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Practice_Midterm_Problems_2011

# Practice_Midterm_Problems_2011 - EE261 Raj Bhatnagar Summer...

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Unformatted text preview: EE261 Raj Bhatnagar Summer 2010-2011 EE 261 The Fourier Transform and its Applications Midterm Practice Problems 1. (10 points) More on evenness and oddness Suppose that a real square integrable function g ( x ) is written as a sum of its even and odd parts as g ( x ) = g e ( x ) + g o ( x ) . (a) Show that integraldisplay ∞ −∞ | g ( x ) | 2 dx = integraldisplay ∞ −∞ | g e ( x ) | 2 dx + integraldisplay ∞ −∞ | g o ( x ) | 2 dx. (b) Also, show that integraldisplay ∞ −∞ g ( x ) g (- x ) dx = integraldisplay ∞ −∞ | g e ( x ) | 2 dx- integraldisplay ∞ −∞ | g o ( x ) | 2 dx. Solution: (a) We start by writing out the required integral: integraldisplay ∞ −∞ | g ( x ) | 2 dx = integraldisplay ∞ −∞ g 2 ( x ) dx (since g ( x ) is real) = integraldisplay ∞ −∞ ( g e ( x ) + g o ( x )) 2 dx = integraldisplay ∞ −∞ ( g 2 e ( x ) + g 2 o ( x ) + 2 g o ( x ) g e ( x ) ) dx = integraldisplay ∞ −∞ | g e ( x ) | 2 dx + integraldisplay ∞ −∞ | g o ( x ) | 2 dx + integraldisplay ∞ −∞ 2 g o ( x ) g e ( x ) dx The last term is a product of an odd function with an even function. Hence the integral from-∞ to ∞ will be zero and our required result is obtained: integraldisplay ∞ −∞ | g ( x ) | 2 dx = integraldisplay ∞ −∞ | g e ( x ) | 2 dx + integraldisplay ∞ −∞ | g o ( x ) | 2 dx. It turns out that this relation is also valid for a complex function g ( x ) . (b) We decompose g ( x ) and g (- x ) into their even and odd components, g e ( x ) and g o ( x ) . integraldisplay ∞ −∞ g ( x ) g (- x ) dx = integraldisplay ∞ −∞ [ g e ( x ) + g o ( x )] [ g e (- x ) + g o (- x )] dx = integraldisplay ∞ −∞ [ g e ( x ) + g o ( x )] [ g e ( x )- g o ( x )] dx = integraldisplay ∞ −∞ g 2 e ( x )- g 2 o ( x ) dx = integraldisplay ∞ −∞ | g e ( x ) | 2 dx- integraldisplay ∞ −∞ | g o ( x ) | 2 dx 1 2. (30 points) Integration and the Fourier Transform Show these formulas by evaluating the integrals: (a) integraldisplay ∞ −∞ sinc 4 ( t ) dt = 2 3 (b) integraldisplay ∞ −∞ 2 1 + (2 πt ) 2 sinc(2 t ) dt = 1- 1 e (c) integraldisplay ∞ −∞ t 2 sinc 4 ( t ) dt = 1 2 π 2 Hint: Use Parseval's Theorem for (a) and (b). For (c), recall the rule of thumb: to simplify an integral of a polynomial times a function, try to use the derivative theorem for the Fourier...
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