set11 - Lecture 30 Laplace Transforms - Dirac delta...

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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 first-order 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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set11 - Lecture 30 Laplace Transforms - Dirac delta...

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