Attenuation for high frequency when using coaxial cable

Question

Can someone tell me why high frequencies (1 MHz sine signals) are getting attenuated in the below case.

1. I have a function generator with 600 Ω output impedence. I connect this output to a digitiser channel input of input resistance 1 MΩ. I connect this via a 30 m BNC cable(https://in.element14.com/belden/9222-004100/triaxial-cable-rg-58a-u-50-ohm/dp/1288489).

I can see that at high frequencies (1 MHz) the voltage of the 1 V input reduced to 200 mV.

So I decided to simulate the above setup in LTspice and got almost the same behaviour. If we replace the transmission line with a wire connection there is no attenuation.

Why does this happen?

Answer 1

A 50 Ω line looks capacitive when driven from 600 Ω, and so they act like a low pass RC filter. The line needs to be matched.

There are three ways you can match the line

• Source matching - drive the line from a 50 Ω resistive impedance. The easiest way to do this is to shunt a 54.5 Ω resistor (exact value, 51 or 56 will be close) across the output of your 600 Ω generator. It will reduce the signal level, but it will be flat with frequency. You can use any resistive load on the far end of the line, including open circuit on the output of the line.
• Load matching - load the line with 50 Ω, (51 will be close). You can drive the line with any resistive impedance, including 600 Ω. The signal will be lower, but flat with frequency.
• Both source and load matching - you'll get half the voltage of the previous two methods.

If you don't want the signal loss that these methods entail, you will need a buffer amplifier on the output of your signal generator, or a 600/50 impedance ratio transformer.

Another alternative is to use a 600 Ω line to convey your signal to where it's needed.

Answer 2

a 50 ohm transmission line acts like a capacitor when it's fed from a 600 ohm source.

To get better performance change R2 and R1 to 56 ohms and R1 to 50 ohms

For your signal generator that would mean placing a 56 ohm resistor in parallel with the start of the coaxial cable, and changing the load at the other end to 50 ohms too.

Answer 3

I have a function generator with 600Ohm output impedance. I connect this output to a digitiser channel input of input resistance 1MOhm. I connect this via a 30m BNC cable

_{Note - this isn't a transmission line problem because the maximum frequency of 1 MHz has a wavelength that is 300 metres}

Your shield (outer) and cable inner are magnetically coupled to each other. However, the coupling isn't perfectly 1:1 so, there is net inductance in series with your signal.

This forms a low pass filter with the distributed capacitance between inner and outer and, higher frequencies get attenuated: -

At 1 MHz you get about 20 dB of attenuation irrespective of the output load resistance you used: -

Different cable types will give slightly less or slightly more attenuation. Your cable: -

If you replaced the lumped inductor/capacitor I used earlier with a transmission line you will see that up to 1 MHz there isn't much difference: -

If you replaced the 600 Ω driving impedance with 50 Ω it would be near-perfect at 1 MHz: -

New circuit with transmission-line and 50 Ω driving impedance: -

Answer 4

Your cable has approximately 100pF of capacitance per meter.

30 meters means 3 nF of capacitance if considered as one lump. In practice the capacitance is not a lump but distributed so effectively it is less.

But still that load capacitance driven with 600 ohms means you have a RC low pass filter which can't pass high frequencies well.

Another reason could be that for high frequency signals, the cable is a 50 ohm load impedance for a 600 ohm source impedance. That is a divide by 13 for the amplitude in itself.

So with 1 MHz sine wave having a wavelength of 300m, the 30m cable starts to look like a transmission line instead of just a piece of wire, so most likely you are seeing some transmission line effects already at 1 MHz.