2_HW2Solutions - EE 211A Fall Quarter 2011 Instructor John...

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1 EE 211A Digital Image Processing I Fall Quarter, 2011 Handout 10 Instructor: John Villasenor Homework 2 Solutions Problems: 1. Textbook Problem 2.4a, b, and f. Let ) , ( 1 n m y and ) , ( 2 n m y be the outputs corresponding to the inputs ) , ( 1 n m x and ). , ( 2 n m x (a) 9 ) , ( 3 ) , ( + = n m x n m y ) , ( 9 ) , ( 3 )] , ( [ 1 1 1 n m y n m x n m x T = + = ) , ( 9 ) , ( 3 )] , ( [ 2 2 2 n m y n m x n m x T = + = ) , ( 9 )) , ( ) , ( ( 3 ] [ 3 2 1 2 1 n m y n m bx n m ax bx ax T = + + = + Since 2 1 3 by ay y + , the system is non-linear. ) , ( 9 ) , ( 3 )] , ( [ 4 1 1 n m y n n m m x n n m m x T = + ʹ′ ʹ′ = ʹ′ ʹ′ Since ) , ( ) , ( 1 4 n n m m y n m y ʹ′ ʹ′ = , the system is shift-invariant. IR = T 9 ) , ( 3 )] , ( [ + = b n a m b n a m δ δ The system is IIR, since the IR is non-zero for all m , n . System (a) is non-linear, shift-invariant, and IIR. (b) ). , ( ) , ( 2 2 n m x n m n m y = ) , ( ) , ( )] , ( [ 1 1 2 2 1 n m y n m x n m n m x T = = ) , ( ) , ( )] , ( [ 2 2 2 2 2 n m y n m x n m n m x T = = ) , ( )) , ( ) , ( ( ] [ 3 2 1 2 2 2 1 n m y n m bx n m ax n m bx ax T = + = + Since 2 1 3 by ay y + = , the system is linear. ) , ( ) , ( )] , ( [ 4 1 2 2 1 n m y n n m m x n m n n m m x T = ʹ′ ʹ′ = ʹ′ ʹ′ Since ) , ( ) , ( 1 4 n n m m y n m y ʹ′
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