HWa-sol-1 - ECE 493 HW #10 Version 1.6 February 18, 2011...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ( )....
View Full Document

Page1 / 6

HWa-sol-1 - ECE 493 HW #10 Version 1.6 February 18, 2011...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online