Derivative of Squared Step Function
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while studying signals & systems I ended up in the following difficult situation after digging up the unit step and delta functions a lot (I wish I hadn't do so, but couldn't help :]). Here it is:
If we define the unit step function as
$$
u(t) =
left{
beginarrayll
1 & mboxif t geq 0 \
0 & mboxif t < 0
endarray
right.
$$
the following holds I presume, $u(t)^2 = u(t)$.
Then, if we dare to take the derivative of both sides, it should be $$2u(t)delta(t)=delta(t)$$ which is also equivalent to $2u(0)delta(t) = 2delta(t)$. So, the RHS and LHS are not equal. But, they were.
So, since we're dealing with generalised functions/derivatives (which I'm not very familiar with) here, I'm making some serious mistakes in the operations above, but don't really know which ones.
derivatives dirac-delta step-function
asked Sep 8 at 11:46
gunes
1943
add a comment |Â
up vote
0
down vote
favorite
while studying signals & systems I ended up in the following difficult situation after digging up the unit step and delta functions a lot (I wish I hadn't do so, but couldn't help :]). Here it is:
If we define the unit step function as
$$
u(t) =
left{
beginarrayll
1 & mboxif t geq 0 \
0 & mboxif t < 0
endarray
right.
$$
the following holds I presume, $u(t)^2 = u(t)$.
Then, if we dare to take the derivative of both sides, it should be $$2u(t)delta(t)=delta(t)$$ which is also equivalent to $2u(0)delta(t) = 2delta(t)$. So, the RHS and LHS are not equal. But, they were.
So, since we're dealing with generalised functions/derivatives (which I'm not very familiar with) here, I'm making some serious mistakes in the operations above, but don't really know which ones.
derivatives dirac-delta step-function
asked Sep 8 at 11:46
gunes
1943
Please do not pretend distributions are functions unless you know what you are doing. Otherwise you get mistakes, such as this one. In particular, multiplication of distributions is often not defined. en.wikipedia.org/wiki/Distribution_(mathematics)
– GEdgar
Sep 8 at 13:06
Well, I guess, square of step function is defined as stated in math.stackexchange.com/questions/442586/…. Do you mean taking derivative of $u(t)^2$ as if it is multiplication of two step functions is a wrong treatment, here? And, if true, does this mean that, once we find that $u(t)^2 = u(t)$, the derivative of this is equal to that of $u(t)$?
– gunes
Sep 8 at 14:46
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
while studying signals & systems I ended up in the following difficult situation after digging up the unit step and delta functions a lot (I wish I hadn't do so, but couldn't help :]). Here it is:
If we define the unit step function as
$$
u(t) =
left{
beginarrayll
1 & mboxif t geq 0 \
0 & mboxif t < 0
endarray
right.
$$
the following holds I presume, $u(t)^2 = u(t)$.
Then, if we dare to take the derivative of both sides, it should be $$2u(t)delta(t)=delta(t)$$ which is also equivalent to $2u(0)delta(t) = 2delta(t)$. So, the RHS and LHS are not equal. But, they were.
So, since we're dealing with generalised functions/derivatives (which I'm not very familiar with) here, I'm making some serious mistakes in the operations above, but don't really know which ones.
derivatives dirac-delta step-function
asked Sep 8 at 11:46
gunes
1943
while studying signals & systems I ended up in the following difficult situation after digging up the unit step and delta functions a lot (I wish I hadn't do so, but couldn't help :]). Here it is:
If we define the unit step function as
$$
u(t) =
left{
beginarrayll
1 & mboxif t geq 0 \
0 & mboxif t < 0
endarray
right.
$$
the following holds I presume, $u(t)^2 = u(t)$.
Then, if we dare to take the derivative of both sides, it should be $$2u(t)delta(t)=delta(t)$$ which is also equivalent to $2u(0)delta(t) = 2delta(t)$. So, the RHS and LHS are not equal. But, they were.
So, since we're dealing with generalised functions/derivatives (which I'm not very familiar with) here, I'm making some serious mistakes in the operations above, but don't really know which ones.
derivatives dirac-delta step-function
derivatives dirac-delta step-function
asked Sep 8 at 11:46
gunes
1943
asked Sep 8 at 11:46
gunes
1943
asked Sep 8 at 11:46
gunes
1943
asked Sep 8 at 11:46
gunes
1943
asked Sep 8 at 11:46
gunes
1943
1943
Please do not pretend distributions are functions unless you know what you are doing. Otherwise you get mistakes, such as this one. In particular, multiplication of distributions is often not defined. en.wikipedia.org/wiki/Distribution_(mathematics)
– GEdgar
Sep 8 at 13:06
Well, I guess, square of step function is defined as stated in math.stackexchange.com/questions/442586/…. Do you mean taking derivative of $u(t)^2$ as if it is multiplication of two step functions is a wrong treatment, here? And, if true, does this mean that, once we find that $u(t)^2 = u(t)$, the derivative of this is equal to that of $u(t)$?
– gunes
Sep 8 at 14:46
add a comment |Â
Please do not pretend distributions are functions unless you know what you are doing. Otherwise you get mistakes, such as this one. In particular, multiplication of distributions is often not defined. en.wikipedia.org/wiki/Distribution_(mathematics)
– GEdgar
Sep 8 at 13:06
Well, I guess, square of step function is defined as stated in math.stackexchange.com/questions/442586/…. Do you mean taking derivative of $u(t)^2$ as if it is multiplication of two step functions is a wrong treatment, here? And, if true, does this mean that, once we find that $u(t)^2 = u(t)$, the derivative of this is equal to that of $u(t)$?
– gunes
Sep 8 at 14:46
Please do not pretend distributions are functions unless you know what you are doing. Otherwise you get mistakes, such as this one. In particular, multiplication of distributions is often not defined. en.wikipedia.org/wiki/Distribution_(mathematics)
– GEdgar
Sep 8 at 13:06
Please do not pretend distributions are functions unless you know what you are doing. Otherwise you get mistakes, such as this one. In particular, multiplication of distributions is often not defined. en.wikipedia.org/wiki/Distribution_(mathematics)
– GEdgar
Sep 8 at 13:06
Well, I guess, square of step function is defined as stated in math.stackexchange.com/questions/442586/…. Do you mean taking derivative of $u(t)^2$ as if it is multiplication of two step functions is a wrong treatment, here? And, if true, does this mean that, once we find that $u(t)^2 = u(t)$, the derivative of this is equal to that of $u(t)$?
– gunes
Sep 8 at 14:46
Well, I guess, square of step function is defined as stated in math.stackexchange.com/questions/442586/…. Do you mean taking derivative of $u(t)^2$ as if it is multiplication of two step functions is a wrong treatment, here? And, if true, does this mean that, once we find that $u(t)^2 = u(t)$, the derivative of this is equal to that of $u(t)$?
– gunes
Sep 8 at 14:46
add a comment |Â
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Please do not pretend distributions are functions unless you know what you are doing. Otherwise you get mistakes, such as this one. In particular, multiplication of distributions is often not defined. en.wikipedia.org/wiki/Distribution_(mathematics)
– GEdgar
Sep 8 at 13:06
Well, I guess, square of step function is defined as stated in math.stackexchange.com/questions/442586/…. Do you mean taking derivative of $u(t)^2$ as if it is multiplication of two step functions is a wrong treatment, here? And, if true, does this mean that, once we find that $u(t)^2 = u(t)$, the derivative of this is equal to that of $u(t)$?
– gunes
Sep 8 at 14:46