Electronic – Characteristic impedance of a D-subminiature connector

This seems the simplest mathematical way to derive characteristic impedance. Consider a "lump" of transmission line connected to the continuation of that transmission line (\$Z_0\$): -

• R is series resistance of cable for a given length
• L is series inductance of cable for a given length
• G is parallel conductance of cable for a given length
• C is parallel capacitance of cable for a given length
• \$Z_0\$ to the right is the continuation of the cable

Therefore the impedance looking into the left is: -

$$Z_0 = R + j\omega L + Z_0||\dfrac{1}{G + j\omega C}$$

$$= R + j\omega L + \dfrac{\frac{Z_0}{G+j\omega C}}{Z_0 + \frac{1}{G+j\omega C}}$$

$$= R + j\omega L + \dfrac{Z_0}{1 + Z_0(G+j\omega C)}$$

$$Z_0[1 + Z_0(G+j\omega C)] = [R+j\omega L][1 + Z_0(G+j\omega C)]+Z_0$$

$$Z_0 + Z_0^2(G+j\omega C) = R+j\omega L + Z_0[(R+j\omega L)(G+j\omega C)]+Z_0$$

$$Z_0^2(G+j\omega C) = R+j\omega L + Z_0[(R+j\omega L)(G+j\omega C)]$$

The important thing next is to recognize that \$(R+j\omega L)(G+j\omega C)\$ is insignificant as the "lump" approaches zero length and we are left with: -

$$Z_0^2(G+j\omega C) = R+j\omega L$$

hence $$Z_0 = \sqrt{\dfrac{R+j\omega L}{G+j\omega C}}$$

High speed cables use shielding, twisted pairs and separation etc to mitigate intereference. Each standard has a maximum length that helps ensure that it can deal with exepected background noise over that distance. So, assuming that you are not using long cable runs, the connector you choose is a low loss type and you try and maintain as much of the original shielding/twists etc( eg do not expose 30cm of cable at your join) it will probably be ok. But until you make the connector you are probably not going to know.