# Mismatched transmission line at DC

The characteristic impedance equation in transmission line theory is this:

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

Take the case of a twisted pair transmission line, say the cat6 cable represented in this link . Cat6 cable info

Either conductor in the pair is well separated by their insulation and an additional plastic spacer, so G is essentially zero. As for R,L, and C, the link above provides that

• R=8.00 ohms/100meters

• L=0 henries

• C=6nF/100meters

If you let omega go to zero, Z_0 goes to infinity. This makes sense if the transmission has no load attached to it, but when a mismatched load is attached things get a little messed up. If you try to calculate the load reflection coefficient, you'll get "-1" as a result. $$\lim_{Z_0\to\infty} \frac{Z_L - Z_0}{Z_L+Z_0} = -1$$ This implies that zero voltage will be measured at the load $$\Gamma_L = \frac {V_L^-}{V_L^+}=-1$$ $$V_L = V_L^+ + V_L^- = V_L^+(1 +\Gamma_L) = 0$$

Now we all now that DC voltage dividers work and that's exactly the circuit that I've explained. If I haven't made any mistakes, could someone tell me the physical significance of this result?

/product_details.aspx?id=33613 "cable info"

• L is not zero!! – winny Aug 17 '16 at 5:47
• The spec sheet does not list L. Plus L being non-zero doesn't affect the end result I'm presenting. – ElecEng2016 Aug 17 '16 at 11:45
• Using $$V_L=V_L^+(1+\Gamma_L )$$ as a steady state solution entails the driver side of the line is terminated to the characteristic impedance. Wherever along the line you look toward the source you see a driver with a finite voltage thru an infinite impedance: output dead. – carloc Aug 17 '16 at 13:01
• My question then is "Why is the line of infinite impedance? " The source will indeed be terminated to the line, some voltage will drop across the source impedance, but the line should act like a resistor spanning between the, say positive pole of the source to the positive pole of the load and also a resistor spanning between the negative poles of the source and load...right? – ElecEng2016 Aug 17 '16 at 13:49
• Who flagged this as low quality? Low quality means quality of the question, if the OP has something wrong then try to get them to correct it not shut it down. It looks like a great question to me as far as length and content – Voltage Spike Aug 17 '16 at 19:30