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How can I calculate the phase of the following complex number?

$ \omega, R ,C $ are some positive constants.

$$ \frac{1-i\omega RC}{2+2i\omega RC} $$

In my book the answer is $ 2\arctan{(\omega\cdot RC}) $

But I cannot see why. Here is my calculation:

$ \frac{1-i\omega RC}{2+2i\omega RC}=\frac{\left(1-i\omega RC\right)\left(1-i\omega RC\right)}{2\left(1+i\omega RC\right)\left(1-i\omega RC\right)}=\frac{1}{2}\frac{1-2i\omega RC-\left(\omega RC\right)^{2}}{\left(1+\left(\omega RC\right)^{2}\right)}=\frac{1-\left(\omega RC\right)^{2}}{2\left(1+\left(\omega RC\right)^{2}\right)}-i\frac{\omega RC}{\left(1+\left(\omega RC\right)^{2}\right)} $

So the phase should be

$ \angle=\arctan\left(\frac{\frac{\omega RC}{1+\left(\omega RC\right)^{2}}}{\left(\frac{1-\left(\omega RC\right)^{2}}{2\left(1+\left(\omega RC\right)^{2}\right)}\right)}\right)=\arctan\left(\frac{2\omega RC}{1-\left(\omega RC\right)^{2}}\right) $

The second way I tried and Im not sure why it is a wrong way, is to subtract the phase of the denominator from the phase on the numerator, so that $ \angle=\angle\left(1-i\omega RC\right)-\angle\left(2+2i\omega RC\right)=\arctan\left(\omega RC\right)-\arctan\left(\omega RC\right) $

Which is obviously wrong.

Any help would be appreciated. Thank in advance.

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Write $z=1-i \omega RC$ then your expression is: $\frac{1}{2} \frac{z}{z^*}$ where $*$ denotes the complex conjugate. Writing $z=r e^{i\theta}$ you have $\frac{1}{2} \frac{z}{z^*}=\frac{1}{2} e^{2i\theta}$.

Since the phase of $z$ is $-\arctan(\omega RC)$, the phase of your expression is : $-2 \arctan(\omega RC)$. To recover this result from yours, you can use the formula $\arctan(x)+\arctan(y)=\arctan\left(\frac{x+y}{1-xy} \right)+k\pi$. You made an error of sign in your last lign, otherwise you get the correct answer.

By the way, one should be careful with the use of $\arctan$ to recover the phase of $x+iy$ since the signs of $x$ and $y$ should be taken into account : $\arctan$ takes values in $]-\pi/2,\pi/2[$ while the phase is generally defined over $]-\pi,\pi]$.

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