This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGN Department of Electrical and Computer Engineering ECE 410 Digital Signal Processing Quiz 2 Thursday, September 18, 2008 Student Name: Prof. Bresler, Prof. Jones INSTRUCTIONS • You may not use any calculators, cell phones, earphones (or other forms of electronic media) for this quiz. You may use one side of one sheet of handwritten notes. • Show all your work to get full credit for your answers. • When you are asked to “ calculate ”, “ determine or “ ﬁnd ”, this means providing closed-form expressions, without summation or integral expressions. Problem 1 (20 points) Let X d ( ω ) = ( | ω | ≤ ω c 1 ω c < | ω | < π where 0 < ω c < π . Determine x [ n ], the inverse DTFT of X d ( ω ). (Specify x [ n ] for all n ∈ Z and show your work step by step for complete credit.) Solution: x [ n ] = 1 2 π Z π-π X d ( ω ) · e jωn dω = 1 2 π •Z-ω c-π e jωn dω + Z π ω c e jωn dω ‚ Now for n = 0, we have x  = 2( π-ω c ) 2 π = 1-ω c π (1) And, for n 6 = 0, we compute x [ n ] = 1 2 πjn e jωn ﬂ ﬂ-ω c-π + 1 2 πjn e jωn ﬂ ﬂ π ω c = 1 2 πjn h e-jω c n-» » » e-jπn + ' '' e jπn-e jω c n i =-ω c π sinc( ω c n ) , n 6 = 0 (2) Combining (1) and (2) we have x [ n ] = δ [ n ]-ω c π sinc( ω c n ) 1 Alternative approach 1: Recall the modulating property of DTFT, using which we can show that x [ n ] = 2 y [ n ] · cos ± π + ω c 2 n ¶ where Y d ( ω ) = ( 1 , | ω | ≤ π-ω c 2 , otherwise ↔ y [ n ] = π-ω c 2 π sinc ± π-ω c 2 n ¶ Therefore, x [ n ] = π-ω c π sinc ± π-ω c 2 n ¶ · cos ± π + ω c 2 n ¶ or equivalently x  = 1-ω c π x [ n ] = 2 πn sin ± π-ω c 2 n ¶...
View Full Document