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Assignment 3-Solutions

# Assignment 3-Solutions - ECE3085 Homework#1 Due Jan 28 2009...

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ECE3085 - Homework #1 Due: Jan. 28, 2009 Problem 1. (5 pts) Compute the inverse Laplace transform of F ( s ) = 1 s ( s + 2) 2 Solution 1. The solution will be obtained by doing the partial fraction expansion of the Laplace domain function F ( s ) then regrouping the numerator terms, F ( s ) = A 1 s + B 11 s + B 12 ( s + 2) 2 = A 1 ( s 2 + 4 s + 4) + B 11 s 2 + B 12 s s ( s + 2) 2 = ( A 1 + B 11 ) s 2 + (4 A 1 + B 12 ) + 4 A 1 s ( s + 2) 2 . Comparing the symbolic partial fraction expansion of F ( s ) to the true F ( s ) leads to the following system of equations 0 = A 1 + B 11 0 = 4 A 1 + B 12 1 = 4 A 1 , whose solution is A 1 = 1 / 4 , B 11 = - 1 / 4 , and B 12 = - 1 . Thus, F ( s ) = 1 4 s + - 1 4 s - 1 ( s + 2) 2 = 1 4 s - s + 4 4( s + 2) 2 = 1 4 s - s + 2 + 2 4( s + 2) 2 = 1 4 s - 1 4( s + 2) - 1 2( s + 2) 2 . Note that you could have done it the other way described in class to arrive at the final form I have above, A 1 = sF ( s ) | s =0 = 1 4 A 3 = ( s + 2) 2 F ( s ) vextendsingle vextendsingle s = - 2 = - 1 2 A 2 = d d s vextendsingle vextendsingle vextendsingle vextendsingle s = - 2 ( ( s + 2) 2 F ( s ) ) = - 1 4 . Either way the inverse Laplace transform is 1 4 1( t ) - 1 4 e - 2 t - 1 2 te - 2 t = 1 4 parenleftbigg 1 - e - 2 t - 2 1 2 te - 2 t parenrightbigg 1( t ) . It is acceptable with or without the 1( t ) factor. Problem 2. (5 pts) The DC gain of a system is the steady-state response of the system to a unit step input. In more mathematical terms, that would mean the limit of the time response as t goes to infinity. What is the the DC gain of the system Y ( s ) = G ( s ) U ( s ) = s + 7 ( s + 1)( s 2 + 4) U ( s ) Solution 2. Oh my! As the discussion in class went, this problem was ill-posed. The imaginary poles imply that the system does not have a steady-state response. That means it gets thrown out and everyone gets full credit. Those who stated that it was ill-posed, get a couple points to boot.

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Problem 3. (10 pts) Solve for the following differential equation using the Laplace transform ¨ y + y = t, y (0) = 1 , ˙ y (0) = - 1 . Solution 3. First step is to transform the equation into the Laplace domain, s 2 Y ( s ) - s ˙ y (0) - y (0) + Y ( s ) = 1 s 2 s 2 Y ( s ) - s + 1 + Y ( s ) = 1 s 2 ( s 2 + 1) Y ( s ) = 1 s 2 + s - 1 s 2 ( s 2 + 1) Y ( s ) = 1 + s 3 - s 2 Y ( s ) = s 3 - s 2 +1 s 2 ( s 2 +1) Looking at the equation, it can be simplified further, Y ( s ) = s 3 - 2 s 2 + s 2 + 1 s 2 ( s 2 + 1) = 1 s 2 + s 2 ( s - 2) s 2 ( s 2 + 1) = 1 s 2 + s s 2 + 1 - 2 1 s 2 + 1 .
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