tut1_solutions_17_01_09

tut1_solutions_17_01_09 - Department of Electrical...

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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 ....
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tut1_solutions_17_01_09 - Department of Electrical...

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