Text from "Fundamentals of Speech Recognition (Rabiner)".

A complex pole of the LPC model spectrum can be expressed as $Z_i = > r_i \cdot e^{j \omega_i} $, where $r_i$ and $\omega_i$ are the radius and frequency of the complex pole, respectively.

The bandwidth of the complex pole is related to the radius by,

$ B_i = - \log(r_i) \cdot f_s/π$

My understanding of bandwidth is that it is the frequency range outside which the signal power drops by 3dB.

What does bandwidth of a complex pole mean? How is the 3dB or any other metric incorporated in the given formula for bandwidth $B_i$ ?

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    $\begingroup$ Note that this is an approximation, valid only for $r_i \approx 1$. I’ll answer shortly $\endgroup$
    – Jdip
    Commented Feb 23 at 14:01

2 Answers 2


My understanding of bandwidth is that it is the frequency range outside which the signal power drops by 3dB.

That is the usual definition, yes. A better phrasing may be "drops by 3dB from the peak power".

What does bandwidth of a complex pole mean?

That, compared to that pole's frequency of least effect, the contribution to the signal's attenuation (or it's loss of gain) of that pole is 3dB down.

How is the 3dB or any other metric incorporated in the given formula for bandwidth $B_i$ ?

It's difficult to explain without dragging in half of DSP theory, so you may find this explanation a bit dense.

I'm going to introduce a bit of common DSP notation here: if you see $H(\omega)$, it denotes the system's response to a single-frequency complex signal: $$H(\omega) = \left . H(z) \right |_{z = e^{j \omega}} \tag 1$$ Yes, the DSP crowd is playing fast and loose with the notation here. Sorry-not-sorry.

In order to isolate the effect of the one pole from the rest of the system, assume that your system transfer function is equal to $$H(z) = H_{tr}(z) H_i(z) = H_{tr}(z) \frac {1 - r_i}{z - Z_i} \tag 2$$ where $H_i(z)$ is the effect of your pole and $H_{tr}(z)$ is the rest of the transfer function.

< Snip some extraneous material here, which means that equation (3) is lost >

Consider $H_i$ by itself, when it is excited by a tone at a frequency offset from $\omega_i$: $$H_i(\omega_i + \Delta \omega) = \frac {1 - r_i} {e^{j(\omega_i + \Delta \omega)} - Z_i} = \frac {1 - r_i} {e^{j(\omega_i + \Delta \omega)} - r_i e^{j \omega_i}} = \frac {1 - r_i} {e^{j\omega_i} \left ( e^{j \Delta \omega} - r_i \right ) } \tag 4$$

Observe that at $\Delta \omega = 0$, $$\left | H_i(\omega_i + \Delta \omega) \right | = 1 \tag 5$$

Observe that for $\Delta \omega \ll 1$, $$e^{j \Delta \omega} \simeq 1 + j \Delta \omega \tag 6$$

Using (6), for sufficiently small $\Delta \omega$, $$\left | H_i(\omega_i + \Delta \omega) \right | \simeq \left | \frac {1 - r_i}{1 - r_i + j\Delta \omega} \right | \tag 7$$

Now assume that $-\log r_i \ll 1$ (note that $0 < r_i < 1$). Then $r_i \simeq 1 + \log r_i$. (7) becomes $$\begin{aligned} \left | H_i(\omega_i + \Delta \omega) \right | &\simeq \left | \frac {1 - (1 + \log r_i)}{1 - (1 + \log r_i) + j\Delta \omega} \right | \\ &= \left | \frac {- \log r_i}{- \log r_i + j\Delta \omega} \right | \\ &= \frac {- \log r_i}{\sqrt{\log^2 r_i + (\Delta \omega)^2}} \end{aligned} \tag 8$$

When $\Delta \omega = \pm \log r_i$, (8) becomes $$\left | H_i(\omega_i + \Delta \omega) \right | \simeq \frac 1 {\sqrt 2} \tag 9$$ i.e., for $\omega = \omega_i \pm \log r_i$, the response is 3dB down from the peak.

To get this into the author's notation, observe that the sampled-time frequency $\omega$ is related to the real-world frequency as $$\omega = 2 \pi \frac f {f_s} \tag {10}$$ Solving for $f$: $$f = \frac {\omega f_s}{2 \pi} \tag {11}$$

To find the real-world bandwidth, plug our 3dB down points into (11): $$B_i = \left( \left(\omega_i + \log r_i \right) - \left( \omega_i - \log r_i \right) \right) \frac {f_s}{2 \pi} = \log r_i \frac{f_s}{\pi} \tag {12}$$

Which is what your book's author so blithely wrote in one sentence.

  • $\begingroup$ Equation 3 is missing. This is such a beautiful answer. The only thing that bothers me is the formulation for Hi(z) (effect of i-th pole). The denominator (z - Zi) makes total sense. At the end of the day whatever denominator H(z) has, it can be broken into (z-Z1)*(z-Z2)*.. and so on ; that is what defines a pole. However, I can't reason (1-ri) in the numerator. Closer the pole is to the origin, greater is the emphasis by the filter? $\endgroup$ Commented Feb 25 at 7:12
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    $\begingroup$ It’s to have unity gain at the resonant frequency (equation 5), and to make it easier to derive 9) from there. Tim’s beautiful answer holds as well with $Hi(z)=\frac{1}{z−Zi}$ but with some tweaks. The normalization makes it easier to follow. $\endgroup$
    – Jdip
    Commented Feb 25 at 9:50
  • $\begingroup$ Nice answer!!!! $\endgroup$
    – Peter K.
    Commented Feb 25 at 15:16
  • $\begingroup$ Equations (7) and (8), are you sure this should be the |magnitude| response? If so, I wouldn't expect the imaginary $j\Delta_{\omega}$ in the denominator. Either you meant to write the frequency response instead, in which case $e^{j\omega_i}$ should still be in the denominator as well (and disappears when the magnitude is taken), or you simply forgot the $||$ on the denominators of (7) and (8)... $\endgroup$
    – Jdip
    Commented Feb 26 at 6:52
  • $\begingroup$ I forgot to indicate the absolute values appropriately. Thanks! $\endgroup$
    – TimWescott
    Commented Feb 26 at 19:44

Tim's answer is a very nice derivation of the approximation in the discrete-time domain.

Two additions to his fantastic answer:

Derivation of the approximation in the continuous-time domain

It's also possible to get to the approximation in the continuous-time domain: see this reference

Calculating the $-3\texttt{dB}$ bandwidth

We can actually calculate the $-3\texttt{dB}$ bandwidth exactly for any value of $r$ and show that the approximation $-\log(r)\frac{f_s}{\pi}$ is valid for $r$ close to $1$.

Starting with the normalized transfer function of a complex one-pole resonator: $$H(z) = \frac{1-r}{1-pz^{-1}}$$ where $p = re^{j\omega_c}$ and $\omega_c = 2\pi\frac{f_c}{f_s}$ is the resonant normalized angular frequency.

The magnitude response can be calculated as: $$ |H(\omega)| = \frac{|1-r|}{|1-re^{j(\omega_c-\omega)}|} = \frac{1-r}{\sqrt{1 - 2r\cos(\omega_c-\omega) + r^2}} $$

As in Tim's answer, note that $|H(\omega_c)| = 1$

The $-3\,\tt{dB}$ bandwidth $\Delta_{\omega}$ is such that $$|H(\omega)| = \frac{1}{\sqrt{2}}$$ Let's solve this: \begin{align} \frac{1-r}{\sqrt{1 - 2r\cos(\frac{\Delta_{\omega}}{2}) + r^2}} &= \frac{1}{\sqrt{2}}\\ 2(1-r)^2 &= 1 - 2r\cos(\frac{\Delta_{\omega}}{2}) + r^2\\ 2\cos^{-1}\left(\frac{r(4-r)-1}{2r}\right) &= \Delta_{\omega}\\ \end{align}

So now we have an exact expression for the bandwidth with respect to $r$: $$ \Delta_{\omega} = 2\cos^{-1}\left(\frac{r(4-r)-1}{2r}\right)$$ Let's verify that this aligns with the approximation $\Delta_{\omega} = -2\log(r)$ for values of $r$ close to $1$:

Note that to get to real-world frequency, these two expressions should be multiplied by $\frac{f_s}{2\pi}$

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    $\begingroup$ I satisfied myself (on the log approximation; area under the graph intuition) with math.stackexchange.com/a/3253689/944228 .But now you have offered exact bandwidth's formulation. Absolutely cool! Thanks. $\endgroup$ Commented Feb 26 at 18:04

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