sol3 - ECE 351, Systems I OSU, Winter 2009 Solutions -...

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ECE 351, Systems I Feb. 11, 2009 OSU, Winter 2009 Solutions - Problem Set 3 Problem 1 1
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Problem 2 0 1 2 0 1 2 1 -1 1 -1 x(t) h(t) = x(2-t) t * t y ( t ) = x ( t ) * v ( t ) = 0 , t < 0 R t 0 1 · - 1 = - t, 0 6 t < 1 R t - 1 0 1 · 1 + R 1 t - 1 1 · - 1 + R t 1 - 1 · - 1 = 3 t - 4 , 1 6 t < 2 R 1 t - 2 1 · 1 + R t - 1 1 1 · - 1 + R 2 t - 1 - 1 · - 1 = - 3 t + 8 , 2 6 t < 3 R 2 t - 2 1 · - 1 = t - 4 , 3 6 t < 4 0 , t > 4 The graph of this convolution is given below. 0 0.5 1 1.5 2 2.5 3 3.5 4 -1 -0.5 0 0.5 1 1.5 2 t y(t) x(t) * h(t) Note that the peak in y ( t ) occurs at time t = 2 when the overlap between x ( t ) and h ( t ) = x (2 - t ) is maximum and the two signals are “matched” to each other, i.e., they have the same shape. This technique is used in a communications receiver to “identify” what shape of signal was 2
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transmitted. If at the transmitter the information to be transmitted is embedded in the shape of the signal, then at the receiver the transmitted information can be identiFed by applying this method. Problem 3 (a) ±irst, write x ( t ) as a sum of complex exponentials, reading from the Fgure left-to-right (note 45 degrees is π/ 4 radians): x ( t ) = 5 e - j 2 π 1050 t e - jπ/ 4 + 5 e - j 2 π 950 t e jπ/ 4 + 5 e j 2 π 950 t e - jπ/ 4 + 5 e j 2 π 1050 t e jπ/ 4 = 10 cos(2 π 950 t - π/ 4) + 10 cos(2 π 1050 t + π/ 4) . In the second line, we have used a form of Euler’s identity:
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sol3 - ECE 351, Systems I OSU, Winter 2009 Solutions -...

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