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ps2sol

# ps2sol - EE 261 The Fourier Transform and its Applications...

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EE 261 The Fourier Transform and its Applications Fall 2011 Solutions to Problem Set Two 1. (30 points) Convolution, Autocorrelation and Fourier Series Recall the convolution of two functions f ( t ) and g ( t ) of period 1 is deﬁned by ( f * g )( t ) Z 1 0 f ( τ ) g ( t - τ ) dτ . (a) (5) Show that f * g is periodic of period 1. (b) (10) Suppose f ( t ) and g ( t ) have Fourier series f ( t ) = X n = -∞ a n e 2 πint g ( t ) = X n = -∞ b n e 2 πint , respectively. Find the Fourier series of ( f * g )( t ) and explain why this implies that f * g = g * f . Let f ( x ) be a real, periodic function of period 1. The autocorrelation of f with itself is the function ( f ? f )( x ) = Z 1 0 f ( y ) f ( y + x ) dy . (c) (5) Show that f ? f is also periodic of period 1. (d) (10) If f ( x ) = X n = -∞ ˆ f ( n ) e 2 πinx show that the Fourier series of ( f ? f )( x ) is ( f ? f )( x ) = X n = -∞ | ˆ f ( n ) | 2 e 2 πinx . See the supplementary problems for more on the autocorrelation function. Solutions 1

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(a) To show that f * g is periodic of period 1 we must show that ( f * g )( t + 1) = ( f * g )( t ) for all t . For this, ( f * g )( t + 1) = Z 1 0 f ( τ ) g ( t + 1 - τ ) = Z 1 0 f ( τ ) g ( t - τ ) (since g is peiodic of period 1) = ( f * g )( t ) Notice that this argument would have worked if we had assumed only that g is periodic of period 1, and the same is true if we had assumed that only f is periodic. That is, f * g is periodic if either f or g is periodic. (b) Next we ﬁnd the Fourier series for f * g . We substitute the Fourier series expansions into the formula for convolution. ( f * g )( t ) = Z 1 0 f ( τ ) g ( t - τ ) = Z 1 0 X n = -∞ a n e 2 πinτ ! X m = -∞ b m e 2 πim ( t - τ ) ! (Notice that we use a diﬀerent summation index for the series for g . This is necessary.) = Z 1 0 X n,m = -∞ a n b m e 2 πinτ e 2 πim ( t - τ ) = Z 1 0 X n,m = -∞ a n b m e 2 πimt e 2 πi ( n - m ) τ = X n,m = -∞ a n b m e 2 πimt Z 1 0 e 2 πi ( n - m ) τ (swapping summation and integration) We’ve seen that integral many times by now: Z 1 0 e 2 πi ( n - m ) τ = ( 1 , n = m 0 , n 6 = m Thus the only terms that survive are when n = m and hence X n,m = -∞ a n b m e 2 πimt Z 1 0 e 2 πi ( n - m ) τ = X n = -∞ a n b n e 2 πint We have shown ( f * g )( t ) = X n = -∞ a n b n e 2 πint . 2
The coeﬃcients of the Fourier series for the convolution are the products of the Fourier coeﬃcients of the individual signals. Symbolically, \ ( f * g )( n ) = ˆ f ( n g ( n ) . (We will see this property for Fourier transforms as well.) From this property of the coeﬃcients it is then clear that f * g = g * f . (c) From the deﬁnition of autocorrelation ( f ? f )( x + 1) = Z 1 0 f ( y ) f ( y + x + 1) dy = Z 1 0 f ( y ) f ( y + x ) dy since f ( y + x + 1) = f ( y + x ) by periodicity of f . This shows rhat f ? f is periodic of period 1. (d) For part (d) we plug the Fourier series of f into the deﬁnition of autocorrelation: ( f ? f )( x ) = Z 1 0 f ( y ) f ( y + x ) dy = Z 1 0 X n = -∞ ˆ f ( n ) e 2 πiny ! X m = -∞ ˆ f ( m ) e 2 πim ( y + x ) ! dy (Again we have to use diﬀerent indices of summation for the two series) = Z 1 0 X n = -∞ ˆ f ( n ) e 2 πiny X m = -∞ ˆ f ( m ) e 2 πimy e 2 πimx !

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ps2sol - EE 261 The Fourier Transform and its Applications...

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