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### Review Problem 8 solution

Course: ME 242, Spring 2008
School: Lehigh
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Word Count: 707

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8.Consider the case where u ( t ) = cos 4t , t 0 and x ( 0 ) = 0 while x ( 0 ) = 0 . Obtain the solution for x ( t ) utilizing the phasor method. Repeat the problem but this time utilize the Laplace transform method. Is the answer the same? Should it be? From Problem 1, X ( s ) = X (s) = U (s) s + 2s + 5 2 U (s) ( s + 2 ) x0 + x0 . As the initial conditions are both zero + 2 s + 2s + 5 s + 2s + 5 2 Phasor Method...

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8.Consider the case where u ( t ) = cos 4t , t 0 and x ( 0 ) = 0 while x ( 0 ) = 0 . Obtain the solution for x ( t ) utilizing the phasor method. Repeat the problem but this time utilize the Laplace transform method. Is the answer the same? Should it be? From Problem 1, X ( s ) = X (s) = U (s) s + 2s + 5 2 U (s) ( s + 2 ) x0 + x0 . As the initial conditions are both zero + 2 s + 2s + 5 s + 2s + 5 2 Phasor Method The phasor method, as shown in Lecture 35, gives the steady state portion of the particular solution when the input is sinusoidal. For this problem, as the initial conditions are zero, the homogeneous solution is zero for all points in time. Thus the phasor solution gives the steady state portion of the total solution. The abrupt start of the input causes a transient solution to occur which is not captured by this method. First find G ( S ) = X (S ) 1 = 2 U ( S ) S + 2S + 5 Recall, for a sinusoidal input, the phasor method gives the general solution, Im G ( j ) , for the case where the input is x p ( t ) = u0 G ( j ) cos ( t + + ) , where tan = Re G ( j ) u0 cos ( t + ) . For Problem 8's sinusoidal inputs S = j and G ( j ) = 1 ( 5 - ) + 2 j 2 . And, for our particular case = 4 , thus G ( j 4 ) = 1 1 1 1 = = while and G ( j 4 ) = -11 + 8 j -11 + 8 j 185 112 + 82 1 -1 8 G ( j 4 ) = = - ( -11 + 8 j ) = - tan . -11 -11 + 8 j 1 8 cos t - tan -1 . 185 -11 As = 0 , and u0 = 1 the steady state solution becomes xss ( t ) = Laplace Transform Method The Laplace Transform Table gives the pair e - at cos t and s+a s . In the case where a = 0 and = 4 the pair becomes pair cos 4t and 2 . 2 2 s + 42 (s + a) + Thus U ( s ) = s s and X ( s ) = 2 . 2 2 s +4 ( s + 4 )( s 2 + 2s + 5) 2 <a href="/keyword/partial-fraction-expansion/" >partial fraction expansion</a> requires the form s as + b cs + d = 2 + 2 . The 2 2 ( s + 4 )( s + 2s + 5) s + 4 s + 2s + 5 2 2 unknown coefficients may be found by multiplying both sides of the expression by the denominator of the left hand side of the previous results to obtain, s = ( as + b ) ( s 2 + 2s + 5 ) + ( cs + d ) ( s 2 + 42 ) . Expanding the expressions and reordering the terms gives s = ( a + c ) s 3 + ( 2a + b + d ) s 2 + ( 5a + 2b + 16c ) s + ( 5b + 16d ) . Equating like terms in the two polynomials in s gives the four equations and four unknowns 0 = a + c , 0 = 2a + b + d , 1 = 5a + 2b + 16c and 0 = 5b + 16d . The first and fourth expressions yield c = -a and d = -5b /16 . Replacing c and d by these equivalences in the second and third expressions 11 11 yields 0 = 2a + b and 1 = -11a + 2b . Solving the first of these to yield a = - b and 16 32 121 + 64 32 substituting these results in the second expression gives 1 = b . Thus b = , 32 185 11 11 10 a=- , c= and d = - . Thus the <a href="/keyword/partial-fraction-expansion/" >partial fraction expansion</a> becomes 185 185 185 1 -11s + 32 11s - 10 X (s) = + 2 . 185 s 2 + 42 s + 2s + 5 We next write these expressions in a more standard form 1 -11s + 8 ( 4 ) 11( s + 1) - 21 X (s) = + and note that the first of these two fractions is 2 185 s 2 + 42 ( s + 1) + 22 associated with the steady state response while the second expression is associated with the transient response. Note that the forcing function caused the transient response. That is if there were no forcing function and the initial conditions were as stated above there would only be the null solution, the transient solution would be zero. Thus the Laplace transform method captures both the transient and steady state solutions caused by the forcing terms whereas the Phasor method only captures the steady state solution. Thus the results are different. But the steady 1 -11s + 8 ( 4 ) state solutions are the same. For the steady state term X ss ( s ) = . 185 s 2 + 42 Conversion from frequency space to time space requires the use of the Laplace Transform table. s Recall earlier we found the pair cos 4t and 2 . Similar methods would result in the pair s + 42 4 1 sin 4t and 2 . Thus xss ( t ) = -11cos ( t ) + 8sin ( t ) . Although these results may 2 s +4 185 look different from those found above they are not as a a cos t + b sin t = a 2 + b 2 cos t - tan -1 . Utilizing this reduction one finds b 1 1 8 8 xss ( t ) = * 121 + 64 cos t - tan -1 cos t - tan -1 = , the same answer. 185 185 -11 -11 Obviously the phaser method is the much easier approach.
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