{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

SolutionsV_1

# SolutionsV_1 - ECE 3250 HOMEWORK V SOLUTIONS Fall 2009 1 In...

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

ECE 3250 HOMEWORK V SOLUTIONS Fall 2009 1. In each case, S ( x ) = Conv h ( x ). (a) S ( x ) = Conv h ( x ), so S ( x )( t ) = Z -∞ h ( t - τ ) x ( τ ) = Z -∞ e - 3( t - τ ) u ( t - τ ) e - 5 τ u ( τ ) = Z 0 e - 3( t - τ ) u ( t - τ ) e - 5 τ = e - 3 t R t 0 e - 2 τ if t 0 0 if t < 0 = (1 / 2)( e - 3 t - e - 5 t ) if t 0 0 if t < 0 . The equality on the third line holds because of the u ( τ ) on the second line. The equality on the fourth line holds because of the u ( t - τ ) on the third line. Note that we can write the final answer as S ( x )( t ) = (1 / 2)( e - 3 t - e - 5 t ) u ( t ) for all t R . (b) S ( x ) = Conv h ( x ), so S ( x )( t ) = Z -∞ h ( t - τ ) x ( τ ) = Z -∞ e - 3( t - τ ) u ( t - τ ) e - 5 | τ | = Z t -∞ e - 3( t - τ ) e - 5 | τ | = ( e - 3 t R t -∞ e 8 τ if t < 0 e - 3 t R 0 -∞ e 8 τ + e - 3 t R t 0 e - 2 τ if t 0 = (1 / 8) e 5 t if t < 0 (1 / 8) e - 3 t + (1 / 2) e - 3 t - (1 / 2) e - 5 t if t 0 . The equality on the third line holds because of the u ( t - τ ) on the second line. Note that we can write the final answer as S ( x )( t ) = ((5 / 8) e - 3 t - (1 / 2) e - 5 t ) u ( t ) + (1 / 8) e 5 t u ( - t ) for all t R . (c) Conv δ ( x ) = x , so Conv Shift 7 ( δ ) ( x ) = Shift 7 ( x ). Since the first term in h is Shift 7 of the impulse response of the system in (a), and since x is the same as in (a), the convolution of the first term in h with x is Shift 7 of the answer to (a). Accordingly, S ( x )( t ) = (1 / 2)( e - 3( t - 7) - e - 5( t - 7) ) u ( t - 7) + e - 5( t - 7) u ( t - 7) for all t R = (1 / 2)( e - 3( t - 7) + e - 5( t - 7) ) u ( t - 7) for all t R . 1

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

View Full Document
2 (d) S ( x ) = Conv h ( x ), so S ( x )( t ) = Z -∞ h ( t - τ ) x ( τ ) = Z -∞ e - 3( t - τ ) u ( t - τ ) u ( - τ ) = Z 0 -∞ e - 3( t - τ ) u ( t - τ ) = ( e - 3 t R 0 -∞ e 3 τ if t 0 e - 3 t R t -∞ e 3 τ if t < 0 . = (1 / 3) e - 3 t if t 0 1 / 3 if t < 0 . The equality on the third line holds because of the u ( - τ ) on the second line. The equality on the fourth line holds because of the u ( t - τ ) on the third line. Note that we can write the final answer as (1 / 3) e - 3 t u ( t ) + 1 / 3 u ( - t ) for all t R . 2. (a) S ( u ) = Conv h ( u ), so S ( u )( t ) = Z -∞ h ( t - τ ) u ( τ ) = Z -∞ 7( e - 3( t - τ ) - e - 5( t - τ ) u ( t - τ ) u ( τ ) = Z 0 7( e - 3( t - τ ) - e - 5( t - τ ) ) u ( t - τ ) = 7 e - 3 t R t 0 e 3 τ - 7 e - 5 t R t 0 e 5 τ if t 0 0 if t < 0 = (7 / 3)(1 - e - 3 t ) - (7 / 5)(1 - e - 5 t ) if t 0 0 if t < 0 The equality on the third line holds because of the u ( τ ) on the second line. The equality on the fourth line holds because of the u ( t - τ ) on the third line. Note that we can write the final answer as S ( x )( t ) = 14 / 15 + ( - (7 / 3) e - 3 t + (7 / 5) e - 5 t ) u ( t ) for all t R .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 6

SolutionsV_1 - ECE 3250 HOMEWORK V SOLUTIONS Fall 2009 1 In...

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

View Full Document
Ask a homework question - tutors are online