springps09_soln

springps09_soln - 18.03 PS9ab Solutions-Spring 09 1a Given...

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18.03 PS9ab Solutions—Spring 09 1a. Given that L ( e at cos ωt ) = s - a ( s - a ) 2 + ω 2 and L ( e at sin ωt ) = ω ( s - a ) 2 + ω 2 , L - 1 ± 1 s ( s 2 - 6 s + 13) ² = L - 1 ± 1 / 13 s + ( - 1 / 13)( s - 3) + (3 / 13) ( s - 3) 2 + 4 ² = 1 13 - 1 13 e 3 t cos 2 t + 3 26 e 3 t sin 2 t. 1b. Using the ordinary coverup method, 1 ( s + 1)( s 2 + 2 s + 5) = 1 / 4 s + 1 + B ( s + 1) + C ( s + 1) 2 + 4 with B and C to be determined. The roots of ( s + 1) 2 + 4 are - 1 ± 2 i . The complex coverup method with s = - 1 + 2 i yields 1 2 i = B (2 i ) + C = B = - 1 / 4 , C = 0 . Therefore L - 1 ± 1 ( s + 1)( s 2 + 2 s + 5) ² = 1 4 e - t - 1 4 e - t cos 2 t. 2a. 1 k P ( D ) x = e st = W ( s ) = k P ( s ) = 5 / 16 2 s 2 + (1 / 2) s + 5 / 16 . The poles are the roots of P ( s ), namely - 1 / 8 ± (3 / 8) i. 2b. The homogeneous equation P ( D ) x = 0 has independent solutions x = e - t/ 8 cos(3 t/ 8) and x = e - t/ 8 sin(3 t/ 8) both having frequency 3 / 8 . 2c. The system is stable because the roots of P ( s ) have negative real part. It is under- damped because the roots are complex. The pseudofrequency ω 1 = 3 / 8 from above. 1 2de. See Figures 1 and 2. The sketches only have to be qualitatively correct. 2f. P ( ) = - 2 ω 2 + (1 / 2) + 5 / 16 = ⇒ | P ( ) | 2 = ( - 2 ω 2 + 5 / 16) 2 + ( ω/ 2) 2 , which simplifies to 4 ω 4 - ω 2 + 25 / 256. The critical points solve 16 ω 3 - 2 ω = 0, so that ω = 0 , ± 1 / 8. The function is symmetric and the minimum achieved at both
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This note was uploaded on 05/06/2010 for the course 18 18.03 taught by Professor Unknown during the Spring '09 term at MIT.

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springps09_soln - 18.03 PS9ab Solutions-Spring 09 1a Given...

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