Limit of The Dirac Comb
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The Dirac comb function with period T is:
$$
f(t,T):=sum_{k=-infty}^{k=infty}delta(t-kT)
$$
What is the limit of:
$$
lim_{Tto0} f(t,T)
$$
?
limits dirac-delta
asked Nov 15 at 16:31
A Slow Learner
388212
add a comment |
up vote
-1
down vote
favorite
The Dirac comb function with period T is:
$$
f(t,T):=sum_{k=-infty}^{k=infty}delta(t-kT)
$$
What is the limit of:
$$
lim_{Tto0} f(t,T)
$$
?
limits dirac-delta
asked Nov 15 at 16:31
A Slow Learner
388212
The limit of f(t,T) is the constant function: f(t) = 1.
– M. Wind
Nov 15 at 18:02
@M.Wind. No, it isn't. For that to happen, you need to multiply it with $T$. That is, $$lim_{T to 0} T sum_{k=-infty}^{k=infty}delta(t-kT) = 1$$ for all $t$. Instead we have $$lim_{Tto0} f(t,T) = infty.$$
– md2perpe
Nov 15 at 18:52
Apologies. I tacitly assumed that the function is properly normalized. There is not much point in considering the properties of a function which obviously diverges.
– M. Wind
Nov 15 at 19:15
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
The Dirac comb function with period T is:
$$
f(t,T):=sum_{k=-infty}^{k=infty}delta(t-kT)
$$
What is the limit of:
$$
lim_{Tto0} f(t,T)
$$
?
limits dirac-delta
asked Nov 15 at 16:31
A Slow Learner
388212
The Dirac comb function with period T is:
$$
f(t,T):=sum_{k=-infty}^{k=infty}delta(t-kT)
$$
What is the limit of:
$$
lim_{Tto0} f(t,T)
$$
?
limits dirac-delta
limits dirac-delta
asked Nov 15 at 16:31
A Slow Learner
388212
asked Nov 15 at 16:31
A Slow Learner
388212
asked Nov 15 at 16:31
A Slow Learner
388212
asked Nov 15 at 16:31
A Slow Learner
388212
asked Nov 15 at 16:31
A Slow Learner
388212
388212
The limit of f(t,T) is the constant function: f(t) = 1.
– M. Wind
Nov 15 at 18:02
@M.Wind. No, it isn't. For that to happen, you need to multiply it with $T$. That is, $$lim_{T to 0} T sum_{k=-infty}^{k=infty}delta(t-kT) = 1$$ for all $t$. Instead we have $$lim_{Tto0} f(t,T) = infty.$$
– md2perpe
Nov 15 at 18:52
Apologies. I tacitly assumed that the function is properly normalized. There is not much point in considering the properties of a function which obviously diverges.
– M. Wind
Nov 15 at 19:15
add a comment |
The limit of f(t,T) is the constant function: f(t) = 1.
– M. Wind
Nov 15 at 18:02
@M.Wind. No, it isn't. For that to happen, you need to multiply it with $T$. That is, $$lim_{T to 0} T sum_{k=-infty}^{k=infty}delta(t-kT) = 1$$ for all $t$. Instead we have $$lim_{Tto0} f(t,T) = infty.$$
– md2perpe
Nov 15 at 18:52
Apologies. I tacitly assumed that the function is properly normalized. There is not much point in considering the properties of a function which obviously diverges.
– M. Wind
Nov 15 at 19:15
The limit of f(t,T) is the constant function: f(t) = 1.
– M. Wind
Nov 15 at 18:02
The limit of f(t,T) is the constant function: f(t) = 1.
– M. Wind
Nov 15 at 18:02
@M.Wind. No, it isn't. For that to happen, you need to multiply it with $T$. That is, $$lim_{T to 0} T sum_{k=-infty}^{k=infty}delta(t-kT) = 1$$ for all $t$. Instead we have $$lim_{Tto0} f(t,T) = infty.$$
– md2perpe
Nov 15 at 18:52
@M.Wind. No, it isn't. For that to happen, you need to multiply it with $T$. That is, $$lim_{T to 0} T sum_{k=-infty}^{k=infty}delta(t-kT) = 1$$ for all $t$. Instead we have $$lim_{Tto0} f(t,T) = infty.$$
– md2perpe
Nov 15 at 18:52
Apologies. I tacitly assumed that the function is properly normalized. There is not much point in considering the properties of a function which obviously diverges.
– M. Wind
Nov 15 at 19:15
Apologies. I tacitly assumed that the function is properly normalized. There is not much point in considering the properties of a function which obviously diverges.
– M. Wind
Nov 15 at 19:15
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
Intuitively it is clear that a very dense set of Dirac functions with constant amplitude, and properly normalized, behaves as the unit constant function. Therefore we expect that the limit of the (normalized) Dirac comb function for $T$ to zero is $f(t) = 1$.
Let us see if we can derive this result a bit more rigorously. Consider a Dirac comb function in the time domain with period $T$. Performing the Fourier transformation yields a Dirac comb function in the frequency domain with period $1/T$. We now take the limit of $T$ to zero. Only the Dirac function at $f=0$ remains, since all the others shift to plus or minus infinity, where they can be assumed to contribute no longer. Now perform the Fourier transformation back to the time regime on the remaining Dirac function. You get the constant function: $f(t) = 1$.
answered Nov 15 at 19:08
M. Wind
2,096715
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Intuitively it is clear that a very dense set of Dirac functions with constant amplitude, and properly normalized, behaves as the unit constant function. Therefore we expect that the limit of the (normalized) Dirac comb function for $T$ to zero is $f(t) = 1$.
Let us see if we can derive this result a bit more rigorously. Consider a Dirac comb function in the time domain with period $T$. Performing the Fourier transformation yields a Dirac comb function in the frequency domain with period $1/T$. We now take the limit of $T$ to zero. Only the Dirac function at $f=0$ remains, since all the others shift to plus or minus infinity, where they can be assumed to contribute no longer. Now perform the Fourier transformation back to the time regime on the remaining Dirac function. You get the constant function: $f(t) = 1$.
answered Nov 15 at 19:08
M. Wind
2,096715
add a comment |
up vote
0
down vote
Intuitively it is clear that a very dense set of Dirac functions with constant amplitude, and properly normalized, behaves as the unit constant function. Therefore we expect that the limit of the (normalized) Dirac comb function for $T$ to zero is $f(t) = 1$.
Let us see if we can derive this result a bit more rigorously. Consider a Dirac comb function in the time domain with period $T$. Performing the Fourier transformation yields a Dirac comb function in the frequency domain with period $1/T$. We now take the limit of $T$ to zero. Only the Dirac function at $f=0$ remains, since all the others shift to plus or minus infinity, where they can be assumed to contribute no longer. Now perform the Fourier transformation back to the time regime on the remaining Dirac function. You get the constant function: $f(t) = 1$.
answered Nov 15 at 19:08
M. Wind
2,096715
add a comment |
up vote
0
down vote
up vote
0
down vote
Intuitively it is clear that a very dense set of Dirac functions with constant amplitude, and properly normalized, behaves as the unit constant function. Therefore we expect that the limit of the (normalized) Dirac comb function for $T$ to zero is $f(t) = 1$.
Let us see if we can derive this result a bit more rigorously. Consider a Dirac comb function in the time domain with period $T$. Performing the Fourier transformation yields a Dirac comb function in the frequency domain with period $1/T$. We now take the limit of $T$ to zero. Only the Dirac function at $f=0$ remains, since all the others shift to plus or minus infinity, where they can be assumed to contribute no longer. Now perform the Fourier transformation back to the time regime on the remaining Dirac function. You get the constant function: $f(t) = 1$.
answered Nov 15 at 19:08
M. Wind
2,096715
Intuitively it is clear that a very dense set of Dirac functions with constant amplitude, and properly normalized, behaves as the unit constant function. Therefore we expect that the limit of the (normalized) Dirac comb function for $T$ to zero is $f(t) = 1$.
Let us see if we can derive this result a bit more rigorously. Consider a Dirac comb function in the time domain with period $T$. Performing the Fourier transformation yields a Dirac comb function in the frequency domain with period $1/T$. We now take the limit of $T$ to zero. Only the Dirac function at $f=0$ remains, since all the others shift to plus or minus infinity, where they can be assumed to contribute no longer. Now perform the Fourier transformation back to the time regime on the remaining Dirac function. You get the constant function: $f(t) = 1$.
answered Nov 15 at 19:08
M. Wind
2,096715
edited Nov 15 at 19:16
answered Nov 15 at 19:08
M. Wind
2,096715
answered Nov 15 at 19:08
M. Wind
2,096715
answered Nov 15 at 19:08
M. Wind
2,096715
2,096715
add a comment |
add a comment |
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The limit of f(t,T) is the constant function: f(t) = 1.
– M. Wind
Nov 15 at 18:02
@M.Wind. No, it isn't. For that to happen, you need to multiply it with $T$. That is, $$lim_{T to 0} T sum_{k=-infty}^{k=infty}delta(t-kT) = 1$$ for all $t$. Instead we have $$lim_{Tto0} f(t,T) = infty.$$
– md2perpe
Nov 15 at 18:52
Apologies. I tacitly assumed that the function is properly normalized. There is not much point in considering the properties of a function which obviously diverges.
– M. Wind
Nov 15 at 19:15