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Unformatted text preview: Lecture 30 Laplace Transforms  Dirac delta function and its uses Periodic application of Dirac delta function Let us once again return to our substance X problem. Left alone in a beaker, substance X decays according to the rate law dx dt = kx, (1) where x ( t ) represents the concentration of X . We then saw how the instantaneous addition of an amount A to the beaker at time t = a can be represented by the Dirac delta function the resulting evolution equation for x ( t ) becomes dx dt = kx + A ( t a ) , x (0) = x . (2) The solution to this problem was obtained using Laplace transforms: x ( t ) = x e kt + Ae k ( t a ) H ( t a ) . (3) As a side remark, the solution to the DE in (2) can also be found using the method of integrating factors for firstorder linear DEs. We leave this as an exercise for the reader. Not surprisingly, the above solution x ( t ) 0 as t . Unless we continue to add more X to the beaker, the concentration of X will decay to zero. So let us consider adding A units of X to the beaker at regular intervals, say at times t n = nT for n = 1 , 2 , , where T > is the time interval between additions. There is now no opportunity for x ( t ) to decay to zero. We wonder, however, whether is is possible that x ( t ) approaches some kind of equilibrium situation. Obviously this equilibrium situation could not be a constant solution, since we are adding an amount A at each time step t n = nT . But after each amount A is added, the concentration x ( t ) decreases. Is it possible that after A is added, and we have an amount y in the beaker, the system will decrease to an amount y A after time T so that when an amount A is added, we start with y again, only to repeat the cycle? The situation is pictured below. We can easily solve for the unknown y for which such a periodic solution would exist. If, at the start of each cycle, the concentration of X at a time t n = nT is y , then at time 1 T 2 T 3 T t y y A t n +1 = ( n + 1) T , i.e. after a time lapse T , the concentration in the beaker will be ye kT . Periodicity of the solution requires that ye kT + A = y, (4) which is satisfied by y = A 1 e kT . (5) Our periodic solution would then be given as follows: For n = 1 , 2 , , x ( t ) = A 1 e kT e k ( t nT ) , for nT t < ( n 1) T. (6) Let us see if this periodic solution is in fact approached as t . We can model this regular addition of X by means of an infinite sum of Dirac delta functions that are positioned at the times t n = nT , i.e., dx dt = kx + f ( t ) , (7) where f ( t ) = summationdisplay n =1 A ( t nT ) . (8) We can solve this DE using LTs in the same way as we did for the DE in (2). Just to recall, take LTs to obtain sX ( s ) x = kX ( s ) + F ( s ) , (9) where F ( s ) = L [ f ]. Then solve for X ( s ): X ( s ) = x k + s + F ( s ) k + s . (10) 2 Then take inverse LTs to give x ( t ) = x e kt + ( g * f )( t ) (11) = x h ( t...
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 Spring '08
 SivabalSivaloganathan

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