solHW3B

# solHW3B - (2 s 2 + s + 2) X ( s ) = e-5 s + ke-t s , or...

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

HW3 solution for Q4-Q7 4. Apply the Laplace transform and the initial conditions, we obtain s 2 X + 4 X = 1 s 2 + 1 - e - 2 πs s 2 + 1 , thus X = 1 - e - 2 πs ( s 2 + 1)( s 2 + 4) = 1 - e - 2 πs 3 ± 1 s 2 + 1 - 1 s 2 + 4 . Thus by Table 5.1 in the lecture note, x ( t ) = 1 6 [1 - u 2 π ( t )](2sin t - sin2 t ) . 5. First rewrite the equation as ¨ x + ˙ x + 5 4 x = sin t - u π ( t )sin t = sin t + u π ( t )sin( t - π ) . Apply the Laplace transform and initial conditions to get ( s 2 + s + 5 4 ) X = 1 + e - πs s 2 + 1 . Hence X = (1 + e - πs ) H ( s ) , x = h ( t ) + u π ( t ) h ( t - π ) , where H ( s ) = 1 ( s 2 + 1)( s 2 + s + 5 / 4) and h ( t ) is the inverse transform of H ( s ) . Rewrite H ( s ) as H ( s ) = 4 - 16 s 17( s 2 + 1) + 16 s + 12 4( s 2 + s + 5 / 4) = - 16 17 s s 2 + 1 + 4 17 1 s 2 + 1 + 16 17 s + 1 / 2 ( s + 1 / 2) 2 + 1 + 4 17 1 ( s + 1 / 2) 2 + 1 , and apply Table 5.1, we see that h ( t ) = 4 17 [ - 4cos t + sin t + 4 e - t/ 2 cos t + e - t/ 2 sin t ] . 1

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

View Full Document
6. Apply the Laplace transform and initial conditions to get ( s 2 + 2 s + 3) X = 1 s 2 + 1 + e - 3 πs , thus X = 1 ( s 2 + 1)( s 2 + 2 s + 3) + e - 3 πs s 2 + 2 s + 3 = 1 - s 4( s 2 + 1) + s + 1 4(( s + 1) 2 + 3) + e - 3 πs ( s + 1) 2 + 3 . We derive that x = 1 4 sin t - 1 4 cos t + 1 4 e - t cos 2 t + 1 2 u 3 π ( t ) e - ( t - 3 π ) sin 2( t - 3 π ) . 7. Consider the equation x + ˙ x + 2 x = δ ( t - 5) + ( t - t 0 ) . Taking the Laplace transform gives that
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: (2 s 2 + s + 2) X ( s ) = e-5 s + ke-t s , or that X ( s ) = e-5 s + ke-t s 2 s 2 + s + 2 = 2 √ 15 ( e-5 s + ke-t s ) 15 / 16 ( s + 1 / 4) 2 + 15 / 16 . Apply Table 5.1 to ﬁnd the inverse transform of X ( s ): x ( t ) = 2 √ 15 ˆ u 5 ( t ) e-( t-5) / 4 sin ˆ √ 15 4 ( t-5) ! + ku t ( t ) e-( t-t ) / 4 sin ˆ √ 15 4 ( t-t ) !! . According to the requirement of the question, we shall set t = 5 + 4 √ 15 × 2 π = 5 + 8 π √ 15 , and set x ( t ) = 0 when t &gt; t , i.e. k = e-( t-5) / 4 = e-2 π/ √ 15 . Putting t and k into x ( t ) gives the answer of (b). 2...
View Full Document

## This note was uploaded on 01/28/2011 for the course MATH 150 taught by Professor T.qian during the Spring '09 term at HKUST.

### Page1 / 2

solHW3B - (2 s 2 + s + 2) X ( s ) = e-5 s + ke-t s , or...

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

View Full Document
Ask a homework question - tutors are online