HW_2_SOLUTIONS - -2 t-e-2 t + 2 te-2 t = e-t-( t + 1) e-2 t...

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

View Full Document Right Arrow Icon
ECE 3085 HW #2 solutions 1. f ( t ) = step ( t ) - 2 step ( t - 1) + step ( t + 2), hence F ( s ) = (1 - 2 exp ( - s ) + exp ( - 2 s )) 1 s = (1 - exp ( - s )) 2 1 s . 2. Let g ( t ) := δ ( t ) - δ ( t - 1). Then f ( t ) = X n =0 g ( t - 2 n ) . Therefore F ( s ) = X n =0 G ( s ) e - 2 ns = G ( s ) 1 1 - exp ( - 2 s ) . But G ( s ) = 1 - exp ( - s ), hence F ( s ) = 1 - exp ( - s ) 1 - exp ( - 2 s ) = 1 1 + exp ( - s ) . 3. x ( t ) X ( s ) dx dt sX ( s ) - 1 d 2 x dt 2 s 2 X ( s ) - s + 2. So, s 2 X ( s ) - s + 2 + 4( sX ( s ) - 1) + 4 X ( s ) = 0, and therefore X ( s ) = s + 2 s 2 + 4 s + 4 = s + 2 ( s + 2) 2 = 1 s + 2 . Thus, x ( t ) = e - et . 4. sX ( s ) - 4 + 3 X ( s ) = 1 s + 1 . ( s + 3) X ( s ) = 4 + 1 s + 1 = 4 s + 5 s + 1 X ( s ) = 4 s + 5 ( s + 1)( s + 3) = A s + 1 + B s + 3 . 4 s + 5 = A ( s + 3) + B ( s + 1) s = - 3 : - 7 = - 2 B B = 3 . 5 s = - 1 : 1 = 2 A A = 0 . 5 X ( s ) = 0 . 5 1 s + 1 + 3 . 5 1 s + 3 x ( t ) = 0 . 5 e - t + 3 . 5 e - 3 t . 1
Background image of page 1

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

View Full DocumentRight Arrow Icon
5. ( s 2 + 3 s + 2) X ( s ) = 1 s + 2 X ( s ) = 1 ( s 2 + 3 s + 2)( s + 2) = 1 ( s + 1)( s + 2) 2 = A s + 1 + B 1 + B 2 s ( s + 2) 2 1 = A ( s + 2) 2 + ( B 1 + B 2 s )( s + 1) s = - 1: 1 = A s = - 2 : 1 = ( B 1 - 2 B 2 )( - 1) = 2 B 2 - B 1 s = 0 : 1 = 4 A + B 1 B 1 = - 3, B 2 = - 1 F ( s ) = 1 s + 1 - 3 + s ( s + 2) 2 Now 1 s 2 t 1 ( s +2) 2 te - 2 t s ( s +2) 2 ( te - 2 t ) 0 = e - 2 t - 2 te - 2 t . So, x ( t ) = e - t - 3 te
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: -2 t-e-2 t + 2 te-2 t = e-t-( t + 1) e-2 t 6. ( s 2 + 4) X ( s ) = 1 s X ( s ) = 1 s ( s 2 + 4) = 1 s ( s-2 j )( s + 2 j ) = A s + B s-2 j + C s + 2 j 1 = A ( s-2 j )( s + 2 j ) + Bs ( s + 2 j ) + Cs ( s-2 j ) s = 0: 1 = 4 A A = 1 4 s = 2 j : 1 = B (2 j )(4 j ) =-8 B , B =-1 8 s =-2 j : 1 =-8 C C =-1 8 F ( s ) = 1 4 1 s-1 8 ( 1 s-2 j + 1 s + 2 j ) f ( t ) = 1 4-1 8 ( e 2 jt + e-2 jt ) = 1 4 (1-cos (2 t )) . 2...
View Full Document

This note was uploaded on 06/09/2009 for the course ECE 3085 taught by Professor Taylor during the Fall '08 term at Georgia Institute of Technology.

Page1 / 2

HW_2_SOLUTIONS - -2 t-e-2 t + 2 te-2 t = e-t-( t + 1) e-2 t...

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