lab09_sol_s08

# lab09_sol_s08 - ECE220 Problem Lab#09 Review for Final Exam...

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ECE220 Problem Lab #09 Review for Final Exam Date: Week of April 21, 2008 The goal of this lab is to review material in preparation for the ﬁnal exam. Make sure you also review the problems we covered in the preparation labs for tests 1, 2, and 3. Work out as many problems as you can. If you have worked on a problem before, you may skip it. Do not worry if you do not ﬁnish all problems in time. ————————————————————————————– 1 Chapter 2 review, signals, their properties and op- erations on signals Question 1. Consider the centered pulse, p τ ( t ), in Equation 1: p τ ( t ) = 1 , - τ/ 2 t < τ/ 2 , 0 , otherwise. (1) Express the signal s ( t ) in Equation 2 in terms of p 1 ( t ), the centered pulse with parameter τ = 1. s ( t ) = 0 , t < 0 , 3 , 0 t < 2 , - 1 , 2 t < 7 , 0 , t 7. (2) Solution: Use time-shifting and time-scaling: 1

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s ( t ) = 3 p 1 ( 1 2 ( t - 1)) - p 1 ( 1 5 ( t - 9 2 )) ————————————————————————————– 2 Chapter 4 review, complex functions Question 2. (10 points) Consider the four complex numbers given by Equation 3: z ( k ) = - ke - j 2 , k = 1 , 2 , 3 , 4 . (3) Accurately place these numbers in the complex plane of ﬁgure 1. Label the four points z (1) ,z (2) ,z (3) and z (4). Solution: See Figure 1. ————————————————————————————– 3 Chapter 5 review, matrices No problems to review in this chapter. ————————————————————————————– 4 Chapter 6 review, systems of linear equations Question 3. Consider the following set of equations: x 1 + x 2 = 0 2 x 1 + x 2 + x 3 = 1 x 1 + bx 3 = 0 b is a real-valued constant. 2
1 2 3 4 30 210 60 240 90 270 120 300 150 330 180 0 z(4) z(2) z(1) z(3) Figure 1: The complex plane. 3

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Determine, if possible, all values for b that will make the system of equations have multiple solutions. Justify your answer. Solution: Consider the matrix B = [ A | b ]: 1 1 0 2 1 1 1 0 b 0 1 0 Subtract 2 times the ﬁrst row from the second row: 1 1 0 0 - 1 1 1 0 b 0 1 0 Subtract the ﬁrst row from the third row: 1 1 0 0 - 1 1 0 - 1 b 0 1 0 Subtract the second row from the third row: 1 1 0 0 - 1 1 0 0 b - 1 0 1 - 1 Looking at the third row, we have the equation ( b - 1) x 3 = - 1. If b = 1, we will have no solution. Any other value for b will cause the system to have only one solution. There are
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lab09_sol_s08 - ECE220 Problem Lab#09 Review for Final Exam...

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