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Could someone help me please. I'm looking for the value of $\omega$

$\text{Argument}(\frac{1,6}{(1+0,004\text{j}\cdot\omega)(1+0,04\text{j}\cdot\omega)})=-135° $

$\text{Argument}({1,6})-\text{Argument}({(1+0,004\text{j}\cdot\omega)(1+0,04\text{j}\cdot\omega)}))=-135° $

$\text{Atan}(\frac{0}{1,6})-\text{Atan}({\frac{0,004\cdot\omega}{1})-\text{Atan}(\frac{0,04\cdot\omega}{1})}=-135° $

$-\text{Atan}({\frac{0,004\cdot\omega}{1})-\text{Atan}(\frac{0,04\cdot\omega}{1})}=-135° $

i'm stucked here :(

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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
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The argument of a complex number is the angle that the complex number makes, when plotted in the complex plane. Therefore, $$ \text{Arg}{\left( z\right)} = -135^{\circ} \Rightarrow z \text{ is in the third quadrant, $\text{Im}{(z)} = \text{Re}(z)<0$} $$ 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): $$ \begin{split} \frac{1}{(1+ai \omega)(1+bi\omega)} &= \frac{(1-ai \omega)(1-bi\omega)}{(1+ai \omega)(1-ai \omega)(1+bi\omega)(1-bi\omega)} \\ &=\frac{(1-ai \omega)(1-bi\omega)}{(1+a^2 \omega^2)(1+b^2\omega^2)} \\ &= \frac{1-bi\omega-ai\omega+abi^2\omega^2}{(1+a^2 \omega^2)(1+b^2\omega^2)} \\ &= \frac{1-ab\omega^2-(a+b)i\omega}{(1+a^2 \omega^2)(1+b^2\omega^2)} \\ \end{split} $$ Now it's easy to separate the imaginary and real part. Equating them, we get $$ \frac{1-ab\omega^2}{(1+a^2 \omega^2)(1+b^2\omega^2)} = \frac{-(a+b)\omega}{(1+a^2 \omega^2)(1+b^2\omega^2)} $$ Multiplying by the denominator (and by $-1$), we obtain $$ ab\omega^2 +(-a-b)\omega -1 = 0 $$ 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 $$ \frac{1.6}{(1+0.004i \omega)(1+0.04i \omega)} \approx -0.0614(1+i) $$

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  • $\begingroup$ Thank you so much $\endgroup$ – omka Jan 13 at 11:28
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The equation directly interpreted, without any argument transforms beyond $\arg(z^{-1})=-\arg(z)+2k\pi j$, 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. $$ 0=1-0.00016⋅ω^2 + 0.044⋅ω\iff (0.04⋅ω-5.5)^2=0.0016⋅ω^2-0.44⋅ω+5.5^2=10+5.5^2 $$ so that $ω>0$ and $ω=25⋅(5.5\pm\sqrt{40.25})\implies ω=25⋅(5.5+\sqrt{40.25})=296.1072192556190..$.

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  • $\begingroup$ Thank you @LutzL $\endgroup$ – omka Jan 13 at 11:29

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