527ex1sl

527ex1sl - 642:527 SOLUTIONS EXAM 1 FALL 2007 1(a What will...

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Unformatted text preview: 642:527 SOLUTIONS: EXAM 1 FALL 2007 1 (a) What will be the form of the partial fraction expansion of G ( s ) = 1 ( s + 2)( s 2 + 9) ? Your answer will involve some constants which you should not evaluate . (Note: There was an error on the exam as distributed; the numerator of G ( s ) was given as s . This was corrected during the exam.) Solution: The expansion will have the form G ( s ) = A s + 2 + Bs + C s 2 + 9 for some constants A , B , and C . (b) Use the Laplace transform to solve for the problem x ′′ + 9 x = e − 2 t , x (0) = 1, x ′ (0) = 3, for x ( t ). Your answer may be left in terms of the constants you introduced in (a). Solution: Taking the Laplace transform of the given equation yields ( s 2 +9) X ( s ) − s − 3 = 1 / ( s +2), so X ( s ) = s + 3 s 2 + 9 + 1 ( s + 2)( s 2 + 9) = s s 2 + 9 + 3 s 2 + 9 + A s + 2 + Bs + C s 2 + 9 = A s + 2 + ( B + 1) s s 2 + 9 + 3 + C s 2 + 9 . From the table of Laplace transforms we find that x ( t ) = Ae − 2 t + ( B + 1)cos 3 t + 3 + C 3 sin 3 t. 2. A function f ( t ) is defined for t ≥ 0 by f ( t ) = braceleftbigg 2 , if 1 ≤ t < 4, , if 0 ≤ t < 1 or t ≥ 4. Express f ( t ) in terms of a single formula using the Heaviside function, then find its Laplace trans- form. Solution: f ( t ) = 2 H ( t − 1) − 2 H ( t − 4), so F ( s ) = 2 e − s /s − 2 e − 4 s /s . 3 (a) Use the Laplace transform to solve the initial value problem for x ( t ): x ′′ + 4 x ′ + 8 x = δ ( t − 3); x (0) = x ′ (0) = 0 . Solution: First, L { δ ( t − 3) } = integraldisplay ∞ δ ( t − 3) e − st dt = e − 3 s , so taking the Laplace transform of the equation we find that X ( s ) = e − 3 s / ( s 2 + 4 s + 8) = e − 3 s / (( s + 2) 2 + 4). Since from the table we have L − 1 { 1 / (( s...
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527ex1sl - 642:527 SOLUTIONS EXAM 1 FALL 2007 1(a What will...

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