{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Solution_Signal

# Solution_Signal - (1 Simplify 3e j 3 4e j 6 Answer 3e j 3...

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

(1) Simplify 6 / 3 / 4 3 π π j j e e - + Answer: 12 . 0 3 4 3 4 3 3 arctan 6 / 3 / 5 5 2 4 3 3 2 3 4 3 ) 6 sin 4 3 sin 3 ( ) 6 cos 3 (cos 3 ) 6 sin 6 (cos 4 ) 3 sin 3 (cos 3 4 3 j j j j e e j j j j e e = = - + + = - + + = - + + = + + - - π π π π π π π π π π (2) Simplify ) 120 cos( 5 ) 30 cos( 5 ) 90 cos( 5 ) ( 0 0 0 - + - + + = t t t t x ϖ ϖ ϖ Answer: ) 45 cos( 59 . 2 ) 75 cos( ) 45 cos( 10 ) 120 cos( 5 ) 30 cos( 5 ) 120 cos( 5 ) 60 cos( ) 30 cos( 10 ) 120 cos( 5 ) 30 cos( 5 ) 90 cos( 5 ) ( 0 0 0 0 0 0 0 0 0 0 0 - = - = - + + = - + + = - + - + + = t t t t t t t t t t x ϖ ϖ ϖ ϖ ϖ ϖ ϖ ϖ ϖ (3) A periodic signal is given by the equation ) 3 / 120 cos( 4 ) 60 sin( 3 ) 5 / 40 cos( 4 2 ) ( π π π π π - + + - + = t t t t x (a) Determine the fundamental frequency and period and the power of x ( t ). (b) Now consider a new signal ). 6 / 50 cos( 10 ) ( ) ( π π - + = t t x t y Is y ( t ) still periodic? If so, what is the period? Answer: (a) ) 3 / 120 cos( 4 ) 2 / 60 cos( 3 ) 5 / 40 cos( 4 2 ) ( π π π π π π - + - + - + = t t t t x

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

View Full Document
For ) 5 / 40 cos( 4 π π - t , the period is 20 1 40 2 2 1 1 1 k k k = = π π ϖ π , N k 1 For ) 2 / 60 cos( 3 π π - t , the period is 30 1 60 2 2 2 2 2 k k k = = π π ϖ π , N k 2 For ) 3 / 120 cos( 4 π π - t , the period is 60 1 120 2 2 3 3 3 k k k = = π π ϖ π , N k 3 When 2 1 = k , 3 2 = k , 6 3 =
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 5

Solution_Signal - (1 Simplify 3e j 3 4e j 6 Answer 3e j 3...

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

View Full Document
Ask a homework question - tutors are online