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solutions_test1_section1_s07-3

# solutions_test1_section1_s07-3 - ECE 220 Section 001,Test 1...

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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, 0 , 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 0 2 4 6 8 10 -4 -3 -2 -1 0 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

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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 -∞ δ ( t - 5) · sin(2 π 10 9 ( t - 5) + π/ 2) · u ( t - 10) dt A = 0 Solution: The δ ( · ) signal is shifted to t 0 = 5. Since t 0 is within the limits of integration, we can apply equation 2.20, page 62, with g ( t ) = sin(2 π 10 9 ( t - 5) + π/ 2) · u ( t - 10) . Therefore, we can write A = g (5) = sin(2 π 10 9 (5 - 5) + π/ 2) · u (5 - 10) = sin( π/ 2) · 0 = 0 3

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Problem 3. (10 points) Determine two periods (call them T 1 and T 2 ) of the signal f ( t ) = 3 cos(2 π 1 3 t + π/ 4) + cos(2 π 1 2 t - π/ 4) Justify your answer. Guesses will not count! T 1 = 6 T 2 = 12 Solution: The period, T 0 , of the signal 3 cos(2 π 1 3 t + π/ 4) is T 0 = 3; other periods would be 6, 9, 12, 15, etc. The period, T 00 , of the signal cos(2 π 1 2 t - π/ 4) is T 00 = 2; other periods would be 4, 6, 8, 10, 12, etc. If we had to guess a period of the signal f ( t ), an intuitive starting point would be to look into the set of values T 0 and T 00 can take. Let’s guess then that, T 1 , a period of the signal f ( t ) would be 6, the first one of the common periods. In order to formally check that T 1 is a period of f ( t ), compute f ( t + T 1 ) = 3 cos(2 π 1 3 ( t + T 1 ) + π/ 4) + cos(2 π 1 2 ( t + T 1 ) - π/ 4) = 3 cos(2 π 1 3 ( t + 6) + π/ 4) + cos(2 π 1 2 ( t + 6) - π/ 4) = 3 cos(2 π 1 3 t + 4 π + π/
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solutions_test1_section1_s07-3 - ECE 220 Section 001,Test 1...

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