solution for problem set 2

solution for problem set 2 - EE 261 The Fourier Transform...

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Unformatted text preview: EE 261 The Fourier Transform and its Applications Fall 2008 Solutions to Problem Set Two 1. (10 points) Whither Rayleigh? What happens to Rayleighs identity if f ( t ) is periodic of period T negationslash = 1? Solution: For a function f with period T , we have the expansion f ( t ) = summationdisplay n = c n e 2 int/T where c n = 1 T integraldisplay T e 2 int/T f ( t ) dt. Define g ( t ) = f ( Tt ). Then g has period 1. We shall derive Rayleighs identity for f from Rayleighs identity for g . First observe that g ( n ) = integraldisplay 1 e 2 int g ( t ) dt = integraldisplay 1 e 2 int f ( Tt ) dt = integraldisplay T e 2 inu/T f ( u ) 1 T du (making the substitution u = Tt ) = 1 T integraldisplay T e 2 inu/T f ( u ) du = c n . Next, integraldisplay 1 | g ( t ) | 2 dt = integraldisplay 1 | f ( Tt ) | 2 dt = integraldisplay T | f ( u ) | 2 1 T du From Rayleighs identity for a function with period 1, integraldisplay 1 | g ( t ) | 2 dt = summationdisplay n = | g ( n ) | 2 . 1 From the equations above, we thus have 1 T integraldisplay T | f ( t ) | 2 dt = summationdisplay n = | c n | 2 , Rayleighs identity for f . Another way to do this is to work with inner products and orthonormal bases in the space L 2 ([0 ,T ]), i.e., for functions of period T . The inner product is ( f,g ) = integraldisplay T f ( t ) g ( t ) dt and the norm (squared) of f is bardbl f bardbl 2 = integraldisplay T | f ( t ) | 2 dt. If e n ( t ) is an orthonormal basis then f = summationdisplay n ( f,e n ) e n and bardbl f bardbl 2 = summationdisplay n | ( f,e n ) | 2 . This is Rayleighs identity always. Lets see how it looks, more explicitly. In the case of functions of period T the orthonormal basis of complex exponentials is e n ( t ) = 1 T e 2 int/T , n = 0 , 1 , 2 ... . In the first version of the solution we wrote c n = 1 T integraldisplay T e 2 int/T f ( t ) dt and the Fourier series as f ( t ) = summationdisplay n c n e 2 int/T . Here ( f,e n ) = 1 T integraldisplay T e 2 int f ( t ) dt = T T integraldisplay T e 2 int f ( t ) dt = Tc n Then bardbl f bardbl 2 = summationdisplay n | ( f,e n ) | 2 . translates to integraldisplay T | f ( t ) | 2 dt = bardbl f bardbl 2 = summationdisplay n | ( f,e n ) | 2 = summationdisplay n | Tc n | 2 = summationdisplay n T | c n | 2 or 1 T integraldisplay T | f ( t ) | 2 dt = summationdisplay n | c n | 2 as we had before. 2 2. (30 points) Convolution, Autocorrelation and Fourier Series The convolution of two functions f ( t ) and g ( t ) of period 1 is defined by ( f g )( t ) integraldisplay 1 f ( ) g ( t ) d ....
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solution for problem set 2 - EE 261 The Fourier Transform...

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