Editing Bilinear transform (section)

== Transformation of a General LTI System ==
A general [[LTI system]] has the transfer function
<math display=block>
    H_a(s) = \frac{b_0 + b_1s + b_2s^2 + \cdots + b_Qs^Q}{a_0 + a_1s + a_2s^2 + \cdots + a_Ps^P}
</math>
The order of the transfer function {{math|''N''}} is the greater of {{math|''P''}} and {{math|''Q''}} (in practice this is most likely {{math|''P''}} as the transfer function must be [[Proper transfer function|proper]] for the system to be stable). Applying the bilinear transform
<math display=block>
    s = K\frac{z - 1}{z + 1}
</math>
where {{math|''K''}} is defined as either {{math|2/''T''}} or otherwise if using [[frequency warping]], gives
<math display=block>
    H_d(z) = \frac{b_0 + b_1\left(K\frac{z - 1}{z + 1}\right) + b_2\left(K\frac{z - 1}{z + 1}\right)^2 + \cdots + b_Q\left(K\frac{z - 1}{z + 1}\right)^Q}
                  {a_0 + a_1\left(K\frac{z - 1}{z + 1}\right) + a_2\left(K\frac{z - 1}{z + 1}\right)^2 + \cdots + b_P\left(K\frac{z - 1}{z + 1}\right)^P}
</math>
Multiplying the numerator and denominator by the largest power of {{math|(''z'' + 1)<sup>−1</sup>}} present, {{math|(''z'' + 1)<sup>−''N''</sup>}}, gives
<math display=block>
   H_d(z) = \frac{b_0(z+1)^N + b_1K(z-1)(z+1)^{N-1} + b_2K^2(z-1)^2(z+1)^{N-2} + \cdots + b_QK^Q(z-1)^Q(z+1)^{N-Q}}
                 {a_0(z+1)^N + a_1K(z-1)(z+1)^{N-1} + a_2K^2(z-1)^2(z+1)^{N-2} + \cdots + a_PK^P(z-1)^P(z+1)^{N-P}}
</math>
It can be seen here that after the transformation, the degree of the numerator and denominator are both {{math|''N''}}.

Consider then the pole-zero form of the continuous-time transfer function
<math display=block>
    H_a(s) = \frac{(s - \xi_1)(s - \xi_2) \cdots (s - \xi_Q)}{(s - p_1)(s - p_2) \cdots (s - p_P)}
</math>
The roots of the numerator and denominator polynomials, {{math|''ξ<sub>i</sub>''}} and {{math|''p<sub>i</sub>''}}, are the [[zeros and poles]] of the system. The bilinear transform is a [[one-to-one mapping]], hence these can be transformed to the z-domain using
<math display=block>
    z = \frac{K + s}{K - s}
</math>
yielding some of the discretized transfer function's zeros and poles {{math|''ξ'<sub>i</sub>''}} and {{math|''p'<sub>i</sub>''}}
<math display=block>
    \begin{aligned}
        \xi'_i &= \frac{K + \xi_i}{K - \xi_i} \quad 1 \leq i \leq Q \\
          p'_i &= \frac{K + p_i}{K - p_i}     \quad 1 \leq i \leq P
    \end{aligned}
</math>
As described above, the degree of the numerator and denominator are now both {{math|''N''}}, in other words there is now an equal number of zeros and poles. The multiplication by {{math|(''z'' + 1)<sup>−''N''</sup>}} means the additional zeros or poles are
<ref>
{{cite web|url=http://www.ee.ic.ac.uk/hp/staff/dmb/courses/DSPDF/00800_TransIIR.pdf
          |last=Bhandari |first=Ayush
          |title=DSP and Digital Filters Lecture Notes
          |access-date=16 August 2022
          |archive-url=https://web.archive.org/web/20220303144755/http://www.ee.ic.ac.uk/hp/staff/dmb/courses/DSPDF/00800_TransIIR.pdf
          |archive-date=3 March 2022}}
</ref>
<math display=block>
    \begin{aligned}
        \xi'_i &= -1 \quad Q < i \leq N \\
          p'_i &= -1 \quad P < i \leq N
    \end{aligned}
</math>
Given the full set of zeros and poles, the z-domain transfer function is then
<math display=block>
    H_d(z) = \frac{(z - \xi'_1)(z - \xi'_2) \cdots (z - \xi'_N)}
                  {(z - p'_1)(z - p'_2) \cdots (z - p'_N)}
</math>