Solving simultaneous equations with complex coefficients using real methods
My circuits analysis textbook teases that there's a way to convert a set of n complex equations into a set of 2n real equations, which can then be solved using any calculator that can solve real simultaneous equations. That is, no capability with complex numbers needed.
e.g.:
$(25 +j100)I_1 - (10+j80)I_2=100angle0^circ\$ (1)
$-(10+j80)I_1+(30+j190)I_2=0$ (2)
I say "teases" because they point me to their website, where after a lengthy sign-up process, I find that the material isn't actually there.
Does anybody know what method they're referring to?
I know how to do this with Cramer's Rule, but that requires a matrix calculator that understands complex numbers (they exist but they're not common).
Here's the page:
https://books.google.com.au/books?id=VLbycoxwas8C&pg=PA959&lpg=PA959&dq=%22Solving+Simultaneous+Equations+with+Complex+Coefficients+Using+Any+Calculator%22&source=bl&ots=Bf9PJRGo3o&sig=B-ssojUUL4fnJXxOFu6VMw0vP_0&hl=en&sa=X&ei=rAekU5_xE5Tr8AWpqoEQ#v=onepage&q=%22Solving%20Simultaneous%20Equations%20with%20Complex%20Coefficients%20Using%20Any%20Calculator%22&f=false
complex-numbers
asked Jun 21 '14 at 1:54
Greg Bell
1615
add a comment |
My circuits analysis textbook teases that there's a way to convert a set of n complex equations into a set of 2n real equations, which can then be solved using any calculator that can solve real simultaneous equations. That is, no capability with complex numbers needed.
e.g.:
$(25 +j100)I_1 - (10+j80)I_2=100angle0^circ\$ (1)
$-(10+j80)I_1+(30+j190)I_2=0$ (2)
I say "teases" because they point me to their website, where after a lengthy sign-up process, I find that the material isn't actually there.
Does anybody know what method they're referring to?
I know how to do this with Cramer's Rule, but that requires a matrix calculator that understands complex numbers (they exist but they're not common).
Here's the page:
https://books.google.com.au/books?id=VLbycoxwas8C&pg=PA959&lpg=PA959&dq=%22Solving+Simultaneous+Equations+with+Complex+Coefficients+Using+Any+Calculator%22&source=bl&ots=Bf9PJRGo3o&sig=B-ssojUUL4fnJXxOFu6VMw0vP_0&hl=en&sa=X&ei=rAekU5_xE5Tr8AWpqoEQ#v=onepage&q=%22Solving%20Simultaneous%20Equations%20with%20Complex%20Coefficients%20Using%20Any%20Calculator%22&f=false
complex-numbers
asked Jun 21 '14 at 1:54
Greg Bell
1615
add a comment |
My circuits analysis textbook teases that there's a way to convert a set of n complex equations into a set of 2n real equations, which can then be solved using any calculator that can solve real simultaneous equations. That is, no capability with complex numbers needed.
e.g.:
$(25 +j100)I_1 - (10+j80)I_2=100angle0^circ\$ (1)
$-(10+j80)I_1+(30+j190)I_2=0$ (2)
I say "teases" because they point me to their website, where after a lengthy sign-up process, I find that the material isn't actually there.
Does anybody know what method they're referring to?
I know how to do this with Cramer's Rule, but that requires a matrix calculator that understands complex numbers (they exist but they're not common).
Here's the page:
https://books.google.com.au/books?id=VLbycoxwas8C&pg=PA959&lpg=PA959&dq=%22Solving+Simultaneous+Equations+with+Complex+Coefficients+Using+Any+Calculator%22&source=bl&ots=Bf9PJRGo3o&sig=B-ssojUUL4fnJXxOFu6VMw0vP_0&hl=en&sa=X&ei=rAekU5_xE5Tr8AWpqoEQ#v=onepage&q=%22Solving%20Simultaneous%20Equations%20with%20Complex%20Coefficients%20Using%20Any%20Calculator%22&f=false
complex-numbers
asked Jun 21 '14 at 1:54
Greg Bell
1615
My circuits analysis textbook teases that there's a way to convert a set of n complex equations into a set of 2n real equations, which can then be solved using any calculator that can solve real simultaneous equations. That is, no capability with complex numbers needed.
e.g.:
$(25 +j100)I_1 - (10+j80)I_2=100angle0^circ\$ (1)
$-(10+j80)I_1+(30+j190)I_2=0$ (2)
I say "teases" because they point me to their website, where after a lengthy sign-up process, I find that the material isn't actually there.
Does anybody know what method they're referring to?
I know how to do this with Cramer's Rule, but that requires a matrix calculator that understands complex numbers (they exist but they're not common).
Here's the page:
https://books.google.com.au/books?id=VLbycoxwas8C&pg=PA959&lpg=PA959&dq=%22Solving+Simultaneous+Equations+with+Complex+Coefficients+Using+Any+Calculator%22&source=bl&ots=Bf9PJRGo3o&sig=B-ssojUUL4fnJXxOFu6VMw0vP_0&hl=en&sa=X&ei=rAekU5_xE5Tr8AWpqoEQ#v=onepage&q=%22Solving%20Simultaneous%20Equations%20with%20Complex%20Coefficients%20Using%20Any%20Calculator%22&f=false
complex-numbers
complex-numbers
asked Jun 21 '14 at 1:54
Greg Bell
1615
asked Jun 21 '14 at 1:54
Greg Bell
1615
edited Nov 18 at 10:16
asked Jun 21 '14 at 1:54
Greg Bell
1615
asked Jun 21 '14 at 1:54
Greg Bell
1615
asked Jun 21 '14 at 1:54
Greg Bell
1615
1615
add a comment |
add a comment |
1 Answer
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The main idea is to split each equation into a real and a complex part.
To easily see how to do this take a look at complex multiplication as a linear transformation.
$(a + bi) * (c + di) = (ac - bd) + (bc + ad)i$ will become
$left(begin{array}{cc}c&-d\d&cend{array}right)
left(begin{array}{c}a\bend{array}right)=
left(begin{array}{c}ac-bd\bc+adend{array}right)$
You can use this pattern to rewrite your example as
$left(begin{array}{cccc|c}
25&-100&-10&80&100\
100&25&-80&-10&0\
-10&80&30&-190&0\
-80&-10&190&30&0
end{array}right)$
answered Dec 17 at 8:29
user10545
111
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
The main idea is to split each equation into a real and a complex part.
To easily see how to do this take a look at complex multiplication as a linear transformation.
$(a + bi) * (c + di) = (ac - bd) + (bc + ad)i$ will become
$left(begin{array}{cc}c&-d\d&cend{array}right)
left(begin{array}{c}a\bend{array}right)=
left(begin{array}{c}ac-bd\bc+adend{array}right)$
You can use this pattern to rewrite your example as
$left(begin{array}{cccc|c}
25&-100&-10&80&100\
100&25&-80&-10&0\
-10&80&30&-190&0\
-80&-10&190&30&0
end{array}right)$
answered Dec 17 at 8:29
user10545
111
add a comment |
The main idea is to split each equation into a real and a complex part.
To easily see how to do this take a look at complex multiplication as a linear transformation.
$(a + bi) * (c + di) = (ac - bd) + (bc + ad)i$ will become
$left(begin{array}{cc}c&-d\d&cend{array}right)
left(begin{array}{c}a\bend{array}right)=
left(begin{array}{c}ac-bd\bc+adend{array}right)$
You can use this pattern to rewrite your example as
$left(begin{array}{cccc|c}
25&-100&-10&80&100\
100&25&-80&-10&0\
-10&80&30&-190&0\
-80&-10&190&30&0
end{array}right)$
answered Dec 17 at 8:29
user10545
111
add a comment |
The main idea is to split each equation into a real and a complex part.
To easily see how to do this take a look at complex multiplication as a linear transformation.
$(a + bi) * (c + di) = (ac - bd) + (bc + ad)i$ will become
$left(begin{array}{cc}c&-d\d&cend{array}right)
left(begin{array}{c}a\bend{array}right)=
left(begin{array}{c}ac-bd\bc+adend{array}right)$
You can use this pattern to rewrite your example as
$left(begin{array}{cccc|c}
25&-100&-10&80&100\
100&25&-80&-10&0\
-10&80&30&-190&0\
-80&-10&190&30&0
end{array}right)$
answered Dec 17 at 8:29
user10545
111
The main idea is to split each equation into a real and a complex part.
To easily see how to do this take a look at complex multiplication as a linear transformation.
$(a + bi) * (c + di) = (ac - bd) + (bc + ad)i$ will become
$left(begin{array}{cc}c&-d\d&cend{array}right)
left(begin{array}{c}a\bend{array}right)=
left(begin{array}{c}ac-bd\bc+adend{array}right)$
You can use this pattern to rewrite your example as
$left(begin{array}{cccc|c}
25&-100&-10&80&100\
100&25&-80&-10&0\
-10&80&30&-190&0\
-80&-10&190&30&0
end{array}right)$
answered Dec 17 at 8:29
user10545
111
answered Dec 17 at 8:29
user10545
111
answered Dec 17 at 8:29
user10545
111
answered Dec 17 at 8:29
user10545
111
111
add a comment |
add a comment |
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