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# homework solution - EE 261 The Fourier Transform and its...

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EE 261 The Fourier Transform and its Applications Fall 2011 Solutions to Problem Set Five 1. (20 points) Windowing functions In signal analysis, it is not realistic to consider a signal f ( t ) from -∞ < t < . Instead, one considers a modiﬁed section of the signal, say from t = - 1 / 2 to t = 1 / 2. The process is referred to as windowing and is achieved by multiplying the signal by a windowing function w ( t ). By the Convolution Theorem (applied in the frequency domain), the Fourier transform of the windowed signal, w ( t ) f ( t ) is the convolution of the Fourier transform of the original signal with the Fourier transform of the window function, F w * F f . The ideal window is no window at all, i.e., it’s just the constant function 1 from -∞ to , which has no eﬀect in either the time domain (multiplying) or in the frequency domain (convolving). Below are three possible windowing functions. Find the Fourier transform of each (you know the ﬁrst two) and plot the log magnitude of each from - 100 < s < 100. Which window is closest to the ideal window? Rectangle Window w ( t ) = Π( t ) Triangular Window w ( t ) = 2Λ(2 t ) Hann Window w ( t ) = ( 2cos 2 ( πt ) , | t | < 1 2 0 , otherwise . Solutions 1

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! 100 ! 80 ! 60 ! 40 ! 20 0 20 40 60 80 100 ! 20 ! 18 ! 16 ! 14 ! 12 ! 10 ! 8 ! 6 ! 4 ! 2 0 log abs sinc ! 100 ! 80 ! 60 ! 40 ! 20 0 20 40 60 80 100 ! 20 ! 18 ! 16 ! 14 ! 12 ! 10 ! 8 ! 6 ! 4 ! 2 0 log abs sinc2 We express w H as w H ( t ) = 2Π( t )cos 2 ( πt ) = Π( t )( 1 2 + 1 2 cos(2 πt )) 2
so the transform is F w H ( s ) = 2(sinc * ( 1 2 δ ( s ) + 1 4 δ ( s - 1) + 1 4 δ ( s + 1))) or F w H ( s ) = sinc( s ) + 1 2 sinc( s + 1) + 1 2 sinc( s - 1) - 100 - 50 0 50 100 - 20 - 15 - 10 - 5 0 log abs Hann From the three plots, the Fourier transform of the Hann window is most like δ , the Fourier transform of 1, in that it drops oﬀ much faster than the other two. The rect is least like the ideal window. 2. (15 points) Let f ( x ) be a signal and for h > 0 let A h f ( x ) be the averaging operator, A h f ( x ) = 1 2 h Z x + h x - h f ( y ) dy (a) Show that we have the alternate expressions for A h f ( x ): A h f ( x ) = 1 2 h Z h - h f ( x + y ) dy = 1 2 h Z h - h f ( x - y ) dy . (b) Show that we should deﬁne A h T for a distribution T by h A h T , ϕ i = h T , A h ϕ i . (c) What is A h δ ? Solutions: 3

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(a) In the integral deﬁning the averaging operator we make the change of variable y = u + x , or u = y - x , to write A h f ( x ) = 1 2 h Z x + h x - h f ( y ) dy = 1 2 h Z h - h f ( u + x ) du. Using the original variable of integration, y , we have the result A h f ( x ) = 1 2 h Z h - h f ( x + y ) dy . Now, in this integral make the substituion v = - y . Then 1 2 h Z h - h f ( x + y ) dy = 1 2 h Z h - h f ( x - v ) dv , so, again using the original variable of inegration, y , A h f ( x ) = 1 2 h Z h - h f ( x + y ) dy = 1 2 h Z h - h f ( x - y ) dy . as desired.
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## This note was uploaded on 01/10/2012 for the course EE 216 taught by Professor Harris,j during the Fall '09 term at Stanford.

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homework solution - EE 261 The Fourier Transform and its...

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