EE110_W08_HW 4 Solution

EE110_W08_HW 4 - Solution to Homework 4 EE 110 Winter 2008 Solution 1(a vs(0.1 =(20 j30 e-2 j50(0.1 =(36.06-56.31o e-0.2 j5 = 36.06e-0.2-56.31o

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Solution to Homework 4, EE 110, Winter 2008 Solution 1. (a) v s (0.1) = (20 – j 30) e (-2 + j 50)(0.1) = (36.06 -56.31 o ) e (-0.2 + j 5) = 36.06 e -0.2 [-56.31 o + j 5(180)/ π ] = 29.52 230.2 o V (or 29.52 -129.8 o V). (b) Re{ v s } = 36.06 e -2 t cos (50 t – 56.31 o ) V. (c) Re{ v s (0.1) } = 29.52 cos (230.2 o ) = -18.89 V. (d) The complex frequency of this waveform is s = -2 + j 50 s -1 (e) s * = (-2 + j 50)* = -2 – j 50 s -1 Solution 2. (a) L { } [ ] - - - - = = + = + 0 0 0 s 5 5 ) ( 3 2 ) ( 3 2 - t t t e dt e dt t u e t u s s s = + - - - t t t t e e s s s 5 lim s 5 lim 0 If the integral is going to converge, then ( 29 finite). be must s (i.e. 0 lim = - t t e s This leads to the first term dropping out (l’Hospital’s rule assures us of this), and so L { } s 5 ) ( 3 2 = + t u (b) L { } + - + - - - + - = = = 0 ) s 8 ( 0 ) s 8 ( 0 s 8 - 8 - 8 s 3 3 3 3 - - t t t t t e dt e dt e e e = + + + - + - + - t t t t e e ) 8 ( 0 ) 8 ( 8 s 3 lim 8 s 3 lim s s = 0 + 8 s 3 + = 8 s 3 + (c) L { } 0 ) 0 ( ) ( ) ( ) ( - - - - 0 0 s 0 0 s 0 s = = - = - = - - - - dt e dt t u e dt t u e t u t t t (d) L { } - - - - - = = = = 0 s 0 s 0 s 0 s s - - t t t t e K dt e K dt e K dt Ke K = + - - - t t t t e K e K s s s lim s lim 0 If the integral is going to converge, then ( 29 finite). be must s (i.e. 0 lim = - t t e s This leads to the first term dropping out (l’Hospital’s rule assures us of this), and so L { } s K K =
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Solution 3. (a) s 1 s 1 = (b) { } s 3 s 1 s 1 s 1 )] ( [ ) ( 1 2 = + + = + + t u t u (c) { } s 3 s 1 3 ) ( 2 - = - t u t (d) { } s 2 s 1 s 1 ) 2 ( ) 1 ( ) ( - 1 - - - + - = - - - + e e t t t δ Solution 4. F ( s ) = 3 2 1 2 4 5 2 π - + + + + s s s s = 2 1 2 ( 2)( 1) - + + + s s s = 2 1 2 ( 2) ( 1) ( 1) a b c - + + + + + + s s s s where 2 2 ( 1) a = - = = + s s 1 ( 2) b = - = = + s s and ( 29 ( 29 1 1 2 2 2 1 1 = = = ( 2) ( 2) ( 2) 1 d d c ds ds = - = - = - = + - - + + + + s s s s s s s s Thus, we may write f ( t ) = 2 ( t ) – u ( t ) + π e –2 t u ( t ) + π te t u ( t ) – π e t u ( t )
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Solution 5. (a) H ( s ) = 1 1 1 2 2 + = - + + s s s , hence h ( t ) = δ ( t ) – e –2 t u ( t ) (b) H ( s ) = ( 29( 29 3 2 1 1 2 1 2 + = - + + + + s s s s s , hence 2 ( ) 2 ( ) t t h t e e u t - - = - (c) We need to perform long division on the second term prior to applying the method of residues (some intermediate steps are not shown): ( 29 3 2 4 2 5 5 7 3 18 32 15 - + + + + + s s s s s s s Thus, H ( s
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This note was uploaded on 03/06/2008 for the course EE 110 taught by Professor Gupta during the Winter '08 term at UCLA.

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EE110_W08_HW 4 - Solution to Homework 4 EE 110 Winter 2008 Solution 1(a vs(0.1 =(20 j30 e-2 j50(0.1 =(36.06-56.31o e-0.2 j5 = 36.06e-0.2-56.31o

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