PS1_soln_fixed

PS1_soln_fixed - EE341 Autumn 2007 Problem Set 1 –...

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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 ].-1-2 2 1 4 3-1-2-3-4-1 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 ]-4-3-1-2 2-1-2 2-2 4-4 4 1-2 2-2 2 2-2 3-1 1 4 1 The signals are now easily plotted: (1 a ) (1 b )-1-2 2 1 4 3-1-2-3-4 5 4 2 3 1 6 5 8 7-3-4-5-1-2 2 1 4 3-2-3 5 4 2 3 1 6 5 2x[-n] 3-x[n] (1 c )-1-2 2 1 4 3-1-2-3-4 4 2 3 1-3-4-5-5-6-7-8 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 time-shifted 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 3-1-2-3-4-1 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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PS1_soln_fixed - EE341 Autumn 2007 Problem Set 1 –...

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