Mixing
Polar plots of $sin(kx)$
$begingroup$
The plots of $sin(kx)$ over the real line are somehow boring and look essentially all the same:
For larger $k$ you cannot easily tell which $k$ it is (not only due to Moiré effects):
But when plotting $sin(kx)$ over the unit circle by
$$x(t) = cos(t) (1 + sin(kt))$$
$$y(t) = sin(t) (1 + sin(kt))$$
interesting patterns emerge, e.g. for $k = 1,2,dots,8$
Interlude: Note that these plots are the stream plots of the complex functions
$$f_k(z)=frac{1}{2i}(z^k - overline{z^k})z $$
on the unit circle (if I didn't make a mistake). Note that $f_k(z)$ is not a holomorphic function.
You may compare this with the stream plot of
$$g_k(z)=frac{1}{2i}(z^k - overline{z^k}) = f_k (z)/z$$
with $g_k(e^{ivarphi}) = sin(kvarphi) $:
[End of the interlude.]
Even for larger $k$ one still could tell $k$:
Furthermore you can see specific effects of rational frequencies $k$ which are invisible in the linear plots. Here are the plots for $k=frac{2n +1}{4}$ with $n = 1,2,dots,8$:
The main advantage of the linear plot of $sin(kx)$ is that it has a simple geometrical interpretation resp. construction: It's the plot of the y-coordinate of a point which rotates with constant speed $k$ on the fixed unit circle:
Alternatively, you can look at the sine as the projection of a helix seen from the side. This was the idea behind one of the earliest depictions of the sine found at Dürer:
Compare this to the cases of cycloids and epicycles. These also have a simple geometrical interpretation - being the plots of the x- and y-coordinates of a point on a circle that rolls on the line
resp. moves on another circle with constant speed
My question is:
By which geometrical interpretation resp. construction (involving circles or
ellipses or whatsoever) can the polar plots of $sin$ be seen resp. generated? Which construction relates to the construction of $sin$ by a rotating point on a circle in the way that the construction of epicycles relates to the construction of cycloids?
Just musing: Might this question have to do with this other question on Hidden patterns in $sin(kx^2)$? (Probably not because you cannot sensibly plot $sin(kx^2)$ radially, since there is no well-defined period.)
geometry trigonometry visualization
asked Jan 30 at 10:36
Hans-Peter StrickerHans-Peter Stricker
6,73543996
$endgroup$
|
show 3 more comments
$begingroup$
The plots of $sin(kx)$ over the real line are somehow boring and look essentially all the same:
For larger $k$ you cannot easily tell which $k$ it is (not only due to Moiré effects):
But when plotting $sin(kx)$ over the unit circle by
$$x(t) = cos(t) (1 + sin(kt))$$
$$y(t) = sin(t) (1 + sin(kt))$$
interesting patterns emerge, e.g. for $k = 1,2,dots,8$
Interlude: Note that these plots are the stream plots of the complex functions
$$f_k(z)=frac{1}{2i}(z^k - overline{z^k})z $$
on the unit circle (if I didn't make a mistake). Note that $f_k(z)$ is not a holomorphic function.
You may compare this with the stream plot of
$$g_k(z)=frac{1}{2i}(z^k - overline{z^k}) = f_k (z)/z$$
with $g_k(e^{ivarphi}) = sin(kvarphi) $:
[End of the interlude.]
Even for larger $k$ one still could tell $k$:
Furthermore you can see specific effects of rational frequencies $k$ which are invisible in the linear plots. Here are the plots for $k=frac{2n +1}{4}$ with $n = 1,2,dots,8$:
The main advantage of the linear plot of $sin(kx)$ is that it has a simple geometrical interpretation resp. construction: It's the plot of the y-coordinate of a point which rotates with constant speed $k$ on the fixed unit circle:
Alternatively, you can look at the sine as the projection of a helix seen from the side. This was the idea behind one of the earliest depictions of the sine found at Dürer:
Compare this to the cases of cycloids and epicycles. These also have a simple geometrical interpretation - being the plots of the x- and y-coordinates of a point on a circle that rolls on the line
resp. moves on another circle with constant speed
My question is:
By which geometrical interpretation resp. construction (involving circles or
ellipses or whatsoever) can the polar plots of $sin$ be seen resp. generated? Which construction relates to the construction of $sin$ by a rotating point on a circle in the way that the construction of epicycles relates to the construction of cycloids?
Just musing: Might this question have to do with this other question on Hidden patterns in $sin(kx^2)$? (Probably not because you cannot sensibly plot $sin(kx^2)$ radially, since there is no well-defined period.)
geometry trigonometry visualization
asked Jan 30 at 10:36
Hans-Peter StrickerHans-Peter Stricker
6,73543996
$endgroup$
$begingroup$
Can you unabbreviate or clarify the meaning of "resp.", please? I'm not familiar with it.
$endgroup$
– timtfj
Jan 30 at 11:08
$begingroup$
(I still upvoted though)
$endgroup$
– timtfj
Jan 30 at 11:16
1
$begingroup$
"Resp." abbreviates "respectively" which just means "or" (with a slightly different connotation).
$endgroup$
– Hans-Peter Stricker
Jan 30 at 11:22
$begingroup$
@HansStricker I think it's more common to use "re" instead of "resp".
$endgroup$
– Jam
Jan 30 at 12:23
$begingroup$
@HansStricker I'm sure some people use "resp" but it's not very common. I'm a native English speaker and I've not seen it more than a couple of times in my life :)
$endgroup$
– Jam
Jan 30 at 12:31
|
show 3 more comments
$begingroup$
The plots of $sin(kx)$ over the real line are somehow boring and look essentially all the same:
For larger $k$ you cannot easily tell which $k$ it is (not only due to Moiré effects):
But when plotting $sin(kx)$ over the unit circle by
$$x(t) = cos(t) (1 + sin(kt))$$
$$y(t) = sin(t) (1 + sin(kt))$$
interesting patterns emerge, e.g. for $k = 1,2,dots,8$
Interlude: Note that these plots are the stream plots of the complex functions
$$f_k(z)=frac{1}{2i}(z^k - overline{z^k})z $$
on the unit circle (if I didn't make a mistake). Note that $f_k(z)$ is not a holomorphic function.
You may compare this with the stream plot of
$$g_k(z)=frac{1}{2i}(z^k - overline{z^k}) = f_k (z)/z$$
with $g_k(e^{ivarphi}) = sin(kvarphi) $:
[End of the interlude.]
Even for larger $k$ one still could tell $k$:
Furthermore you can see specific effects of rational frequencies $k$ which are invisible in the linear plots. Here are the plots for $k=frac{2n +1}{4}$ with $n = 1,2,dots,8$:
The main advantage of the linear plot of $sin(kx)$ is that it has a simple geometrical interpretation resp. construction: It's the plot of the y-coordinate of a point which rotates with constant speed $k$ on the fixed unit circle:
Alternatively, you can look at the sine as the projection of a helix seen from the side. This was the idea behind one of the earliest depictions of the sine found at Dürer:
Compare this to the cases of cycloids and epicycles. These also have a simple geometrical interpretation - being the plots of the x- and y-coordinates of a point on a circle that rolls on the line
resp. moves on another circle with constant speed
My question is:
By which geometrical interpretation resp. construction (involving circles or
ellipses or whatsoever) can the polar plots of $sin$ be seen resp. generated? Which construction relates to the construction of $sin$ by a rotating point on a circle in the way that the construction of epicycles relates to the construction of cycloids?
Just musing: Might this question have to do with this other question on Hidden patterns in $sin(kx^2)$? (Probably not because you cannot sensibly plot $sin(kx^2)$ radially, since there is no well-defined period.)
geometry trigonometry visualization
asked Jan 30 at 10:36
Hans-Peter StrickerHans-Peter Stricker
6,73543996
$endgroup$
The plots of $sin(kx)$ over the real line are somehow boring and look essentially all the same:
For larger $k$ you cannot easily tell which $k$ it is (not only due to Moiré effects):
But when plotting $sin(kx)$ over the unit circle by
$$x(t) = cos(t) (1 + sin(kt))$$
$$y(t) = sin(t) (1 + sin(kt))$$
interesting patterns emerge, e.g. for $k = 1,2,dots,8$
Interlude: Note that these plots are the stream plots of the complex functions
$$f_k(z)=frac{1}{2i}(z^k - overline{z^k})z $$
on the unit circle (if I didn't make a mistake). Note that $f_k(z)$ is not a holomorphic function.
You may compare this with the stream plot of
$$g_k(z)=frac{1}{2i}(z^k - overline{z^k}) = f_k (z)/z$$
with $g_k(e^{ivarphi}) = sin(kvarphi) $:
[End of the interlude.]
Even for larger $k$ one still could tell $k$:
Furthermore you can see specific effects of rational frequencies $k$ which are invisible in the linear plots. Here are the plots for $k=frac{2n +1}{4}$ with $n = 1,2,dots,8$:
The main advantage of the linear plot of $sin(kx)$ is that it has a simple geometrical interpretation resp. construction: It's the plot of the y-coordinate of a point which rotates with constant speed $k$ on the fixed unit circle:
Alternatively, you can look at the sine as the projection of a helix seen from the side. This was the idea behind one of the earliest depictions of the sine found at Dürer:
Compare this to the cases of cycloids and epicycles. These also have a simple geometrical interpretation - being the plots of the x- and y-coordinates of a point on a circle that rolls on the line
resp. moves on another circle with constant speed
My question is:
By which geometrical interpretation resp. construction (involving circles or
ellipses or whatsoever) can the polar plots of $sin$ be seen resp. generated? Which construction relates to the construction of $sin$ by a rotating point on a circle in the way that the construction of epicycles relates to the construction of cycloids?
Just musing: Might this question have to do with this other question on Hidden patterns in $sin(kx^2)$? (Probably not because you cannot sensibly plot $sin(kx^2)$ radially, since there is no well-defined period.)
geometry trigonometry visualization
geometry trigonometry visualization
asked Jan 30 at 10:36
Hans-Peter StrickerHans-Peter Stricker
6,73543996
asked Jan 30 at 10:36
Hans-Peter StrickerHans-Peter Stricker
6,73543996
edited Jan 30 at 20:02
Hans-Peter Stricker
asked Jan 30 at 10:36
Hans-Peter StrickerHans-Peter Stricker
6,73543996
asked Jan 30 at 10:36
Hans-Peter StrickerHans-Peter Stricker
6,73543996
asked Jan 30 at 10:36
Hans-Peter StrickerHans-Peter Stricker
6,73543996
6,73543996
$begingroup$
Can you unabbreviate or clarify the meaning of "resp.", please? I'm not familiar with it.
$endgroup$
– timtfj
Jan 30 at 11:08
$begingroup$
(I still upvoted though)
$endgroup$
– timtfj
Jan 30 at 11:16
1
$begingroup$
"Resp." abbreviates "respectively" which just means "or" (with a slightly different connotation).
$endgroup$
– Hans-Peter Stricker
Jan 30 at 11:22
$begingroup$
@HansStricker I think it's more common to use "re" instead of "resp".
$endgroup$
– Jam
Jan 30 at 12:23
$begingroup$
@HansStricker I'm sure some people use "resp" but it's not very common. I'm a native English speaker and I've not seen it more than a couple of times in my life :)
$endgroup$
– Jam
Jan 30 at 12:31
|
show 3 more comments
$begingroup$
Can you unabbreviate or clarify the meaning of "resp.", please? I'm not familiar with it.
$endgroup$
– timtfj
Jan 30 at 11:08
$begingroup$
(I still upvoted though)
$endgroup$
– timtfj
Jan 30 at 11:16
1
$begingroup$
"Resp." abbreviates "respectively" which just means "or" (with a slightly different connotation).
$endgroup$
– Hans-Peter Stricker
Jan 30 at 11:22
$begingroup$
@HansStricker I think it's more common to use "re" instead of "resp".
$endgroup$
– Jam
Jan 30 at 12:23
$begingroup$
@HansStricker I'm sure some people use "resp" but it's not very common. I'm a native English speaker and I've not seen it more than a couple of times in my life :)
$endgroup$
– Jam
Jan 30 at 12:31
$begingroup$
Can you unabbreviate or clarify the meaning of "resp.", please? I'm not familiar with it.
$endgroup$
– timtfj
Jan 30 at 11:08
$begingroup$
Can you unabbreviate or clarify the meaning of "resp.", please? I'm not familiar with it.
$endgroup$
– timtfj
Jan 30 at 11:08
$begingroup$
(I still upvoted though)
$endgroup$
– timtfj
Jan 30 at 11:16
$begingroup$
(I still upvoted though)
$endgroup$
– timtfj
Jan 30 at 11:16
1
1
$begingroup$
"Resp." abbreviates "respectively" which just means "or" (with a slightly different connotation).
$endgroup$
– Hans-Peter Stricker
Jan 30 at 11:22
$begingroup$
"Resp." abbreviates "respectively" which just means "or" (with a slightly different connotation).
$endgroup$
– Hans-Peter Stricker
Jan 30 at 11:22
$begingroup$
@HansStricker I think it's more common to use "re" instead of "resp".
$endgroup$
– Jam
Jan 30 at 12:23
$begingroup$
@HansStricker I think it's more common to use "re" instead of "resp".
$endgroup$
– Jam
Jan 30 at 12:23
$begingroup$
@HansStricker I'm sure some people use "resp" but it's not very common. I'm a native English speaker and I've not seen it more than a couple of times in my life :)
$endgroup$
– Jam
Jan 30 at 12:31
$begingroup$
@HansStricker I'm sure some people use "resp" but it's not very common. I'm a native English speaker and I've not seen it more than a couple of times in my life :)
$endgroup$
– Jam
Jan 30 at 12:31
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Thanks to Yves Danoust's hint to Grandi's roses I found this "answer without words" best fitting to my question:
– even though I'm not quite sure how to choose the parameters (radius of the circle, rotation speeds of line and circle) to exactly reproduce my plot:
– and even though I don't see clearly what happens. As Robert Ferréol describes it on his web page on Grandi's roses:
The roses can also be obtained as the trajectories of the second intersection point between a line and a circle in uniform rotation
around one of their points.
The difference between Grandi's rose and my plot is the order, in which the curve is drawn: not lobe by lobe, but as a rose. This difference vanishes, when we plot $sin(e^{ikvarphi})$ not over the circle (as the base line), but from the origin. Here for $0 leq varphi < pi$ (left) and $pi < varphi < 2pi$ (right):
answered Jan 30 at 14:47
Hans-Peter StrickerHans-Peter Stricker
6,73543996
$endgroup$
$begingroup$
better you take just a ray from the origin: then it is easier to controll the parameters
$endgroup$
– G Cab
Jan 30 at 14:58
$begingroup$
@GCab: Instead of what? Don't I need the whole line through the origin? And how would it help me to control the parameters? (Note that it was not me who created the animated gif.)
$endgroup$
– Hans-Peter Stricker
Jan 30 at 15:01
$begingroup$
judging by eye from the animation, that's constructed by crossing the entire line ($pm rho$) with the circle. If you take instead only the ray, then the ratio of the two angular speeds will govern (better) the number of lobes.
$endgroup$
– G Cab
Jan 30 at 15:50
add a comment |
$begingroup$
I did not grasp exactly what you are asking, however it might be of interest to know that in "old times" electrical engineers were used to visualize phase and frequency of a sinusoidal wave by feeding it to the $x$ axis of an oscilloscope in combination to a known and tunable signal (sinusoidal, triangular , ..) fed to the $y$ axis and produce a Lissajous figure.
answered Jan 30 at 12:38
G CabG Cab
20.4k31341
$endgroup$
$begingroup$
No, the question is not related to the Lissajous curves.
$endgroup$
– Yves Daoust
Jan 30 at 13:58
$begingroup$
@YvesDaoust: now that the question has been edited, it is more clear.
$endgroup$
– G Cab
Jan 30 at 15:00
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
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$begingroup$
Thanks to Yves Danoust's hint to Grandi's roses I found this "answer without words" best fitting to my question:
– even though I'm not quite sure how to choose the parameters (radius of the circle, rotation speeds of line and circle) to exactly reproduce my plot:
– and even though I don't see clearly what happens. As Robert Ferréol describes it on his web page on Grandi's roses:
The roses can also be obtained as the trajectories of the second intersection point between a line and a circle in uniform rotation
around one of their points.
The difference between Grandi's rose and my plot is the order, in which the curve is drawn: not lobe by lobe, but as a rose. This difference vanishes, when we plot $sin(e^{ikvarphi})$ not over the circle (as the base line), but from the origin. Here for $0 leq varphi < pi$ (left) and $pi < varphi < 2pi$ (right):
answered Jan 30 at 14:47
Hans-Peter StrickerHans-Peter Stricker
6,73543996
$endgroup$
$begingroup$
better you take just a ray from the origin: then it is easier to controll the parameters
$endgroup$
– G Cab
Jan 30 at 14:58
$begingroup$
@GCab: Instead of what? Don't I need the whole line through the origin? And how would it help me to control the parameters? (Note that it was not me who created the animated gif.)
$endgroup$
– Hans-Peter Stricker
Jan 30 at 15:01
$begingroup$
judging by eye from the animation, that's constructed by crossing the entire line ($pm rho$) with the circle. If you take instead only the ray, then the ratio of the two angular speeds will govern (better) the number of lobes.
$endgroup$
– G Cab
Jan 30 at 15:50
add a comment |
$begingroup$
Thanks to Yves Danoust's hint to Grandi's roses I found this "answer without words" best fitting to my question:
– even though I'm not quite sure how to choose the parameters (radius of the circle, rotation speeds of line and circle) to exactly reproduce my plot:
– and even though I don't see clearly what happens. As Robert Ferréol describes it on his web page on Grandi's roses:
The roses can also be obtained as the trajectories of the second intersection point between a line and a circle in uniform rotation
around one of their points.
The difference between Grandi's rose and my plot is the order, in which the curve is drawn: not lobe by lobe, but as a rose. This difference vanishes, when we plot $sin(e^{ikvarphi})$ not over the circle (as the base line), but from the origin. Here for $0 leq varphi < pi$ (left) and $pi < varphi < 2pi$ (right):
answered Jan 30 at 14:47
Hans-Peter StrickerHans-Peter Stricker
6,73543996
$endgroup$
$begingroup$
better you take just a ray from the origin: then it is easier to controll the parameters
$endgroup$
– G Cab
Jan 30 at 14:58
$begingroup$
@GCab: Instead of what? Don't I need the whole line through the origin? And how would it help me to control the parameters? (Note that it was not me who created the animated gif.)
$endgroup$
– Hans-Peter Stricker
Jan 30 at 15:01
$begingroup$
judging by eye from the animation, that's constructed by crossing the entire line ($pm rho$) with the circle. If you take instead only the ray, then the ratio of the two angular speeds will govern (better) the number of lobes.
$endgroup$
– G Cab
Jan 30 at 15:50
add a comment |
$begingroup$
Thanks to Yves Danoust's hint to Grandi's roses I found this "answer without words" best fitting to my question:
– even though I'm not quite sure how to choose the parameters (radius of the circle, rotation speeds of line and circle) to exactly reproduce my plot:
– and even though I don't see clearly what happens. As Robert Ferréol describes it on his web page on Grandi's roses:
The roses can also be obtained as the trajectories of the second intersection point between a line and a circle in uniform rotation
around one of their points.
The difference between Grandi's rose and my plot is the order, in which the curve is drawn: not lobe by lobe, but as a rose. This difference vanishes, when we plot $sin(e^{ikvarphi})$ not over the circle (as the base line), but from the origin. Here for $0 leq varphi < pi$ (left) and $pi < varphi < 2pi$ (right):
answered Jan 30 at 14:47
Hans-Peter StrickerHans-Peter Stricker
6,73543996
$endgroup$
Thanks to Yves Danoust's hint to Grandi's roses I found this "answer without words" best fitting to my question:
– even though I'm not quite sure how to choose the parameters (radius of the circle, rotation speeds of line and circle) to exactly reproduce my plot:
– and even though I don't see clearly what happens. As Robert Ferréol describes it on his web page on Grandi's roses:
The roses can also be obtained as the trajectories of the second intersection point between a line and a circle in uniform rotation
around one of their points.
The difference between Grandi's rose and my plot is the order, in which the curve is drawn: not lobe by lobe, but as a rose. This difference vanishes, when we plot $sin(e^{ikvarphi})$ not over the circle (as the base line), but from the origin. Here for $0 leq varphi < pi$ (left) and $pi < varphi < 2pi$ (right):
answered Jan 30 at 14:47
Hans-Peter StrickerHans-Peter Stricker
6,73543996
edited Jan 31 at 12:45
answered Jan 30 at 14:47
Hans-Peter StrickerHans-Peter Stricker
6,73543996
answered Jan 30 at 14:47
Hans-Peter StrickerHans-Peter Stricker
6,73543996
answered Jan 30 at 14:47
Hans-Peter StrickerHans-Peter Stricker
6,73543996
6,73543996
$begingroup$
better you take just a ray from the origin: then it is easier to controll the parameters
$endgroup$
– G Cab
Jan 30 at 14:58
$begingroup$
@GCab: Instead of what? Don't I need the whole line through the origin? And how would it help me to control the parameters? (Note that it was not me who created the animated gif.)
$endgroup$
– Hans-Peter Stricker
Jan 30 at 15:01
$begingroup$
judging by eye from the animation, that's constructed by crossing the entire line ($pm rho$) with the circle. If you take instead only the ray, then the ratio of the two angular speeds will govern (better) the number of lobes.
$endgroup$
– G Cab
Jan 30 at 15:50
add a comment |
$begingroup$
better you take just a ray from the origin: then it is easier to controll the parameters
$endgroup$
– G Cab
Jan 30 at 14:58
$begingroup$
@GCab: Instead of what? Don't I need the whole line through the origin? And how would it help me to control the parameters? (Note that it was not me who created the animated gif.)
$endgroup$
– Hans-Peter Stricker
Jan 30 at 15:01
$begingroup$
judging by eye from the animation, that's constructed by crossing the entire line ($pm rho$) with the circle. If you take instead only the ray, then the ratio of the two angular speeds will govern (better) the number of lobes.
$endgroup$
– G Cab
Jan 30 at 15:50
$begingroup$
better you take just a ray from the origin: then it is easier to controll the parameters
$endgroup$
– G Cab
Jan 30 at 14:58
$begingroup$
better you take just a ray from the origin: then it is easier to controll the parameters
$endgroup$
– G Cab
Jan 30 at 14:58
$begingroup$
@GCab: Instead of what? Don't I need the whole line through the origin? And how would it help me to control the parameters? (Note that it was not me who created the animated gif.)
$endgroup$
– Hans-Peter Stricker
Jan 30 at 15:01
$begingroup$
@GCab: Instead of what? Don't I need the whole line through the origin? And how would it help me to control the parameters? (Note that it was not me who created the animated gif.)
$endgroup$
– Hans-Peter Stricker
Jan 30 at 15:01
$begingroup$
judging by eye from the animation, that's constructed by crossing the entire line ($pm rho$) with the circle. If you take instead only the ray, then the ratio of the two angular speeds will govern (better) the number of lobes.
$endgroup$
– G Cab
Jan 30 at 15:50
$begingroup$
judging by eye from the animation, that's constructed by crossing the entire line ($pm rho$) with the circle. If you take instead only the ray, then the ratio of the two angular speeds will govern (better) the number of lobes.
$endgroup$
– G Cab
Jan 30 at 15:50
add a comment |
$begingroup$
I did not grasp exactly what you are asking, however it might be of interest to know that in "old times" electrical engineers were used to visualize phase and frequency of a sinusoidal wave by feeding it to the $x$ axis of an oscilloscope in combination to a known and tunable signal (sinusoidal, triangular , ..) fed to the $y$ axis and produce a Lissajous figure.
answered Jan 30 at 12:38
G CabG Cab
20.4k31341
$endgroup$
$begingroup$
No, the question is not related to the Lissajous curves.
$endgroup$
– Yves Daoust
Jan 30 at 13:58
$begingroup$
@YvesDaoust: now that the question has been edited, it is more clear.
$endgroup$
– G Cab
Jan 30 at 15:00
add a comment |
$begingroup$
I did not grasp exactly what you are asking, however it might be of interest to know that in "old times" electrical engineers were used to visualize phase and frequency of a sinusoidal wave by feeding it to the $x$ axis of an oscilloscope in combination to a known and tunable signal (sinusoidal, triangular , ..) fed to the $y$ axis and produce a Lissajous figure.
answered Jan 30 at 12:38
G CabG Cab
20.4k31341
$endgroup$
$begingroup$
No, the question is not related to the Lissajous curves.
$endgroup$
– Yves Daoust
Jan 30 at 13:58
$begingroup$
@YvesDaoust: now that the question has been edited, it is more clear.
$endgroup$
– G Cab
Jan 30 at 15:00
add a comment |
$begingroup$
I did not grasp exactly what you are asking, however it might be of interest to know that in "old times" electrical engineers were used to visualize phase and frequency of a sinusoidal wave by feeding it to the $x$ axis of an oscilloscope in combination to a known and tunable signal (sinusoidal, triangular , ..) fed to the $y$ axis and produce a Lissajous figure.
answered Jan 30 at 12:38
G CabG Cab
20.4k31341
$endgroup$
I did not grasp exactly what you are asking, however it might be of interest to know that in "old times" electrical engineers were used to visualize phase and frequency of a sinusoidal wave by feeding it to the $x$ axis of an oscilloscope in combination to a known and tunable signal (sinusoidal, triangular , ..) fed to the $y$ axis and produce a Lissajous figure.
answered Jan 30 at 12:38
G CabG Cab
20.4k31341
answered Jan 30 at 12:38
G CabG Cab
20.4k31341
answered Jan 30 at 12:38
G CabG Cab
20.4k31341
answered Jan 30 at 12:38
G CabG Cab
20.4k31341
20.4k31341
$begingroup$
No, the question is not related to the Lissajous curves.
$endgroup$
– Yves Daoust
Jan 30 at 13:58
$begingroup$
@YvesDaoust: now that the question has been edited, it is more clear.
$endgroup$
– G Cab
Jan 30 at 15:00
add a comment |
$begingroup$
No, the question is not related to the Lissajous curves.
$endgroup$
– Yves Daoust
Jan 30 at 13:58
$begingroup$
@YvesDaoust: now that the question has been edited, it is more clear.
$endgroup$
– G Cab
Jan 30 at 15:00
$begingroup$
No, the question is not related to the Lissajous curves.
$endgroup$
– Yves Daoust
Jan 30 at 13:58
$begingroup$
No, the question is not related to the Lissajous curves.
$endgroup$
– Yves Daoust
Jan 30 at 13:58
$begingroup$
@YvesDaoust: now that the question has been edited, it is more clear.
$endgroup$
– G Cab
Jan 30 at 15:00
$begingroup$
@YvesDaoust: now that the question has been edited, it is more clear.
$endgroup$
– G Cab
Jan 30 at 15:00
add a comment |
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$begingroup$
Can you unabbreviate or clarify the meaning of "resp.", please? I'm not familiar with it.
$endgroup$
– timtfj
Jan 30 at 11:08
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(I still upvoted though)
$endgroup$
– timtfj
Jan 30 at 11:16
1
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"Resp." abbreviates "respectively" which just means "or" (with a slightly different connotation).
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– Hans-Peter Stricker
Jan 30 at 11:22
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@HansStricker I think it's more common to use "re" instead of "resp".
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– Jam
Jan 30 at 12:23
$begingroup$
@HansStricker I'm sure some people use "resp" but it's not very common. I'm a native English speaker and I've not seen it more than a couple of times in my life :)
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– Jam
Jan 30 at 12:31