MidtermSol-ECE205-F16.pdf - University of Waterloo ECE 205...

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Unformatted text preview: University of Waterloo ECE 205 – Fall 2016 Midterm Exam, Solutions Friday October 28, 10:00-11:20 a.m. Name (print): ________________________________________________________ ID Number: ____________________________________ Signature: _____________________________________ Instructor: Shahla Aliakbari Instructions: Use the back of the preceding page if more space is required. Marking Scheme: Question 1 Marks available 10 2 10 3 10 4 10 5 10 Total 50 Marks Awarded 1. A source voltage V (t ) t is connected in series with a resistance R 10 and an inductance L 0 .1 H . The differential equation for the current in the circuit, i (t ) , is given by: di 0.1 10 i t dt Find the current i (t ) , assuming that i (0 ) 0 . Solutions: We divide both sides of the DE by 0.1, 100 10 The integrating factor is: Multiplying both sides of DE by this integrating factor, we have 100 10 The left hand side can be written as the derivative of the integrating factor times the dependent variable: 10 Integrating from both sides, we have 10 The integral can be evaluated using the method of integration by parts, 1 10 1 1000 We obtain, 1 10 1 1000 Applying the initial condition, we have: 0 This gives 0 1 0 10 1 1000 0.001. 2. Solve the differential equation x t dx t x 2 dt 0 . [Hint: This equation becomes exact if you multiply both sides by t ]. Solutions: We multiply both sides of the DE by , 0 We check if the equation is exact, 2 This is an exact equation, so the general solution is , . We have: Integrating these equations: 2 , 2 , 3 2 We set these two equations equal: 2 3 → 2 3 So the solution is: , → 3 2 3 2 3 2 3. Consider the differential equation x x cos( t ) . (a) Find the general solution subject to the initial conditions x(0) 0 and x(0) 0 . Solutions: First, we find the complementary solution to the homogeneous equation. The characteristic equation is: 1 0 → We have two complex roots, so the complementary solution is: cos sin To obtain a particular solution, we use the trial solution cosω sinω Taking derivatives and substituting into the DE: ωsinω " ω cosω 1 ωcosω ω cosω ω sinω ω cosω Then we obtain: ω sinω cosω 1 sinω ω sinω 1 , 1 ω So a particular solution is: cosω cosω 0 1 cosω 1 ω The general solution is the sum of the complementary solution and the and particular solution: cos Applying the initial conditions, 0 0 → cos0 sin0 sin 1 cosω0 1 ω 1 cosω 1 ω 0 → 1 1 ω 3 0 0 → sin0 Thus, we have: cos0 1 cos 1 ω 1 ω sin 0 1 cos 1 ω 0 → 0 1 (b) Describe the solution of the differential equation for 1 . 1, we have beats. Solutions: If 1 1 Using the formula cos a cos b 2 sin b a sin b a , the solution (1) can be also 2 2 written as: 2 1 1 sin 1 t sin 1 t 2 1 2 2 x(t ) Note that for the case → 1 the solution is t sin t/2. This can be obtained by trying the particular solution cos sin , or by taking the limit of the solution (1) as → 1 (using the L’Hopital’s rule). 4. (a) Find two linearly independent solutions to the differential equation y 2 y y 0 . Show that your solutions are indeed linearly independent. Solutions: First we find the solution to the homogeneous equation. The characteristic equation is: 2 1 0 → 1 0 → 1 We have a repeated root, so the solution is: x The solutions and are linearly independent if the Wronskian determinant is not equal to zero: 0 So these two solutions are indeed linearly independent. (b) Find the general solution to y 2 y y 2 e x . Solutions: From part (a), we obtained the complementary solution: x To find a particular solution, we use Taking the derivatives and substituting into the DE: 2 " 2 2 4 2 2 4 2 2 2 2 4 2 2 → 1 So a particular solution is: The general solution is: x 5. (a) Find L t cos( 2t ) 3e t . Solutions: cos 2t cos 2t 3 cos 2t 3 4 cos 2t 4 2 4 4 4 Using the formula: f t 1 3 cos 2t 1 3 1 4 4 3 3 1 (b) Find L1 tan 1 s . Solutions: We know that 1 tan 1 Also, we have (n = 1) 1 Setting F s tan , we have 1 tan 1 Taking the inverse of Laplace transform, 1 1 sin sin 5 Formula Sheet Table of integrals dx x C e x n x dx dx e x C cos( x)dx sin( x) C 2 sec ( x)dx tan( x) C dx ln | x | C x sin( x)dx cos( x) C 2 csc ( x)dx cot( x) C dx x cos 1 ( ) C a a2 x2 dx 1 1 x a 2 x 2 a tan ( a ) C x n 1 C,n 1 n 1 dx x sin 1 ( ) C a a2 x2 x a dx ax C (a 0,a 1) ln a Table of Laplace Transforms L{e f (t )} F ( s a) at d n F ( s) , L{t f (t )} (1) ds n n n L{ f (t )} sF ( s) f (0) 6 ...
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