102_1_W08-102-Week2-Recap

102_1_W08-102-Week2-Recap - 5 WEEK 2 RECAP 5 U and...

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5 WEEK 2 RECAP 5. U ( · ) and δ ( · ) : Continued Recall that δ ( t ) is 0 everywhere, except at t = 0, where it is equal to 1 ! Therefore one says that δ ( t ) sits at t = 0 . Note that δ ( t ) = δ ( t ) , t R δ ( · ) is an even function. In general you have, for t and τ both taking real values δ ( t τ ) := 0 , t 6 = τ , (5.1) := 1 , t = τ , (5.2) and Z 1 −1 δ ( t τ ) dt = 1 , (5.3) where δ ( t τ ) is taken to be a function of t . You also have f ( t ) δ ( t τ ) = f ( τ ) δ ( t τ ) , (5.4) Z 1 −1 f ( t ) δ ( t τ ) dt = Z 1 −1 f ( τ ) δ ( t τ ) dt = f ( τ ) Z 1 −1 δ ( t τ ) dt, (5.5) = f ( τ ) . (5.6) You therefore conclude from (5.4) that: “The product f ( · ) δ ( · ) is equal to the product of the value of f ( · ) calculated at the place where delta sits — and delta.” Similarly, it follows from (5.5) and (5.6) that: “The integral of the product f ( · ) δ ( · ) is equal to the value of f ( · ) com- puted at the place where δ ( · ) sits.” Remark d [ f ( t ) U ( t )] dt = df ( t ) dt U ( t ) + f ( t ) dU ( t ) dt , (5.7) = df ( t ) dt U ( t ) + f ( t ) δ ( t ) = df ( t ) dt U ( t ) + f (0) δ ( t ) . (5.8) Now interchanging t and τ in (5.5) and (5.6) you get (5.9) f ( t ) = Z 1 −1 f ( τ ) δ ( t τ ) d τ , t R . This is an important result! It means that a signal f ( · ) can be considered as the limit of the sum of its “resolutions” into its values at each τ , i.e., f ( τ ), multiplying by the delta function sitting at τ , i.e., δ ( t τ ) .
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6 NHAN LEVAN 6. Impulse Response Function (IRF) h ( t ; τ ) , t, τ R , of S IMPULSE := δ ( · ) Recall that := means “equal by definition”. Let S be a system with IPOP Description: t R : x ( t ) −→ [ S ] −→ y ( t ) = T [ x ( t )] Picture 1 of the Week : t R : δ ( t ) −→ [ S ] −→ h ( t ; 0 ) = T [ δ ( t 0 )] The input δ ( t ) — sits at t = 0 — and it is equal to 0 at t 6 = 0. The corresponding OP is written as h ( t ; 0) to remind you that it is the response of S to an impulse applied at 0.
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