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Sifting Properties of Delta

Module by: Richard Baraniuk. E-mail the author

Summary: A module concerning the sifting properties of delta.

f⁢n⁢δ⁢n−k=f⁢k⁢δ⁢n−k f n δ n k f k δ n k Sum both sides over K K ∑ K f⁢n⁢δ⁢n−k=∑ K f⁢k⁢δ⁢n−k K f n δ n k K f k δ n k f⁢n⁢∑ K δ⁢n−k=∑ K f⁢k⁢δ⁢n−k f n K δ n k K f k δ n k where (∑ K δ⁢n−k={1  if  k=n0  otherwise  )→1 K δ n k 1 k n 0 1

f⁢n=∑ K f⁢k⁢δ⁢n−k f n K f k δ n k
(1)
Interpret sifting as building f⁢n f n from weighted linear combos of δ⁢n−k δ n k eg: n= n 0, 1, ..., N−1 N 1 . f⁢n=f⁢0⁢δ⁢n+f⁢1⁢δ⁢n−1+...+f⁢N−1⁢δ⁢n−N+1 f n f 0 δ n f 1 δ n 1 ... f N 1 δ n N 1 ( f⁢0 f⁢1 ... f⁢N−1 )=f⁢0⁢( 1 0 ... 0 )+f⁢1⁢( 0 1 ... 0 )+...+f⁢N−1⁢( 0 ... 0 1 ) f 0 f 1 ... f N 1 f 0 1 0 ... 0 f 1 0 1 ... 0 ... f N 1 0 ... 0 1 ie: δ⁢n−k δ n k are the "canonical basis vectors" or ℝN ℝ N .

Interpret sifting propertiy as a matrix / vector product ( f⁢0 f⁢1 ... f⁢N−1 )=( )=( f⁢0 f⁢1 ... f⁢N−1 ) f 0 f 1 ... f N 1 f 0 f 1 ... f N 1

Unit IMPULSE "Function"

aka Dirac Delta "Function" super duper useful!!!

Definition 1:
δ⁢t=0 δ t 0 , t≠0 t 0 ∫−∞∞δ⁢td t =1 t δ t 1
To integrate to 1, δ⁢t δ t must be "infinitely tall but infinitely skinny" when viewed as a function (actually δ⁢0 δ 0 is undefined).
Figure 1
Figure 1 (fig1.png)
This is no problem, however, becuase δ δ functions always end up hanging around integrals.

Can think of δ⁢t δ t as limit of pulse functions as we compress

Figure 2
Figure 2 (fig2.png)
scaled impulse: k⁢δ⁢t k δ t k⁢δ⁢t=0 k δ t 0 , t≠0 t 0 ∫k⁢δ⁢td t =k t k δ t k
Figure 3
Figure 3 (fig3.png)
scaled-delta product

Assume φ⁢t φ t is continuous at t=0 t 0 . Then

φ⁢t⁢δ⁢t=φ⁢0⁢δ⁢t φ t δ t φ 0 δ t
(2)
Figure 4
Figure 4 (fig4.png)
more generally:
φ⁢t⁢δ⁢t−T=φ⁢0⁢δ⁢t−T φ t δ t T φ 0 δ t T
(3)

Sifting Properties of Impulse

Recall that f⁢t⁢δ⁢t−T=f⁢T⁢δ⁢t−T f t δ t T f T δ t T Thus ∫−∞∞f⁢t⁢δ⁢t−Td t =∫−∞∞f⁢T⁢δ⁢t−Td t t f t δ t T t f T δ t T pulling out f⁢T f T =f⁢T⁢∫−∞∞δ⁢t−Td t f T t δ t T =f⁢T⁢∫−∞∞dδ⁢tdd t′ f T t δ t =f⁢T f T a.k.a. "impulse sampling"

∫−∞∞f⁢t⁢δ⁢t−Td t =f⁢T t f t δ t T f T
(4)
This operation "sifts" thru the entire signal and picks out the value f⁢T f T

Note:

Also ∫−∞∞f⁢t⁢δ⁢T−td t =f⁢T t f t δ T t f T Why?

Exercise 1

Is δ⁢t δ t in any L p L p ?

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