7 1 30 - Find the inverse Laplace transform of the function F(s) = 9 + s 4 − s2 Solution - If we break up this fraction we get: L−1 ( 9 + s 4 − s2 ) = −92L−1 ( 2
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[PDF] Chapter 7 Laplace Transforms Section 74 Inverse Laplace
Example 1 Determine the inverse Laplace transform of the given function (a) F(s ) = 2 s3 SOLUTION L−1 { 2 s3 } = L−1 {2 s3 } = t2 (b) F(s) = 2 s2+4
[PDF] The Inverse Laplace Transform 1 If L{f(t)} = F(s), then the inverse
1 2 L−1{ 2 s3 } + 3L −1{ 2 s2 + 4 } = s2 2 + 3 sin 2t (4) 3 Example: Suppose you want to find the inverse Laplace transform x(t) of X(s) = 1 (s + 1)4 + s − 3
[PDF] 63 Inverse Laplace Transforms
(s − 1)2 + 4 ] = ex £−1[ s s2 + 4 ] = ex cos 2x (using property 1 of Theorem 6 17 in reverse) The inverse Laplace transform is a linear operator Theorem 6 27
[PDF] The Inverse Laplace Transform - UAH
Given a function F(s), the inverse Laplace transform of F , denoted by L−1[F], B = 2 and C = 8 Hence Y(s) = A s − 4 + Bs + C s2 + 4 = 1 s − 4 + 2s + 8
[PDF] Section 74: Inverse Laplace Transform A natural question to ask
We now ask this question about the Laplace transform: given a function F(s), 2s2 + 8s + 10 } = 5多 -1{ 1 s - 6 } -6多 -1{ s s2 + 9 } + 3 2 多 -1 { 1 s2 + 4s + 5 }
[PDF] Problem 1 Find the inverse Laplace transform of the following function
Problem 1 Find the inverse Laplace transform of the following function (a) s - 1 s2 - 4s + 14 (b) s + 4 (s2 + 2s - 3)(s - 2) e -4s (c) 1 (s - 4)5 Problem 2
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the “Cover-up Rule” and “Keily's Method”; see Unit 1 9) EXAMPLES 1 Determine the Inverse Laplace Transform of F(s) = 3 s3 + 4 s − 2 Solution f(t) = 3 2
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Use the roots command to check the poles obtained in part a 28 Find the inverse Laplace transform of : s2 + 4s + 7 s + 2 ( )
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(s + 4)2 Inverse LT gives y(t) = te-4t - 1 2 cos 4t (e) Laplace transform of the equation leads to (s2 - 2s + 2)Y (s) = s s2 + 1 + (s - 2) Solving for Y (s) and taking
[PDF] Math 2280 - Assignment 9
7 1 30 - Find the inverse Laplace transform of the function F(s) = 9 + s 4 − s2 Solution - If we break up this fraction we get: L−1 ( 9 + s 4 − s2 ) = −92L−1 ( 2
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Math 2280 - Assignment 9
Dylan Zwick
Summer 2013
Section 7.1- 1, 6, 20, 30, 36
Section 7.2- 1, 4, 15, 20, 29
Section 7.3- 3, 8, 19, 24, 30, 33
1Section7.1-LaplaceTransformsandInverseTrans-forms7.1.1- Calculate the Laplace transform off(t) =tusing the definition of
the Laplace transform.Solution- The Laplace transform is:
L(f(t)) =?
0 te-stdt =-te-st s????∞0+1s? 0 e-st. The move from the first line to the second requires an integration by parts. Ifs >0then the limits above converge, and we get: =-e-st s2????∞0=1s2. 27.1.6- Calculate the Laplace transform off(t) = sin2tusing the definition
of the Laplace transform.Solution- The Laplace transform is:
L(f(t)) =?
0 e-stsin2(t)dt =-e-stsin2(t) s????∞0+2s? 0 e-stsin(t)cos(t)dt. Again, the move from the first line to the second required an inte- gration by parts, and ifs >0the limits above converge and we get (after another integration by parts):2sin(t)cos(t)e-st
s2????∞0+2s2? 0 e-st(cos2(t)-sin2(t))dt.Noting
cos2(t)-sin2(t) = 1-2sin2(t)
we get: 2 s2? 0 e-st(1-2sin2(t))dt.Combining the string of equalities above we get:
1 +4 s2? 0 e-stsin2(t)dt=2s2? 0 e-stdt=2s3. So,L(f(t)) =?
0 e-stsin2(t)dt=?s2 s2+ 4?? 2s3? =2s(s2+ 4). 37.1.20- Find the Laplace transform of the functionf(t) =tet.
Solution- Using the definition of the Laplace transform we get:L(f(t)) =?
0 e-sttetdt=? 0 te(1-s)tdt.After an integration by parts this becomes:
te (1-s)t1-s????∞0-11-s?
0 e(1-s)tdt.Ifs >1the limits converge, and we get:
=-e(1-s)t (1-s)2????∞0=1(1-s)2. 47.1.30- Find the inverse Laplace transform of the functionF(s) =9 +s4-s2.
Solution- If we break up this fraction we get:
L -1?9 +s 4-s2? =-92L-1?2s2-4? - L -1?ss2-4? =-92sinh(2t)-cosh(2t).
57.1.36- Show that the functionf(t) = sin(et2)is of exponential order as
t→ ∞but that its derivative is not.Solution- We have
fort >0. So,f(t)is of exponential order withM= 1,c= 1,T= 0. On the other hand the derivative isf(t)isf?(t) = 2tet2cos(et2). Sup- pose there exist positive constantsM,candTsuch that:2tet2cos(et2)< Mect
for allt > T. Now, for any positivecwe knowet2> ectwhent > c, and in factet2> Mectwhent >c+? c2+ 4ln(M)2.1Furthermore,
the cosine function is periodic with maximum 1, so for any positiveTthere exists at > Tsuch that
2tet2cos(et2) = 2tet2.
And, fort >1
2we haveet2<2tet2. So, we can see that, no matter
what values ofM,c, andTwe pick there will always be at > Tsuch thatf?(t)> Mect, which meansf?(t)is not of exponential order.1We can assumeM >1.
6Section 7.2 - Transformation of Initial Value Prob-lems7.2.1- Use Laplace transforms to solve the initial value problem below.
x ??+ 4x= 0;x(0) = 5;x?(0) = 0. Solution- The Laplace transform of the left side is:L(x??+ 4x) =s2X(s)-sx(0)-x?(0) + 4X(s)
=s2X(s)-5s+ 4X(s) = (s2+ 4)X(s)-5s. The Laplace transform of the right side is0, and equating the Laplace transforms we get: (s2+ 4)X(s)-5s= 0 ?X(s) =5s s2+ 4= 5?ss2+ 4? From this we see the inverse Laplace transform, our solution, is: x(t) = 5cos(2t). 77.2.4- Use Laplace transforms to solve the initial value problem below.
x ??+ 8x?+ 15x= 0;x(0) = 2;x?(0) =-3. Solution-The Laplace transform of the left side of the above equation is: L(x??+8x?+15x) =s2X(S)-sx(0)-x?(0)+8sX(s)-8x(0)+15X(s) = (s2+ 8s+ 15)X(s)-2s-13. The Laplace transform of the right side is0, and equating these twoLaplace transforms we get:
(s2+ 8s+ 15)X(s)-2s-13 = 0 ?X(s) =2s+ 13 s2+ 8s+ 15=2s+ 13(s+ 5)(s+ 3). If we take a partial fraction decomposition of the rational function we get:2s+ 13
(s+ 5)(s+ 3)=As+ 5+Bs+ 3==A(s+ 3) +B(s+ 5)(s+ 5)(s+ 3).So, we get the linear equations:
A+B= 2
3A+ 5B= 13.
8Solving forAandBwe getA=-32,B=72. So,
X(s) =-3
2?1s+ 5?
+72?1s+ 3?
From this we get the inverse Laplace transform, our solution, is: x(t) =-32e-5t+72e-3t.
97.2.15- Use Laplace transforms to solve the initial value problem below.
x ??+x?+y?+ 2x-y= 0, y ??+x?+y?+ 4x-2y= 0; x(0) =y(0) = 1;x?(0) =y?(0) = 0. Solution- If we take the Laplace transforms of both equations we get: s2X(s)-s+sX(s)-1 +sY(s)-1 + 2X(s)-Y(s) = 0,
s2Y(s)-s+sX(s)-1 +sY(s)-1 + 4X(s)-2Y(s) = 0.
We can simplify this system as:
(s2+s+ 2)X(s) + (s-1)Y(s) =s+ 2, (s+ 4)X(s) + (s2+s-2)Y(s) =s+ 2.Solving for forY(s)in the second equation we get:
Y(s) =s+ 2
s2+s-2-(s+ 4)X(s)s2+s-2=1s-1-(s+ 4)X(s)(s+ 2)(s-1). If we plug this in forY(s)in the first equation we get: (s2+s+ 2)X(s) + 1-?s+ 4 s+ 2?X(s) =s+ 2.
Solving forX(s)we get:
10X(s) =(s+ 1)(s+ 2)s(s2+ 3s+ 3).
Plugging this into our equation forY(s)we get:
Y(s) =1
s-1-(s+ 4)(s+ 1)s(s2+ 3s+ 3)(s-1)=s3+ 2s2-2s-4s(s-1)(s2+ 3s+ 3).A partial fraction decomposition onX(s)gives us:
X(s) =(s+ 1)(s+ 2)
s(s2+ 3s+ 3)=As+Bs+Cs2+ 3s+ 3. If we solve forA,B,Cusing our usual procedure we get:X(s) =2
3? 1s? +13s+12?s+3
2?2+34+1
2?s+3 2? 2+34, from which we get: x(t) =23+13e-3
2tcos?
⎷3 2t? +⎷3sin? ⎷3 2t?Doing the same thing withY(s)we get:
Y(s) =s3+ 2s2-2s-4
s(s-1)(s2+ 3s+ 3)=As+Bs-1+Cs+ds2+ 3s+ 3 2821?
1s? -921? 1s-1? +2