑠2𝑡 = 𝑠2 + 4 ∗ 2 − 2𝑠 ∗ 2𝑠 (𝑠2 + 4)2 = 2𝑠2 + 8 − 4𝑠2 (𝑠2 + 4)2 = −12𝑠 (𝑠2+4)2 − 8−2𝑠2 𝑠2+4 2 = −2𝑠2−12𝑠+8 (𝑠2+4)2
23. 23. 23 5.Time Derivative: ℒ 𝑑𝑓 𝑡 𝑑𝑡 = 𝑠𝑓 𝑠 − 𝑓(0) Where 𝑓(0) is the initial value of 𝑓(𝑡) ,evaluated as the one - side limit of 𝑓(𝑡) as 𝑡 → 0 from positive valued. ℒ 𝑓 𝑡 ′ = 0 ∞ 𝑓 𝑡 ′ 𝑒−𝑠𝑡 . 𝑑𝑡 Using 𝑢𝑑𝑣 = 𝑢𝑣 − 𝑣𝑑𝑢 𝑢 = 𝑒−𝑠𝑡 , 𝑑𝑢 = −𝑠𝑒−𝑠𝑡, 𝑑𝑣 = 𝑓 𝑡 ′ , 𝑣 = 𝑓(𝑡) = 𝑒−𝑠𝑡 𝑓 𝑡 𝑤𝑖𝑡ℎ 𝑙𝑖𝑚𝑖𝑡𝑠 𝑓𝑟𝑜𝑚 0 − ∞ − 0 ∞ 𝑓(𝑡)(−𝑠𝑒−𝑠𝑡 ). 𝑑𝑡 = 0 − 𝑓 0 + 𝑠𝑓(𝑠) Then ℒ 𝑑2𝑓 𝑡 𝑑𝑡2 = 𝑠2 𝑓 𝑠 − 𝑠𝑓 0 − 𝑓(0)′ and in general: ℒ 𝑑𝑛 𝑓 𝑡 𝑑𝑡𝑛 = 𝑠𝑛 𝑓 𝑠 − 𝑖=1 𝑛 𝑓(0)(𝑖−1) 𝑠𝑛−𝑖
24. 24. 24 - Example: 𝑓 𝑡 = 𝑡 , 𝐹𝑖𝑛𝑑 ℒ 1 𝑢𝑠𝑖𝑛𝑔 𝑡𝑖𝑚𝑒 − 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 Since ℒ 1 = 1 𝑠 , 𝑓 0 = 0 ℒ 𝑑𝑓 𝑡 𝑑𝑡 = 𝑠𝑓 𝑠 − 𝑓 0 ℒ 𝑑𝑓 1 𝑑𝑡 = 𝑠𝑓 𝑠 − 𝑓 0 ℒ 1 = 𝑠 ∗ 1 𝑠2 − 0 = 1 𝑠
25. 25. 25 - Example: Use Laplace transform in solving for the current in an electric circuit. Consider the RL-circuit in the following figure, where 𝑉 is constant. The loop equation for this circuit is given by: 𝐿. 𝑑𝑖(𝑡) 𝑑𝑡 + 𝑅𝑖 𝑡 = 𝑉𝑢 𝑡 𝑓𝑜𝑟 𝑡 > 0
26. 26. 26 Since the switch is closed at 𝑡 = 0 . The Laplace transform of this equation yields: 𝐿 𝑠𝐼 𝑠 − 𝑖 0 + 𝑅𝐼 𝑠 = 𝑉 𝑠 The initial current is zero 0 = 0 , 𝑖(𝑡) is zero for negative time since the switch is open for 𝑡 < 0 and the current in an inductance cannot change instantaneously. 𝐼 𝑠 = 𝑉 𝑠 𝐿𝑠 + 𝑅 = 𝑉 𝐿 𝑠 𝑠 + 𝑅 𝐿 𝑑𝑖𝑣𝑖𝑑𝑖𝑛𝑔 𝑏𝑦 𝐿 Solve with partial fraction: 𝐼 𝑠 = 𝑉 𝐿 𝑠 𝑠 + 𝑅 𝐿 = 𝑎 𝑠 + 𝑏 𝑠 + 𝑅 𝐿 = 𝑎𝑠 + 𝑎 𝑅 𝐿 + 𝑏𝑠 𝑠 𝑠 + 𝑅 𝐿
27. 27. 27 = 𝑎 + 𝑏 𝑠 + 𝑎 𝑅 𝐿 𝑠(𝑠 + 𝑅 𝐿) 𝑎 + 𝑏 = 0 → 𝑎 = −𝑏 𝑎 𝑅 𝐿 = 𝑉 𝐿 → 𝑎 = 𝑉 𝑅 𝐼 𝑠 = 𝑉 𝑅 𝑠 − 𝑉 𝑅 𝑠 + 𝑅 𝐿 𝐴 = 𝑠 . 𝑉 𝑅 𝑠.(𝑠+𝑅 𝐿 ) 𝑤ℎ𝑒𝑛 𝑠 = 0 𝐴 = 𝑉 𝑅 𝐵 = 𝑠 + 𝑅 𝐿 . 𝑉 𝐿 𝑠. 𝑠+𝑅 𝐿 𝑤ℎ𝑒𝑛 𝑠 = −𝑅 𝐿 𝐵 = 𝑉 𝐿 − 𝑅 𝐿 = − 𝑉 𝑅 ℒ𝐼(𝑠)−1 = 𝑖 𝑡 = 𝑉 𝑅 − 𝑉 𝑅 𝑒− 𝑅 𝐿𝑡 𝑖 𝑡 = 𝑉 𝑅 1 − 𝑒−( 𝑅 𝐿)𝑡 𝑡 > 0 The initial condition 𝑖 0 = 0 is satisfied by 𝑖 𝑡 also substitution of 𝑖 𝑡 into the differential equation satisfies that equation.
28. 28. 28 - Example: Solve the following differential equation: 𝑦′′ − 4𝑦′ + 5𝑦 = 𝑥 𝑡 𝑦 0 = 0 , 𝑦 0 ′ 𝑠2 𝑦 𝑠 − 𝑠𝑦 0 − 𝑦′ 0 − 4 𝑠𝑦 𝑠 − 𝑦 0 + 5𝑦 𝑠 = 𝑥 𝑠 𝑠2𝑦 𝑠 − 4𝑠𝑦 𝑠 + 5𝑦 𝑠 = 𝑥 𝑠 𝑦 𝑠 𝑠2 − 4𝑠 + 5 = 𝑥 𝑠 𝑦(𝑠) 𝑥(𝑠) = 1 𝑠2 − 4𝑠 + 5 - Home Work : 1. 𝑦′′ − 3𝑦′ + 2𝑦 = 𝑢 𝑡 2. 𝑦′′ + 4𝑦′ + 20𝑦 = 𝑢(𝑡)
29. 29. 29 6. Real Shifting: 𝑓 𝑡 − 𝑡0 𝑢 𝑡 − 𝑡0 = 𝑓 𝑡 − 𝑡0 𝑡 > 𝑡0 1. 𝑡 < 𝑡0 ℒ 𝑓 𝑡 − 𝑡0 𝑢 𝑡 − 𝑡0 = 0 ∞ 𝑓 𝑡 − 𝑡0 𝑢 𝑡 − 𝑡0 𝑒−𝑠𝑡. 𝑑𝑡 = 0 ∞ 𝑓(𝑡 − 𝑡0) 𝑒−𝑠𝑡 . 𝑑𝑡 We make the change of variable 𝑡 − 𝑡0 = 𝜏 , hence 𝑡 = (𝜏 + 𝑡0), 𝑑𝑡 = 𝑑𝜏 and ℒ 𝑓 𝑡 − 𝑡0 𝑢 𝑡 − 𝑡0 = 0 ∞ 𝑓 𝜏 𝑒−𝑠(𝜏+𝑡0). 𝑑𝜏 = 𝑒−𝑡0𝑠 0 ∞ 𝑓 𝜏 𝑒−𝑠𝜏 . 𝑑𝜏 = 𝑒−𝑡0𝑠 . 𝑓(𝑠)
30. 30. 30 Since 𝜏 is the variable of integration and can be replaced with , the integral on the right side is 𝑓(𝑠) , hence the Laplace transform of the shifted time function is given by: ℒ 𝑓 𝑡 − 𝑡0 𝑢 𝑡 − 𝑡0 = 𝑒−𝑡0𝑠𝑓(𝑠) 𝑡0 ≥ 0 And ℒ 𝑓 𝑡 = 𝑓(𝑠). This relationship called the real translation, properly applies for a function of the type in the figure: It is necessary that the function to be zero in time less than 𝑡0 , the amount of the shift.
31. 31. 31 - Example: Find the Laplace transform of a delayed exponential function: 𝑓 𝑡 = 5𝑒−0.3 𝑡−2 𝑢(𝑡 − 2) 𝑤ℎ𝑒𝑟𝑒 𝑡 = 2 , ℒ𝑓 𝑡 = 𝑒−2𝑠 𝑓 𝑠 = 5𝑒−2𝑠 𝑠 + 0.3
32. 32. 32 7. Time Integral: 𝑔 𝑡 = 0 𝑡 𝑓 𝜏 . 𝑑𝜏 ℒ𝑔 𝑡 = ℒ[ 0 𝑡 𝑓 𝜏 . 𝑑𝜏] ∴ 0 ∞ [ 0 𝑡 𝑓 𝜏 . 𝑑𝜏]𝑒−𝑠𝑡 . 𝑑𝑡 𝑢 = 0 𝑡 𝑓 𝜏 . 𝑑𝜏 𝑑𝑣 = 𝑒−𝑠𝑡 𝑑𝑢 = 𝑓 𝑡 . 𝑑𝑡 𝑣 = 𝑒−𝑠𝑡 −𝑠 𝑢𝑑𝑣 = 𝑢𝑣 − 𝑣𝑑𝑢 𝑒−𝑠𝑡 −𝑠 . 0 𝑡 𝑓 𝜏 . 𝑑𝜏 𝑤𝑖𝑡ℎ 𝑙𝑖𝑚𝑖𝑡𝑠 𝑓𝑟𝑜𝑚 0 𝑡𝑜 ∞ − 0 ∞ 𝑒−𝑠𝑡 −𝑠 𝑓 𝑡 . 𝑑𝑡 = 1 𝑠 0 − 0 + 1 𝑠 0 ∞ 𝑓(𝑡)𝑒−𝑠𝑡 . 𝑑𝑡
33. 33. 33 ∴ ℒ[ 0 𝑡 𝑓 𝜏 . 𝑑𝜏] = 1 𝑠 0 ∞ 𝑓 𝑡 𝑒−𝑠𝑡 . 𝑑𝑡 ∴ 𝑓(𝑠) 𝑠 - Example: Find the Laplace transform for 𝑓 𝑡 = 0 ∞ 𝑠𝑖𝑛𝑡. 𝑑𝑡 By integration the sine the result will be cosine and have the Laplace 𝑠 𝑠2+1 By using the Laplace transform properties, the result will be ℒ 1 𝑠2+1 Then the Laplace transform for integration the sine is 1 𝑠2+1 𝑠 Which leads the result to 𝑠 𝑠2+1
34. 34. 34 8.Initial Value Theorem The initial value 𝑓(0) of the function 𝑓(𝑡) whose L.T is 𝑓(𝑠) is: 𝑓 0 = lim 𝑡→0 𝑓(𝑡) = lim 𝑠→∞ 𝑠𝑓(𝑠) - Example: For the function 𝑓 𝑡 = 3𝑒−2𝑡, prove the initial – value theorem: ℒ3𝑒−2𝑡 = 3 𝑠2 + 2 𝑓 0 = lim 𝑡→0 3𝑒−2𝑡 = lim 𝑠→∞ 𝑠 3 𝑠2 + 2 3 = 3 According to Lobital rule in limits that states when there is ∞ ∞ then the limit is taking the derivative of the function of both the nominators and denominators which is in this case 3 1
35. 35. 35 9. Final value theorem: The final value of the function 𝑓(𝑡) whose L.T is𝑓 ∞ = lim 𝑡→∞ 𝑓(𝑡) = lim 𝑠→0 𝑠𝑓(𝑠) - Example: For the function 𝑓 𝑡 = 3𝑒−2𝑡 , prove the final – value theorem: 𝑓 ∞ = lim 𝑡→∞ 3𝑒−2𝑡 = lim 𝑠→0 𝑠 3 𝑠2 + 2 0 = 0
36. 36. THANK YOU 36