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Unformatted text preview: 18.03 Problem Set 8: Solutions 29. [Part I problems: 0] (a)  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)  (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 ....
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This note was uploaded on 02/06/2009 for the course 18 18.03 taught by Professor Unknown during the Spring '09 term at MIT.
- Spring '09