# Chap02 - Chapter 2 Time-Domain Representation of Linear...

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Chapter 2: Time-Domain Representation of Linear Time-Invariant Systems H. F. Francis Lu 2011 Spring: Signal and Systems Ver. 2011.03.06 H. F. Francis Lu 1 / 93
Sec. 2.2 Convolution Sum Sec 2.3 Convolution Sum Evaluation Procedure 2011 Spring: Signal and Systems Ver. 2011.03.06 H. F. Francis Lu 2 / 93
Definition 1 (Discrete Convolution Sum) Given two discrete time signals x [ n ] and y [ n ] , their convolution sum is defined as ( x y )[ n ] = x [ n ] y [ n ] = k = -∞ x [ k ] y [ n - k ] 2011 Spring: Signal and Systems Ver. 2011.03.06 H. F. Francis Lu 3 / 93
Polynomial View of Convolution Sum Given two discrete time signals x [ n ] and y [ n ] , we define x ( ζ ) = n = -∞ x [ n ] ζ n and y ( ζ ) = n = -∞ y [ n ] ζ n For example x [ n ] = δ [ n ] + 2 δ [ n - 1 ] x ( ζ ) = 1 + 2 ζ y [ n ] = δ [ n - 1 ] + 3 δ [ n - 2 ] y ( ζ ) = ζ + 3 ζ 2 Clearly we have x ( ζ ) y ( ζ ) = ( 1 + 2 ζ ) ζ + 3 ζ 2 = ζ + 5 ζ 2 + 6 ζ 3 On the other hand, it is easy to show ( x y )[ n ] = δ [ n - 1 ] + 5 δ [ n - 2 ] + 6 δ [ n - 3 ] we see ( x y )( ζ ) = x ( ζ ) y ( ζ ) 2011 Spring: Signal and Systems Ver. 2011.03.06 H. F. Francis Lu 4 / 93
Convolution sum is just Polynomial Multiplication Theorem 1 (Polynomial Interpretation of Convolution Sum) ( x y )( ζ ) = x ( ζ ) y ( ζ ) Proof. ( x y )( ζ ) = n = -∞ ( x y )[ n ] ζ n = n = -∞ k = -∞ x [ k ] y [ n - k ] ζ n = n = -∞ k = -∞ x [ k ] y [ n - k ] ζ k ζ n - k = k = -∞ x [ k ] ζ k n = -∞ y [ n - k ] ζ n - k = x ( ζ ) y ( ζ ) 2011 Spring: Signal and Systems Ver. 2011.03.06 H. F. Francis Lu 5 / 93
Example 1 (cf. Example 2.2 of the book) Let x [ n ] = u [ n ] and y [ n ] = 3 4 n u [ n ] . Compute ( x y )[ n ] . Sol. x ( ζ ) = n = -∞ u [ n ] ζ n = n = 0 ζ n = 1 1 - ζ y ( ζ ) = n = -∞ 3 4 n ζ n u [ n ] = n = 0 3 ζ 4 n = 1 1 - 3 ζ 4 Using Partial Fractions we can show x ( ζ ) y ( ζ ) = 1 ( 1 - ζ )( 1 - 3 ζ 4 ) = 4 1 - ζ - 3 1 - 3 ζ 4 = 4 n = -∞ u [ n ] ζ n - 3 n = -∞ 3 4 n ζ n u [ n ] 2011 Spring: Signal and Systems Ver. 2011.03.06 H. F. Francis Lu 6 / 93
x ( ζ ) y ( ζ ) = 4 n = -∞ u [ n ] ζ n - 3 n = -∞ 3 4 n ζ n u [ n ] Hence ( x y )[ n ] = 4 - 3 3 4 n u [ n ] For example: ( x y )[ 5 ] = 4 - 3 3 4 5 u [ 5 ] = 3 . 288 ( x y )[ 10 ] = 4 - 3 3 4 10 u [ 10 ] = 3 . 831 2011 Spring: Signal and Systems Ver. 2011.03.06 H. F. Francis Lu 7 / 93
Compute by Multiplication Worktable Example 2 x [ n ] = 1 , n = 0 2 , n = 1 0 , otherwise y [ n ] = 1 , n = 0 2 , n = 1 3 , n = 2 0 , otherwise 2011 Spring: Signal and Systems Ver. 2011.03.06 H. F. Francis Lu 8 / 93
Sol. 1 2 × 1 2 3 1 2 2 4 3 6 1 4 7 6 Hence ( x y )[ n ] = 1 , n = 0 4 , n = 1 7 , n = 2 6 , n = 3 0 , otherwise 2011 Spring: Signal and Systems Ver. 2011.03.06 H. F. Francis Lu 9 / 93
Example 3 (cf. Example 2.1 of the book) x [ n ] = 1 , n = 0 1 2 , n = 1 0 , otherwise y [ n ] = 2 , n = 0 4 , n = 1 - 2 , n = 2 0 , otherwise 2011 Spring: Signal and Systems Ver. 2011.03.06 H. F. Francis Lu 10 / 93
Sol. 1 1 2 × 2 4 - 2 2 1 4 2 - 2 - 1 2 5 0 - 1 Hence ( x y )[ n ] = 2 , n = 0 5 , n = 1 0 , n = 3 - 1 , n = 4 0 , otherwise 2011 Spring: Signal and Systems Ver. 2011.03.06 H. F. Francis Lu 11 / 93
Multiplication Notation 1 1 2 × 2 4 - 2 2 1 4 2 - 2 - 1 2 5 0 - 1 For simplicity we will write the above tabular multiplication as 1 , 1 2 ( 2 , 4 , - 2 ) = ( 2 , 5 , 0 , - 1 ) where denotes the place of time n = 0. 2011 Spring: Signal and Systems Ver. 2011.03.06 H. F. Francis Lu 12 / 93
Discrete Time Linear Time Invariant System Definition 2 (Discrete Time Linear Time Invariant System) Let H { } be a discrete time system. We say H is a Discrete Time Linear Time Invariant (DT-LTI) System if 1 H is linear H { ax 1 [ n ] + bx 2 [ n ]} = aH { x 1 [ n ]} + bH { x 2 [ n ]} 2 H is time invariant y [ n ] = H { x [ n ]} y [ n - n 0 ] = H { x [ n - n 0 ]} 2011 Spring: Signal and Systems Ver. 2011.03.06 H. F. Francis Lu 13 / 93
Impulse Response Definition 3 Let H be a DT-LTI system. Then the impulse response of H is h [ n ] = H { δ [ n ]} where δ [ n ] is the Kronecker delta.