ps3sol - 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 Three 1. (5 points) Equivalent width: Still another reciprocal relationship The equivalent width of a signal f ( t ), with f (0) 6 = 0, is the width of a rectangle having height f (0) and area the same as under the graph of f ( t ). Thus W f = 1 f (0) Z -∞ f ( t ) dt. This is a measure for how spread out a signal is. Show that W f W F f = 1. Thus, the equivalent widths of a signal and its Fourier transform are reciprocal. From the Internet Encyclopedia of Science: Equivalent width A measure of the strength of a spectral line. On a plot of intensity against wavelength, a spectral line appears as a curve with a shape defined by the line profile. The equivalent width is the width of a rectangle centered on a spectral line that, on a plot of intensity against wavelength, has the same area as the line. Solution: The things we need to note are F f (0) = Z -∞ e - 2 πi 0 · t f ( t ) dt = Z -∞ f ( t ) dt, and the corresponding statement using the inverse Fourier transform, f (0) = Z -∞ e 2 πis · 0 F f ( s ) ds = Z -∞ F f ( s ) ds. 1
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Then we compute W f = 1 f (0) Z -∞ f ( t ) dt = F f (0) f (0) W F f = 1 F f (0) Z -∞ F f ( s ) ds = f (0) F f (0) W f W F f = F f (0) f (0) f (0) F f (0) W f W F f = 1 2. (20 points) Reversals, Shifts and Stretches If f ( t ) is a signal the corresponding reversed signal is defined to be f - ( t ) = f ( - t ) . Define the shift operator τ b f and the stretch operator σ a f by ( τ b f )( t ) = f ( t - b ) , ( σ a f )( t ) = f ( at ) . (a) Express f (2 t + 3) as σ a ( τ b f ) and as τ b 0 ( σ a 0 f ) for suitable shifts and stretches. Throwing in a reversal (using f - instead of f ), do the same for f ( - 2 t + 3). Find the Fourier transform of each. (b) Find the Fourier transforms of the function shown in the graph (a shifted sinc) Solution: (a) From a signal f ( t ) we want to end up with f (2 t + 3). As much as anything else, this problem is a worthwhile exercise in parentheses. In general, we have ( σ a ( τ b f ))( t ) = ( τ b f )( at ) = f ( at - b ) . 2
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Hence to wind up with f (2 t + 3) we take ( σ 2 ( τ - 3 f ))( t ). On the other hand ( τ b ( σ a f ))( t ) = ( σ a f )( t - b ) = f ( a ( t - b )) = f ( at - ab ) . So to wind up with f (2 t + 3) we take ( τ - 3 / 2 ( σ 2 f ))( t ). To work in reversal, we see that ( σ a ( τ b ( f - )))( t ) = ( τ b ( f - ))( at ) = f - ( at - b ) = f ( - ( at - b )) = f ( - at + b ) . Thus f ( - 2 t + 3) = ( σ 2 ( τ 3 ( f - )))( t ) . We could also do this the other way; ( τ b ( σ a ( f - )))( t ) = ( σ a ( f - ))( t - b ) = f - ( a ( t - b )) = f ( - a ( t - b )) = f ( - at + ab ) . So to get f ( - 2 t + 3) this way, f ( - 2 t + 3) = ( τ 3 / 2 ( σ 2 ( f - )))( t ) . To find the Fourier transforms, my advice is to go right to the definition and compute from there: F ( f (2 t + 3)) = Z -∞ e - 2 πist f (2 t + 3) dt = Z -∞ e - 2 πis ( u - 3) / 2 f ( u ) du (using u = 2 t + 3) = Z -∞ e - 2 πi ( s/ 2) u e 2 πi (3 s/ 2) f ( u )( du/ 2) = 1 2 e 3 πis Z -∞ e - 2 πi ( s/ 2) u f ( u ) du = 1 2 e 3 πis F f ( s/ 2) Thus f ( t ) ± F ( s ) = f (2 t + 3) ± 1 2 e 3 πis F ( s/ 2) And throwing in a reversal: F ( f ( - 2 t + 3)) = Z -∞ e - 2 πist f ( - 2 t + 3) dt (substitute u = - 2 t + 3) = Z -∞ + e - 2 πis ( u - 3) / ( - 2) f ( u - du/ 2) (note that the limits of integration are reversed) = Z -∞ e - 2 πis ( u - 3) / ( - 2) f ( u du/ 2) (and note how we reversed them back because of the - du ) = 1 2 e - 3 πis Z -∞ e - 2 πi ( - s/ 2) u f ( u ) du = 1 2 e - 3 πis F f ( - s/ 2) 3
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Thus f ( t ) ± F ( s ) = f ( - 2 t + 3) ±
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ps3sol - EE 261 The Fourier Transform and its Applications...

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