Is there a simpler way to calculate the amplitude response of the following filter

Question

I would like to calculate the amplitude response $|H(z)|$, $z=e^{j\omega}$, of the following filter:

$$H(z)=\frac{\frac{b}{2}+z^{-2}}{2+bz^{-2}}$$

and I would like to avoid using Euler's formula and complex numbers cartesian form as much as possible.

Answer 1

The given transfer function has a specific form. With the denominator polynomial $A(z)=2+bz^{-2}$, you can write the transfer function as

$$H(z)=\frac{z^{-2}}{2}\frac{A\left(\frac{1}{z}\right)}{A(z)}\tag{1}$$

Since for $z=e^{j\omega}$ (i.e., on the unit circle, where we evaluate the frequency response), we have $|A(z)|=|A(z^{-1})|$, the magnitude of $H(e^{j\omega})$ is given by

$$\left|H(e^{j\omega})\right|=\frac12\tag{2}$$

The given filter is a (scaled) all-pass filter. It's easy to recognize all-pass filters by noting that their numerator and denominator polynomials have mirrored coefficients.

Answer 2

I am not sure how one would avoid complex math or why one would want to given the simplification complex math provides (an oxymoron due to poor naming or perhaps to scare those that don't know yet how easy it is away) but perhaps the graphical explanation can help provide further intuitive insight.

First some necessary background: $z$ is the domain of all complex numbers, meaning $z$ itself can take on any complex number (therefore it is the plane of all complex numbers). But the frequency response is found when you only use the unit circle for $z$. Meaning a circle on the complex plane, centered on the origin, that has a magnitude of 1. Mathematically, this is given as $z= e^{j\omega}$ where $\omega$ represents frequencies from $0$ to $2\pi$, or if we prefer $\pm\pi$, where the sampling rate $f_s= 2\pi$ (in units of radians per sample, hence the reason for $\pi$ showing up. Each point on this unit circle represents a different frequency between $f=0$ or "DC" and the sampling rate with $f=2\pi$.

The term $e^{j\omega}$ is a phasor that has a magnitude of 1 always, and an angle given by $\omega$.

So we use $z= e^{j\omega}$ to get the frequency response which will include both the magnitude and phase for all points $\omega$. Thus the magnitude response is often expressed as $|H(e^{j\omega})|$, showing that we are making this substitution for $z$. Likewise $z^{-1}= e^{-j\omega}$ and $z^{-2}= e^{-j2\omega}$ etc. Notice what is occuring here: raising $z$ to the $-1$ power results in a phasor that has a magnitude of 1 and rotates clockwise as the frequency goes from $0$ to $2\pi$ (as the angle is increasingly negative), and raising $z$ to the $-2$ power results in the phasor rotating twice.

So now observe the numerator in the OP's expression for $H(z)$, which is $b/2+ z^{-2}$. This is graphically depicted below on the left, on its own as a phasor of magnitude 1 rotating clocwise twice (the $z^{-2}$ term) added to a phasor of magnitude $b/2$ and angle 0. (The $b/2$ does not rotate as we sweep the frequency but the $z^{-2}$ term does.) The net result of this summation is the phasor with the dashed line, here shown when the frequency is approximately at $\pi/8$. That dashed line has the magnitude and phase of the numerator at that particular frequency, and for purpose of computing the magnitude of $H(e^{j\omega})$, the phase will be of no consequence. Since the rotating phasor in the picture is given by $z^{-2}$, it will have rotated twice as the frequency is swept from $0$ to $2\pi$. For example, when the frequency is at half the sampling rate, which is given as $f=\pi$, the rotating phasor will have rotated $-2\pi$ or be back at angle 0 (for a numerator magnitude of $1+b/2$ at that frequency).

The denominator is formed the same way, thus overall resulting in a numerator mangitude divided by a denominator magnitude to give the final magnitude of $H(e^{j\omega})$ for each frequency value $\omega$ between $0$ and $2\pi$.

Answer 3

Use fvtool and freqz in Matlab.

b=[0.5b 0 1];
a=[2 0 b];
freqz(b,a);