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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 $H(e^{... 1 Use fvtool and freqz in Matlab. b=[0.5b 0 1]; a=[2 0 b]; freqz(b,a); 1 I am not sure how one would avoid complex math or why one would want to given the simplification complex math provides (an oxymoron due to poor naming or perhaps to scare those that don't know yet how easy it is away) but perhaps the graphical explanation can help provide further intuitive insight. First some necessary background:$z$is the domain of all ... 1 First of all, your filter coefficients are given by b = g * [1, b1, b2] a = [1,a1,a2] And second, I believe that it is the phase delay that you're interested in, not the group delay. The phase delay at frequency$\omega_0>0$is given by $$\tau_p(\omega_0)=-\frac{\phi(\omega_0)}{\omega_0}\tag{1}$$ where$\phi(\omega)\$ is the (unwrapped) phase of the ...