finding phase and group delay from frequency response (From Schaum's Outlines, DSP, second edition, problem 5.25, second part of problem)

What's the procedure to find the Phase and Group Delay of:

$$
H(e^{j\omega}) = e^{-j\theta}\left( \frac{e^{-j(\omega-\theta)}- |\alpha|
}{1-|\alpha|\ e^{-j(\omega - \theta)}} \right)
$$

Book says that group delay should be:

$$
\tau(\omega) = \tau_0(\omega - \theta) = \frac{1-\alpha^2}{|1-\alpha\ e^{-j(\omega-\theta)}|^2}
$$

I want to say the phase is $\phi(\omega) = (\omega - \theta)$ and group delay is: $\tau(\omega) = - \phi ' (\omega)= 1$... but i know that's not correct because the book gives a more complicated group delay.  it looks like book version has something to do with "$\tau_0$", which i'm guessing is the initial group delay at time zero??  actually, i have no idea... its not explained in the book. (Any ideas how to do this same problem using the method in second part of problem?)

I feel there's a 50% chance i need to do this the complicated algebra and calculus way using this expresss as start:

$$
\phi(\omega) = \arg\Big\{ H_R(e^{j\omega})+jH_I(e^{j\omega}) \Big\}
$$

but then, i'm not really sure if there's a shortcut to just convert it to polar form instead of slugging it out in complex Cartesian form alegbra. like it seems the phase is $\phi(\omega) = (\omega - \theta)$...

with a lot of algebra I obtained this result:

$$
\tau(\omega) = \frac{(-\alpha^2_I - \alpha^2_R + 1)}{1 + \alpha_{R}^{2} + \alpha_{I}^{2} -2 \alpha_{R} \cos(\omega) - 2 \alpha_{I} \sin(\omega)}
$$

still would be nice if there were a simpler way..like the way the second part of the problem suggests...but doesn't really explain...

Fancy factoring to match book result:

$$1 + \alpha_{R}^{2} + \alpha_{I}^{2} -2 \alpha_{R} \cos(\omega) - 2 \alpha_{I} \sin(\omega)$$

$$= (1-\alpha e^{-j\omega}) (1- \alpha^{*} e^{j\omega})$$

$$=|1-\alpha\ e^{-j\omega}|^2$$

other factor:

$$|z|^2 = z_I^2 + z_R^2 $$

$$(-\alpha^2_I - \alpha^2_R + 1)$$

$$=(1 -(\alpha^2_I + \alpha^2_R) )$$

$$=(1 - |\alpha|^2)$$

can remove absolute value as long as $\alpha$ remains squared.

$$=(1 - \alpha^2)$$
Define a frequency response

$$\tilde{H}(e^{j\omega})=\frac{e^{-j\omega}-|\alpha|}{1-|\alpha|e^{-j\omega}}\tag{1}$$

and note that the group delay $\tilde{\tau}(\omega)$ corresponding to $(1)$ is related to the group delay $\tau(\omega)$ of the original frequency response $H(e^{j\omega})$ by

$$\tau(\omega)=\tilde{\tau}(\omega-\theta)\tag{2}$$

because $(1)$ is just a scaled and frequency-shifted version of the original $H(e^{j\omega})$.

$\tilde{H}(e^{j\omega})$ is the frequency response of a first-order all-pass filter with a real-valued pole at $z_{\infty}=|\alpha|&lt;1$, and a real-valued zero at $z_0=1/|\alpha|$. The original frequency response $H(e^{j\omega})$ describes a first-order all-pass filter with a complex-valued pole at $z_{\infty}=\alpha=|\alpha|e^{j\theta}$ ($|\alpha|&lt;1$), and a zero at $z_0=1/\alpha^*$.

Now write

$$\tilde{H}(e^{j\omega})=\frac{B(e^{j\omega})}{A(e^{j\omega})}\tag{3}$$

and note that

$$\tilde{\phi}(\omega)=\phi_B(\omega)-\phi_A(\omega)\tag{4}$$

where $\tilde{\phi}(\omega)$ is the phase of $\tilde{H}(e^{j\omega})$, and $\phi_A(\omega)$ and $\phi_B(\omega)$ are the phases of $A(e^{j\omega})$ and $B(e^{j\omega})$, respectively. Consequently,

$$\tilde{\tau}(\omega)=\tau_B(\omega)-\tau_A(\omega)\tag{5}$$

where $\tau_A(\omega)$ and $\tau_B(\omega)$ are the group delays of $A(e^{j\omega})$ and $B(e^{j\omega})$, respectively.

We have

$$\phi_B(\omega)=-\arctan\left(\frac{\sin(\omega)}{\cos(\omega)-|\alpha|}\right)\tag{6}$$

and

$$\phi_A(\omega)=\arctan\left(\frac{|\alpha|\sin(\omega)}{1-|\alpha|\cos(\omega)}\right)\tag{7}$$

where I've ignored any ambiguities by $\pm\pi$ due to the principal value of $\arctan()$ because they are irrelevant for computing the group delay.

Taking the negative derivative of $(6)$ and $(7)$ with respect to $\omega$ gives

$$\tau_B(\omega)=\frac{1-|\alpha|\cos(\omega)}{1-2|\alpha|\cos(\omega)+|\alpha|^2}\tag{8}$$

and

$$\tau_A(\omega)=\frac{|\alpha|^2-|\alpha|\cos(\omega)}{1-2|\alpha|\cos(\omega)+|\alpha|^2}\tag{9}$$

From $(5)$ we get

$$\tilde{\tau}(\omega)=\frac{1-|\alpha|^2}{1-2|\alpha|\cos(\omega)+|\alpha|^2}=\frac{1-|\alpha|^2}{\big|1-|\alpha|e^{-j\omega}\big|^2}\tag{10}$$

From $(2)$ the group delay of the complex first-order all-pass is given by

$$\tau(\omega)=\frac{1-|\alpha|^2}{1-2|\alpha|\cos(\omega-\theta)+|\alpha|^2}=\frac{1-|\alpha|^2}{\big|1-|\alpha|e^{-j(\omega-\theta)}\big|^2}\tag{11}$$

The difference with the given solution in Schaum's outline is the fact that in $(11)$ you have $|\alpha|$ instead of the complex-valued $\alpha$. Note that $(11)$ is correct and Schaum's formula is wrong because a complex-valued group delay doesn't make much sense.