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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.

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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).

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  • $\begingroup$ 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 $\pi - \omega$? $\endgroup$ – Narendra Deconda Oct 5 '18 at 12:38
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    $\begingroup$ @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. $\endgroup$ – Matt L. Oct 5 '18 at 14:06

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