527ex1sl

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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This note was uploaded on 09/29/2009 for the course 642 527 taught by Professor Speer during the Fall '07 term at Rutgers.

Page1 / 3

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online