(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.

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

  • $\begingroup$ The complex $\arg\{\cdot\}$ is not always that $\arctan(\cdot)$ function you had. You should look it up. $\endgroup$ Jan 18, 2019 at 21:08
  • $\begingroup$ what is $\alpha_R$ and $\alpha_I$? isn't $\alpha$ real? and i think that $\theta$ needs to be in the expression somewhere, doesn't it? $\endgroup$ Jan 18, 2019 at 21:13
  • $\begingroup$ $\alpha_R=Re\{\alpha\}$ and $\alpha_I=Im\{\alpha\}$ $\endgroup$
    – Bill Moore
    Jan 19, 2019 at 1:05
  • $\begingroup$ but Bill, there is no place in your transfer function (which totally describes the input-output relationship of the LTI system) where there are separated real and imaginary parts to $\alpha$. in fact, it is only shown as $|\alpha|$ which must be real and non-negative. $$H(e^{j\omega}) = e^{-j\theta}\left( \frac{e^{-j(\omega-\theta)}- |\alpha|}{1-|\alpha|\ e^{-j(\omega - \theta)}} \right)$$ nowhere, anywhere, can there be a meaningful imaginary part to the value of $\alpha$. you can just replace $|\alpha|$ with $\beta$ and require that $\beta$ is real and non-negative. $\endgroup$ Jan 19, 2019 at 4:17
  • 1
    $\begingroup$ sorry, i'm just showing part II of question, H(z) is really an all-pass filter: $$H(z)=\frac{z^-1 - \alpha}{1-\alpha^{*} z^{-1}}$$. first part of problem is a little bit of a simplification because they just assume that $\alpha$ is always real. $$H(z)=\frac{z^-1 - \alpha}{1-\alpha z^{-1}}$$. but then they fix it in part II. so, I was trying to solve the complex $\alpha$ using the same method as part I by assuming that ($\alpha = \Re{\alpha} + j\Im{\alpha})$, but really they tell you a better way in part II to use polar for $\alpha$ and factor out the phase offet rather than using cartesian. $\endgroup$
    – Bill Moore
    Jan 19, 2019 at 20:07

2 Answers 2


Define a frequency response


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


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 $z_{\infty}=|\alpha|<1$, 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 $z_{\infty}=\alpha=|\alpha|e^{j\theta}$ ($|\alpha|<1$), and a zero at $z_0=1/\alpha^*$.

Now write


and note that


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,


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




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




From $(5)$ we get


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


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.

  • $\begingroup$ Thanks. the $\tilde{H}$ trick makes it about 10 times easier. $\endgroup$
    – Bill Moore
    Jan 19, 2019 at 15:38
  • $\begingroup$ I can follow it to the end of the problem, but i'm still a little confused about the minus sign for $\tilde{\phi}(\omega)=\phi_B(\omega)-\phi_A(\omega)\tag{4}$ I would have thought its positive? where does the minus sign come from? $$\tilde{H}(e^{j\omega})=(e^{-j\omega}-|\alpha|) \left(\frac{1}{1-|\alpha|e^{-j\omega}}\tag{1}\right)$$ $\endgroup$
    – Bill Moore
    Jan 19, 2019 at 16:34
  • $\begingroup$ maybe i lost a sign somewhere... $\endgroup$
    – Bill Moore
    Jan 19, 2019 at 16:39
  • $\begingroup$ @BillMoore: The minus sign is there because $\phi_A$ is the phase of the denominator, so you have to subtract it from the phase of the numerator. The phase of $u=v/w$ is the phase of $v$ minus the phase of $w$. $\endgroup$
    – Matt L.
    Jan 19, 2019 at 17:06

Just apply the definitions.

$$\begin{align} H(e^{j\omega}) &= \Re\Big\{H(e^{j\omega})\Big\} + j \Im\Big\{H(e^{j\omega})\Big\} \\ \\ &= \Big| H(e^{j\omega}) \Big| e^{j \arg\{H(e^{j\omega})\}} \\ \\ &= \Big| H(e^{j\omega}) \Big| e^{j \phi(\omega)} \\ \end{align}$$


$$ \phi(\omega)\triangleq \arg\Big\{ H(e^{j\omega}) \Big\} $$


$$\begin{align} \arg\Big\{H(e^{j\omega})\Big\} &= \operatorname{atan2}(\Im\Big\{H(e^{j\omega})\Big\},\, \Re\Big\{H(e^{j\omega})\Big\}) \\ \\ &=\begin{cases} \arctan\left(\frac{\Im\{H(e^{j\omega})\}}{\Re\{H(e^{j\omega})\}}\right) &\text{if } \Re\Big\{H(e^{j\omega})\Big\} > 0 \\ \\ \frac{\pi}{2} - \arctan\left(\frac{\Re\{H(e^{j\omega})\}}{\Im\{H(e^{j\omega})\}}\right) &\text{if } \Im\Big\{H(e^{j\omega})\Big\} > 0 \\ \\ -\frac{\pi}{2} - \arctan\left(\frac{\Re\{H(e^{j\omega})\}}{\Im\{H(e^{j\omega})\}}\right) &\text{if } \Im\Big\{H(e^{j\omega})\Big\} < 0 \\ \\ \arctan\left(\frac{\Im\{H(e^{j\omega})\}}{\Re\{H(e^{j\omega})\}}\right) \pm \pi &\text{if } \Re\Big\{H(e^{j\omega})\Big\} < 0 \\ \\ \text{undefined} &\text{if } H(e^{j\omega}) = 0 \end{cases} \end{align}$$

Phase delay is the delay in time of a sinusoid at normalized angular frequency of $\omega$ passed through this LTI system and turns out to be this ratio ($\phi(\omega)$ must be "unwrapped" in this phase delay definition):

$$ \tau_\phi(\omega) \triangleq -\frac{\phi(\omega)}{\omega} $$

Group delay is the delay in time of an envelope of a sinusoid at normalized angular frequency of $\omega$ passed through this LTI system and turns out to be this derivative:

$$ \tau_\mathrm{g}(\omega) \triangleq -\frac{\mathrm{d}\phi(\omega)}{\mathrm{d}\omega} $$

Just apply the definitions and slug it out.

  • $\begingroup$ how to get the $\phi(\omega)$ to start with? $\endgroup$
    – Bill Moore
    Jan 18, 2019 at 21:03
  • $\begingroup$ You will have to split $H(e^{j\omega})$ into real and imaginary parts. then apply the definition of the complex $\arg\{\cdot\}$. $\endgroup$ Jan 18, 2019 at 21:12

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