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Unformatted text preview: EE341, Autumn 2007 Problem Set 1 – SOLUTION (updated 8 th October) Reading for this problem set: Chapter 9 of Signals, Systems, and Transforms . 1. The following is the graph of a signal x [ n ].12 2 1 4 312341 6 5 4 2 3 1 Assuming that the signal is zero except as shown, plot the following: (a) 2 x [ n ] (b) 3 x [ n ] (c) 2 x [ n ] 4 solution It may be helpful to first make a table of x [ n ], x [ n ], and x [ n ]: x x [ n ] x [ n ] x [ n ]4312 212 22 44 4 12 22 2 22 31 1 4 1 The signals are now easily plotted: (1 a ) (1 b )12 2 1 4 31234 5 4 2 3 1 6 5 8 734512 2 1 4 323 5 4 2 3 1 6 5 2x[n] 3x[n] (1 c )12 2 1 4 31234 4 2 3 13455678 2x[n]4 2. Sometimes signals can be decomposed into combinations of simple unit sequences, such as this one: y [ n ] = 2 u [ n 1] 2 u [ n 5] 2 u [1 n ] + 2 u [3 n ] Sketch the signal y [ n ], and then plot the following: (a) 2 3 y [ n ] (b) 3 y [ n 2] (c) 1 + 2 y [ 2 + n ] 2 SOLUTION One approach to drawing y [ n ] is to notice that two terms are of the form u [ n ], and two are similar to u [ n ]. Grouping the terms: y [ n ] = 2 u [ n 1] 2 u [ n 5] 2 u [1 n ] + 2 u [3 n ] = 2 · ( u [ n 1] u [ n 5]) + ( 2) · ( u [1 n ] u [3 n ]) = 2 · ( u [ n 1] u [ n 5]) + ( 2) · ( u [ ( n 1)] u [ ( n 3)]) Now it’s a bit clearer that this signal y [ n ] is composed of two timeshifted slices of the unit step function. Both slices are scaled in amplitude by 2, and there’s positive reinforcement where they overlap. The plot of y [ n ] looks like this:1 2 1 312341 6 5 4 2 3 1 4 The three signals are all scaled combinations of y [ n ] and y [ n 2], so you can first make a table of...
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 Spring '08
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 Period, Fundamental frequency, 1, solution i. memoryless

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