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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.

fnδnk=fkδnk f n δ n k f k δ n k Sum both sides over K K K fnδnk= K fkδnk K f n δ n k K f k δ n k fn K δnk= K fkδnk f n K δ n k K f k δ n k where ( K δnk={1  if  k=n0  otherwise  )1 K δ n k 1 k n 0 1

fn= K fkδnk f n K f k δ n k
(1)
Interpret sifting as building fn f n from weighted linear combos of δnk δ n k eg: n= n 0, 1, ..., N1 N 1 . fn=f0δn+f1δn1+...+fN1δnN+1 f n f 0 δ n f 1 δ n 1 ... f N 1 δ n N 1 ( f0 f1 ... fN1 )=f0( 1 0 ... 0 )+f1( 0 1 ... 0 )+...+fN1( 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: δnk δ n k are the "canonical basis vectors" or N N .

Interpret sifting propertiy as a matrix / vector product ( f0 f1 ... fN1 )=( )=( f0 f1 ... fN1 ) 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:
δt=0 δ t 0 , t0 t 0 δtd t =1 t δ t 1
To integrate to 1, δt δ t must be "infinitely tall but infinitely skinny" when viewed as a function (actually δ0 δ 0 is undefined). This is no problem, however, becuase δ δ functions always end up hanging around integrals.

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

scaled impulse: kδt k δ t kδt=0 k δ t 0 , t0 t 0 kδtd t =k t k δ t k scaled-delta product

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

φtδt=φ0δt φ t δ t φ 0 δ t
(2)
more generally:
φtδtT=φ0δtT φ t δ t T φ 0 δ t T
(3)

## Sifting Properties of Impulse

Recall that ftδtT=fTδtT f t δ t T f T δ t T Thus ftδtTd t =fTδtTd t t f t δ t T t f T δ t T pulling out fT f T =fTδtTd t f T t δ t T =fTdδtdd t f T t δ t =fT f T a.k.a. "impulse sampling"

ftδtTd t =fT t f t δ t T f T
(4)
This operation "sifts" thru the entire signal and picks out the value fT f T

### Note:

Also ftδTtd t =fT t f t δ T t f T Why?

### Exercise 1

Is δt δ t in any L p L p ?

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