Separate real and imaginary part of a transfer function
$begingroup$
So I have this fourth order transfer function
$$ G(s) = frac{2}{(1+itauomega)^4} $$
I have figured out that multiplying with the complex conjugate let's you separate the real and imaginary part.
However, I don't seem to find the right answer according to my book.
I end up with
$$ Re = frac{2 - 12tau^2omega^2 + 2tau^4omega^4}{(1+tau^2omega^2)^4} $$
$$ Im = frac{-8tauomega + 8tau^3omega^3}{(1+tau^2omega^2)^4} $$
Calculating the phase angle leaves me with
$$ arctan(frac{-8tauomega + 8tau^3omega^3}{2 - 12tau^2omega^2+2tau^4omega^4}) $$
While the book says the solution is
$$ -4arctan(tauomega) $$
Comparing the solution of the modulus suggests that my approach is correct, since the denominator is the same.. Am I missing something? I don't see how I can ever reduce the fraction to the one from the solutions.
complex-numbers
asked Jan 20 '17 at 22:08
boortmansboortmans
1012
$endgroup$
add a comment |
$begingroup$
So I have this fourth order transfer function
$$ G(s) = frac{2}{(1+itauomega)^4} $$
I have figured out that multiplying with the complex conjugate let's you separate the real and imaginary part.
However, I don't seem to find the right answer according to my book.
I end up with
$$ Re = frac{2 - 12tau^2omega^2 + 2tau^4omega^4}{(1+tau^2omega^2)^4} $$
$$ Im = frac{-8tauomega + 8tau^3omega^3}{(1+tau^2omega^2)^4} $$
Calculating the phase angle leaves me with
$$ arctan(frac{-8tauomega + 8tau^3omega^3}{2 - 12tau^2omega^2+2tau^4omega^4}) $$
While the book says the solution is
$$ -4arctan(tauomega) $$
Comparing the solution of the modulus suggests that my approach is correct, since the denominator is the same.. Am I missing something? I don't see how I can ever reduce the fraction to the one from the solutions.
complex-numbers
asked Jan 20 '17 at 22:08
boortmansboortmans
1012
$endgroup$
add a comment |
$begingroup$
So I have this fourth order transfer function
$$ G(s) = frac{2}{(1+itauomega)^4} $$
I have figured out that multiplying with the complex conjugate let's you separate the real and imaginary part.
However, I don't seem to find the right answer according to my book.
I end up with
$$ Re = frac{2 - 12tau^2omega^2 + 2tau^4omega^4}{(1+tau^2omega^2)^4} $$
$$ Im = frac{-8tauomega + 8tau^3omega^3}{(1+tau^2omega^2)^4} $$
Calculating the phase angle leaves me with
$$ arctan(frac{-8tauomega + 8tau^3omega^3}{2 - 12tau^2omega^2+2tau^4omega^4}) $$
While the book says the solution is
$$ -4arctan(tauomega) $$
Comparing the solution of the modulus suggests that my approach is correct, since the denominator is the same.. Am I missing something? I don't see how I can ever reduce the fraction to the one from the solutions.
complex-numbers
asked Jan 20 '17 at 22:08
boortmansboortmans
1012
$endgroup$
So I have this fourth order transfer function
$$ G(s) = frac{2}{(1+itauomega)^4} $$
I have figured out that multiplying with the complex conjugate let's you separate the real and imaginary part.
However, I don't seem to find the right answer according to my book.
I end up with
$$ Re = frac{2 - 12tau^2omega^2 + 2tau^4omega^4}{(1+tau^2omega^2)^4} $$
$$ Im = frac{-8tauomega + 8tau^3omega^3}{(1+tau^2omega^2)^4} $$
Calculating the phase angle leaves me with
$$ arctan(frac{-8tauomega + 8tau^3omega^3}{2 - 12tau^2omega^2+2tau^4omega^4}) $$
While the book says the solution is
$$ -4arctan(tauomega) $$
Comparing the solution of the modulus suggests that my approach is correct, since the denominator is the same.. Am I missing something? I don't see how I can ever reduce the fraction to the one from the solutions.
complex-numbers
complex-numbers
asked Jan 20 '17 at 22:08
boortmansboortmans
1012
asked Jan 20 '17 at 22:08
boortmansboortmans
1012
asked Jan 20 '17 at 22:08
boortmansboortmans
1012
asked Jan 20 '17 at 22:08
boortmansboortmans
1012
asked Jan 20 '17 at 22:08
boortmansboortmans
1012
1012
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
begin{eqnarray*}
tan^{-1}(A)+tan^{-1}(B) &=& tan^{-1}(frac{A+B}{1-AB})
end{eqnarray*}
Is that helpful ?
begin{eqnarray*}
-tan^{-1}(A) &=& tan^{-1}(-A).
end{eqnarray*}
Maybe they are the same !
answered Jan 20 '17 at 22:16
Donald SplutterwitDonald Splutterwit
22.8k21446
$endgroup$
add a comment |
$begingroup$
$$arg(1+irw)=arctanfrac{rw}{1}=arctan rw$$
$$argfrac{sqrt[4]{2}}{1+irw}=-arctan rw$$
because numbers doesn't effect on arg,
$$argfrac{2}{(1+irw)^4}=argBig(frac{sqrt[4]{2}}{1+irw}Big)^4=4argfrac{sqrt[4]{2}}{1+irw}=-4arctan rw$$
answered Jan 20 '17 at 22:19
NosratiNosrati
26.6k62354
$endgroup$
add a comment |
$begingroup$
$$argfrac {4}{(1+iromega)^4}=arg 4-4arg(1+iromega)=0-4arctan romega$$
answered Jan 20 '17 at 22:23
David QuinnDavid Quinn
24.1k21141
$endgroup$
add a comment |
$begingroup$
Observe that
$$tan 4theta=frac{2tan2theta}{1-tan^22theta}=frac{2dfrac{2tantheta}{1-tan^2theta}}{1-left(dfrac{2tantheta}{1-tan^2theta}right)^2}=frac{4tantheta-4tan^3theta}{1-6tan^2theta+tan^4theta}$$
and compare the two solutions.
answered Oct 12 '18 at 21:09
Yves DaoustYves Daoust
129k676227
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
begin{eqnarray*}
tan^{-1}(A)+tan^{-1}(B) &=& tan^{-1}(frac{A+B}{1-AB})
end{eqnarray*}
Is that helpful ?
begin{eqnarray*}
-tan^{-1}(A) &=& tan^{-1}(-A).
end{eqnarray*}
Maybe they are the same !
answered Jan 20 '17 at 22:16
Donald SplutterwitDonald Splutterwit
22.8k21446
$endgroup$
add a comment |
$begingroup$
begin{eqnarray*}
tan^{-1}(A)+tan^{-1}(B) &=& tan^{-1}(frac{A+B}{1-AB})
end{eqnarray*}
Is that helpful ?
begin{eqnarray*}
-tan^{-1}(A) &=& tan^{-1}(-A).
end{eqnarray*}
Maybe they are the same !
answered Jan 20 '17 at 22:16
Donald SplutterwitDonald Splutterwit
22.8k21446
$endgroup$
add a comment |
$begingroup$
begin{eqnarray*}
tan^{-1}(A)+tan^{-1}(B) &=& tan^{-1}(frac{A+B}{1-AB})
end{eqnarray*}
Is that helpful ?
begin{eqnarray*}
-tan^{-1}(A) &=& tan^{-1}(-A).
end{eqnarray*}
Maybe they are the same !
answered Jan 20 '17 at 22:16
Donald SplutterwitDonald Splutterwit
22.8k21446
$endgroup$
begin{eqnarray*}
tan^{-1}(A)+tan^{-1}(B) &=& tan^{-1}(frac{A+B}{1-AB})
end{eqnarray*}
Is that helpful ?
begin{eqnarray*}
-tan^{-1}(A) &=& tan^{-1}(-A).
end{eqnarray*}
Maybe they are the same !
answered Jan 20 '17 at 22:16
Donald SplutterwitDonald Splutterwit
22.8k21446
answered Jan 20 '17 at 22:16
Donald SplutterwitDonald Splutterwit
22.8k21446
answered Jan 20 '17 at 22:16
Donald SplutterwitDonald Splutterwit
22.8k21446
answered Jan 20 '17 at 22:16
Donald SplutterwitDonald Splutterwit
22.8k21446
22.8k21446
add a comment |
add a comment |
$begingroup$
$$arg(1+irw)=arctanfrac{rw}{1}=arctan rw$$
$$argfrac{sqrt[4]{2}}{1+irw}=-arctan rw$$
because numbers doesn't effect on arg,
$$argfrac{2}{(1+irw)^4}=argBig(frac{sqrt[4]{2}}{1+irw}Big)^4=4argfrac{sqrt[4]{2}}{1+irw}=-4arctan rw$$
answered Jan 20 '17 at 22:19
NosratiNosrati
26.6k62354
$endgroup$
add a comment |
$begingroup$
$$arg(1+irw)=arctanfrac{rw}{1}=arctan rw$$
$$argfrac{sqrt[4]{2}}{1+irw}=-arctan rw$$
because numbers doesn't effect on arg,
$$argfrac{2}{(1+irw)^4}=argBig(frac{sqrt[4]{2}}{1+irw}Big)^4=4argfrac{sqrt[4]{2}}{1+irw}=-4arctan rw$$
answered Jan 20 '17 at 22:19
NosratiNosrati
26.6k62354
$endgroup$
add a comment |
$begingroup$
$$arg(1+irw)=arctanfrac{rw}{1}=arctan rw$$
$$argfrac{sqrt[4]{2}}{1+irw}=-arctan rw$$
because numbers doesn't effect on arg,
$$argfrac{2}{(1+irw)^4}=argBig(frac{sqrt[4]{2}}{1+irw}Big)^4=4argfrac{sqrt[4]{2}}{1+irw}=-4arctan rw$$
answered Jan 20 '17 at 22:19
NosratiNosrati
26.6k62354
$endgroup$
$$arg(1+irw)=arctanfrac{rw}{1}=arctan rw$$
$$argfrac{sqrt[4]{2}}{1+irw}=-arctan rw$$
because numbers doesn't effect on arg,
$$argfrac{2}{(1+irw)^4}=argBig(frac{sqrt[4]{2}}{1+irw}Big)^4=4argfrac{sqrt[4]{2}}{1+irw}=-4arctan rw$$
answered Jan 20 '17 at 22:19
NosratiNosrati
26.6k62354
answered Jan 20 '17 at 22:19
NosratiNosrati
26.6k62354
answered Jan 20 '17 at 22:19
NosratiNosrati
26.6k62354
answered Jan 20 '17 at 22:19
NosratiNosrati
26.6k62354
26.6k62354
add a comment |
add a comment |
$begingroup$
$$argfrac {4}{(1+iromega)^4}=arg 4-4arg(1+iromega)=0-4arctan romega$$
answered Jan 20 '17 at 22:23
David QuinnDavid Quinn
24.1k21141
$endgroup$
add a comment |
$begingroup$
$$argfrac {4}{(1+iromega)^4}=arg 4-4arg(1+iromega)=0-4arctan romega$$
answered Jan 20 '17 at 22:23
David QuinnDavid Quinn
24.1k21141
$endgroup$
add a comment |
$begingroup$
$$argfrac {4}{(1+iromega)^4}=arg 4-4arg(1+iromega)=0-4arctan romega$$
answered Jan 20 '17 at 22:23
David QuinnDavid Quinn
24.1k21141
$endgroup$
$$argfrac {4}{(1+iromega)^4}=arg 4-4arg(1+iromega)=0-4arctan romega$$
answered Jan 20 '17 at 22:23
David QuinnDavid Quinn
24.1k21141
answered Jan 20 '17 at 22:23
David QuinnDavid Quinn
24.1k21141
answered Jan 20 '17 at 22:23
David QuinnDavid Quinn
24.1k21141
answered Jan 20 '17 at 22:23
David QuinnDavid Quinn
24.1k21141
24.1k21141
add a comment |
add a comment |
$begingroup$
Observe that
$$tan 4theta=frac{2tan2theta}{1-tan^22theta}=frac{2dfrac{2tantheta}{1-tan^2theta}}{1-left(dfrac{2tantheta}{1-tan^2theta}right)^2}=frac{4tantheta-4tan^3theta}{1-6tan^2theta+tan^4theta}$$
and compare the two solutions.
answered Oct 12 '18 at 21:09
Yves DaoustYves Daoust
129k676227
$endgroup$
add a comment |
$begingroup$
Observe that
$$tan 4theta=frac{2tan2theta}{1-tan^22theta}=frac{2dfrac{2tantheta}{1-tan^2theta}}{1-left(dfrac{2tantheta}{1-tan^2theta}right)^2}=frac{4tantheta-4tan^3theta}{1-6tan^2theta+tan^4theta}$$
and compare the two solutions.
answered Oct 12 '18 at 21:09
Yves DaoustYves Daoust
129k676227
$endgroup$
add a comment |
$begingroup$
Observe that
$$tan 4theta=frac{2tan2theta}{1-tan^22theta}=frac{2dfrac{2tantheta}{1-tan^2theta}}{1-left(dfrac{2tantheta}{1-tan^2theta}right)^2}=frac{4tantheta-4tan^3theta}{1-6tan^2theta+tan^4theta}$$
and compare the two solutions.
answered Oct 12 '18 at 21:09
Yves DaoustYves Daoust
129k676227
$endgroup$
Observe that
$$tan 4theta=frac{2tan2theta}{1-tan^22theta}=frac{2dfrac{2tantheta}{1-tan^2theta}}{1-left(dfrac{2tantheta}{1-tan^2theta}right)^2}=frac{4tantheta-4tan^3theta}{1-6tan^2theta+tan^4theta}$$
and compare the two solutions.
answered Oct 12 '18 at 21:09
Yves DaoustYves Daoust
129k676227
answered Oct 12 '18 at 21:09
Yves DaoustYves Daoust
129k676227
answered Oct 12 '18 at 21:09
Yves DaoustYves Daoust
129k676227
answered Oct 12 '18 at 21:09
Yves DaoustYves Daoust
129k676227
129k676227
add a comment |
add a comment |
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