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Can we combine convolution and higher powers for locally maximising a function?
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Can we somehow find local maximum function value (for strictly positive functions) using a convolution?
My idea is based on the result that $$ lim_pto inftyleft[frac1Nsum_k=1^N (a_k) ^pright]^frac 1 p = max(a_k)$$
Similarly we might be able to use the result that
$$ min(a_k) = max(a_k)-max(max(a_k)-a_k)$$
Could we perhaps use this for building envelope detection? It seems theoretically sound to me, but would it be practically feasible?
limits approximation signal-processing computational-mathematics
asked 2 days ago
mathreadler
13.5k71857
add a comment |Â
up vote
0
down vote
favorite
Can we somehow find local maximum function value (for strictly positive functions) using a convolution?
My idea is based on the result that $$ lim_pto inftyleft[frac1Nsum_k=1^N (a_k) ^pright]^frac 1 p = max(a_k)$$
Similarly we might be able to use the result that
$$ min(a_k) = max(a_k)-max(max(a_k)-a_k)$$
Could we perhaps use this for building envelope detection? It seems theoretically sound to me, but would it be practically feasible?
limits approximation signal-processing computational-mathematics
asked 2 days ago
mathreadler
13.5k71857
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Can we somehow find local maximum function value (for strictly positive functions) using a convolution?
My idea is based on the result that $$ lim_pto inftyleft[frac1Nsum_k=1^N (a_k) ^pright]^frac 1 p = max(a_k)$$
Similarly we might be able to use the result that
$$ min(a_k) = max(a_k)-max(max(a_k)-a_k)$$
Could we perhaps use this for building envelope detection? It seems theoretically sound to me, but would it be practically feasible?
limits approximation signal-processing computational-mathematics
asked 2 days ago
mathreadler
13.5k71857
Can we somehow find local maximum function value (for strictly positive functions) using a convolution?
My idea is based on the result that $$ lim_pto inftyleft[frac1Nsum_k=1^N (a_k) ^pright]^frac 1 p = max(a_k)$$
Similarly we might be able to use the result that
$$ min(a_k) = max(a_k)-max(max(a_k)-a_k)$$
Could we perhaps use this for building envelope detection? It seems theoretically sound to me, but would it be practically feasible?
limits approximation signal-processing computational-mathematics
asked 2 days ago
mathreadler
13.5k71857
asked 2 days ago
mathreadler
13.5k71857
asked 2 days ago
mathreadler
13.5k71857
asked 2 days ago
mathreadler
13.5k71857
13.5k71857
add a comment |Â
add a comment |Â
1 Answer
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Yes, we can do this.
If we consider the function $$f(t)=casest$$
We can with the convolution
- If we are in continuous time:
$$(g*f)(t)=int_-infty^infty g(tau)cdot f(t-tau) dtau$$
- If we are in discrete time:
$$(g*f)(t)=sum_tau=-infty^infty g(tau)cdot f(t-tau)$$
Define the non-linear operation $$O(g,p) = (g^p*f)(t)^1/p$$
Where exponent means multiplicative power.
For example: if $g(t) = sin(t), O(g,4) = ((sin(x)^4*f))(t)^1/4$
In practice we usually want much higher $p$ exponents, maybe 32 or 1024.
We now investigate the test-function
$$g(t) = sin(60t^2)^2cdot (sin(16t)^2 + cos(8t)^2)$$
The square is to get same level of minimum values: 0.
If we analyze $g(t)$, we can see that we have a chirp function modulated by a slower wave. This is to investigate behaviors for different frequencies.
We see that for higher frequencies we get reasonable envelope behavior. It is also possible to observe that as we change parameter $N$ above, which frequency range we can calculate maxima for are affected - but also the rectangular shape of local maxima will grow. We may need to smooth these out somehow. But how to do so is outside scope of question.
answered yesterday
mathreadler
13.5k71857
Can you explain what you have plotted here? The red "max-envelope" curve can't be simply $g*f$, since that would give you the average value in a window rather than the maximum.
– Rahul
yesterday
@Rahul yes you are right, I will correct.
– mathreadler
yesterday
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Yes, we can do this.
If we consider the function $$f(t)=casest$$
We can with the convolution
- If we are in continuous time:
$$(g*f)(t)=int_-infty^infty g(tau)cdot f(t-tau) dtau$$
- If we are in discrete time:
$$(g*f)(t)=sum_tau=-infty^infty g(tau)cdot f(t-tau)$$
Define the non-linear operation $$O(g,p) = (g^p*f)(t)^1/p$$
Where exponent means multiplicative power.
For example: if $g(t) = sin(t), O(g,4) = ((sin(x)^4*f))(t)^1/4$
In practice we usually want much higher $p$ exponents, maybe 32 or 1024.
We now investigate the test-function
$$g(t) = sin(60t^2)^2cdot (sin(16t)^2 + cos(8t)^2)$$
The square is to get same level of minimum values: 0.
If we analyze $g(t)$, we can see that we have a chirp function modulated by a slower wave. This is to investigate behaviors for different frequencies.
We see that for higher frequencies we get reasonable envelope behavior. It is also possible to observe that as we change parameter $N$ above, which frequency range we can calculate maxima for are affected - but also the rectangular shape of local maxima will grow. We may need to smooth these out somehow. But how to do so is outside scope of question.
answered yesterday
mathreadler
13.5k71857
Can you explain what you have plotted here? The red "max-envelope" curve can't be simply $g*f$, since that would give you the average value in a window rather than the maximum.
– Rahul
yesterday
@Rahul yes you are right, I will correct.
– mathreadler
yesterday
add a comment |Â
up vote
0
down vote
Yes, we can do this.
If we consider the function $$f(t)=casest$$
We can with the convolution
- If we are in continuous time:
$$(g*f)(t)=int_-infty^infty g(tau)cdot f(t-tau) dtau$$
- If we are in discrete time:
$$(g*f)(t)=sum_tau=-infty^infty g(tau)cdot f(t-tau)$$
Define the non-linear operation $$O(g,p) = (g^p*f)(t)^1/p$$
Where exponent means multiplicative power.
For example: if $g(t) = sin(t), O(g,4) = ((sin(x)^4*f))(t)^1/4$
In practice we usually want much higher $p$ exponents, maybe 32 or 1024.
We now investigate the test-function
$$g(t) = sin(60t^2)^2cdot (sin(16t)^2 + cos(8t)^2)$$
The square is to get same level of minimum values: 0.
If we analyze $g(t)$, we can see that we have a chirp function modulated by a slower wave. This is to investigate behaviors for different frequencies.
We see that for higher frequencies we get reasonable envelope behavior. It is also possible to observe that as we change parameter $N$ above, which frequency range we can calculate maxima for are affected - but also the rectangular shape of local maxima will grow. We may need to smooth these out somehow. But how to do so is outside scope of question.
answered yesterday
mathreadler
13.5k71857
Can you explain what you have plotted here? The red "max-envelope" curve can't be simply $g*f$, since that would give you the average value in a window rather than the maximum.
– Rahul
yesterday
@Rahul yes you are right, I will correct.
– mathreadler
yesterday
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Yes, we can do this.
If we consider the function $$f(t)=casest$$
We can with the convolution
- If we are in continuous time:
$$(g*f)(t)=int_-infty^infty g(tau)cdot f(t-tau) dtau$$
- If we are in discrete time:
$$(g*f)(t)=sum_tau=-infty^infty g(tau)cdot f(t-tau)$$
Define the non-linear operation $$O(g,p) = (g^p*f)(t)^1/p$$
Where exponent means multiplicative power.
For example: if $g(t) = sin(t), O(g,4) = ((sin(x)^4*f))(t)^1/4$
In practice we usually want much higher $p$ exponents, maybe 32 or 1024.
We now investigate the test-function
$$g(t) = sin(60t^2)^2cdot (sin(16t)^2 + cos(8t)^2)$$
The square is to get same level of minimum values: 0.
If we analyze $g(t)$, we can see that we have a chirp function modulated by a slower wave. This is to investigate behaviors for different frequencies.
We see that for higher frequencies we get reasonable envelope behavior. It is also possible to observe that as we change parameter $N$ above, which frequency range we can calculate maxima for are affected - but also the rectangular shape of local maxima will grow. We may need to smooth these out somehow. But how to do so is outside scope of question.
answered yesterday
mathreadler
13.5k71857
Yes, we can do this.
If we consider the function $$f(t)=casest$$
We can with the convolution
- If we are in continuous time:
$$(g*f)(t)=int_-infty^infty g(tau)cdot f(t-tau) dtau$$
- If we are in discrete time:
$$(g*f)(t)=sum_tau=-infty^infty g(tau)cdot f(t-tau)$$
Define the non-linear operation $$O(g,p) = (g^p*f)(t)^1/p$$
Where exponent means multiplicative power.
For example: if $g(t) = sin(t), O(g,4) = ((sin(x)^4*f))(t)^1/4$
In practice we usually want much higher $p$ exponents, maybe 32 or 1024.
We now investigate the test-function
$$g(t) = sin(60t^2)^2cdot (sin(16t)^2 + cos(8t)^2)$$
The square is to get same level of minimum values: 0.
If we analyze $g(t)$, we can see that we have a chirp function modulated by a slower wave. This is to investigate behaviors for different frequencies.
We see that for higher frequencies we get reasonable envelope behavior. It is also possible to observe that as we change parameter $N$ above, which frequency range we can calculate maxima for are affected - but also the rectangular shape of local maxima will grow. We may need to smooth these out somehow. But how to do so is outside scope of question.
answered yesterday
mathreadler
13.5k71857
edited yesterday
answered yesterday
mathreadler
13.5k71857
answered yesterday
mathreadler
13.5k71857
answered yesterday
mathreadler
13.5k71857
13.5k71857
Can you explain what you have plotted here? The red "max-envelope" curve can't be simply $g*f$, since that would give you the average value in a window rather than the maximum.
– Rahul
yesterday
@Rahul yes you are right, I will correct.
– mathreadler
yesterday
add a comment |Â
Can you explain what you have plotted here? The red "max-envelope" curve can't be simply $g*f$, since that would give you the average value in a window rather than the maximum.
– Rahul
yesterday
@Rahul yes you are right, I will correct.
– mathreadler
yesterday
Can you explain what you have plotted here? The red "max-envelope" curve can't be simply $g*f$, since that would give you the average value in a window rather than the maximum.
– Rahul
yesterday
Can you explain what you have plotted here? The red "max-envelope" curve can't be simply $g*f$, since that would give you the average value in a window rather than the maximum.
– Rahul
yesterday
@Rahul yes you are right, I will correct.
– mathreadler
yesterday
@Rahul yes you are right, I will correct.
– mathreadler
yesterday
add a comment |Â
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