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Unformatted text preview: ECE 493 HW #10 Version 1.6 February 18, 2011 Spring 2011 Univ. of Illinois Due Th, Feb 10 Prof. Allen Topic of this homework: Hilbert transform; Functions classes; Positive definite functions; Deliverable: Show your work. 1 Hilbert transforms Analyze the real impulse response h ( t ) = e t/ u ( t ) , with = 10 [ms], in terms of its Hilbert transform (integral) relations. 1. Find H ( s ), the Laplace transform of h ( t ). Solution: Let a = 1 / . Then e at u ( t ) 1 s + a . 2. Find the real and imaginary parts of H ( ) H ( s )  s = i and plot each. Solution: First rationalize the denominator: H ( s ) = s + a ( s + a )( s + a ) = i + a 2 + 2 + a 2 . Next take the real and imaginary parts, and evaluate s on the axis (set = 0): H ( ) = a/ ( a 2 + 2 ) , H ( ) = / ( a 2 + 2 ) . The real part is constant below the cutoff (resonant) frequency = a, and goes at 12 dB/oct above the cutoff. The imaginary part is bandpass with 6 dB/Oct above and below the resonance frequency. 3. The even and odd parts of a function h ( t ) are defined as h e ( t ) h ( t ) + h ( t ) 2 and h o ( t ) h ( t ) h ( t ) 2 . Plot the symmetric h e ( t ) and antisymmetric h o ( t ) functions. Solution: 2 h e ( t ) = h ( t ) + h ( t ) e at u ( t ) + e at u ( t ), while 2 h o ( t ) = h ( t ) h ( t ) e at u ( t ) e at u ( t ). It trivially follows that h ( t ) = h e ( t ) + h o ( t ). 4. Find the Fourier transforms of h e ( t ) H e ( ) and h o ( t ) H o ( ). Solution: Since h ( t ) H ( ), a symmetric time function is real in the frequency domain, 2 h e ( t ) = h ( t ) + h ( t ) H ( ) + H ( ) = 2 H ( ) , thus h e ( t ) H ( ). In a similar fashion, an antisymmetric time function is pure imaginary h o ( t ) i H ( ) . Again with a 1 / : H e ( ) = H ( ) = a 2 + a 2 , 1 H o ( ) = i H ( ) = i 2 + a 2 , thus H ( ) = H e ( ) + H o ( ) h ( t ) = h e ( t ) + h o ( t ) . The inverse Fourier transform of H o ( ) is zero at t = 0, which makes it very different from the inverse Laplace transform, which is not defined at t = 0. What is the inverse FT of H e ( )? Be sure to discuss what happens at t = 0. 5. Determine the Hilbert (integral) relations between H e ( ) and H o ( ) by use of the Fourier transform relations 1 2 sgn( t ) 1 i and/or u ( t ) ( ) + 1 i . Note: sgn( t ) = t/  t  . Solution: This follows simply from the following obvious relationship h o ( t ) = h e ( t )sgn( t ) 1 2 H e ( ) parenleftbigg 2 i parenrightbigg (1) since sgn( t ) 2 i . Thus H o ( ) = 1 i H e ( ) (2) 6. Find the Hilbert (integral) relations between H r H ( ) (real part) and H i H ( ) (imag part) of H ( )....
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 Spring '11
 JontB.Allen

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