Frequency Warping

It is easy to check that the bilinear transform gives a one-to-one, order-preserving, conformal map [57] between the analog frequency axis $s=j\omega_a$ and the digital frequency axis $z=e^{j\omega_d T}$ , where $T$ is the sampling interval. Therefore, the amplitude response takes on exactly the same values over both axes, with the only defect being a frequency warping such that equal increments along the unit circle in the $z$ plane correspond to larger and larger bandwidths along the $j\omega$ axis in the $s$ plane [88]. Some kind of frequency warping is obviously unavoidable in any one-to-one map because the analog frequency axis is infinite while the digital frequency axis is finite. The relation between the analog and digital frequency axes may be derived immediately from Eq.(I.9) as

$$\begin{eqnarray*} j\omega_a &=& c\frac{1-e^{-j\omega_d T}}{1+e^{-j\omega_d T}} = c\frac{e^{j\omega_d T}-1}{e^{j\omega_d T}+1}\\ &=& c\frac{e^{j\omega_d T/2}\left(e^{j\omega_d T/2}-e^{-j\omega_d T/2}\right)}{e^{j\omega_d T/2}\left(e^{j\omega_d T/2}+e^{-j\omega_d T/2}\right)}\\ &=& jc\frac{\sin(\omega_dT/2)}{\cos(\omega_dT/2)}\\ &=& jc\tan(\omega_dT/2). \end{eqnarray*}$$

Given an analog cut-off frequency $\omega_a=\omega_c$ , to obtain the same cut-off frequency in the digital filter, we set

$$c = \omega_c\cot(\omega_cT/2)$$