EE110_W08_HW 4 Solution - Solution to Homework 4 EE 110...

Info icon This preview shows pages 1–4. Sign up to view the full content.

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 =
Image of page 1

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

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 δ δ Solution 4. F ( s ) = 3 2 1 2 4 5 2 π - + + + + s s s s = 2 1 2 ( 2)( 1) π - + + + s s s = 2 1 2 ( 2) ( 1) ( 1) a b c - + + + + + + s s s s where 2 2 ( 1) a π π = - = = + s s 1 ( 2) b π π = - = = + s s and ( 29 ( 29 1 1 2 2 2 1 1 = = = ( 2) ( 2) ( 2) 1 d d c ds ds π π π π = - = - = - = + - - + + + + s s s s s s s s Thus, we may write f ( t ) = 2 δ ( t ) – u ( t ) + π e –2 t u ( t ) + π te t u ( t ) – π e t u ( t )
Image of page 2
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 ) = 2 2 18 32 15 3 5 ( 1) ( 3) + - + - + + s s s s s s + 1 = ( 29 2 2 6 1 3 1 A B C + + + + + + + s s s s where A = –1/2, B = 9/4, and C = –81/4.
Image of page 3

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

Image of page 4
This is the end of the preview. Sign up to access the rest of the document.
  • Winter '08
  • Gupta
  • Cos, lim, ribosomal RNA

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern