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102_1_vdis3 - EE102 Discussion 3 vasiliy karasev 1 LTI...

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EE102 Discussion 3 2010-10-11 vasiliy karasev 1. LTI System Response δ(t) t 1 0 1 h(t) = e u(t) -at t 1 0 1 *h(t) impulse impulse response x(t) t 1 0 1 *h(t) y(t) = ? t 1 0 1 complex input complex output Linearity allows representing x [ n ] as a sum of weighted δ functions: x [ n ] = ... + x [ - 1] δ [ n + 1] + x [0] δ [ n ] + x [1] δ [ n - 1] + ... Time Invariance asserts that output to delayed delta function is the delayed impulse response h ( t ). δ [ n ] 7→ h [ n ] δ [ n - 1] 7→ h [ n - 1] ... Linearity claims that weighted, delayed impulse responses can be summed to produce y [ n ]: y [ n ] = ... + x [ - 1] h [ n + 1] + x [0] h [ n ] + x [1] h [ n - 1] + ... This can also be summarized (in continuous time) as: x ( t ) = Z -∞ x ( τ ) δ ( t - τ ) y ( t ) = Z -∞ x ( τ ) h ( t - τ ) = ( x * h )( t ) = ( h * x )( t ) Exercise 1. f ( t ) = u ( t ), g ( t ) = δ ( t ) - δ ( t - 1). Find ( f * g )( t ). Solution It’s easy to do this graphically. Alternatively, recall that δ ( t ) is the identity element of convolution, so f ( t ) * δ ( t ) = f ( t ). Similarly, δ ( t - 1) only delays the input signal. Hence, ( f * g )( t ) = u ( t ) - u ( t - 1) = rect( t - 1 2 ).
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