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# PS3_sol - EE341 Autumn 2007 Problem Set 3 SOLUTION Reading...

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EE341, Autumn 2007 Problem Set 3 SOLUTION Reading for this problem set: Chapter 10 of Signals, Systems, and Transforms . 1. Show that, for any function g [ n ], g [ n ] * δ [ n ] = g [ n ] solution From the definition of the convolution sum, g [ n ] * δ [ n ] = k = -∞ g [ k ] δ [ n - k ] however δ [ n - k ] = 0 only at k = n, therefore: = . . . + g k - 1 [ n ] + g k [ n ] + g k +1 [ n ] + . . . ...which is the impulse representation of the discrete time signal: = g [ n ] 2. Show that the convolution of three signals can be performed in any order, by showing that: ( f [ n ] * g [ n ]) * h [ n ] = f [ n ] * ( g [ n ] * h [ n ]) solution for ( f [ n ] * g [ n ]) * h [ n ] : for f [ n ] * ( g [ n ] * h [ n ]) : let P [ n ] = f [ n ] * g [ n ] let g [ n ] * h [ n ] = Q [ n ] P [ n ] = k = -∞ f [ k ] g [ n - k ] k = -∞ h [ k ] g [ n - k ] = Q [ n ] now let i = n - k so k = n - i now let i = n - k so k = n - i P [ n ] = i = -∞ g [ i ] f [ n - i ] i = -∞ g [ i ] h [ n - i ] = Q [ n ] so ( f [ n ] * g [ n ]) * h [ n ] = P [ n ] * h [ n ] so f [ n ] * Q [ n ] = f [ n ] * ( g [ n ] * h [ n ]) = k = -∞ h [ k ] P [ n - k ] k = -∞ f [ k ] Q [ n - k ] = 1

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but P [ n - k ] = P [ i ] , but Q [ n - k ] = Q [ i ] , = k = -∞ h [ k ]( g [ i ] f [ n - i ]) k = -∞ f [ k ]( g [ i ] h [ n - i ]) = k = -∞ h [ k ] f [ k ] g [ n - k ] = k = -∞ f [ k ] g [ n - k ] h [ k ] therefore ( f [ n ] * g [ n ]) * h [ n ] = f [ n ] * ( g [ n ] * h [ n ]) 3. Find the system output, y [ n
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