102_1_lecture6

102_1_lecture6 - UCLA Fall 2011 Systems and Signals Lecture...

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UCLA Fall 2011 Systems and Signals Lecture 6: Convolution October 12, 2011 EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 1
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Agenda Homework 2 due in class at 10:00 am. During grace period (after class and before 5:00 pm): TA office hours room (Engineering IV, 67-112) EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 2
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Convolution Evaluation and Properties Today’s topics: Review: Convolution Representation of convolution Graphical interpretation Examples Properties of convolution EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 3
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Response of LTI System A linear system is completely characterized by its impulse response h ( t,τ ) . If the system is also time invariant, the impulse response h ( t ) of impulse at time 0 can be used to find all other impulse responses. For a linear system with an input signal x ( t ) , the output is given by the superposition integral y ( t ) = Z -∞ x ( τ ) h ( ) If the system is also time invariant, the superposition integral simplifies to y ( t ) = Z -∞ x ( τ ) h ( t - τ ) = Z -∞ x ( t - τ ) h ( τ ) which is in the form of a convolution integral . EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 4
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Convolution Integral The convolution of an input signal x ( t ) with and impulse response h ( t ) is y ( t ) = Z -∞ x ( τ ) h ( t - τ ) = ( x * h )( t ) or y = x * h. This is also often written as y ( t ) = x ( t ) * h ( t ) EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 5
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Convolution Integral for Causal Systems For a causal system h ( t ) = 0 for t < 0 , y ( t ) = Z -∞ x ( τ ) h ( t - τ ) | {z } =0 if t - τ< 0 = y ( t ) = Z t -∞ x ( τ ) h ( t - τ ) Only past and present values of x ( τ ) contribute to y ( t ) . EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 6
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t 0 t 0 y ( t ) τ x ( t ) x ( t )= Z - x ( τ ) δ ( t - τ ) d τ τ>t τ<t y ( t Z t -∞ x ( τ ) h ( t - τ ) Does not contribute to y(t) If x ( t ) is also causal, x ( t ) = 0 for t < 0 , and the integral further simplifies y ( t ) = Z t 0 x ( τ ) h ( t - τ ) dτ. t 0 t 0 y ( t ) τ x ( t ) Does not contribute to y(t) Does not contribute to y(t) x ( t Z t 0 x ( τ ) δ ( t - τ ) y ( t Z t 0 x ( τ ) h ( t - τ ) EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 7
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Graphical Interpretation An increment in input x ( τ ) δ ( t - τ ) produces an impulse response x ( τ ) h ( t - τ ) . The output is the integral of all of these responses y ( t ) = Z -∞ x ( τ ) h ( t - τ ) Another perspective is just to look at the integral. h ( t - τ ) is the impulse response delayed to time τ If we consider h ( t - τ ) to be a function of τ , then h ( t - τ ) is delayed to time t , and reversed . τ t h ( t - τ ) τ t h ( t - τ ) EE102: Systems and Signals; Fall 2011, Jin Hyung Lee 8
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This is multiplied point by point with the input, τ t h ( t - τ ) x ( τ ) τ t x ( τ ) h ( t - τ ) Then integrate over τ to find y ( t ) for this t .
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This note was uploaded on 12/13/2011 for the course EE 102 taught by Professor Levan during the Fall '08 term at UCLA.

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102_1_lecture6 - UCLA Fall 2011 Systems and Signals Lecture...

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