Mixing
argument of complex number for bode plot
$begingroup$
Could someone help me please. I'm looking for the value of $omega$
$text{Argument}(frac{1,6}{(1+0,004text{j}cdotomega)(1+0,04text{j}cdotomega)})=-135° $
$text{Argument}({1,6})-text{Argument}({(1+0,004text{j}cdotomega)(1+0,04text{j}cdotomega)}))=-135° $
$text{Atan}(frac{0}{1,6})-text{Atan}({frac{0,004cdotomega}{1})-text{Atan}(frac{0,04cdotomega}{1})}=-135° $
$-text{Atan}({frac{0,004cdotomega}{1})-text{Atan}(frac{0,04cdotomega}{1})}=-135° $
i'm stucked here :(
complex-analysis complex-numbers complex-geometry
asked Jan 8 at 12:41
omkaomka
102
$endgroup$
add a comment |
$begingroup$
Could someone help me please. I'm looking for the value of $omega$
$text{Argument}(frac{1,6}{(1+0,004text{j}cdotomega)(1+0,04text{j}cdotomega)})=-135° $
$text{Argument}({1,6})-text{Argument}({(1+0,004text{j}cdotomega)(1+0,04text{j}cdotomega)}))=-135° $
$text{Atan}(frac{0}{1,6})-text{Atan}({frac{0,004cdotomega}{1})-text{Atan}(frac{0,04cdotomega}{1})}=-135° $
$-text{Atan}({frac{0,004cdotomega}{1})-text{Atan}(frac{0,04cdotomega}{1})}=-135° $
i'm stucked here :(
complex-analysis complex-numbers complex-geometry
asked Jan 8 at 12:41
omkaomka
102
$endgroup$
$begingroup$
What have you tried? The first step would be the interpret the definition of complex argument.
$endgroup$
– Matti P.
Jan 8 at 12:50
$begingroup$
Hint: $tan{-135^{circ}} = 1$
$endgroup$
– Matti P.
Jan 8 at 12:58
$begingroup$
Please @MattiP. Can i write -0,004.w-0,04.w=1 then w=22,72? thank you
$endgroup$
– omka
Jan 10 at 17:14
$begingroup$
I made a plot and it seems like the answer is something like $omega approx 296.1$
$endgroup$
– Matti P.
Jan 11 at 7:11
add a comment |
$begingroup$
Could someone help me please. I'm looking for the value of $omega$
$text{Argument}(frac{1,6}{(1+0,004text{j}cdotomega)(1+0,04text{j}cdotomega)})=-135° $
$text{Argument}({1,6})-text{Argument}({(1+0,004text{j}cdotomega)(1+0,04text{j}cdotomega)}))=-135° $
$text{Atan}(frac{0}{1,6})-text{Atan}({frac{0,004cdotomega}{1})-text{Atan}(frac{0,04cdotomega}{1})}=-135° $
$-text{Atan}({frac{0,004cdotomega}{1})-text{Atan}(frac{0,04cdotomega}{1})}=-135° $
i'm stucked here :(
complex-analysis complex-numbers complex-geometry
asked Jan 8 at 12:41
omkaomka
102
$endgroup$
Could someone help me please. I'm looking for the value of $omega$
$text{Argument}(frac{1,6}{(1+0,004text{j}cdotomega)(1+0,04text{j}cdotomega)})=-135° $
$text{Argument}({1,6})-text{Argument}({(1+0,004text{j}cdotomega)(1+0,04text{j}cdotomega)}))=-135° $
$text{Atan}(frac{0}{1,6})-text{Atan}({frac{0,004cdotomega}{1})-text{Atan}(frac{0,04cdotomega}{1})}=-135° $
$-text{Atan}({frac{0,004cdotomega}{1})-text{Atan}(frac{0,04cdotomega}{1})}=-135° $
i'm stucked here :(
complex-analysis complex-numbers complex-geometry
complex-analysis complex-numbers complex-geometry
asked Jan 8 at 12:41
omkaomka
102
asked Jan 8 at 12:41
omkaomka
102
edited Jan 8 at 13:21
omka
asked Jan 8 at 12:41
omkaomka
102
asked Jan 8 at 12:41
omkaomka
102
asked Jan 8 at 12:41
omkaomka
102
102
$begingroup$
What have you tried? The first step would be the interpret the definition of complex argument.
$endgroup$
– Matti P.
Jan 8 at 12:50
$begingroup$
Hint: $tan{-135^{circ}} = 1$
$endgroup$
– Matti P.
Jan 8 at 12:58
$begingroup$
Please @MattiP. Can i write -0,004.w-0,04.w=1 then w=22,72? thank you
$endgroup$
– omka
Jan 10 at 17:14
$begingroup$
I made a plot and it seems like the answer is something like $omega approx 296.1$
$endgroup$
– Matti P.
Jan 11 at 7:11
add a comment |
$begingroup$
What have you tried? The first step would be the interpret the definition of complex argument.
$endgroup$
– Matti P.
Jan 8 at 12:50
$begingroup$
Hint: $tan{-135^{circ}} = 1$
$endgroup$
– Matti P.
Jan 8 at 12:58
$begingroup$
Please @MattiP. Can i write -0,004.w-0,04.w=1 then w=22,72? thank you
$endgroup$
– omka
Jan 10 at 17:14
$begingroup$
I made a plot and it seems like the answer is something like $omega approx 296.1$
$endgroup$
– Matti P.
Jan 11 at 7:11
$begingroup$
What have you tried? The first step would be the interpret the definition of complex argument.
$endgroup$
– Matti P.
Jan 8 at 12:50
$begingroup$
What have you tried? The first step would be the interpret the definition of complex argument.
$endgroup$
– Matti P.
Jan 8 at 12:50
$begingroup$
Hint: $tan{-135^{circ}} = 1$
$endgroup$
– Matti P.
Jan 8 at 12:58
$begingroup$
Hint: $tan{-135^{circ}} = 1$
$endgroup$
– Matti P.
Jan 8 at 12:58
$begingroup$
Please @MattiP. Can i write -0,004.w-0,04.w=1 then w=22,72? thank you
$endgroup$
– omka
Jan 10 at 17:14
$begingroup$
Please @MattiP. Can i write -0,004.w-0,04.w=1 then w=22,72? thank you
$endgroup$
– omka
Jan 10 at 17:14
$begingroup$
I made a plot and it seems like the answer is something like $omega approx 296.1$
$endgroup$
– Matti P.
Jan 11 at 7:11
$begingroup$
I made a plot and it seems like the answer is something like $omega approx 296.1$
$endgroup$
– Matti P.
Jan 11 at 7:11
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The argument of a complex number is the angle that the complex number makes, when plotted in the complex plane. Therefore,
$$
text{Arg}{left( zright)} = -135^{circ} Rightarrow z text{ is in the third quadrant, $text{Im}{(z)} = text{Re}(z)<0$}
$$
This is because $tan{-135^{circ}} = 1$. Next, let's do some algebra (note that multiplying by $1.6$ doesn't change the argument, therefore we can leave it out):
$$
begin{split}
frac{1}{(1+ai omega)(1+biomega)} &=
frac{(1-ai omega)(1-biomega)}{(1+ai omega)(1-ai omega)(1+biomega)(1-biomega)} \
&=frac{(1-ai omega)(1-biomega)}{(1+a^2 omega^2)(1+b^2omega^2)} \
&= frac{1-biomega-aiomega+abi^2omega^2}{(1+a^2 omega^2)(1+b^2omega^2)} \
&= frac{1-abomega^2-(a+b)iomega}{(1+a^2 omega^2)(1+b^2omega^2)} \
end{split}
$$
Now it's easy to separate the imaginary and real part. Equating them, we get
$$
frac{1-abomega^2}{(1+a^2 omega^2)(1+b^2omega^2)} = frac{-(a+b)omega}{(1+a^2 omega^2)(1+b^2omega^2)}
$$
Multiplying by the denominator (and by $-1$), we obtain
$$
abomega^2 +(-a-b)omega -1 = 0
$$
or
$$
omega = frac{ (a+b) pm sqrt{(a+b)^2 + 4 ab} }{ 2ab }
$$
Now we can plug in the values $a=0.004$ and $b=0.04$ to get
$$
omega approx frac{ 0.044 pm 0.0508 }{ 0.00032 }
$$
$ Rightarrow omega approx 21.1 $ or $omega approx 296.1 $ .
In addition, we had the condition that the real and imaginary parts have to be negative. Inserting these values for $omega$ into the original equation, we see that
$$
omega approx 296.1077
$$
is the solution. Just to check, plugging in this value results in
$$
frac{1.6}{(1+0.004i omega)(1+0.04i omega)} approx -0.0614(1+i)
$$
answered Jan 11 at 6:48
Matti P.Matti P.
1,900413
$endgroup$
$begingroup$
Thank you so much
$endgroup$
– omka
Jan 13 at 11:28
add a comment |
$begingroup$
The equation directly interpreted, without any argument transforms beyond $arg(z^{-1})=-arg(z)+2kpi j$, gives
$$
(1+0,004j⋅ω)(1+0,04j⋅ω)=re^{j⋅135°}=r'(-1+i)
$$
for some $r=sqrt2 r'>0$. One reads off that the sum of real part and imaginary part on the left side has to be zero, with the imaginary part positive, as it is the case on the right.
$$
0=1-0.00016⋅ω^2 + 0.044⋅ωiff (0.04⋅ω-5.5)^2=0.0016⋅ω^2-0.44⋅ω+5.5^2=10+5.5^2
$$
so that $ω>0$ and $ω=25⋅(5.5pmsqrt{40.25})implies ω=25⋅(5.5+sqrt{40.25})=296.1072192556190..$.
answered Jan 11 at 11:53
LutzLLutzL
57.6k42054
$endgroup$
$begingroup$
Thank you @LutzL
$endgroup$
– omka
Jan 13 at 11:29
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The argument of a complex number is the angle that the complex number makes, when plotted in the complex plane. Therefore,
$$
text{Arg}{left( zright)} = -135^{circ} Rightarrow z text{ is in the third quadrant, $text{Im}{(z)} = text{Re}(z)<0$}
$$
This is because $tan{-135^{circ}} = 1$. Next, let's do some algebra (note that multiplying by $1.6$ doesn't change the argument, therefore we can leave it out):
$$
begin{split}
frac{1}{(1+ai omega)(1+biomega)} &=
frac{(1-ai omega)(1-biomega)}{(1+ai omega)(1-ai omega)(1+biomega)(1-biomega)} \
&=frac{(1-ai omega)(1-biomega)}{(1+a^2 omega^2)(1+b^2omega^2)} \
&= frac{1-biomega-aiomega+abi^2omega^2}{(1+a^2 omega^2)(1+b^2omega^2)} \
&= frac{1-abomega^2-(a+b)iomega}{(1+a^2 omega^2)(1+b^2omega^2)} \
end{split}
$$
Now it's easy to separate the imaginary and real part. Equating them, we get
$$
frac{1-abomega^2}{(1+a^2 omega^2)(1+b^2omega^2)} = frac{-(a+b)omega}{(1+a^2 omega^2)(1+b^2omega^2)}
$$
Multiplying by the denominator (and by $-1$), we obtain
$$
abomega^2 +(-a-b)omega -1 = 0
$$
or
$$
omega = frac{ (a+b) pm sqrt{(a+b)^2 + 4 ab} }{ 2ab }
$$
Now we can plug in the values $a=0.004$ and $b=0.04$ to get
$$
omega approx frac{ 0.044 pm 0.0508 }{ 0.00032 }
$$
$ Rightarrow omega approx 21.1 $ or $omega approx 296.1 $ .
In addition, we had the condition that the real and imaginary parts have to be negative. Inserting these values for $omega$ into the original equation, we see that
$$
omega approx 296.1077
$$
is the solution. Just to check, plugging in this value results in
$$
frac{1.6}{(1+0.004i omega)(1+0.04i omega)} approx -0.0614(1+i)
$$
answered Jan 11 at 6:48
Matti P.Matti P.
1,900413
$endgroup$
$begingroup$
Thank you so much
$endgroup$
– omka
Jan 13 at 11:28
add a comment |
$begingroup$
The argument of a complex number is the angle that the complex number makes, when plotted in the complex plane. Therefore,
$$
text{Arg}{left( zright)} = -135^{circ} Rightarrow z text{ is in the third quadrant, $text{Im}{(z)} = text{Re}(z)<0$}
$$
This is because $tan{-135^{circ}} = 1$. Next, let's do some algebra (note that multiplying by $1.6$ doesn't change the argument, therefore we can leave it out):
$$
begin{split}
frac{1}{(1+ai omega)(1+biomega)} &=
frac{(1-ai omega)(1-biomega)}{(1+ai omega)(1-ai omega)(1+biomega)(1-biomega)} \
&=frac{(1-ai omega)(1-biomega)}{(1+a^2 omega^2)(1+b^2omega^2)} \
&= frac{1-biomega-aiomega+abi^2omega^2}{(1+a^2 omega^2)(1+b^2omega^2)} \
&= frac{1-abomega^2-(a+b)iomega}{(1+a^2 omega^2)(1+b^2omega^2)} \
end{split}
$$
Now it's easy to separate the imaginary and real part. Equating them, we get
$$
frac{1-abomega^2}{(1+a^2 omega^2)(1+b^2omega^2)} = frac{-(a+b)omega}{(1+a^2 omega^2)(1+b^2omega^2)}
$$
Multiplying by the denominator (and by $-1$), we obtain
$$
abomega^2 +(-a-b)omega -1 = 0
$$
or
$$
omega = frac{ (a+b) pm sqrt{(a+b)^2 + 4 ab} }{ 2ab }
$$
Now we can plug in the values $a=0.004$ and $b=0.04$ to get
$$
omega approx frac{ 0.044 pm 0.0508 }{ 0.00032 }
$$
$ Rightarrow omega approx 21.1 $ or $omega approx 296.1 $ .
In addition, we had the condition that the real and imaginary parts have to be negative. Inserting these values for $omega$ into the original equation, we see that
$$
omega approx 296.1077
$$
is the solution. Just to check, plugging in this value results in
$$
frac{1.6}{(1+0.004i omega)(1+0.04i omega)} approx -0.0614(1+i)
$$
answered Jan 11 at 6:48
Matti P.Matti P.
1,900413
$endgroup$
$begingroup$
Thank you so much
$endgroup$
– omka
Jan 13 at 11:28
add a comment |
$begingroup$
The argument of a complex number is the angle that the complex number makes, when plotted in the complex plane. Therefore,
$$
text{Arg}{left( zright)} = -135^{circ} Rightarrow z text{ is in the third quadrant, $text{Im}{(z)} = text{Re}(z)<0$}
$$
This is because $tan{-135^{circ}} = 1$. Next, let's do some algebra (note that multiplying by $1.6$ doesn't change the argument, therefore we can leave it out):
$$
begin{split}
frac{1}{(1+ai omega)(1+biomega)} &=
frac{(1-ai omega)(1-biomega)}{(1+ai omega)(1-ai omega)(1+biomega)(1-biomega)} \
&=frac{(1-ai omega)(1-biomega)}{(1+a^2 omega^2)(1+b^2omega^2)} \
&= frac{1-biomega-aiomega+abi^2omega^2}{(1+a^2 omega^2)(1+b^2omega^2)} \
&= frac{1-abomega^2-(a+b)iomega}{(1+a^2 omega^2)(1+b^2omega^2)} \
end{split}
$$
Now it's easy to separate the imaginary and real part. Equating them, we get
$$
frac{1-abomega^2}{(1+a^2 omega^2)(1+b^2omega^2)} = frac{-(a+b)omega}{(1+a^2 omega^2)(1+b^2omega^2)}
$$
Multiplying by the denominator (and by $-1$), we obtain
$$
abomega^2 +(-a-b)omega -1 = 0
$$
or
$$
omega = frac{ (a+b) pm sqrt{(a+b)^2 + 4 ab} }{ 2ab }
$$
Now we can plug in the values $a=0.004$ and $b=0.04$ to get
$$
omega approx frac{ 0.044 pm 0.0508 }{ 0.00032 }
$$
$ Rightarrow omega approx 21.1 $ or $omega approx 296.1 $ .
In addition, we had the condition that the real and imaginary parts have to be negative. Inserting these values for $omega$ into the original equation, we see that
$$
omega approx 296.1077
$$
is the solution. Just to check, plugging in this value results in
$$
frac{1.6}{(1+0.004i omega)(1+0.04i omega)} approx -0.0614(1+i)
$$
answered Jan 11 at 6:48
Matti P.Matti P.
1,900413
$endgroup$
The argument of a complex number is the angle that the complex number makes, when plotted in the complex plane. Therefore,
$$
text{Arg}{left( zright)} = -135^{circ} Rightarrow z text{ is in the third quadrant, $text{Im}{(z)} = text{Re}(z)<0$}
$$
This is because $tan{-135^{circ}} = 1$. Next, let's do some algebra (note that multiplying by $1.6$ doesn't change the argument, therefore we can leave it out):
$$
begin{split}
frac{1}{(1+ai omega)(1+biomega)} &=
frac{(1-ai omega)(1-biomega)}{(1+ai omega)(1-ai omega)(1+biomega)(1-biomega)} \
&=frac{(1-ai omega)(1-biomega)}{(1+a^2 omega^2)(1+b^2omega^2)} \
&= frac{1-biomega-aiomega+abi^2omega^2}{(1+a^2 omega^2)(1+b^2omega^2)} \
&= frac{1-abomega^2-(a+b)iomega}{(1+a^2 omega^2)(1+b^2omega^2)} \
end{split}
$$
Now it's easy to separate the imaginary and real part. Equating them, we get
$$
frac{1-abomega^2}{(1+a^2 omega^2)(1+b^2omega^2)} = frac{-(a+b)omega}{(1+a^2 omega^2)(1+b^2omega^2)}
$$
Multiplying by the denominator (and by $-1$), we obtain
$$
abomega^2 +(-a-b)omega -1 = 0
$$
or
$$
omega = frac{ (a+b) pm sqrt{(a+b)^2 + 4 ab} }{ 2ab }
$$
Now we can plug in the values $a=0.004$ and $b=0.04$ to get
$$
omega approx frac{ 0.044 pm 0.0508 }{ 0.00032 }
$$
$ Rightarrow omega approx 21.1 $ or $omega approx 296.1 $ .
In addition, we had the condition that the real and imaginary parts have to be negative. Inserting these values for $omega$ into the original equation, we see that
$$
omega approx 296.1077
$$
is the solution. Just to check, plugging in this value results in
$$
frac{1.6}{(1+0.004i omega)(1+0.04i omega)} approx -0.0614(1+i)
$$
answered Jan 11 at 6:48
Matti P.Matti P.
1,900413
edited Jan 11 at 8:15
answered Jan 11 at 6:48
Matti P.Matti P.
1,900413
answered Jan 11 at 6:48
Matti P.Matti P.
1,900413
answered Jan 11 at 6:48
Matti P.Matti P.
1,900413
1,900413
$begingroup$
Thank you so much
$endgroup$
– omka
Jan 13 at 11:28
add a comment |
$begingroup$
Thank you so much
$endgroup$
– omka
Jan 13 at 11:28
$begingroup$
Thank you so much
$endgroup$
– omka
Jan 13 at 11:28
$begingroup$
Thank you so much
$endgroup$
– omka
Jan 13 at 11:28
add a comment |
$begingroup$
The equation directly interpreted, without any argument transforms beyond $arg(z^{-1})=-arg(z)+2kpi j$, gives
$$
(1+0,004j⋅ω)(1+0,04j⋅ω)=re^{j⋅135°}=r'(-1+i)
$$
for some $r=sqrt2 r'>0$. One reads off that the sum of real part and imaginary part on the left side has to be zero, with the imaginary part positive, as it is the case on the right.
$$
0=1-0.00016⋅ω^2 + 0.044⋅ωiff (0.04⋅ω-5.5)^2=0.0016⋅ω^2-0.44⋅ω+5.5^2=10+5.5^2
$$
so that $ω>0$ and $ω=25⋅(5.5pmsqrt{40.25})implies ω=25⋅(5.5+sqrt{40.25})=296.1072192556190..$.
answered Jan 11 at 11:53
LutzLLutzL
57.6k42054
$endgroup$
$begingroup$
Thank you @LutzL
$endgroup$
– omka
Jan 13 at 11:29
add a comment |
$begingroup$
The equation directly interpreted, without any argument transforms beyond $arg(z^{-1})=-arg(z)+2kpi j$, gives
$$
(1+0,004j⋅ω)(1+0,04j⋅ω)=re^{j⋅135°}=r'(-1+i)
$$
for some $r=sqrt2 r'>0$. One reads off that the sum of real part and imaginary part on the left side has to be zero, with the imaginary part positive, as it is the case on the right.
$$
0=1-0.00016⋅ω^2 + 0.044⋅ωiff (0.04⋅ω-5.5)^2=0.0016⋅ω^2-0.44⋅ω+5.5^2=10+5.5^2
$$
so that $ω>0$ and $ω=25⋅(5.5pmsqrt{40.25})implies ω=25⋅(5.5+sqrt{40.25})=296.1072192556190..$.
answered Jan 11 at 11:53
LutzLLutzL
57.6k42054
$endgroup$
$begingroup$
Thank you @LutzL
$endgroup$
– omka
Jan 13 at 11:29
add a comment |
$begingroup$
The equation directly interpreted, without any argument transforms beyond $arg(z^{-1})=-arg(z)+2kpi j$, gives
$$
(1+0,004j⋅ω)(1+0,04j⋅ω)=re^{j⋅135°}=r'(-1+i)
$$
for some $r=sqrt2 r'>0$. One reads off that the sum of real part and imaginary part on the left side has to be zero, with the imaginary part positive, as it is the case on the right.
$$
0=1-0.00016⋅ω^2 + 0.044⋅ωiff (0.04⋅ω-5.5)^2=0.0016⋅ω^2-0.44⋅ω+5.5^2=10+5.5^2
$$
so that $ω>0$ and $ω=25⋅(5.5pmsqrt{40.25})implies ω=25⋅(5.5+sqrt{40.25})=296.1072192556190..$.
answered Jan 11 at 11:53
LutzLLutzL
57.6k42054
$endgroup$
The equation directly interpreted, without any argument transforms beyond $arg(z^{-1})=-arg(z)+2kpi j$, gives
$$
(1+0,004j⋅ω)(1+0,04j⋅ω)=re^{j⋅135°}=r'(-1+i)
$$
for some $r=sqrt2 r'>0$. One reads off that the sum of real part and imaginary part on the left side has to be zero, with the imaginary part positive, as it is the case on the right.
$$
0=1-0.00016⋅ω^2 + 0.044⋅ωiff (0.04⋅ω-5.5)^2=0.0016⋅ω^2-0.44⋅ω+5.5^2=10+5.5^2
$$
so that $ω>0$ and $ω=25⋅(5.5pmsqrt{40.25})implies ω=25⋅(5.5+sqrt{40.25})=296.1072192556190..$.
answered Jan 11 at 11:53
LutzLLutzL
57.6k42054
answered Jan 11 at 11:53
LutzLLutzL
57.6k42054
answered Jan 11 at 11:53
LutzLLutzL
57.6k42054
answered Jan 11 at 11:53
LutzLLutzL
57.6k42054
57.6k42054
$begingroup$
Thank you @LutzL
$endgroup$
– omka
Jan 13 at 11:29
add a comment |
$begingroup$
Thank you @LutzL
$endgroup$
– omka
Jan 13 at 11:29
$begingroup$
Thank you @LutzL
$endgroup$
– omka
Jan 13 at 11:29
$begingroup$
Thank you @LutzL
$endgroup$
– omka
Jan 13 at 11:29
add a comment |
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$begingroup$
What have you tried? The first step would be the interpret the definition of complex argument.
$endgroup$
– Matti P.
Jan 8 at 12:50
$begingroup$
Hint: $tan{-135^{circ}} = 1$
$endgroup$
– Matti P.
Jan 8 at 12:58
$begingroup$
Please @MattiP. Can i write -0,004.w-0,04.w=1 then w=22,72? thank you
$endgroup$
– omka
Jan 10 at 17:14
$begingroup$
I made a plot and it seems like the answer is something like $omega approx 296.1$
$endgroup$
– Matti P.
Jan 11 at 7:11