HW1_solution

HW1_solution - ECEN 314 Signals and Systems Solution to HW...

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

View Full Document Right Arrow Icon
ECEN 314: Signals and Systems Solution to HW 1 Problem 1.3 (a) x 1 ( t ) = e - 2 t u ( t ) E = Z 0 e - 4 t dt = 1 4 . P = 0, because E = 1 4 < (b) x 2 ( t ) = e j (2 t + π 4 ) Noting that | x 2 ( t ) | = 1, we have E = Z -∞ | x 2 ( t ) | 2 dt = Z -∞ dt = P = lim T →∞ 1 2 T Z T - T | x 2 ( t ) | 2 dt = lim T →∞ 1 2 T Z T - T 1 dt = 1 . (c) x 3 ( t ) = cos( t ) E = Z -∞ | x 3 ( t ) | 2 dt = Z -∞ cos 2 t dt = P = lim T →∞ 1 2 T Z T - T cos 2 tdt = lim T →∞ 1 2 T Z T - T 1 + cos(2 t ) 2 dt = 1 2 . (d) x 1 [ n ] = ( 1 2 ) n u [ n ] E = X n = -∞ | x 1 [ n ] | 2 = X n =0 ± 1 4 n = 1 1 - 1 / 4 = 4 / 3 . We have P = 0, because E < . (e) x 2 [ n ] = e j ( π n 2 + π n 8 ) Noting that | x 2 [ n ] | 2 = 1, we have E = X n = -∞ 1 = . 1
Background image of page 1

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

View Full Document Right Arrow Icon
P = lim N →∞ 1 2 N + 1 N X n = - N | x 2 [ n ] | 2 = lim N →∞ 1 2 N + 1 N X n = - N 1 = 1 . (f)
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 4

HW1_solution - ECEN 314 Signals and Systems Solution to HW...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online