solut3 - # = L-1 " 4 e-3( s-2) e 6 ( s-2) 2 +...

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Some solutions for Homework 3 3.6: 6. The answer is - 6 + 6 e 5 t - 30 t - 75 t 2 - 125 t 3 2750 . 3.6: 10. We have Y ( s ) = F ( s ) ( s + 4)( s + 6) + 10 + s ( s + 4)( s + 6) . L - 1 " 1 ( s + 4)( s + 6) # = - 1 2 e - 6 t + 1 2 e - 4 t . Next, 10 + s ( s + 4)( s + 6) = 3 s + 4 - 2 s + 6 , and L - 1 " s + 10 ( s + 4)( s + 6) # = 3 e - 4 t - 2 e - 6 t . Therefore, we get y ( t ) = f * ( - 1 2 e - 6 t + 1 2 e - 4 t ) + 3 e - 4 t - 2 e - 6 t . 3.4: 18. The equation is f = - t + f * sin t. Therefore, F = - 1 s 2 + F µ 1 s 2 + 1 . F ( s ) = - s 2 + 1 s 4 = - 1 s 2 - 1 s 4 . f ( t ) = - t - t 3 / 6 . 1
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3.5: 2. (13 - 4 s + s 2 ) Y = 4 e - 3 s . Y ( s ) = 4 e - 3 s s 2 - 4 s + 13 = 4 e - 3 s ( s - 2) 2 + 9 . L - 1 " 4 e - 3 s ( s - 2) 2 + 9
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Unformatted text preview: # = L-1 " 4 e-3( s-2) e 6 ( s-2) 2 + 9 # = 4 e 6 e 2 t L-1 " e-3 s 2 2 + 9 # , by frst shiFting theorem. Also, L-1 " e-3 s 2 2 + 9 # = 1 3 sin[3( t-3)] H ( t-3) , by second shiFting theorem. ThereFore, y ( t ) = 4 e 6+2 t 3 sin[3( t-3)] H ( t-3) . 2...
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solut3 - # = L-1 " 4 e-3( s-2) e 6 ( s-2) 2 +...

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