Mixing
boxcar function integral
$begingroup$
What is the following integral?
$$int f(x) , Pi left(x, a - frac{w}{2}, a + frac{w}{2} right) dx$$
where $Pi(x, a - frac{w}{2},a+ frac{w}{2})$ is a rectangular pulse centered at $x=a$ with width $w$. I know that if $w$ is sufficiently small, then it approximates a Dirac pulse, and then the integral is known as $f(a)$. But how about this case?
calculus
asked Jan 19 at 17:13
Vitor RibeiroVitor Ribeiro
11
$endgroup$
add a comment |
$begingroup$
What is the following integral?
$$int f(x) , Pi left(x, a - frac{w}{2}, a + frac{w}{2} right) dx$$
where $Pi(x, a - frac{w}{2},a+ frac{w}{2})$ is a rectangular pulse centered at $x=a$ with width $w$. I know that if $w$ is sufficiently small, then it approximates a Dirac pulse, and then the integral is known as $f(a)$. But how about this case?
calculus
asked Jan 19 at 17:13
Vitor RibeiroVitor Ribeiro
11
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$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience.
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– dantopa
Jan 19 at 17:18
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What is the height of this rectangular pulse? It will only approximate a delta function if it grows in height as it gets narrower.
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– John Barber
Jan 19 at 17:28
$begingroup$
The height is some constant real number h. Lets say h=1.
$endgroup$
– Vitor Ribeiro
Jan 19 at 17:50
$begingroup$
What the integral is would depend on $f$.
$endgroup$
– David C. Ullrich
Jan 19 at 18:23
$begingroup$
If h=1 then the integrand is $f(x)$ if $a-frac{w}{2}<x<a+frac{w}{2}$ and 0 otherwise. No?
$endgroup$
– Vitor Ribeiro
Jan 19 at 19:50
add a comment |
$begingroup$
What is the following integral?
$$int f(x) , Pi left(x, a - frac{w}{2}, a + frac{w}{2} right) dx$$
where $Pi(x, a - frac{w}{2},a+ frac{w}{2})$ is a rectangular pulse centered at $x=a$ with width $w$. I know that if $w$ is sufficiently small, then it approximates a Dirac pulse, and then the integral is known as $f(a)$. But how about this case?
calculus
asked Jan 19 at 17:13
Vitor RibeiroVitor Ribeiro
11
$endgroup$
What is the following integral?
$$int f(x) , Pi left(x, a - frac{w}{2}, a + frac{w}{2} right) dx$$
where $Pi(x, a - frac{w}{2},a+ frac{w}{2})$ is a rectangular pulse centered at $x=a$ with width $w$. I know that if $w$ is sufficiently small, then it approximates a Dirac pulse, and then the integral is known as $f(a)$. But how about this case?
calculus
calculus
asked Jan 19 at 17:13
Vitor RibeiroVitor Ribeiro
11
asked Jan 19 at 17:13
Vitor RibeiroVitor Ribeiro
11
edited Jan 19 at 19:41
Vitor Ribeiro
asked Jan 19 at 17:13
Vitor RibeiroVitor Ribeiro
11
asked Jan 19 at 17:13
Vitor RibeiroVitor Ribeiro
11
asked Jan 19 at 17:13
Vitor RibeiroVitor Ribeiro
11
11
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience.
$endgroup$
– dantopa
Jan 19 at 17:18
$begingroup$
What is the height of this rectangular pulse? It will only approximate a delta function if it grows in height as it gets narrower.
$endgroup$
– John Barber
Jan 19 at 17:28
$begingroup$
The height is some constant real number h. Lets say h=1.
$endgroup$
– Vitor Ribeiro
Jan 19 at 17:50
$begingroup$
What the integral is would depend on $f$.
$endgroup$
– David C. Ullrich
Jan 19 at 18:23
$begingroup$
If h=1 then the integrand is $f(x)$ if $a-frac{w}{2}<x<a+frac{w}{2}$ and 0 otherwise. No?
$endgroup$
– Vitor Ribeiro
Jan 19 at 19:50
add a comment |
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience.
$endgroup$
– dantopa
Jan 19 at 17:18
$begingroup$
What is the height of this rectangular pulse? It will only approximate a delta function if it grows in height as it gets narrower.
$endgroup$
– John Barber
Jan 19 at 17:28
$begingroup$
The height is some constant real number h. Lets say h=1.
$endgroup$
– Vitor Ribeiro
Jan 19 at 17:50
$begingroup$
What the integral is would depend on $f$.
$endgroup$
– David C. Ullrich
Jan 19 at 18:23
$begingroup$
If h=1 then the integrand is $f(x)$ if $a-frac{w}{2}<x<a+frac{w}{2}$ and 0 otherwise. No?
$endgroup$
– Vitor Ribeiro
Jan 19 at 19:50
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience.
$endgroup$
– dantopa
Jan 19 at 17:18
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience.
$endgroup$
– dantopa
Jan 19 at 17:18
$begingroup$
What is the height of this rectangular pulse? It will only approximate a delta function if it grows in height as it gets narrower.
$endgroup$
– John Barber
Jan 19 at 17:28
$begingroup$
What is the height of this rectangular pulse? It will only approximate a delta function if it grows in height as it gets narrower.
$endgroup$
– John Barber
Jan 19 at 17:28
$begingroup$
The height is some constant real number h. Lets say h=1.
$endgroup$
– Vitor Ribeiro
Jan 19 at 17:50
$begingroup$
The height is some constant real number h. Lets say h=1.
$endgroup$
– Vitor Ribeiro
Jan 19 at 17:50
$begingroup$
What the integral is would depend on $f$.
$endgroup$
– David C. Ullrich
Jan 19 at 18:23
$begingroup$
What the integral is would depend on $f$.
$endgroup$
– David C. Ullrich
Jan 19 at 18:23
$begingroup$
If h=1 then the integrand is $f(x)$ if $a-frac{w}{2}<x<a+frac{w}{2}$ and 0 otherwise. No?
$endgroup$
– Vitor Ribeiro
Jan 19 at 19:50
$begingroup$
If h=1 then the integrand is $f(x)$ if $a-frac{w}{2}<x<a+frac{w}{2}$ and 0 otherwise. No?
$endgroup$
– Vitor Ribeiro
Jan 19 at 19:50
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You're right in that if we let $tilde{f}!(x) = f(x)Pi(x,a - w/2,a+w/2),$ $mathscr{D}$ denote the region where $Pi$ is nonzero, and $f$ be integrable on $mathscr{D}$, then
$$
tilde{f}!(x) =
begin{cases}
begin{align}
& f(x) & x in mathscr{D} \
& 0 &text{otherwise}.
end{align}
end{cases}
$$
It seems like it would make sense at first if it was true that
$$
int !tilde{f}!(x) , dx =
begin{cases}
begin{align}
& !!int ! f(x) , dx & x in mathscr{D}\
& 0 & text{otherwise},
end{align}
end{cases}
$$
but this equality is generally not true. To see this claim, let's rewrite $Pi(x, cdot, cdot)$ as
$$
Pi ! left(x, a - frac{w}{2}, a + frac{w}{2} right) equiv h ! left(x - a + frac{w}{2} right) - h ! left(x - a - frac{w}{2} right).
$$
Then
$$
int !tilde{f}!(x) , dx = ! int ! f(x) , h ! left(x - a + frac{w}{2} right) ! dx - int ! f(x) , h ! left(x - a - frac{w}{2} right) ! dx.
$$
For $x < a - w/2,$ both integrands are exactly zero and we've so far accumulated zero area under $tilde{f}!$. For $x in mathscr{D},$ $f$ is integrated normally on the left, and the right integrand is still zero. We've now accumulated some nonzero result from the left integral scanning across $mathscr{D},$ but the area previous was zero, so we must begin keeping track of the area from zero. For $x > a + w/2,$ both the left and right integrands are again zero, and we're not going to accumulate any more area under $tilde{f}!.$ But what's been covered already doesn't just disappear, so we must have that, given
$$
F(x) = int ! f(x) , dx, $$
$$
int ! tilde{f}!(x) , dx =
begin{cases}
begin{align}
&0 & x &< minmathscr{D} \
& F(x) - lim_{x downarrow min mathscr{D}} F(x) & x &in mathscr{D} \
&lim_{x downarrow max mathscr{D}} F(x) - lim_{x downarrow min mathscr{D}} F(x) & x &> max mathscr{D}
end{align}
end{cases}
$$
This argument is simply a continuity argument using the fact that the indefinite integral always comes with a free constant, even when integrating zero.
If we use this continuity, we can rid of the limits and explicitly write, using $eta = a - w/2$ and $xi = a + w/2$ as shorthand,
$$
int ! tilde{f}!(x) , dx = Big( F(x) - F(eta) Big) , h ( x - eta) - Big( F(x) - F(xi) Big) , h( x - xi).
$$
As correctly pointed out, this result may be written as
begin{align}
int tilde{f} ! (x) , dx &= h(x - eta) int_eta^x f(s) , ds - h(x - xi)int_xi^x f(s) , ds \
&= h(x - eta) int_eta^x f(s) , ds + h(x - xi)int_x^xi f(s) , ds.
end{align}
answered Jan 20 at 10:11
AEngineerAEngineer
1,5441317
$endgroup$
1
$begingroup$
Thanks. The last equation is really a good shorthand.
$endgroup$
– Vitor Ribeiro
Jan 21 at 17:13
$begingroup$
I think the last equation can be written as $int_eta^x{f(x)} dx h(x-eta)-int_psi^x{f(x)} dx h(x-psi)$
$endgroup$
– Vitor Ribeiro
Jan 31 at 15:42
$begingroup$
Yea you can write it like that too if you'd like. I think I like your expression better; I'll add it in.
$endgroup$
– AEngineer
Jan 31 at 17:49
add a comment |
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1 Answer
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active
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votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
You're right in that if we let $tilde{f}!(x) = f(x)Pi(x,a - w/2,a+w/2),$ $mathscr{D}$ denote the region where $Pi$ is nonzero, and $f$ be integrable on $mathscr{D}$, then
$$
tilde{f}!(x) =
begin{cases}
begin{align}
& f(x) & x in mathscr{D} \
& 0 &text{otherwise}.
end{align}
end{cases}
$$
It seems like it would make sense at first if it was true that
$$
int !tilde{f}!(x) , dx =
begin{cases}
begin{align}
& !!int ! f(x) , dx & x in mathscr{D}\
& 0 & text{otherwise},
end{align}
end{cases}
$$
but this equality is generally not true. To see this claim, let's rewrite $Pi(x, cdot, cdot)$ as
$$
Pi ! left(x, a - frac{w}{2}, a + frac{w}{2} right) equiv h ! left(x - a + frac{w}{2} right) - h ! left(x - a - frac{w}{2} right).
$$
Then
$$
int !tilde{f}!(x) , dx = ! int ! f(x) , h ! left(x - a + frac{w}{2} right) ! dx - int ! f(x) , h ! left(x - a - frac{w}{2} right) ! dx.
$$
For $x < a - w/2,$ both integrands are exactly zero and we've so far accumulated zero area under $tilde{f}!$. For $x in mathscr{D},$ $f$ is integrated normally on the left, and the right integrand is still zero. We've now accumulated some nonzero result from the left integral scanning across $mathscr{D},$ but the area previous was zero, so we must begin keeping track of the area from zero. For $x > a + w/2,$ both the left and right integrands are again zero, and we're not going to accumulate any more area under $tilde{f}!.$ But what's been covered already doesn't just disappear, so we must have that, given
$$
F(x) = int ! f(x) , dx, $$
$$
int ! tilde{f}!(x) , dx =
begin{cases}
begin{align}
&0 & x &< minmathscr{D} \
& F(x) - lim_{x downarrow min mathscr{D}} F(x) & x &in mathscr{D} \
&lim_{x downarrow max mathscr{D}} F(x) - lim_{x downarrow min mathscr{D}} F(x) & x &> max mathscr{D}
end{align}
end{cases}
$$
This argument is simply a continuity argument using the fact that the indefinite integral always comes with a free constant, even when integrating zero.
If we use this continuity, we can rid of the limits and explicitly write, using $eta = a - w/2$ and $xi = a + w/2$ as shorthand,
$$
int ! tilde{f}!(x) , dx = Big( F(x) - F(eta) Big) , h ( x - eta) - Big( F(x) - F(xi) Big) , h( x - xi).
$$
As correctly pointed out, this result may be written as
begin{align}
int tilde{f} ! (x) , dx &= h(x - eta) int_eta^x f(s) , ds - h(x - xi)int_xi^x f(s) , ds \
&= h(x - eta) int_eta^x f(s) , ds + h(x - xi)int_x^xi f(s) , ds.
end{align}
answered Jan 20 at 10:11
AEngineerAEngineer
1,5441317
$endgroup$
1
$begingroup$
Thanks. The last equation is really a good shorthand.
$endgroup$
– Vitor Ribeiro
Jan 21 at 17:13
$begingroup$
I think the last equation can be written as $int_eta^x{f(x)} dx h(x-eta)-int_psi^x{f(x)} dx h(x-psi)$
$endgroup$
– Vitor Ribeiro
Jan 31 at 15:42
$begingroup$
Yea you can write it like that too if you'd like. I think I like your expression better; I'll add it in.
$endgroup$
– AEngineer
Jan 31 at 17:49
add a comment |
$begingroup$
You're right in that if we let $tilde{f}!(x) = f(x)Pi(x,a - w/2,a+w/2),$ $mathscr{D}$ denote the region where $Pi$ is nonzero, and $f$ be integrable on $mathscr{D}$, then
$$
tilde{f}!(x) =
begin{cases}
begin{align}
& f(x) & x in mathscr{D} \
& 0 &text{otherwise}.
end{align}
end{cases}
$$
It seems like it would make sense at first if it was true that
$$
int !tilde{f}!(x) , dx =
begin{cases}
begin{align}
& !!int ! f(x) , dx & x in mathscr{D}\
& 0 & text{otherwise},
end{align}
end{cases}
$$
but this equality is generally not true. To see this claim, let's rewrite $Pi(x, cdot, cdot)$ as
$$
Pi ! left(x, a - frac{w}{2}, a + frac{w}{2} right) equiv h ! left(x - a + frac{w}{2} right) - h ! left(x - a - frac{w}{2} right).
$$
Then
$$
int !tilde{f}!(x) , dx = ! int ! f(x) , h ! left(x - a + frac{w}{2} right) ! dx - int ! f(x) , h ! left(x - a - frac{w}{2} right) ! dx.
$$
For $x < a - w/2,$ both integrands are exactly zero and we've so far accumulated zero area under $tilde{f}!$. For $x in mathscr{D},$ $f$ is integrated normally on the left, and the right integrand is still zero. We've now accumulated some nonzero result from the left integral scanning across $mathscr{D},$ but the area previous was zero, so we must begin keeping track of the area from zero. For $x > a + w/2,$ both the left and right integrands are again zero, and we're not going to accumulate any more area under $tilde{f}!.$ But what's been covered already doesn't just disappear, so we must have that, given
$$
F(x) = int ! f(x) , dx, $$
$$
int ! tilde{f}!(x) , dx =
begin{cases}
begin{align}
&0 & x &< minmathscr{D} \
& F(x) - lim_{x downarrow min mathscr{D}} F(x) & x &in mathscr{D} \
&lim_{x downarrow max mathscr{D}} F(x) - lim_{x downarrow min mathscr{D}} F(x) & x &> max mathscr{D}
end{align}
end{cases}
$$
This argument is simply a continuity argument using the fact that the indefinite integral always comes with a free constant, even when integrating zero.
If we use this continuity, we can rid of the limits and explicitly write, using $eta = a - w/2$ and $xi = a + w/2$ as shorthand,
$$
int ! tilde{f}!(x) , dx = Big( F(x) - F(eta) Big) , h ( x - eta) - Big( F(x) - F(xi) Big) , h( x - xi).
$$
As correctly pointed out, this result may be written as
begin{align}
int tilde{f} ! (x) , dx &= h(x - eta) int_eta^x f(s) , ds - h(x - xi)int_xi^x f(s) , ds \
&= h(x - eta) int_eta^x f(s) , ds + h(x - xi)int_x^xi f(s) , ds.
end{align}
answered Jan 20 at 10:11
AEngineerAEngineer
1,5441317
$endgroup$
1
$begingroup$
Thanks. The last equation is really a good shorthand.
$endgroup$
– Vitor Ribeiro
Jan 21 at 17:13
$begingroup$
I think the last equation can be written as $int_eta^x{f(x)} dx h(x-eta)-int_psi^x{f(x)} dx h(x-psi)$
$endgroup$
– Vitor Ribeiro
Jan 31 at 15:42
$begingroup$
Yea you can write it like that too if you'd like. I think I like your expression better; I'll add it in.
$endgroup$
– AEngineer
Jan 31 at 17:49
add a comment |
$begingroup$
You're right in that if we let $tilde{f}!(x) = f(x)Pi(x,a - w/2,a+w/2),$ $mathscr{D}$ denote the region where $Pi$ is nonzero, and $f$ be integrable on $mathscr{D}$, then
$$
tilde{f}!(x) =
begin{cases}
begin{align}
& f(x) & x in mathscr{D} \
& 0 &text{otherwise}.
end{align}
end{cases}
$$
It seems like it would make sense at first if it was true that
$$
int !tilde{f}!(x) , dx =
begin{cases}
begin{align}
& !!int ! f(x) , dx & x in mathscr{D}\
& 0 & text{otherwise},
end{align}
end{cases}
$$
but this equality is generally not true. To see this claim, let's rewrite $Pi(x, cdot, cdot)$ as
$$
Pi ! left(x, a - frac{w}{2}, a + frac{w}{2} right) equiv h ! left(x - a + frac{w}{2} right) - h ! left(x - a - frac{w}{2} right).
$$
Then
$$
int !tilde{f}!(x) , dx = ! int ! f(x) , h ! left(x - a + frac{w}{2} right) ! dx - int ! f(x) , h ! left(x - a - frac{w}{2} right) ! dx.
$$
For $x < a - w/2,$ both integrands are exactly zero and we've so far accumulated zero area under $tilde{f}!$. For $x in mathscr{D},$ $f$ is integrated normally on the left, and the right integrand is still zero. We've now accumulated some nonzero result from the left integral scanning across $mathscr{D},$ but the area previous was zero, so we must begin keeping track of the area from zero. For $x > a + w/2,$ both the left and right integrands are again zero, and we're not going to accumulate any more area under $tilde{f}!.$ But what's been covered already doesn't just disappear, so we must have that, given
$$
F(x) = int ! f(x) , dx, $$
$$
int ! tilde{f}!(x) , dx =
begin{cases}
begin{align}
&0 & x &< minmathscr{D} \
& F(x) - lim_{x downarrow min mathscr{D}} F(x) & x &in mathscr{D} \
&lim_{x downarrow max mathscr{D}} F(x) - lim_{x downarrow min mathscr{D}} F(x) & x &> max mathscr{D}
end{align}
end{cases}
$$
This argument is simply a continuity argument using the fact that the indefinite integral always comes with a free constant, even when integrating zero.
If we use this continuity, we can rid of the limits and explicitly write, using $eta = a - w/2$ and $xi = a + w/2$ as shorthand,
$$
int ! tilde{f}!(x) , dx = Big( F(x) - F(eta) Big) , h ( x - eta) - Big( F(x) - F(xi) Big) , h( x - xi).
$$
As correctly pointed out, this result may be written as
begin{align}
int tilde{f} ! (x) , dx &= h(x - eta) int_eta^x f(s) , ds - h(x - xi)int_xi^x f(s) , ds \
&= h(x - eta) int_eta^x f(s) , ds + h(x - xi)int_x^xi f(s) , ds.
end{align}
answered Jan 20 at 10:11
AEngineerAEngineer
1,5441317
$endgroup$
You're right in that if we let $tilde{f}!(x) = f(x)Pi(x,a - w/2,a+w/2),$ $mathscr{D}$ denote the region where $Pi$ is nonzero, and $f$ be integrable on $mathscr{D}$, then
$$
tilde{f}!(x) =
begin{cases}
begin{align}
& f(x) & x in mathscr{D} \
& 0 &text{otherwise}.
end{align}
end{cases}
$$
It seems like it would make sense at first if it was true that
$$
int !tilde{f}!(x) , dx =
begin{cases}
begin{align}
& !!int ! f(x) , dx & x in mathscr{D}\
& 0 & text{otherwise},
end{align}
end{cases}
$$
but this equality is generally not true. To see this claim, let's rewrite $Pi(x, cdot, cdot)$ as
$$
Pi ! left(x, a - frac{w}{2}, a + frac{w}{2} right) equiv h ! left(x - a + frac{w}{2} right) - h ! left(x - a - frac{w}{2} right).
$$
Then
$$
int !tilde{f}!(x) , dx = ! int ! f(x) , h ! left(x - a + frac{w}{2} right) ! dx - int ! f(x) , h ! left(x - a - frac{w}{2} right) ! dx.
$$
For $x < a - w/2,$ both integrands are exactly zero and we've so far accumulated zero area under $tilde{f}!$. For $x in mathscr{D},$ $f$ is integrated normally on the left, and the right integrand is still zero. We've now accumulated some nonzero result from the left integral scanning across $mathscr{D},$ but the area previous was zero, so we must begin keeping track of the area from zero. For $x > a + w/2,$ both the left and right integrands are again zero, and we're not going to accumulate any more area under $tilde{f}!.$ But what's been covered already doesn't just disappear, so we must have that, given
$$
F(x) = int ! f(x) , dx, $$
$$
int ! tilde{f}!(x) , dx =
begin{cases}
begin{align}
&0 & x &< minmathscr{D} \
& F(x) - lim_{x downarrow min mathscr{D}} F(x) & x &in mathscr{D} \
&lim_{x downarrow max mathscr{D}} F(x) - lim_{x downarrow min mathscr{D}} F(x) & x &> max mathscr{D}
end{align}
end{cases}
$$
This argument is simply a continuity argument using the fact that the indefinite integral always comes with a free constant, even when integrating zero.
If we use this continuity, we can rid of the limits and explicitly write, using $eta = a - w/2$ and $xi = a + w/2$ as shorthand,
$$
int ! tilde{f}!(x) , dx = Big( F(x) - F(eta) Big) , h ( x - eta) - Big( F(x) - F(xi) Big) , h( x - xi).
$$
As correctly pointed out, this result may be written as
begin{align}
int tilde{f} ! (x) , dx &= h(x - eta) int_eta^x f(s) , ds - h(x - xi)int_xi^x f(s) , ds \
&= h(x - eta) int_eta^x f(s) , ds + h(x - xi)int_x^xi f(s) , ds.
end{align}
answered Jan 20 at 10:11
AEngineerAEngineer
1,5441317
edited Jan 31 at 17:52
answered Jan 20 at 10:11
AEngineerAEngineer
1,5441317
answered Jan 20 at 10:11
AEngineerAEngineer
1,5441317
answered Jan 20 at 10:11
AEngineerAEngineer
1,5441317
1,5441317
1
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Thanks. The last equation is really a good shorthand.
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– Vitor Ribeiro
Jan 21 at 17:13
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I think the last equation can be written as $int_eta^x{f(x)} dx h(x-eta)-int_psi^x{f(x)} dx h(x-psi)$
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– Vitor Ribeiro
Jan 31 at 15:42
$begingroup$
Yea you can write it like that too if you'd like. I think I like your expression better; I'll add it in.
$endgroup$
– AEngineer
Jan 31 at 17:49
add a comment |
1
$begingroup$
Thanks. The last equation is really a good shorthand.
$endgroup$
– Vitor Ribeiro
Jan 21 at 17:13
$begingroup$
I think the last equation can be written as $int_eta^x{f(x)} dx h(x-eta)-int_psi^x{f(x)} dx h(x-psi)$
$endgroup$
– Vitor Ribeiro
Jan 31 at 15:42
$begingroup$
Yea you can write it like that too if you'd like. I think I like your expression better; I'll add it in.
$endgroup$
– AEngineer
Jan 31 at 17:49
1
1
$begingroup$
Thanks. The last equation is really a good shorthand.
$endgroup$
– Vitor Ribeiro
Jan 21 at 17:13
$begingroup$
Thanks. The last equation is really a good shorthand.
$endgroup$
– Vitor Ribeiro
Jan 21 at 17:13
$begingroup$
I think the last equation can be written as $int_eta^x{f(x)} dx h(x-eta)-int_psi^x{f(x)} dx h(x-psi)$
$endgroup$
– Vitor Ribeiro
Jan 31 at 15:42
$begingroup$
I think the last equation can be written as $int_eta^x{f(x)} dx h(x-eta)-int_psi^x{f(x)} dx h(x-psi)$
$endgroup$
– Vitor Ribeiro
Jan 31 at 15:42
$begingroup$
Yea you can write it like that too if you'd like. I think I like your expression better; I'll add it in.
$endgroup$
– AEngineer
Jan 31 at 17:49
$begingroup$
Yea you can write it like that too if you'd like. I think I like your expression better; I'll add it in.
$endgroup$
– AEngineer
Jan 31 at 17:49
add a comment |
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– dantopa
Jan 19 at 17:18
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What is the height of this rectangular pulse? It will only approximate a delta function if it grows in height as it gets narrower.
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– John Barber
Jan 19 at 17:28
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The height is some constant real number h. Lets say h=1.
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– Vitor Ribeiro
Jan 19 at 17:50
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What the integral is would depend on $f$.
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– David C. Ullrich
Jan 19 at 18:23
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If h=1 then the integrand is $f(x)$ if $a-frac{w}{2}<x<a+frac{w}{2}$ and 0 otherwise. No?
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– Vitor Ribeiro
Jan 19 at 19:50