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EE200_Weber_9-23

# EE200_Weber_9-23 - EE 200 Impulse Response Consider the...

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1 EE 200 Impulse Response Consider the zero-state output response What happens if our input is the Kronecker delta function? n Integers, δ (n) = 1 if n = 0 = 0 if n 0 The resulting output is the impulse response of the system, h(n). y ( n ) = ca n " 1 " m bx ( m ) m = 0 n " 1 # \$ % ( ) + dx ( n )

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2 EE 200 Impulse Response Letting x(n) = δ (n) For n 1, only the first term of the sum (when m=0) is non-zero. This results in h ( n ) = ca n " 1 " m b # ( m ) m = 0 n " 1 \$ % ( ) * + d ( n ) h ( n ) = 0 if n < 0 d if n = 0 ca n " 1 b if n # 1 \$ % &
3 EE 200 Impulse Response For our earlier example of a SISO system where the output was equal to 2/3 of the current input plus 1/3 of the previous input, what is the impulse response? We found that y(n) = cs(n) + dx(n) = (1/3)s(n) + (2/3)x(n) c = 1/3 and d = 2/3. and s(n+1) = as(n) + bx(n) = x(n) a = 0 and b = 1.

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4 EE 200 Impulse Response The impulse response is given by Putting in the values for a, b, c and d gives h ( n ) = 0 if n < 0 d if n = 0 ca n " 1 b if n # 1 \$ % h ( n ) = 0 i f n < 0 2/3 if n = 0 1/3 if n = 1 0 if n > 1 " # \$ \$ % \$ \$ 1 1 2 3 4 h(n) n
5 EE 200 Impulse Response The expression for the impulse response can be put in the equation for the zero-state output response, y(n), to remove the a, b, c and d terms. For the terms inside the summation and we can change the summation to be y ( n ) = ca n " 1 " m bx ( m ) m = 0 n " 1 # \$ % ( ) + dx ( n ) h ( n ) = 0 d ca n " 1 b # \$ % % if n < 0 if n = 0 if n " 1 ca n " 1 " m b = ca ( n " m ) " 1 b = h ( n " m ) y ( n ) = h ( n " m ) x ( m ) m = 0 n " 1 # \$ % ( ) + dx ( n )

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6 EE 200 Impulse Response We can eliminate the dx(n) term by including it in the summation. If we make the summation go one value higher to n, then it includes in the sum. The zero-state output response is then given by The impulse response is ALL the information we need in order to determine the response of the system to ANY input, assuming an initial state of zero. " n # 0 y ( n ) = h ( n \$ m ) x ( m ) m = 0 n % h ( n " n ) x ( n ) = h (0) x ( n ) = dx ( n )
7 EE 200 Convolution Sum Since x(n) = 0 for n < 0, and h(n) = 0 for n < 0, we can make the summation run from - to and have the same result. The other terms now included in the summation are all zero and don’t affect the result. This is called a convolution sum and is written as " n # 0 y ( n ) = h ( n \$ m ) x ( m ) m = \$% % y = h " x

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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. 2
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EE200_Weber_9-23 - EE 200 Impulse Response Consider the...

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