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How do you find the function delta(f(t))?

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purnapragna

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query:dirac impulse

how do you find the function \[\delta(f(t))\]

please help me. I was solving a problem in Bracewell book about Fourier Transforms where I encoutered the problem like find \[\delta(t^2-1)\]

help me

thanx

purna!
 

Re: query:dirac impulse

Well, in terms of delta(t^2 - 1),

I believe the delta function is defined as such:

delta(0) = 1

delta(anything else) = 0

So if we have delta(t^2 - 1),
What would make the inside term (t^2 - 1) = 0?
answer ==> t = 1 or -1

so delta(t^2 - 1) = 1 for t = 1 or -1
delta(t^2 - 1) = 0 for t = anything else
 

Re: query:dirac impulse

check this : https://mathworld.wolfram.com/DeltaFunction.html
especially equation(6)
regards,
a.safwat

Added after 6 minutes:

rtarbell said:
Well, in terms of delta(t^2 - 1),

I believe the delta function is defined as such:

delta(0) = 1

delta(anything else) = 0

So if we have delta(t^2 - 1),
What would make the inside term (t^2 - 1) = 0?
answer ==> t = 1 or -1

so delta(t^2 - 1) = 1 for t = 1 or -1
delta(t^2 - 1) = 0 for t = anything else

delta(t^2 - 1) = 0.5 for t = 1 or -1
i think the main idea is that the integration of any delta from -∞ to ∞ must be 1 , so the weight added is to asure that area is 1 and not 2 if we do the integration on ur fromula.
 

Re: query:dirac impulse

remember this is the general formula for impulse.

δ(f(x))=∑ over κ [1/mod(derivative of f(Θκ)]δ(Θ-Θκ).here Θκ are the roots of f(Θ)=0.
 

Re: query:dirac impulse

delta function represents impulse of action at the origin of system. For example when we push a key it is like an impulse of force at that instant(say t=0, key pushing time). When this delta is multiplied with some other function it makes that function an impulse function at zeros of that function i. e given function shifts origin of impulse. For example key pressed at t=0, t=1,t=2 can be represented by

delte*F, (delta-1)*F, (delta-2)*F.
 

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