lecture6n - Stanford University Winter 2009-2010 Signal...

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Stanford University Winter 2009-2010 Signal Processing and Linear Systems I Lecture 6: Convolution January 14, 2010 EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 1 Convolution Evaluation and Properties Today’s topics: Review: response of an LTI system Representation of convolution Graphical interpretation Examples Properties of convolution EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 2

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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 is only a function of t , and is written h ( t ) . For a linear system with an input signal x ( t ) , the output is given by the superposition integral y ( t ) = −∞ x ( τ ) h ( t, τ ) d τ If the system is also time invariant, the superposition integral simplifies to y ( t ) = −∞ x ( τ ) h ( t τ ) d τ = −∞ x ( t τ ) h ( τ ) d τ which is in the form of a convolution integral . EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 3 The block diagrams for a convolution system with an impulse response h ( t ) : y ( t ) x ( t ) h ( t ) y ( t ) x ( t ) h ( t ) EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 4
From last time, this was illustrated: t 0 t h ( t ) 0 t 0 t 0 ! ( t ) t 0 t 0 t 0 t 0 x ( t ) y ( t ) ( x ( ! ) d ! ) " ( t ! ) ( x ( ! ) d ! ) h ( t ! ) ! ( t " ) ! ! ! ! ! x ( t ) x ( t ) = Z ! ! x ( " ) # ( t " ) d " y ( t ) = Z ! ! x ( " ) h ( t " ) d " Input Output h ( t ! ) EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 5 Convolution Integral The convolution of an input signal x ( t ) with and impulse response h ( t ) is y ( t ) = −∞ x ( τ ) h ( t τ ) d τ = ( x h )( t ) or y = x h. This is also often written as y ( t ) = x ( t ) h ( t ) which is potentially confusing, since the t ’s have di ff erent interpretations on the left and right sides of the equation (your book does this). EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 6

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Convolution Integral for Causal Systems For a causal system h ( t ) = 0 for t < 0 , y ( t ) = −∞ x ( τ ) h ( t τ ) =0 if t τ < 0 d τ = y ( t ) = t −∞ x ( τ ) h ( t τ ) d τ Only past and present values of x ( τ ) contribute to y ( t ) . EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 7 t 0 t 0 y ( t ) ! x ( t ) x ( t ) = Z ! ! x ( " ) # ( t " ) d " τ > t τ < t y ( t ) = t −∞ x ( τ ) h ( t τ ) d 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 ) = t 0 x ( τ ) h ( t τ ) d τ . t 0 t 0 y ( t ) ! x ( t ) τ > t τ < t Does not contribute to y(t) Does not contribute to y(t) x ( t ) = t 0 x ( τ ) δ ( t τ ) d τ y ( t ) = t 0 x ( τ ) h ( t τ ) d τ EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 8
Graphical Interpretation An increment in input x ( τ ) δ ( t τ ) d τ produces an impulse response x ( τ ) h ( t τ ) d τ . The output is the integral of all of these responses

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