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Unformatted text preview: ECE 301 Detailed Solutions  Set 2: Impulse Response 1 21. Let a linear system with input x t and output y t be described by the differential equation d 2 dt 2 y t 7 d dt y t 12 y t = 6 x t (a) Compute the mathematical form of the impulse response h t for this system. (b) Plot h t over the range 1 t 2. DETAILED SOLUTION: (a) Compute the mathematical form of the impulse response h t . First find the transfer function H s and then compute the inverse Laplace transform h t . To do this, take the Laplace transform of the differential equation in the problem statement : s 2 Y s 7 sY s 12 Y s = 6 X s a If x t equals the impulse δ t , then the output y t is by definition the impulse response : If x t = δ t ; then y t = h t b Note that b gives the following sdomain result as well : If x t = δ t ; then Y s = H s c If x t equals the impulse, then X s is the Laplace transform of the impulse, which has previously been found, and is repeated below for reference : x t = δ t X s = 1 d Therefore, substitute Y s = H s from c and X s = 1 from d into equation a , obtaining s 2 H s 7 sH s 12 H s = 6 e Solving e for H s then gives H s = 6 s 2 7 s 12 = 6 s 3 s 4 f We can expand the far right term in f using partial fraction expansion. (You can see the Review Notes on Laplace Transforms for a worked partial fraction example). The partial fraction expansion produces 6 s 3 s 4 = 6 h 1 s 3 1 s 4 i g Substituting g into f thus gives H s = 6 h 1 s 3 1 s 4 i h Finally, taking the inverse transform of h produces the desired answer for the impulse response: h t = 6 h e 3 t e 4 t i u t j ECE 301 Detailed Solutions  Set 2: Impulse Response 2 21 (cont.) (b) Plot h t over the range 1 t 2 . Here is an easy way to quickly plot the function in equation j : ¯ Plot the first exponential 6 e 3 t over approximately 4 time constants. This is a plot from t = 0 out to t = 4 ¢ : 333 = 1 : 333 seconds, shown in the top dashed graph in the figure below. ¯ On the same graph and using the same scale, plot the second exponential 6 e 4 t over approximately 4 time constants. This is a plot from t = 0 out to t = 4 ¢ : 25 = 1 : 0 seconds, shown in the bottom dashed graph in the figure below. t 6 4 2 0.33 t= t = 0642 e1 6 = 2.21 0.25 t= 1.00 t= e16 = 2.21 e4t 6 e3t 6 Make sure you use the appropriate time constant for each exponential. Once you have plotted both functions, now “visually add them” to produce the final plot for g t . This is shown on the next page. ECE 301 Detailed Solutions  Set 2: Impulse Response 3 21 (cont.) The upper dashed curve in the figure below is the plot of 6 e 3 t and the lower dashed curve is the plot of the function 6 e 4 t . The solid line is the sum, which is the solution curve for h t in equation...
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This note was uploaded on 01/25/2010 for the course ECE 301 taught by Professor Hoozturk during the Spring '08 term at N.C. State.
 Spring '08
 Hoozturk

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