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HW4Soln - ECE 301 Homework 4 SOLUTION Landis Huffman Purdue...

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Unformatted text preview: ECE 301: Homework 4 SOLUTION Landis Huffman Purdue University, Summer 2009 Due Thursday July 16, 2009 in class 1. Chapter 4 problems 4.21(a,g), 4.26(a)(i), 4.33(a,b) 4.21 Solution: (a) We have x ( t ) = [ e- αt cos ω t ] u ( t ) = e- αt [ e jω t + e- jω t ) ] / 2 u ( t ) = 1 2 [ e t (- α + jω ) + e t (- α- jω ) ] u ( t ) X ( ω ) = integraldisplay ∞ 1 2 [ e t (- α + jω ) + e t (- α- jω ) ] e- jωt dt = 1 2 integraldisplay ∞ e- t ( α- jω + jω ) + e- t ( α + jω + jω ) dt = 1 2( α- jω + jω ) + 1 2( α + jω + jω ) 1 (g) We have X ( jω ) = integraldisplay ∞-∞ x ( t ) e- jωt dt = integraldisplay ∞-∞ x ( t )(- j sin( ωt )) dt ( ∵ x ( t ) is odd) = 2 integraldisplay ∞ x ( t )(- j sin( ωt )) dt = 2 integraldisplay 1 t (- j sin( ωt )) dt + 2 integraldisplay 2 1 (- j sin( ωt )) dt = 2 j bracketleftbigg t cos( ωt ) ω bracketrightbigg 1- 2 j integraldisplay 1 cos( ωt ) ω dt + 2 j bracketleftbigg cos( ωt ) ω bracketrightbigg 2 1 = 2 j cos( ω ) ω-...
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HW4Soln - ECE 301 Homework 4 SOLUTION Landis Huffman Purdue...

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