Lemma 82 suppose that ℜ α then if we set e α t e

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Lemma 8.2. Suppose that α < 0 . Then, if we set e α ( t ) = e αt H ( t ) , we obtain ˆ e α ( λ ) = 1 α . (Some applied mathematicians would leave out the condition α < 0 in the lemma just given and most would write ˆ H ( λ ) = 1 / ( ). The study of Laplace transforms reveals why this reckless behaviour does not lead to disaster.) Lemma 8.3. The solution F = K f of F ′′ ( t ) + ( a + b ) F ( t ) + abF ( t ) = f ( t ) (where a, b > 0 ), subject to the boundary condition F ( t ) , F ( t ) 0 as t → −∞ , is given by K f = K⋆f where K ( t ) = e bt e at a b H ( t ) . 20
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Observe that K ( t ) = 0 for t 0 and so, if f ( t ) = 0 for t 0, we have K f ( t ) = K⋆f ( t ) = 0 for t 0 , K f ( t ) = K⋆f ( t ) = integraldisplay t 0 f ( s ) K ( t s ) ds for t> 0 . Thus K is indeed causal. There is another way of analysising black boxes. Let g n be a sequence of functions such that (i) g n ( t ) 0 for all t , (ii) integraldisplay −∞ g n ( t ) dt = 1, (iii) g n ( t ) = 0 for | t | > 1 /n . In some sense, the g n ‘converge’ towards the ‘idealised impulse function’ δ whose defining property runs as follows. Definition 8.4. If f : R R is a well behaved function then integraldisplay −∞ f ( t ) δ ( t ) dt = f (0) . If the black box is well behaved we expect K f n to converge to some function E . We write K δ = E and say that the response of the black box to the delta function is the ele- mentary solution E . Note that, since our black box is causal, K ( t ) = 0 for t< 0. If f is a ordinary function, we define its translate by some real number a to be f a where f a ( t ) = f ( t a ). In the same way, we define the translate by a of the delta function by a to be δ a where δ a ( t ) = δ ( t a ) or, more formally, by integraldisplay −∞ f ( t ) δ a ( t ) dt = integraldisplay −∞ f ( t ) δ ( t a ) dt = f ( a ) . Since our black box is time invariant, we have K δ a = E a and, since it is linear, K n summationdisplay j =1 λ j δ a j ( t ) = n summationdisplay j =1 λ j E a j ( t ) . 21
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In particular, if F is a well behaved function, K MN summationdisplay j = MN N 1 F ( j/N ) δ j/N ( t ) = MN summationdisplay j = MN N 1 F ( j/N ) E j/N ( t ) = MN summationdisplay j = MN N 1 F ( j/N ) E ( t j/N ) . Crossing our fingers and allowing M and N to tend to infinity, we obtain K F ( t ) = integraldisplay −∞ F ( s ) E ( t s ) ds, so K F = F E. Thus the response of the black box to a signal F is obtained by convolving F with the response of the black box to the delta function. (This is why the acoustics of concert halls are tested by letting off starting pistols.) We now understand the importance of convolution, delta functions and elementary solutions in signal processing and the study of partial differential equations. To see what happens in our specific example, we use Fourier transform methods find the elementary solution of equation . Lemma 8.5. The solution E = K δ of E ′′ ( t ) + ( a + b ) E ( t ) + abE ( t ) = δ ( t ) (where, a, b > 0 ), subject to the boundary condition E ( t ) , E ( t ) 0 as t → −∞ , is given by E ( t ) = e bt e at a b H ( t ) .
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  • Fall '08
  • Groah
  • Math, Analytic function, Q7, Cauchy, Lemma

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