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The Delta Function∫t–∞f( τ)δ( τ) dτ=∫t–∞f0 ( )δ( τ) dτ+∫t–∞[ f( τ) – f0 ( )]δ( τ) dτ(1.39)The first integral on the right side of (1.39) contains the constant termoutside the integral, that is,∫t–∞f0 ( )δ( τ) dτThe second integral of the right side of (1.39) is always zero because∫t= f0 ( ) δ( τ) dτ–∞f0 ( ); this can be written(1.40)andTherefore, (1.39) reduces to∫δ() t = 0 for t < 0 and t > 0[ f( τ) – f0 ( )] τ = 0= f0 ( ) – f0 ( ) = 0t–∞f( τ)δ( τ) dτ∫t= f0 ( ) δ( τ) dτ–∞(1.41)Differentiating both sides of (1.41), and replacingτwith t , we obtainf()δtt= f0 ( )δ()tSampling Property of δ()t(1.42)1.4.2 The Sifting Property of the Delta FunctionThe sifting property of the delta function states thatδ()t∫∞–∞ft ()δt ( – α) dt=f( α)(1.43)that is, if we multiply any function ft () by δt ( – α) , and integrate from – ∞ to +∞ , we will obtainthe value of ft () evaluated at t = α .Proof:Let us consider the integral∫abft ()δt ( – α) dt where a < α < b(1.44)We will use integration by parts to evaluate this integral. We recall from the derivative of productsthatdxy ( ) = xdy + ydx or xdy = dxy ( ) – ydx(1.45)and integrating both sides we obtainSignals and Systems with MATLAB ® Computing and Simulink ® Modeling, Fourth EditionCopyright © Orchard Publications1−13