{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# HW8sol - ECEN 314 Signals and Systems Solutions to HW 8 1...

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

ECEN 314: Signals and Systems Solutions to HW 8 1. Note that H ( s ) = 10 s +10 . The steady state response is then: y ( t ) = 2 | H ( j 10) | cos(10 t + H ( j 10)) = 1 . 414 cos(10 t - 45 ) 2. Note that the period of x ( t ) is T = 1. The simplest way to do this problem is as follows: 2 cos 10 πt + sin 6 πt = e j 10 πt + e - j 10 πt + 1 2 j e j 6 πt - e - j 6 πt = e j 2 π (5) t + e j 2 π ( - 5) t + 1 2 j e j 2 π (3) t - e j 2 π ( - 3) t One can easily see that x 5 = x - 5 = 1 and x 3 = - x - 3 = - j/ 2. By direct inspection of x ( t ) one easily gets a v = 0, a 5 = 2 and b 3 = 1. We can also verify the above result by using the standard integrals as follows: x k = 1 / 2 Z - 1 / 2 x ( t ) e - j 2 πkt dt = 1 / 2 Z - 1 / 2 e j 10 πt + e - j 10 πt + 1 2 j ( e j 6 πt - e - j 6 πt ) e - j 2 πkt dt = sin π (5 - k ) π (5 - k ) + sin π (5 + k ) π (5 + k ) + j 2 sin π (3 + k ) π (3 + k ) - sin π (3 - k ) π (3 - k ) = 1 k = ± 5 j/ 2 k = - 3 - j/ 2 k = 3 0 else We will now look at the alternative form of the Fourier series. a v = 1 / 2 Z - 1 / 2 ( 2 cos 10 πt + sin 6 πt ) dt = 0 .

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 / 4

HW8sol - ECEN 314 Signals and Systems Solutions to HW 8 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