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ps8sol

# ps8sol - 18.03 Problem Set 8 Solutions 29[Part I problems 0...

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18.03 Problem Set 8: Solutions 29. [Part I problems: 0] (a) [9] 1 ( s - 1)( s 2 + 4) has poles at s = 1 and at s = ± 2 i . The rightmost pole has real part 1, so the inverse Laplace transform f ( t ) of any function with the same pole diagram grows (for large t ) with exponential rate 1, like e t . More precisely, for large t , | f ( t ) | < e bt for any b > 1, and e at < | f ( t ) | for any a < 1. The numerator of e - s s 2 + 2 s + 2 is never zero and never infinity, so the poles occur at the roots of the denominator, which are - 1 ± i . All functions f ( t ) whose Laplace transform have this pole diagram die off like e - t (that is, more quickly than e - at for any a < 1 and less quickly than e - at for any a > 1), and oscillate with circular frequency 1. The constant function 3 never becomes infinite, so it has no poles; the pole diagram is empty. Functions whose Laplace transform has this pole diagram die off faster than any exponential as t → ∞ . For example any function which is zero for large t has Laplace transform with empty pole diagram. Another example is the function e - t 2 . (b) [9] (i) W ( s ) = L e - t/ 2 sin ( 3 t 2 ) = 3 / 2 ( s + 1 2 ) 2 + 9 4 . p ( s ) = 1 W ( s ) = 2 3 s 2 + s + 5 2 . (ii) The poles occur at the roots of p ( s ), which are at s = - 1 2 ± 3 i 2 .

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