Unformatted text preview: ECE 2704 Homework Number 6 Due at the start of class on Wednesday, 19 March 2008 1. Exercise 4.13 (page 479, a, b, f) Solutions (a) Thus (b) The poles (roots of the denominator polynomial) are complex conjugates, thus we use 10c in Table 4.1 directly. Thus, (f) = Multiple both sides by and collect like terms of , . Equate coefficients for powers of on each side, Thus , and and 2. Consider the differential equation with initial conditions Solutions and . Find the Laplace transform of . Computing the Laplace transform of both sides of the of differential equation yields which is expressed Thus, To get the preferred final form, we multiply both the numerator and denominator by , 3. Exercise 4.35 (page 481, a and c only) Solutions (a) (c) , , 4. Exercise 4.36 (a and b only). Hint: simply do the previous problem in reverse order to arrive at a differential equation. Since we are considering a transfer function, it is assumed that initial conditions are zero. You do not need to be concerned with controllability and observability, as these terms are slightly misused in this problem. The author is trying to let you know that you can assume that there is no cancellation between roots in the numerator (zeros) and roots in the denominator (zeros). Solutions (a) (b) 5. Exercise 4.37 (a only). Solutions Since both inputs are simple transformations of a single unit step, let us compute the output for the input , and We will use the Laplace transform pairs (2 and 10d in Table 4.1), where . For this problem, partial fraction techniques, , using , , and . To find , we can use standard (1) To compute and , we multiple the left and right side of (1) by yields . By equating like powers of on both sides of the equality, Thus and This is the output when the input is (i) For (i) For , , . ...
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This note was uploaded on 03/29/2008 for the course ECE 2074 taught by Professor Stilwell during the Spring '08 term at Virginia Tech.
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