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SolSet5 - EE 562a Homework Solutions 5 1 1 Well denote a...

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EE 562a Homework Solutions 5 October 27, 2009 1 1. We’ll denote a point in the abstract space of deterministic sequences (i.e. discrete time signals) by ˜ x = { x ( n ) : n ∈ T } and ˜ y = { y ( n ) : n ∈ T } , where the index set is T = Z , the set of integers. The sequence represented by ˜ y is the output of the system G when ˜ x is the input (i.e. ˜ y = G ˜ x ) if and only if ay ( n ) + by ( n - 1) + cy ( n - 2) = αx ( n ) + βx ( n - 1) . (a) To obtain the block diagram from the difference equation, it is useful to express the output in terms of shifted versions of the input and previous outputs. y ( n ) = 1 a [( αx ( n ) + βx ( n - 1)) - ( by ( n - 1) + cy ( n - 2))] . The realization follows from this equation. x ( n ) fi fl 1/ a c b z -1 z -1 z -1 z -1 = unit delay device realization of system zeros realization of system poles + - + - + + y ( n ) It is possible to realize this system with fewer delay devices - can you see how? (b) & (c) It is actually easier to do (c) first. ˜ y = G ˜ x ⇐⇒ a T 0 ˜ y + b T - 1 ˜ y + c T - 2 ˜ y = α T 0 ˜ x + β T - 1 ˜ x. Using the fact that T k = ( T 1 ) k ( k unit shifts is the same as one k shift), the solution for (c) is a ( T 1 ) 0 ˜ y + b ( T 1 ) - 1 ˜ y + c ( T 1 ) - 2 ˜ y = α ( T 1 ) 0 ˜ x + β ( T 1 ) - 1 ˜ x. Now we can use the linearity of the shift operator to check that G is linear. You can check homogeneity and additivity separately or just check superposition. Let’s check the superposition property. We want to show that ˜ y 1 = G ˜ x 1 , ˜ y 2 = G ˜ x 2 γ 1 ˜ y 1 + γ 2 ˜ y 2 = G ( γ 1 ˜ x 1 + γ 2 ˜ x 2 ) , where ˜ x i is an arbitrary signal in the space and γ i is an arbitrary scalar for i = 1 , 2. Now we’ll claim that G has the superposition property, γ 1 ˜ y 1 + γ 2 ˜ y 2 = G ( γ 1 ˜ x 1 + γ 2 ˜ x 2 ), and show that this is indeed true. G has this superposition property if and only if a T 0 ( γ 1 ˜ y 1 + γ 2 ˜ y 2 ) + b T - 1 ( γ 1 ˜ y 1 + γ 2 ˜ y 2 ) + c T - 2 ( γ 1 ˜ y 1 + γ 2 ˜ y 2 ) = α T 0 ( γ 1 ˜ x 1 + γ 2 ˜ x 2 ) + β T - 1 ( γ 1 ˜ x 1 + γ 2 ˜ x 2 ) .
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2 EE 562a Homework Solutions 5 October 27, 2009 Since T k is linear, we know that it has the superposition property. T k ( γ 1 ˜ x 1 + γ 2 ˜ x 2 ) = γ 1 T k ˜ x 1 + γ 1 T k ˜ x 2 . Applying this to both sides of the input/output relation yields γ 1 ( a T 0 ˜ y 1 + b T - 1 ˜ y 1 + c T - 2 ˜ y 1 ) + γ 2 ( a T 0 ˜ y 2 + b T - 1 ˜ y 2 + c T - 2 ˜ y 2 ) = γ 1 ( α T 0 ˜ x 1 + β T - 1 ˜ x 1 ) + γ 2 ( α T 0 ˜ x 2 + β T - 1 ˜ x 2 ) . So G obeys the superposition principle if and only if the above equation is true. This equation is true since we know that, by definition ˜ y 1 = G ˜ x 1 , ˜ y 2 = G ˜ x 2 a T 0 ˜ y i + b T - 1 ˜ y i + c T - 2 ˜ y i = α T 0 ˜ x i + β T - 1 ˜ x i , for i = 1 , 2. It follows that G has the superposition property and therefore is linear. Notice that we have also proven homogeneity (i.e. take γ 2 = 0) and additivity (i.e. take γ 1 = γ 2 = 1). To show G is time-invariant we need to show that it commutes with the shift operator: GT k = T k G . This occurs ⇐⇒ ˜ y = G ˜ x implies that G ( T k ˜ x ) = T k G ˜ x = T k ˜ y . Again we will we will assume this is true and show that it is consistent. So we claim that GT k ˜ x = T k ˜ y when G ˜ x = ˜ y , or equivalently a T 0 ( T k ˜ y ) + b T - 1 ( T k ˜ y ) + c T - 2 ( T k ˜ y ) = α T 0 ( T k ˜ x ) + β T - 1 ( T k ˜ x ) .
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