s10_ps10_sol

# s10_ps10_sol - Unified Engineering II Spring 2004 Problem...

This preview shows pages 1–3. Sign up to view the full content.

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Unified Engineering II Spring 2004 Problem S10 (Signals and Systems) Solution 1. Because the numerator is the same order as the denominator, the partial fraction expansion will have a constant term: 3 s 2 + 3 s − 10 G ( s ) = 2 s − 4 3 s 2 + 3 s − 10 = ( s − 2)( s + 2) b c = a + + s + 2 s − 2 To find a , b , and c , use coverup method: a = G ( s ) | = 3 s = ∞ 3 s 2 + 3 s − 10 b = s + 2 = 2 s =2 3 s 2 + 3 s − 10 1 c = = s − 2 s = − 2 So 2 1 G ( s ) = 3 + s − 2 + s + 2 , Re[ s ] > 2 We can take the inverse LT by simple pattern matching. The result is that g ( t ) = 3 δ ( t ) + 2 e 2 t + e − 2 t σ ( t ) 2. 6 s 2 + 26 s + 26 G ( s ) = ( s + 1)( s + 2)( s + 3) a b c = + + s + 1 s + 2 s + 3 Using partial fraction expansions, 6 s 2 + 26 s + 26 ( s + 2)( s + 3) 3 a = = s = − 1 6 s 2 + 26 s + 26 b = 2 = ( s + 1)( s + 3) s = − 2 6 s 2 + 26 s + 26 c = = 1 ( s + 1)( s + 2) s = − 3 So 3 2 1 G ( s ) = + + , Re[ s ] > − 1 s + 1 s + 2 s + 3 The inverse LT is given by...
View Full Document

## This note was uploaded on 01/28/2012 for the course AERO 16.01 taught by Professor Markdrela during the Fall '05 term at MIT.

### Page1 / 4

s10_ps10_sol - Unified Engineering II Spring 2004 Problem...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online