20082ee102_1_HW_5_solution

# 20082ee102_1_HW_5_solution - HW5 solution SPRING 2008 EE...

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

HW5 solution, SPRING 2008, EE 102 1. Y ( s ) = - 7 s + 3 + 8 s + 4 = 1 s + 3 × s - 4 s + 4 X ( s ) = 1 s + 3 x ( t ) = e - 3 t U ( t ) H ( s ) = s - 4 s + 4 = 1 - 8 s + 4 h ( t ) = δ ( t ) - 8 e - 4 t U ( t ) 2. For S 1 z ( t ) = Z t 0 e - ( t - τ ) x ( τ ) = e - t * x ( t ) Z ( s ) = 1 s + 1 X ( s ) H 1 ( s ) = 1 s + 1 For S 2 g 2 ( t ) = te - t U ( t ) G 2 ( s ) = 1 ( s + 1) 2 = H 2 ( s ) 1 s H 2 ( s ) = s ( s + 1) 2 x ( t ) = δ ( t ) + 2 costU ( t ) X ( s ) = 1 + 2 s s 2 + 1 = ( s + 1) 2 s 2 + 1 For the cascaded system, we have Y ( s ) = X ( s ) H 1 ( s ) H 2 ( s ) = s ( s + 1)( s 2 + 1) = - 1 / 2 s + 1 + (1 / 2) s + 1 / 2 s 2 + 1 y ( t ) = 1 2 ( - e - t + cost + sint ) U ( t ) 3. Note the identities: Z te t dt = ( t - 1) e t + C 1

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

View Full Document
Z t 2 e t dt = ( t 2 - 2 t + 2) e t + C Z t 3 e t dt = ( t 3 - 3 t 2 + 6 t - 6) e t + C The ﬁrst integral = Z -∞ ( t - τ ) e - ( t - τ ) U ( t - τ )2( δ ( τ ) - τ 2 U ( τ )) = 2 te - t U ( t ) - 2 e - t tU ( t ) Z t 0 τ 2 e τ
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 5

20082ee102_1_HW_5_solution - HW5 solution SPRING 2008 EE...

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

View Full Document
Ask a homework question - tutors are online