HW8_solution - EEL3105 Fall2011 Homework8 1.

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EEL 3105 Fall 2011 Homework 8 1. Solve the following problems for both of these the differential equations 2 2 2 2 61 6( ) 12 16 ( ) dy d y yu t dt dt d y t dt dt  To get ready to solve the problems below, let us first calculate the roots of the corresponding polynomials: 2 2 6 12 16. and  The roots are 37 j  [labeled as 12 , ] and 10.47, 1.53 [labeled as 34 , ] respectively. Thus, the solution the homogeneous part of the first differential equation [I will label it (i)] has the form: ( 3 2.65) ( 3 2.65) 3 [ cos(2.65 ) sin(2.65 )] jt t Ae Be or eM tN t   where A,B or M,N are unknown constants that depend on the initial conditions. Note that this function is a decaying sinusoid. Please plot it using MATLAB to understand this important point. Similarly, the solution to the homogeneous part of the second differential equation [labeled as (ii)] has the form 10.47 1.53 tt Ae Be This function, by contrast, has no oscillations but simply decays exponentially to zero. The slower time constant 1.53 determines, to a dominant extent, the rate of decay to zero. Again, please plot using MATLAB to get a sense of this function. We are now ready to attack the problems below. A. Suppose y(0)=0, (0) 0 dy dt and u=U(t) where U(t) is the unit step function. Find y(t).
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Solution: I will do this first the Equation (i). Since initial conditions are zero, we can directly try to calculate the solution. Using the general form of solution for step functions, the solution has the form 3 1 ( ) ( ) [ cos(2.65 ) sin(2.65 )] t A yt CU t e M t N t  I am using subscript A to denote that this is solution to Problem A. We will reuse this solution to solve other parts below. In the sequel, I will use subscript to denote solutions to various parts. We need to find the three constants C 1 , M, N. This will be done using two facts: y must satisfy the differential equation and it must satisfy the initial conditions on y and dy/dt. Let’s plug y into the differential equation (i). Note that since U(t) is a constant for all t> 0, its derivatives are all zero for t>0. Moreover, the second term in the solution form for y(t) will render the left hand side zero since it is a solution to the homogeneous equation. Therefore, when we plug y into the Equation (i), we get 1 16 1. C Therefore C 1 =1/16. Let’s now apply the initial conditions: (0) 1/16 0, (0) 3 2.65 0. yM dy MN dt   Thus, M= 1/16 and N= 31 . 1 6 2.65*16 16  . Thus, the solution is 3 11 ( ) [cos(2.65 ) 1.13sin(2.65 )] ( ) 16 16 t A yt e t t U t     . We can repeat this process for Equation (ii). In this case, the general form of solution is 10.47 1.53 1 () , 0 . tt A y CU t Ae Be t  Plugging it into the differential equation yields 1 16 1. C Application of initial conditions yields 1 0 10.47 1.53 0 CA B AB Solving for A and B gives rise to A=0.0107, B= .073. Thus, the solution for differential equation (ii) is 10.47 1.53 1 0.0107 0.073 ( ) 16 A ye e U t   
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B. Suppose y(0) =1, (0) 0 dy dt and u=unit impulse function. Find y(t).
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This note was uploaded on 12/27/2011 for the course EEL 3105 taught by Professor Boykins during the Fall '10 term at University of Florida.

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HW8_solution - EEL3105 Fall2011 Homework8 1.

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