Phase plot of $H(e^{j\omega})=(1-re^{-j\omega})\left(1-\dfrac{1}{r}e^{-j\omega}\right)$, $0<r<1$

Question

I want to find the phase plot of $H(e^{j\omega})=(1-re^{-j\omega})\left(1-\dfrac{1}{r}e^{-j\omega}\right)$, $0<r<1$ for the interval $0\leq \omega \leq \pi$.

Method 1:

$H(e^{j\omega})=1-\left(r+\dfrac{1}{r}\right)e^{-j\omega}+e^{-2j\omega}=e^{-j\omega}\left[2\cos \omega-\left(r+\dfrac{1}{r}\right)\right]$

$\angle H(e^{j\omega})=-\omega+\angle \left[2\cos \omega-(r+\dfrac{1}{r})\right]=-\omega+\pi$

Method 2:

$\angle H(e^{j\omega})=\angle(1-re^{-j\omega})+\angle(1-\dfrac{1}{r}e^{-j\omega})$

$$\angle H(e^{j\omega})=\begin{cases} \tan^{-1}\bigg(\dfrac{r\sin \omega}{1-r\cos\omega}\bigg)+\pi-\tan^{-1}\bigg(\dfrac{\sin \omega}{\cos \omega-r}\bigg),0\leq \omega\leq \cos^{-1}(r)\\ \tan^{-1}\bigg(\dfrac{r\sin\omega}{1-r\cos\omega}\bigg)+\tan^{-1}\bigg(\dfrac{\sin\omega}{r-\cos \omega}\bigg),\cos^{-1}(r)\leq \omega\leq \pi\\ \end{cases}$$

I tried applying the formula $\tan^{-1}x+\tan^{-1}y=\tan^{-1}\bigg(\dfrac{x+y}{1-xy}\bigg)$, but I am confused when to use this formula and unable to proceed forward. Please help in solving this.

Answer 1

In fact you've already proved that the two expressions must be identical. But if you want to show it the hard way, you can indeed use the formula you suggested. The argument of the combined arctangent function becomes

$$\begin{align}\frac{\frac{r\sin(\omega)}{1-r\cos(\omega)}+\frac{\sin(\omega)}{r-\cos(\omega)}}{1-\frac{rsin^2(\omega)}{(1-r\cos(\omega))(r-\cos(\omega))}}&=\frac{r\sin(\omega)(r-\cos(\omega))+\sin(\omega)(1-r\cos(\omega))}{(1-r\cos(\omega))(r-\cos(\omega))-r\sin^2(\omega)}\\&=\frac{\sin(\omega)(1-2r\cos(\omega)+r^2)}{-\cos(\omega)(1-r\cos(\omega)+r^2)+r(\underbrace{1-\sin^2(\omega)}_{\cos^2(\omega)})}\\&=\frac{\sin(\omega)(1-2r\cos(\omega)+r^2)}{-\cos(\omega)(1-2r\cos(\omega)+r^2)}=-\frac{\sin(\omega)}{\cos(\omega)}=-\tan(\omega)\end{align}$$

Now you just have to be careful with the principal value and you'll obtain the same result as with your first method (which is of course much smarter).