Tuto5_sol - EE3210 Tutorial 5 (Solution) 1. Since the input...

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

View Full Document Right Arrow Icon
EE3210 Tutorial 5 (Solution) 1. Since the input is δ [ n ] and the system is initially at rest, we can conclude that h [ n ] = 0 for n < 0. Now, ] 1 [ 2 ] [ ] [ = n y n x n y . Therefore, 8 ] 2 [ 2 ] 3 [ ] 3 [ 4 ] 1 [ 2 ] 2 [ ] 2 [ 2 ] 0 [ 2 ] 1 [ ] 1 [ 1 ] 1 [ 2 ] 0 [ ] 0 [ = = = = = = = = h x h h x h h x h h x h and so on. Combining the two cases, we can write the impulse response in closed form as follows: ] [ ) 2 ( ] [ n u n h n = . 2. Since the system is initially at rest, we must have h [ n ] = 0 for n < 0. Next, we consider the following polynomial equation: 0 16 . 0 6 . 0 2 = r r . Factorize it, 0 ) 8 . 0 )( 2 . 0 ( = + r r . The two roots are -0.2 and 0.8. Hence, for n 0, we have ( ) n n C C n h ) 8 . 0 ( 2 . 0 ] [ + = 2 1 . We determine two initial conditions recursively as follows: 3 ] 1 [ 16 . 0 ] 0 [ 6 . 0 ] 1 [ 5 ] 1 [ 5 ] 2 [ 16 . 0 ] 1 [ 6 . 0 ] 0 [ 5 ] 0 [ = + + = = + + = h h h h h h Now we solve for C 1 and C 2 : 2 1 2 1 8 . 0 2 . 0 3 5 C C C C + = + = Therefore, C 1 = 1 and C 2 = 4. Hence, ( ) ] [ ] ) 8 . 0 ( 4 2 . 0 [ ] [ n u n h n n + = .
Background image of page 1

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

View Full DocumentRight Arrow Icon
3. The Fourier transform can be found as follows: ω j j e dt e dt e t u e j X t j t j t j t + = + = = = + + 2 1
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 11/22/2009 for the course ELECTRONIC EE3210 taught by Professor Sungchiwan during the Spring '09 term at École Normale Supérieure.

Page1 / 2

Tuto5_sol - EE3210 Tutorial 5 (Solution) 1. Since the input...

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