tut1_solutions_17_01_09

# tut1_solutions_17_01_09 - Department of Electrical...

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

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.

Unformatted text preview: Department of Electrical Engineering EE 308: Communication Systems (Spring 2009) Course Instructor: Prof. Abhay Karandikar Solutions to Problem Set 1 1. What is the Fourier transform of a function that is (a) Real and even? (b) Real and odd? (c) Imaginary and even? (d) Imaginary and odd? Solution: Let G ( f ) be the Fourier transform of the function g ( t ). Using De Moivre’s formula, G ( f ) = Z ∞-∞ g ( t )exp(- j 2 πft ) dt = Z ∞-∞ g ( t )cos2 πftdt- j Z ∞-∞ g ( t )sin2 πftdt (1) If g(t) is even, R ∞-∞ g ( t )sin2 πftdt = 0 and if g(t) be odd, R ∞-∞ g ( t )cos2 πftdt = 0. Using these properties, the proof of (a)-(d) is trivial. 2. For k > 0, find the Fourier transform of g ( t ) = 2 k k 2 +(2 πt ) 2 and use it to evaluate R ∞-∞ dx (1+ x 2 ) 2 . Solution: The Fourier transform of g ( t ) can be expressed as G ( f ) = 2 k Z ∞-∞ exp(- j 2 πft ) k 2 + (2 πt ) 2 dt (2) = 2 k Z ∞-∞ exp( j 2 πft ) k 2 + (2 πt ) 2 dt (3) Using the contour integral approach and evaluating the residues in the upper half plane (see the chapters on Complex Analysis in Kreyszig), from (2) and (3), we obtain G ( f ) = exp(-| k | f ). Alternatively, one may show that g ( f ) is the Fourier transform of G ( t ) and then use duality. Duality: If g ( t ) F ←→ G ( f ), then G ( t ) F ←→ g (- f ). 1 Using Parseval’s theorem, we have Z ∞-∞ 2 k k 2 + (2 πt ) 2 2 dt = Z ∞-∞ exp(- 2 k | f | ) df = 1 k (4) Substituting k = 2 π in the above and simplifying, we obtain Z ∞-∞ dt (1 + t 2 ) 2 = π 2 (5) Parseval’s Theorem: Z ∞-∞ | g ( t ) | 2 dt = Z ∞-∞ | G ( f ) | 2 df (6) 3. Let g ( t ) = e- a | t | sgn( t ) ,a > 0. (a) Find G ( f ), the Fourier transform of g ( t ). (b) What happens to G ( f ) as a → 0? (c) Can you now evaluate the Fourier transform of the step function u ( t )? Solution: (a) The Fourier transform of g ( t ) can be expressed as G ( f ) =- Z-∞ e ( a- j 2 πf ) t dt + Z ∞ e- ( a + j 2 πf ) t dt = 1 a + j 2 πf- 1 a- j 2 πf =- j 4 πf a 2 + (2 πf ) 2 (7) (b) lim a → G ( f ) = 1 jπf (8) Thus, we have obtained sgn( t ) F ←→ 1 jπf ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 7

tut1_solutions_17_01_09 - Department of Electrical...

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

View Full Document
Ask a homework question - tutors are online