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How to find the phase-shift from the sum of two sines?
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I want to find out how interference of two sine waves can affect the output-phase of the interfered wave.
Consider two waves,
$$ E_1 = sin(x) \
E_2 = 2 sin(x + delta)
$$
First off, I don't know how to prove it, but I can see visually (plotting numerically) that the sum of these waves looks a new sine wave.
I want to find out what the phase of $E_1 + E_2 $ looks like. First I tried finding using functions like ArcSin() and Ln() but ran into trouble for both methods. For example when I try ArcSin(Sin[x] + 2*Sin(x - $delta$)), I get answers that disagree with my numerical answers.
Numerically, I solve for zeros and find the one with a positive derivative in a 2-pi region.
Now I plot the phase-shift of $E_3$ as a function of $delta$ (in blue) and compare it to $E_2$ (in purple):
Is there a "formula" I can use to find this answer without having to trace through one of the zeros? I think the key when using something like ArcSin is using the right normalization (I think it only works for sine of amplitude 1), but I'm not sure exactly the proper way of doing it.
trigonometry wave-equation
asked Aug 6 at 2:06
Steven Sagona
1257
add a comment |Â
up vote
2
down vote
favorite
I want to find out how interference of two sine waves can affect the output-phase of the interfered wave.
Consider two waves,
$$ E_1 = sin(x) \
E_2 = 2 sin(x + delta)
$$
First off, I don't know how to prove it, but I can see visually (plotting numerically) that the sum of these waves looks a new sine wave.
I want to find out what the phase of $E_1 + E_2 $ looks like. First I tried finding using functions like ArcSin() and Ln() but ran into trouble for both methods. For example when I try ArcSin(Sin[x] + 2*Sin(x - $delta$)), I get answers that disagree with my numerical answers.
Numerically, I solve for zeros and find the one with a positive derivative in a 2-pi region.
Now I plot the phase-shift of $E_3$ as a function of $delta$ (in blue) and compare it to $E_2$ (in purple):
Is there a "formula" I can use to find this answer without having to trace through one of the zeros? I think the key when using something like ArcSin is using the right normalization (I think it only works for sine of amplitude 1), but I'm not sure exactly the proper way of doing it.
trigonometry wave-equation
asked Aug 6 at 2:06
Steven Sagona
1257
1
I think this link might be useful for you: scipp.ucsc.edu/~haber/ph5B/addsine.pdf
– Karn Watcharasupat
Aug 6 at 2:22
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I want to find out how interference of two sine waves can affect the output-phase of the interfered wave.
Consider two waves,
$$ E_1 = sin(x) \
E_2 = 2 sin(x + delta)
$$
First off, I don't know how to prove it, but I can see visually (plotting numerically) that the sum of these waves looks a new sine wave.
I want to find out what the phase of $E_1 + E_2 $ looks like. First I tried finding using functions like ArcSin() and Ln() but ran into trouble for both methods. For example when I try ArcSin(Sin[x] + 2*Sin(x - $delta$)), I get answers that disagree with my numerical answers.
Numerically, I solve for zeros and find the one with a positive derivative in a 2-pi region.
Now I plot the phase-shift of $E_3$ as a function of $delta$ (in blue) and compare it to $E_2$ (in purple):
Is there a "formula" I can use to find this answer without having to trace through one of the zeros? I think the key when using something like ArcSin is using the right normalization (I think it only works for sine of amplitude 1), but I'm not sure exactly the proper way of doing it.
trigonometry wave-equation
asked Aug 6 at 2:06
Steven Sagona
1257
I want to find out how interference of two sine waves can affect the output-phase of the interfered wave.
Consider two waves,
$$ E_1 = sin(x) \
E_2 = 2 sin(x + delta)
$$
First off, I don't know how to prove it, but I can see visually (plotting numerically) that the sum of these waves looks a new sine wave.
I want to find out what the phase of $E_1 + E_2 $ looks like. First I tried finding using functions like ArcSin() and Ln() but ran into trouble for both methods. For example when I try ArcSin(Sin[x] + 2*Sin(x - $delta$)), I get answers that disagree with my numerical answers.
Numerically, I solve for zeros and find the one with a positive derivative in a 2-pi region.
Now I plot the phase-shift of $E_3$ as a function of $delta$ (in blue) and compare it to $E_2$ (in purple):
Is there a "formula" I can use to find this answer without having to trace through one of the zeros? I think the key when using something like ArcSin is using the right normalization (I think it only works for sine of amplitude 1), but I'm not sure exactly the proper way of doing it.
trigonometry wave-equation
asked Aug 6 at 2:06
Steven Sagona
1257
asked Aug 6 at 2:06
Steven Sagona
1257
asked Aug 6 at 2:06
Steven Sagona
1257
asked Aug 6 at 2:06
Steven Sagona
1257
1257
1
I think this link might be useful for you: scipp.ucsc.edu/~haber/ph5B/addsine.pdf
– Karn Watcharasupat
Aug 6 at 2:22
add a comment |Â
1
I think this link might be useful for you: scipp.ucsc.edu/~haber/ph5B/addsine.pdf
– Karn Watcharasupat
Aug 6 at 2:22
1
1
I think this link might be useful for you: scipp.ucsc.edu/~haber/ph5B/addsine.pdf
– Karn Watcharasupat
Aug 6 at 2:22
I think this link might be useful for you: scipp.ucsc.edu/~haber/ph5B/addsine.pdf
– Karn Watcharasupat
Aug 6 at 2:22
add a comment |Â
2 Answers
2
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oldest
votes
up vote
1
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accepted
beginalign
& sin x + 2sin(x+delta) \[10pt]
= & sin x + 2sin xcosdelta + 2cos x sindelta \[10pt]
= & (1+2cosdelta) sin x + (2sindelta) cos x \[10pt]
= & Asin x + Bcos x \[10pt]
= & sqrtA^2+B^2,, left( frac A sqrtA^2+B^2,, ,sin x + frac B sqrtA^2+B^2,, , cos x right) \[10pt]
= & sqrtA^2+B^2,, big( cosvarphi sin x + sinvarphi cos x big) \
& qquad textThese coefficients can be written as the cosine and sine \
& qquad textof some angle $varphi$ because the sum of their squares is $1.$ \[10pt]
= & sqrtA^2+B^2,, , sin(x+varphi).
endalign
In this case we have
$$
A^2+B^2 = (1+4cosdelta + 4cos^2delta) + 4sin^2delta = 5 + 4cosdelta.
$$
answered Aug 6 at 4:23
Michael Hardy
204k23186463
add a comment |Â
up vote
2
down vote
We can write $E_1$ and $E_2$ as the imaginary parts of $e^ix$ and $2 e^i(x+delta)$, and by linearity $E_1 + E_2$ will be the imaginary part of the sum of these two complex numbers:
$$e^ix + 2e^i(x+delta) = e^ix(1 + 2 e^idelta) = e^ix(1 + 2 cos delta + i 2 sin delta) = e^ixleft(sqrt1+ 4 cosdelta + 4 cos^2delta + 4 sin^2 deltae^ioperatornameatan2(2sindelta,1+2cosdelta)right) $$
$$ = sqrt5+ 4 cosdelta e^i[x + operatornameatan2(2sindelta,1+2cosdelta)].$$
Taking the imaginary part, we therefore find
$$ E_1 + E_2 = left(sqrt5+ 4 cosdeltaright) sinleft(x + operatornameatan2 left( 2sindelta, 1+2cosdeltaright) right),$$
so the phase-shift is given by
$$operatornameatan2left(2sindelta,1+2cosdeltaright).$$
answered Aug 6 at 2:27
Matthew Kvalheim
671416
Note the conspicuous typographical difference between $itextatan2(cdots)$ and $ioperatornameatan2(cdots).$ With operatornameatan2 the amount of space to the left and right of $operatornameatan2$ depends on the context. I edited this answer accordingly. $qquad$
– Michael Hardy
Aug 6 at 4:46
@MichaelHardy Thanks for the tip.
– Matthew Kvalheim
Aug 6 at 5:32
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
beginalign
& sin x + 2sin(x+delta) \[10pt]
= & sin x + 2sin xcosdelta + 2cos x sindelta \[10pt]
= & (1+2cosdelta) sin x + (2sindelta) cos x \[10pt]
= & Asin x + Bcos x \[10pt]
= & sqrtA^2+B^2,, left( frac A sqrtA^2+B^2,, ,sin x + frac B sqrtA^2+B^2,, , cos x right) \[10pt]
= & sqrtA^2+B^2,, big( cosvarphi sin x + sinvarphi cos x big) \
& qquad textThese coefficients can be written as the cosine and sine \
& qquad textof some angle $varphi$ because the sum of their squares is $1.$ \[10pt]
= & sqrtA^2+B^2,, , sin(x+varphi).
endalign
In this case we have
$$
A^2+B^2 = (1+4cosdelta + 4cos^2delta) + 4sin^2delta = 5 + 4cosdelta.
$$
answered Aug 6 at 4:23
Michael Hardy
204k23186463
add a comment |Â
up vote
1
down vote
accepted
beginalign
& sin x + 2sin(x+delta) \[10pt]
= & sin x + 2sin xcosdelta + 2cos x sindelta \[10pt]
= & (1+2cosdelta) sin x + (2sindelta) cos x \[10pt]
= & Asin x + Bcos x \[10pt]
= & sqrtA^2+B^2,, left( frac A sqrtA^2+B^2,, ,sin x + frac B sqrtA^2+B^2,, , cos x right) \[10pt]
= & sqrtA^2+B^2,, big( cosvarphi sin x + sinvarphi cos x big) \
& qquad textThese coefficients can be written as the cosine and sine \
& qquad textof some angle $varphi$ because the sum of their squares is $1.$ \[10pt]
= & sqrtA^2+B^2,, , sin(x+varphi).
endalign
In this case we have
$$
A^2+B^2 = (1+4cosdelta + 4cos^2delta) + 4sin^2delta = 5 + 4cosdelta.
$$
answered Aug 6 at 4:23
Michael Hardy
204k23186463
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
beginalign
& sin x + 2sin(x+delta) \[10pt]
= & sin x + 2sin xcosdelta + 2cos x sindelta \[10pt]
= & (1+2cosdelta) sin x + (2sindelta) cos x \[10pt]
= & Asin x + Bcos x \[10pt]
= & sqrtA^2+B^2,, left( frac A sqrtA^2+B^2,, ,sin x + frac B sqrtA^2+B^2,, , cos x right) \[10pt]
= & sqrtA^2+B^2,, big( cosvarphi sin x + sinvarphi cos x big) \
& qquad textThese coefficients can be written as the cosine and sine \
& qquad textof some angle $varphi$ because the sum of their squares is $1.$ \[10pt]
= & sqrtA^2+B^2,, , sin(x+varphi).
endalign
In this case we have
$$
A^2+B^2 = (1+4cosdelta + 4cos^2delta) + 4sin^2delta = 5 + 4cosdelta.
$$
answered Aug 6 at 4:23
Michael Hardy
204k23186463
beginalign
& sin x + 2sin(x+delta) \[10pt]
= & sin x + 2sin xcosdelta + 2cos x sindelta \[10pt]
= & (1+2cosdelta) sin x + (2sindelta) cos x \[10pt]
= & Asin x + Bcos x \[10pt]
= & sqrtA^2+B^2,, left( frac A sqrtA^2+B^2,, ,sin x + frac B sqrtA^2+B^2,, , cos x right) \[10pt]
= & sqrtA^2+B^2,, big( cosvarphi sin x + sinvarphi cos x big) \
& qquad textThese coefficients can be written as the cosine and sine \
& qquad textof some angle $varphi$ because the sum of their squares is $1.$ \[10pt]
= & sqrtA^2+B^2,, , sin(x+varphi).
endalign
In this case we have
$$
A^2+B^2 = (1+4cosdelta + 4cos^2delta) + 4sin^2delta = 5 + 4cosdelta.
$$
answered Aug 6 at 4:23
Michael Hardy
204k23186463
answered Aug 6 at 4:23
Michael Hardy
204k23186463
answered Aug 6 at 4:23
Michael Hardy
204k23186463
answered Aug 6 at 4:23
Michael Hardy
204k23186463
204k23186463
add a comment |Â
add a comment |Â
up vote
2
down vote
We can write $E_1$ and $E_2$ as the imaginary parts of $e^ix$ and $2 e^i(x+delta)$, and by linearity $E_1 + E_2$ will be the imaginary part of the sum of these two complex numbers:
$$e^ix + 2e^i(x+delta) = e^ix(1 + 2 e^idelta) = e^ix(1 + 2 cos delta + i 2 sin delta) = e^ixleft(sqrt1+ 4 cosdelta + 4 cos^2delta + 4 sin^2 deltae^ioperatornameatan2(2sindelta,1+2cosdelta)right) $$
$$ = sqrt5+ 4 cosdelta e^i[x + operatornameatan2(2sindelta,1+2cosdelta)].$$
Taking the imaginary part, we therefore find
$$ E_1 + E_2 = left(sqrt5+ 4 cosdeltaright) sinleft(x + operatornameatan2 left( 2sindelta, 1+2cosdeltaright) right),$$
so the phase-shift is given by
$$operatornameatan2left(2sindelta,1+2cosdeltaright).$$
answered Aug 6 at 2:27
Matthew Kvalheim
671416
Note the conspicuous typographical difference between $itextatan2(cdots)$ and $ioperatornameatan2(cdots).$ With operatornameatan2 the amount of space to the left and right of $operatornameatan2$ depends on the context. I edited this answer accordingly. $qquad$
– Michael Hardy
Aug 6 at 4:46
@MichaelHardy Thanks for the tip.
– Matthew Kvalheim
Aug 6 at 5:32
add a comment |Â
up vote
2
down vote
We can write $E_1$ and $E_2$ as the imaginary parts of $e^ix$ and $2 e^i(x+delta)$, and by linearity $E_1 + E_2$ will be the imaginary part of the sum of these two complex numbers:
$$e^ix + 2e^i(x+delta) = e^ix(1 + 2 e^idelta) = e^ix(1 + 2 cos delta + i 2 sin delta) = e^ixleft(sqrt1+ 4 cosdelta + 4 cos^2delta + 4 sin^2 deltae^ioperatornameatan2(2sindelta,1+2cosdelta)right) $$
$$ = sqrt5+ 4 cosdelta e^i[x + operatornameatan2(2sindelta,1+2cosdelta)].$$
Taking the imaginary part, we therefore find
$$ E_1 + E_2 = left(sqrt5+ 4 cosdeltaright) sinleft(x + operatornameatan2 left( 2sindelta, 1+2cosdeltaright) right),$$
so the phase-shift is given by
$$operatornameatan2left(2sindelta,1+2cosdeltaright).$$
answered Aug 6 at 2:27
Matthew Kvalheim
671416
Note the conspicuous typographical difference between $itextatan2(cdots)$ and $ioperatornameatan2(cdots).$ With operatornameatan2 the amount of space to the left and right of $operatornameatan2$ depends on the context. I edited this answer accordingly. $qquad$
– Michael Hardy
Aug 6 at 4:46
@MichaelHardy Thanks for the tip.
– Matthew Kvalheim
Aug 6 at 5:32
add a comment |Â
up vote
2
down vote
up vote
2
down vote
We can write $E_1$ and $E_2$ as the imaginary parts of $e^ix$ and $2 e^i(x+delta)$, and by linearity $E_1 + E_2$ will be the imaginary part of the sum of these two complex numbers:
$$e^ix + 2e^i(x+delta) = e^ix(1 + 2 e^idelta) = e^ix(1 + 2 cos delta + i 2 sin delta) = e^ixleft(sqrt1+ 4 cosdelta + 4 cos^2delta + 4 sin^2 deltae^ioperatornameatan2(2sindelta,1+2cosdelta)right) $$
$$ = sqrt5+ 4 cosdelta e^i[x + operatornameatan2(2sindelta,1+2cosdelta)].$$
Taking the imaginary part, we therefore find
$$ E_1 + E_2 = left(sqrt5+ 4 cosdeltaright) sinleft(x + operatornameatan2 left( 2sindelta, 1+2cosdeltaright) right),$$
so the phase-shift is given by
$$operatornameatan2left(2sindelta,1+2cosdeltaright).$$
answered Aug 6 at 2:27
Matthew Kvalheim
671416
We can write $E_1$ and $E_2$ as the imaginary parts of $e^ix$ and $2 e^i(x+delta)$, and by linearity $E_1 + E_2$ will be the imaginary part of the sum of these two complex numbers:
$$e^ix + 2e^i(x+delta) = e^ix(1 + 2 e^idelta) = e^ix(1 + 2 cos delta + i 2 sin delta) = e^ixleft(sqrt1+ 4 cosdelta + 4 cos^2delta + 4 sin^2 deltae^ioperatornameatan2(2sindelta,1+2cosdelta)right) $$
$$ = sqrt5+ 4 cosdelta e^i[x + operatornameatan2(2sindelta,1+2cosdelta)].$$
Taking the imaginary part, we therefore find
$$ E_1 + E_2 = left(sqrt5+ 4 cosdeltaright) sinleft(x + operatornameatan2 left( 2sindelta, 1+2cosdeltaright) right),$$
so the phase-shift is given by
$$operatornameatan2left(2sindelta,1+2cosdeltaright).$$
answered Aug 6 at 2:27
Matthew Kvalheim
671416
edited Aug 7 at 5:22
answered Aug 6 at 2:27
Matthew Kvalheim
671416
answered Aug 6 at 2:27
Matthew Kvalheim
671416
answered Aug 6 at 2:27
Matthew Kvalheim
671416
671416
Note the conspicuous typographical difference between $itextatan2(cdots)$ and $ioperatornameatan2(cdots).$ With operatornameatan2 the amount of space to the left and right of $operatornameatan2$ depends on the context. I edited this answer accordingly. $qquad$
– Michael Hardy
Aug 6 at 4:46
@MichaelHardy Thanks for the tip.
– Matthew Kvalheim
Aug 6 at 5:32
add a comment |Â
Note the conspicuous typographical difference between $itextatan2(cdots)$ and $ioperatornameatan2(cdots).$ With operatornameatan2 the amount of space to the left and right of $operatornameatan2$ depends on the context. I edited this answer accordingly. $qquad$
– Michael Hardy
Aug 6 at 4:46
@MichaelHardy Thanks for the tip.
– Matthew Kvalheim
Aug 6 at 5:32
Note the conspicuous typographical difference between $itextatan2(cdots)$ and $ioperatornameatan2(cdots).$ With operatornameatan2 the amount of space to the left and right of $operatornameatan2$ depends on the context. I edited this answer accordingly. $qquad$
– Michael Hardy
Aug 6 at 4:46
Note the conspicuous typographical difference between $itextatan2(cdots)$ and $ioperatornameatan2(cdots).$ With operatornameatan2 the amount of space to the left and right of $operatornameatan2$ depends on the context. I edited this answer accordingly. $qquad$
– Michael Hardy
Aug 6 at 4:46
@MichaelHardy Thanks for the tip.
– Matthew Kvalheim
Aug 6 at 5:32
@MichaelHardy Thanks for the tip.
– Matthew Kvalheim
Aug 6 at 5:32
add a comment |Â
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1
I think this link might be useful for you: scipp.ucsc.edu/~haber/ph5B/addsine.pdf
– Karn Watcharasupat
Aug 6 at 2:22