solutions_test1_section1_s07-3

solutions_test1_section1_s07-3 - ECE 220, Section 001,Test...

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Unformatted text preview: ECE 220, Section 001,Test 1 Solutions 15 February 2007 Problem 1. (15 points) Consider the graph of the signal s ( t ) shown in figure 1. The graph is accurate. Write down a mathematical expression for this signal, using (time-shifted and scaled, if need be) versions of the signal v ( t ), defined as v ( t ) = 1 ,- 3 t < 3, , otherwise. s ( t ) = 2 v (2 t + 9) + 3 v (3 t + 6) + v (3 t- 6)- 2 v (2 t- 9)-10-8-6-4-2 2 4 6 8 10-4-3-2-1 1 2 3 4 time t The signal s(t) s(t) 1 3-1-3 Figure 1: The signal s ( t ) in problem 1. 1 Solution: The signal v ( t ) is shown in figure 6, page 18. It has a width of 6 time units. Observe that there are four pulses in the signal s ( t ), shown in figures 7 through 10 (at the end of this document) . To express the signal s 1 ( t ) in terms of v ( t ), we need to: time-scale the signal v ( t ) by a factor of 2, to compress its width to 3 time units; we can then write y ( t ) = v (2 t ) time-shift the signal y ( t ) by 4.5 time units, to align the rising edges of the pulse; we can then write q ( t ) = y ( t + 4 . 5) = v (2( t + 4 . 5) = v (2 t + 9) amplify the signal q ( t ) by a factor of 2, to match the amplitudes; we can then write s 1 ( t ) = 2 q ( t ) = 2 v (2 t + 9) In a similar fashion, then the following three signals would match the pulses in t [- 3 ,- 1], t [1 , 3], and t [3 , 6] respectively: s 2 ( t ) = 3 v (3 t + 6) s 3 ( t ) = v (3 t- 6) s 4 ( t ) =- 2 v (2 t- 9) We can then write s ( t ) = s 1 ( t ) + s 2 ( t ) + s 3 ( t ) + s 4 ( t ) = 2 v (2 t + 9) + 3 v (3 t + 6) + v (3 t- 6)- 2 v (2 t- 9) 2 Problem 2. (10 points) Calculate the integral A = Z - ( t- 5) sin(2 10 9 ( t- 5) + / 2) u ( t- 10) dt A = 0 Solution: The ( ) signal is shifted to t = 5. Since t is within the limits of integration, we can apply equation 2.20, page 62, with g ( t ) = sin(2 10 9 ( t- 5) + / 2) u ( t- 10) . Therefore, we can write A = g (5) = sin(2 10 9 (5- 5) + / 2) u (5- 10) = sin( / 2) = 0 3 Problem 3. (10 points) Determine two periods (call them T 1 and T 2 ) of the signal f ( t ) = 3cos(2 1 3 t + / 4) + cos(2 1 2 t- / 4) Justify your answer. Guesses will not count! T 1 = 6 T 2 = 12 Solution: The period, T , of the signal 3cos(2 1 3 t + / 4) is T = 3; other periods would be 6, 9, 12, 15, etc. The period, T 00 , of the signal cos(2 1 2 t- / 4) is T 00 = 2; other periods would be 4, 6, 8, 10, 12, etc....
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This note was uploaded on 08/24/2009 for the course ECE 220 taught by Professor Nilson during the Summer '08 term at N.C. State.

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solutions_test1_section1_s07-3 - ECE 220, Section 001,Test...

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