# HWa-sol-1 - ECE 493 HW#10 – Version 1.6 Spring 2011 Univ...

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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/τ u ( t ) , with τ = 10 [ms], in terms of its Hilbert transform (integral) relations. 1. Find H ( s ), the Laplace transform of h ( t ). Solution: Let a = 1 /τ . Then e − at u ( t ) ↔ 1 s + a . 2. Find the real and imaginary parts of H ( ω ) ≡ H ( s ) | s = iω and plot each. Solution: First rationalize the denominator: H ( s ) = s ∗ + a ( s + a )( s ∗ + a ) = σ − iω + a σ 2 + ω 2 + a 2 . Next take the real ℜ and imaginary ℑ parts, and evaluate s on the ω axis (set σ = 0): ℜ H ( ω ) = a/ ( a 2 + ω 2 ) , ℑ H ( ω ) = − ω/ ( a 2 + ω 2 ) . The real part is constant below the cutoff (resonant) frequency ω = 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 ( ω ) and h o ( t ) ↔ H o ( ω ). Solution: Since h ( − t ) ↔ H ∗ ( ω ), a symmetric time function is real in the frequency domain, 2 h e ( t ) = h ( t ) + h ( − t ) ↔ H ( ω ) + H ∗ ( ω ) = 2 ℜ H ( ω ) , thus h e ( t ) ↔ ℜ H ( ω ). In a similar fashion, an antisymmetric time function is pure imaginary h o ( t ) ↔ i ℑ H ( ω ) . Again with a ≡ 1 /τ : H e ( ω ) = ℜ H ( ω ) = a ω 2 + a 2 , 1 H o ( ω ) = i ℑ H ( ω ) = − iω ω 2 + a 2 , thus H ( ω ) = H e ( ω ) + H o ( ω ) ↔ h ( t ) = h e ( t ) + h o ( t ) . The inverse Fourier transform of H o ( ω ) 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 ( ω )? Be sure to discuss what happens at t = 0. 5. Determine the Hilbert (integral) relations between H e ( ω ) and H o ( ω ) by use of the Fourier transform relations 1 2 sgn( t ) ↔ 1 iω and/or u ( t ) ↔ πδ ( ω ) + 1 iω . Note: sgn( t ) = t/ | t | . Solution: This follows simply from the following obvious relationship h o ( t ) = h e ( t )sgn( t ) ↔ 1 2 π H e ( ω ) ⋆ parenleftbigg 2 iω parenrightbigg (1) since sgn( t ) ↔ 2 iω . Thus H o ( ω ) = 1 πiω ⋆ H e ( ω ) (2) 6. Find the Hilbert (integral) relations between H r ≡ ℜ H ( ω ) (real part) and H i ≡ ℑ H ( ω ) (imag part) of H ( ω )....
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## This note was uploaded on 03/28/2011 for the course ECE ECE 493 taught by Professor Jontb.allen during the Spring '11 term at University of Illinois at Urbana–Champaign.

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HWa-sol-1 - ECE 493 HW#10 – Version 1.6 Spring 2011 Univ...

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