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hw2sol - ECE220 Signals and Information Spring 2007...

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ECE220 Signals and Information Spring 2007 Homework 2 Solutions Problem 1 Consider the signal x ( t ) = 2 + 4 cos(10 t + 2) - cos(120 t - 2) 1. Sketch the spectrum of this signal as in Figure 3.1 in the textbook. Show your calculations. From Euler’s formula we have x ( t ) = 2 e j · 0 + 2 e j(10 t +2) + e - j(10 t +2) - 1 2 e j(120 t - 2) + e - j(120 t - 2) = 2 e j · 0 + 2 e 2j e j(2 π ) 5 π t + 2 e - 2j e - j(2 π ) 5 π t - 1 2 e - 2j e j(2 π ) 60 π t - 1 2 e 2j e - j(2 π ) 60 π t The above form allows us to sketch the spectrum in the form as in Figure 3.1 in the textbook. The spectrum is shown in Figure 1. -5/ pi 5/ pi - 60/pi 60/pi 2 1 2 exp(-2j) 1 2 exp(2j) 2exp(2j) 2exp(-2j) - - Hz Figure 1. 2. Is this signal periodic? If so, what is its fundamental period? Let us consider the two cosine waves individally. We need to show that there exists T s > 0 such that for all t cos(10 t + 2) = cos(10( t + T s ) + 2) cos(120 t - 2) = cos(120( t + T s ) - 2) As the cosine function is periodic with the period 2 π , we want to find T s such that 10 T s = 2 πk 120 T s = 2 πl for some integers k and l . It is clear that T s = π 5 is the smallest positive number which satisfies the two conditions. Thus T s = π 5 is the fundamental period.
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3. Now consider the signal y ( t ) = x ( t ) + 5 sin(15 t - 1 2 ) and answer questions (1) and (2) for this signal. How do the answers change? Clearly we need to express 5 sin(15 t - 1 2 ) as a linear combination of complex exponentials, 5 sin(15 t - 1 2 ) = 5 2j e j((2 π ) 7 . 5 π t - 1 2 ) - e - j((2 π ) 7 . 5 π t - 1 2 ) = 5 2 e j 3 π - 1 2 e j(2 π ) 7 . 5 π t - 5 2 e - j 3 π - 1 2 e - j(2 π ) 7 . 5 π t Thus y ( t ) = 2 e j · 0 + 2 e 2j e j(2 π ) 5 π t + 2 e - 2j e - j(2 π ) 5 π t - 1 2 e - 2j e j(2 π ) 60 π t - 1 2 e 2j e - j(2 π ) 60 π t + 5 2 e j 3 π - 1 2 e j(2 π ) 7 . 5 π t - 5 2 e - j 3 π - 1 2 e - j(2 π ) 7 . 5 π t and the fundamental period grows to T s ( y ) = 2 π 5 .
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