{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

10Fall_hw08soln

10Fall_hw08soln - ECE210 Fall 2010 Homework 08 Solutions 1...

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

ECE210 - Fall 2010 - Homework 08 Solutions 1. Find the exponential and trigonometric Fourier series coefficients of the periodic signal f ( t ) = e - 2 t , 0 t < 3 . The period is T = 3 . Solution: F n = 1 T Z T 0 f ( t ) e - jnω 0 t dt = 1 3 Z 3 0 e - 2 t e - jnω 0 t dt = 1 3 Z 3 0 e - ( jnω 0 +2) t dt = - 1 j 3 0 + 6 e - ( jnω 0 +2) t 3 0 = - 1 j 3 0 + 6 ( e - j 3 0 - 6 - 1) Substituting for ω 0 = 2 π T = 2 π 3 F n = 1 j 2 πn + 6 ( 1 - e - jn 2 π - 6 ) = 1 2 1 - e - 6 3 + jπn = 1 2 1 - e - 6 q 9 + ( πn ) 2 e - j arctan ( n π 3 ) = 1 2 1 - e - 6 9 + ( πn ) 2 ! (3 - jπn ) Then a n = F n + F - n = 3 1 - e - 6 9 + ( πn ) 2 ! b n = j ( F n - F - n ) = ( 1 - e - 6 ) 9 + ( πn ) 2 2. Problem 6.3 from text. Solution: There is not one unique approach to deriving this solution, but most approaches should follow one of these two methods: 1 2 2 e - jnπ - 0 - jn π 2 - 1 2 e - jn π 2 t ( - jn π 2 ) 2 2 0 = e - jnπ - jn π 2 - 1 2 e - jnπ ( - jn π 2 ) 2 - 1 ( - jn π 2 ) 2 evaluate the right side at bounds = e - jnπ - jn π 2 + 2(1 - e - jnπ ) j 2 n 2 π 2 simplify terms = e - jnπ - jn π 2 + 4 e - jn π 2 jn 2 π 2 1 2 j ( e jn π 2 - e - jn π 2 ) pull-out sin( n π 2 ) terms = e - jnπ - jn π 2 + 4 e - jn π 2 jn 2 π 2 sin( n π 2 ) convert to sin( n π 2 ) = -

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.

{[ snackBarMessage ]}

Page1 / 7

10Fall_hw08soln - ECE210 Fall 2010 Homework 08 Solutions 1...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online