# How does twisted pair turn rate affect interference rejection bandwidth?

I'm curious to know how the turn rate of a twisted pair affects the frequency of crosstalk interference rejection.

It's clear that an untwisted pair has no inherent rejection. Alexander Graham Bell figured out in 1881 that twisting is good, so we now twist wire to reduce crosstalk.

As twisted pair communication has evolved, cabling has started to be specified with more and more twisting (cat5 is around 50-70 twists/meter). Ostensibly this is to ensure we can reduce crosstalk at higher frequencies of communication.

I understand the principle of why twisting is good. I'm looking for a mathematical or rule-of-thumb explanation for the relationship between the twist ratio and the bandwidth I get from crosstalk rejection.

The Internet is not yielding such an explanation for this. Am I missing something?

• I visited Bell's Museum in PEI this summer. The phone was such a small part of his legacy but with excellent records at > 15yrs of litigation , he got the patent. The speech therapy, sign language, speed boats and planes were more interesting. – Tony Stewart EE75 Dec 13 '19 at 19:56

As I recall from doing this 40 yrs ago, twisted pair was about 240 to 120 Ohms differential impedance depending mostly on the insulation thickness. The closer the conductors the more capacitance and lower impedance while increasing the number of twists per unit length increases both the capacitance and inductance so the change is less significant, where $$\Z= \sqrt \frac{L}{C}\$$ .

Meanwhile the crosstalk is the result of a coupled input signal. The attenuation ratio of E field in free space to impedance of the cable and the tolerance of wire mismatch will be effectively cause poor CMRR similar to a say 1% resistor (-40dB CMRR) unlike a -100 dB CMRR in an INA (INstrument Amp). As you go to "far field" this imbalance can improves.

When that cable has a differential signal it is easy to understand the induced crosstalk is greater when the load impedance is mismatched and thus $$\V(F)=I(f)*ΔZ(f)\$$.

Wire Inductance per m depends slightly on L/D ratio from 0.5uH to 1 uH /m Wire Pair Capacitance per m length depends on gap of insulations. Both L,C increase with twists per cm or inch.

I would expect (but never measured) insulation thickness of 2 strands is approx equal to wire diameter to get the rated impedance of the twisted pair.

Ribbon cable I recall was 200 to 220 Ohms.

But it is harder to visualize say a 0.2% mismatch in coupling capacitance or inductance unless you have perfect uniformity and it the twists are far away from the interfering source.

For an E field in [V/m] the coupling is capacitive and for a B field it is inductive [A/m] while the bandwidth of the signal is limit by the 0~90% risetime $$\BW_{-3dB} = 0.35 /T_r\$$

If your source/load impedances are low and somewhat matched then your signal BW is increased and CM attenuation from impedance ratio is improved. Thus adding more twists may not be as significant.

Since CAT 5 uses matched impedance applications, I suspect the number of twists is far more tolerant than the insulation thickness to wire thickness ratio.

Beyond 1GHz you need a shielded pair for adequate SNR even with the best BALUNs. That's my hand-waving argument for today.

The idea behind a twisted pair (TP) is that any external signal, EMI, is coupled equally to both wires, so that the interfering source becomes a common mode noise on the twisted pair. Common mode noise is then rejected by the differential receiver at the destination. This is one of the major benefits of a differential interface.

If something happens that causes the external noise to couple differently onto each wires of the, then the noise seen by the receiver is no longer common mode, but rather is differential noise and, depending on the amplitude and frequency of the differential noise, may be received by the receiver.

Because the two wires in the diff pairs are slightly separated from one another, the external noise induces slightly different voltages into each wire at any place along their run. By twisting the wires, this difference in induced voltage tends to average out, with the end result being that the same voltage is induces on both wires across the run.

At higher interfering frequencies, there is, because of the shorter wavelength, more likelihood that the coupled noise will be differential instead of common mode. Hence more twists per length of run are better at higher frequencies.

In twisted pair bundles, it is common for the individual pairs to have different twists per length. This is so that inside the bundle, wire A+ of twisted pair A+/A- will not always lie next to wire B+ or pair B+/B-. Another way of saying this is it reduces crosstalk between the pairs.

I thought there were some tech articles out there that discussed this in more detail. I will look through some of my references and see what I can find.