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EE200_Weber_10-5

# EE200_Weber_10-5 - EE 200 Convolution Sum Example A system...

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8 EE 200 Convolution Sum Example: A system has the impulse response h(n) below and we apply the input x(n) to it. " n # 0 y ( n ) = h ( k ) x ( n \$ k ) k = 0 3 % 2 1 1 2 3 4 h(n) n 2 1 1 2 3 4 x(n) n h (0) = 1 h (1) = 2 h (2) = 2 h (3) = 1 x (0) = 2 x (1) = " 1 x (2) = 1

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9 EE 200 Convolution Sum Evaluate the convolution sum for values of n: y (0) = h ( k ) x (0 " k ) k = 0 3 # = h (0) x (0 " 0) + h (1) x (0 " 1) + h (2) x (0 " 2) + h (3) x (0 " 3) = h (0) x (0) + h (1) x ( " 1) + h (2) x ( " 2) + h (3) x ( " 3) = (1 \$ 2) + (2 \$ 0) + (2 \$ 0) + (1 \$ 0) = 2 + 0 + 0 + 0 = 2 2 1 1 2 3 4 h(k) k 2 1 1 2 3 4 h(k) k y (1) = h ( k ) x (1 " k ) k = 0 3 # = h (0) x (1 " 0) + h (1) x (1 " 1) + h (2) x (1 " 2) + h (3) x (1 " 3) = h (0) x (1) + h (1) x (0) + h (2) x ( " 1) + h (3) x ( " 2) = (1 \$" 1) + (2 \$ 2) + (2 \$ 0) + (1 \$ 0) = " 1 + 4 + 0 + 0 = 3 x(0-k) x(1-k)
10 EE 200 Convolution Sum Evaluate the convolution sum for values of n: y (2) = h ( k ) x (2 " k ) k = 0 3 # = h (0) x (2 " 0) + h (1) x (2 " 1) + h (2) x (2 " 2) + h (3) x (2 " 3) = h (0) x (2) + h (1) x (1) + h (2) x (0) + h (3) x ( " 1) = (1 \$ 1) + (2 \$" 1) + (2 \$ 2) + (1 \$ 0) = 1 " 2 + 4 + 0 = 3 2 1 1 2 3 4 h(k) k 2 1 1 2 3 4 h(k) k y (3) = h ( k ) x (3 " k ) k = 0 3 # = h (0) x (3 " 0) + h (1) x (3 " 1) + h (2) x (3 " 2) + h (3) x (3 " 3) = h (0) x (3) + h (1) x (2) + h (2) x (1) + h (3) x (0) = (1 \$ 0) + (2 \$ 1) + (2 \$" 1) + (1 \$ 2) = 0 + 2 " 2 + 2 = 2 x(2-k) x(3-k)

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11 EE 200 Convolution Sum Evaluate the convolution sum for values of n: y (4) = h ( k ) x (4 " k ) k = 0 3 # = h (0) x (4 " 0) + h (1) x (4 " 1) + h
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