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

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

\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}
• Thanks. One question. You've taken $tan(\arctan(\dfrac{rsin\omega}{1-rcos\omega})+\arctan(\dfrac{sin\omega}{r-cos\omega}))$ and got $-tan(\omega)$. Now, how to take the inverse of $tan$ to get original $\arctan(\dfrac{rsin\omega}{1-rcos\omega})+\arctan(\dfrac{sin\omega}{r-cos\omega})$? I feel, the answer will contain multpile cases, and will the answer be just simple as $\pi - \omega$? Oct 5, 2018 at 12:38
• @NarendraDeconda: That's what my last sentence referred to. From the result you see that the phase is a linear function. The offset is most easily seen from the original transfer function by checking the points $\omega=0$ and/or $\omega=\pi$. E.g., for $\omega=\pi$, the frequency response is real-valued and positive, so the phase must be zero (or a multiple of $2\pi$), this uniquely determines the offset you need. Oct 5, 2018 at 14:06